Transcripts
1. Electric Circuits Promo: Hi, and welcome everyone to our course for
electric circuits. This course is designed
for anyone who would like to learn about electric
circuits from scratch. Even if he doesn't know what does an
electricity even mean? So if you don't know about electricity or you
would like to learn about electric circuits
from the very beginning. Then this course is for you. I am a madman and
electrical power engineer. Let's start by learning what are you going to get
from this course? In the beginning of the course, we will start
learning gap out with that basic concepts
of electric circuits, such as Zach, current, voltage, and power, and so on. Then we are going
to discuss that basic laws of electric circuits, such as what is the
meaning of Ohm's law? And what is the
meaning of resistance, Zach, meaning of
conductance, and so on. Then we are going
to start learning how exact methods of
analysis in which we are going to learn about mesh analysis and
nodal analysis. Then we are going to start
learning about with Zan, different circuit theorems such as super position theorem, Norton's theorem, that social
transformation theorem, and that maximum power transfer. Then we are going to learn about a very important component
inside electric circuits, which is the
operational amplifiers, or abbreviated as x. Then we are going to start
learning about Zack capacitors and inductors which are used in the electrical power systems. Then we are going
to start learning about that first order circuits, which is format of resistor and inductor or a resistance
and capacitance. In both cases, our source of free gifts and step
response gaze. Then we are going to start learning about the
exact AC circuits. And we will start to learn
what's the difference between DC and AC circuits. And we will understand also that different concepts related
to the AC circuits, such as that phasor
representation or as a phasor diagram. We will understand that
meaning of admittance and impedance in electric
circuits and much more. Then in the next section
we will discuss as a sinusoidal AC power analysis. All we are going to start
learning how can we apply that different circuit theorems in DC circuits to the AC circuits? Then we are going
to start learning about with Zach AC
power analysis, which are representing
different types of power, such as the active power, reactive power, and
apparent power. We will understand
what is the meaning of this concepts which are used in the electrical
power system. So z are really, really important in understanding their
electric circuits. We are going to discuss
an important phenomena which occurs in
electric power systems, which is that electrical
resonance or electric resonance. We will understand
what is the meaning of resonance and what
is the meaning of a series resonant circuit and
the parallel resonant sack. Finally, in the scores, we have an additional component
or an additional section, which is not found
in any other course. You'll learn about what is the electric circuit
simulations. We are going to simulate most of the electric circuits that
we learned in the course. In the Matlab program. Matlab is an important program
for electrical simulation. We will use MATLAB Simulink
to start simulating this different electrical
circuits or electric sockets. So if you are looking for one course that
will help you learn all the basics of
electric circuits are from scratch without
any previous knowledge, then z-scores is for you. So I hope to see you in our
course for electric circuits. And for any question, you can send me a message. Thank you and see you in
our course for electric.
2. Introduction to Electrical Systems: Hi and welcome everyone to our course for
electric circuits. So in this course, we are going to learn how to do the analysis for
electric circuits. We are going to understand is that definition of the voltage, current, electricity in general. And how can we do KVL, KCL, or the different
circuit theorems, okay? In this lesson, we would like to understand an introduction
about electricity. So what happens in the
electrical power system? So for us that we have, in our electrical power system, we have three main stages. We have that generation stage
or the generation phase. We have the transmission system, we have the distribution system. So first, what are
we going to do in our electrical system for us? So we start generating
our electricity. We generate electricity
from different sources. It can be renewable
energy sources or non-renewable energy sources. Renewable energy sources
such as solar energy, wind energy, and so on. For the non-renewable sources, we are generating electricity from fossil fuels as an example. So for example, we generate
our electricity at a voltage and we will
understand in this course. So what does the
voltage you mean? Ten kilo volt. This is how we generate
electricity as an example, okay, at this voltage. Now, the next step
is that we are connected to a
transmission system, okay? Is this transformation system, what is the benefit of
this transmission system? It transmits or transport that electrical power or
generated power from the generating station to
the distribution network. Okay? So that's the transmission
system has a voltage, for example, 220 kilo volt. It has 500 kilo volt and
so on, different voltages. So as you can see,
there is a difference between this voltage, the generation of voltage, and the transmission voltage. Then we reach that
distribution system. This distribution
system, its function is to distribute the
electrical power. It can distribute at 3.3
kilo volt, for example. Then finally at 080 volts. So the generation
of electricity, transmitting electricity, then distributing electricity
to our electrical load, such as in our home. If we look at this system again, we have generation,
then substation. What is a function
of substation? It has many important functions. As an example is a
substation is used to provide protection to our
electrical power system. It contains a
protection system such as circuit breakers, relays, and so also the substation
has something which is called the transformer runs former. What does a transform
or do the transform, our function is to
change the voltage. So as an example, we
said before we have here ten kilovolt and we would like to have 220 kilovolt
adds a transmission system. So how can we change the
voltage using the transformer? The transformer is
used to step up or increase the voltage and
can be used to step-downs, such as in here from the
transmission system, shows a distribution system
or step down the voltage. Okay? So this is an overview about
the electrical power system. So what are we going
in this course to do? We are going to learn forest
that main definitions. We need to understand
what does that mean? What does the voltage
amine and so on. So we need to understand what
does electricity even mean? Okay, so let's start first. If you look at any atom, okay? So for example, that metal
or a non-metal materials, each of them is
consisting of atoms. Atoms. Okay? So the atom contains inside it. Here is a nucleus. This is called the nucleus. It contains part of it
is called Z neutrons, also part called the protons, and we have around
18 orbits electrons. So the neutron is not outpost of the charge
or a negative charge. It is not a positive
or negative. It is a neutron or two, we can say it is a
neutrally charged. So it is nice, nicer net. Protons are positively charged. And around the atom we have electrons which
are negatively charged. Now, as you can see
from this figure, you can see we have
electrons which are negative and we have
positive protons. So since the z I have
different signs in nature, what happens is that this negative charge
would like to go to the post of a charge
supposed to surge would like to go to the negative z, would like to attract each
other because they have the same charge as they
have different signs. One positive, one negative. But if this happened, there will be no atom in nature. So what happens is that you will find that the new electrons, electrons, these
electrons are rotating around the nucleus with
a very high speed. So if we look at
that atom in 3D, it will be something like this. You can see the nucleus, which contains
protons and neutrons. And around it in orbits, there are electrons
which are rotating or moving in a very high-speed. As you can see, this particles which are moving are
called electrons. In atoms, in atoms
of that metals, they have something which is
called the free electrons. So the electron, this
electron like this, which are not
connected to an orbit, they are free to move
inside the material itself. Like this. So we have, this is an atom
which has its own electrons. In the metal atoms, we have something which is
called the free electrons. They are not attached
to any atoms. In the metal atoms. They have very large
amount of free electrons. Metals such as, for
example iron or copper or aluminium
have free electrons. So this is free electrons or
the cost of the electricity. So when they are exposed to
a difference in voltage, as we will learn or does a
voltage means of course, when they are exposed to
all different and voltage, they are moving in a
certain direction. So as you can see here, they are scattered and
moving randomly here when we are exposing it
to a certain voltage. So the OnStar to moving
in a certain direction. So this motion in a
certain direction form is the electric current. Okay? So the motion of electrons
inside the material itself or the metal is the cause
of the electric current. Now we have to know that here, similar to the magnets, if they have, are both poles, they attract each other. The North and South
attract each other. If we have cells and
South or North and North, they repel from
each other as they would like to go away
from each other. So you can think about with Zao voltage such as a
battery, for example, the battery has a
bold step terminal and a negative terminal. So you can think about
the positive terminal as a large amount of
positive charges. And negative term as our large
amount of negative chores. You can think about
it like this. Okay? Now what happens here? We have a metal here, this brown line
representing a metal. And this metal has what
has free electrons. Free electrons. So when we connect 11 here
and another one here, or one wire here and
another wire here. In a closed circuit. What will happen is that
you will find that here we have negative electrons
here, and so it's a wire. Negative electrons. Now, betweens and
negative electrons ends up positive of the battery. What will happen? They are different signs so
they attract each other. So the electron, so the
negative electrons tries to go to the positive
terminal of the battery. All the negative electron goes to the positive terminal
of the battery. Now what about the
negative betweens and negative terminal and
the negative electrons, they are same sign. So they are rebelling
the ripple between them. So what will happen is that
it starts to go away from it. So it starts to go away
from it, away from it. So you will see that
from this figure, what you can see is that
when we connect a wire here, negative and positive two, the two terminals
of the battery, you will find that
the electrons are moving in a certain direction. As you can see, they are going
into the positive terminal from the negative going
into as opposed of Turner. Okay? So since it's the outgoing
in a certain direction, this is an electric current. The electric current is
the motion of electrons. Now when the electric current
passes through any load, such as a lamp or a
refrigerator or anything. When it passes through
it, it operates it. Okay. So as you can see, when
the electric current is passing through it. You can see that the bulb
is giving light. Okay? So that's what happens in
real life when we have an electrical source such
as a generator for example, this generator produces
a difference in voltage. So this difference
in voltage such as like a battery, okay? This difference is
volts causes the motion of electrons or the
free electrons. The motion of free
electrons leads to formation of
electric current. The electric current
leads the two operation of our electrical device. Okay. Now what about the voltage? So if we have a battery, battery like this, battery, one was a positive terminal
and negative terminal. How can we or what does that difference
in voltage you mean? Okay. So you will sometimes see
a battery with a 12-volt, battery with a 24 volt. So the higher the voltage, the higher the electrical current passing
through the system. As if we are increasing the force of attraction
by the positive terminal, or worse or repelled by
the negative terminal. Okay? They sometimes say is
that you can imagine the voltage or the difference involves such as
the 24 or at 12, as if we have two tanks, Bank a and Bank B. Okay? So as this elevation increases, more pressure, more pressure, you can see that the water is false and due to
the force of gravity, it tries to go from the upper direction going
through thank beam. So this difference in height, difference and heist in height representing a
difference in voltage. So when we have high
difference in height, we have high voltage. So the height difference in
height produces high flowing of water exists two
tanks were like this. There will be no flow of water because they
have the same height. Or here we can say same voltage. So the difference
between them is 0 volts as if they are
on the same height. However, if it is like this, then the water will flow
from the higher location. Location as if we have high difference between
these two voltages. Okay? Now, here is how it happens. So when we have an
electric conductor such as, for example, a copper wire or aluminium, you will find that it contains a large number of
free electrons. This free electrons, as you
can see, are moving randomly. They are moving everywhere. However, in this case, we don't have any
potential difference or no source is applied or
no voltage is applied. However, in this case, you will find that when we apply a difference in voltages
such as for example, when you have a wire and
connecting it to a battery, for example, with a positive
and the negative terminal, you will find that
this electron will start moving from the
negative terminal, going into the
positive terminal. Okay? So as you can see, here's
the electrons itself start moving in a
certain direction. Why due to the presence of a potential difference
between two points. Here, without any potential, they are moving randomly. Here when we have a
potential difference or a difference in voltage, they start to moving in
a certain direction, which means that we have
an electric current. Okay? So in the next lesson, we are going to
start learning more about electricity or
electric current, the voltage, current, voltage, energy, power and so on. Okay. So don't worry, we are going
to start learning about each of these definitions in
details with their equations.
3. Electric Circuits, Charges, and Current: Hi, and welcome everyone to our lesson in our course
for electric circuits. In this lesson, we will discuss some basic concepts are
powered electric sockets. So first, we need to
understand what does an electric circuit mean or
what is an electric socket? So simply, an
electric circuit is an interconnection of
electrical elements. So a basic electric circuit
consisting of a battery, which is a source of
electricity, electrical loads, and wires connecting between them. So what does this mean? If you look at this
figure, this figure here, is this representing
an electric circuit or a very basic electric socket? You can see it's consisting
of several elements. Number one, we have the
source of electricity, which is our battery. So our battery is a
source of electricity. Now, other sources
of electricity may be an electrical generator, such as easy
synchronous generator and many other types of
electrical generators. That second part, which is
the electrical load, e.g. we have here a lamp which
consumes the electricity. So we have sub-patterns
which will provide electricity or electrical power. And the lamp is our
electrical load, which are consumes electricity. Then in order to connect
between the load, which is our lamp, and the source of
electricity, we need wires. You can see we connect
the wires between the lamp and our
source of electricity. So these are the
three elements which involves or form is
an electric socket. You can represent this figure in the form of a
schematic diagram, a 2D schematic
diagram like this. You can see we have
our pattering, we have our lamp, we have wires. Can see wires which are
connectors between, between zap positive of the battery and negative
of the battery. You can see any battery has positive terminal
and a negative term. We will understand
what does this mean later inside the course. Then we have a switch. Now as an example,
as an example, if we have in our
home we have switch. So when we close the switch, that lamp or electrical
equipment starts operating. When we open the switch, the electrical lamp
will turn off. Okay. So here's the
switch is like this. So this switch, when we open the switch like
this, it is open. You can see is this
wires are disconnected. Can see this connected here, which means that this lamp will not have electrical power. So when we close the switch, when we close the
switch will exist, e.g. it will be closed like this. And the electrical current
will flow through the lamp. And the lamp will
start giving light. Okay. So here, if you look
at this figure, we have the pottery, Okay? You can see we have
all stiff terminal and we have negative terminal. Okay? So the battery is format of different chemical
composition or different chemical has
different chemical reactions. And in the end, we have our one side
of the battery, one side, one terminal here, which is a positive, and another terminal
which is negative. Now what does this mean? It means that one side, one side of the battery, one side of the battery,
which is negative. It means it has large
number of electrons, very large number of electrons. And the positive side it
means has large number of balls they've holds,
holds like this. You can think of the slides, or it has very low
amount of electrons. So we have one side, which is a four-step, have, has very low amount of electrons and another side which has large amount of electrons. Now what does the pottery
would like to have? The battery would like
to be in balance. We would like to have
the neutral state. So what does this mean? It means that the electrons. Would like to go and
fill this holds. Each one of these
electrons would like to go and fill this holds. Okay? So how does this even happen? You will find that here that this wire also
contains electrons. This wire contains
electrons like this. Okay? Now since we have very large
number of electrons here, and we have electrons inside
the wires itself, e.g. in copper or aluminium. What happened is
that we will have a very large repulsion
force since we have very large amount of
negative electrons here. And we have electrons
inside that conductor. Than zero will be
a repulsion force. The electrons in the wire
will start to go hoeing away and go to the
positive term. And this holds. Now if you have this
circuit opened, as you can see here, open circuit or this wire is cut off. What will happen in this case? In this case, you will find that the electrons can
travel from here, from this site and go to the battery so there
will be no current. Okay? So we say is that the motion of these electrons due to the
presence of repulsion force. And at the same time these
electrons wants to go and develop all of these
positive holes to be in neutral state. This motion when the
electric salts moving like this and going
into the other side, this motion is known as
the electric current. Okay? So we have here
several elements. We have forest, the pottery
which is measured in volts. We have voltage. We will understand
what does this mean. That we have the electrons
or the electric charges. And then we have that flow of electrons or the
motion of electrons, which is the electric current. So we need to understand
all of these elements. Okay? So first, what is
an electric charge? A charge is an
electrical property of the atomic particles of which
matter consists is off. It is measured in columns. Columns is the unit
of measurement of electrons or
electric charges. In general. Each atom, as you already
know from physics, that each atom consisting of electrons, protons,
and neutrons. If you look at the
structure of any atom, we have the nucleus
and we have orbits around this nucleus.
This all bits. So we have electrons,
as you can see here, electrons, which are
negative charges. And inside the
nucleus of the atom, we have neutrons
and two protons. We have all stupid charges, which is a proton
supposed the charges. And we have neutrons which is neither positive nor negative. It has no charge. And the electrons, electrons
are negatively charged. So we have negative charges. We have both individuals
and we have neutrons. Okay? Now, this is how it looks like. A very simplified
look of this add-on. You will find that the
electrons are rotating around the nucleus at a
very high-speed. Okay? Okay. So the electric charges, or is called the fall
or the electrons, neutrons and protons are
called electric charges. Here what is the
important part which is essential for us
is the electron. Electron is negatively
charged and it has a value of 1.602 multiplied by ten to the power
negative 19 coulombs. This is the amount of the electron is equal
to one electrons lost. One electron has this
amount of columns, which is a measurement of
the electrical charges. Okay? Okay. So what does ten to
the power negative 19 mean? It means that we have, this means 1.602 multiplied by ten to the power negative 19th means
1/10 to the power nine. Or what does this even mean? It means that we will have one. And beside it, 1900s, we have one like this. All of these zeros are 19. So you can see that
the amount of columns of one electron is
very, very small. Okay? Now, in one column, just one column, how many electrons are
required to form one column? You can know, you know
that one electron. Gibbs, this value 1.602 multiplied by ten to
the power negative 19. Okay? Now, what if I would
like one column? This is column, we
fought like one column. So how many electrons required? From this equation,
you will find that x, or the amount of electrons
required is 1/1, 0.609 multiplied by ten
to the power negative 19. What does this mean? It means that we need 6.24
multiplied by ten to the power 18 electrons in order to
form one column of charge. Okay? Okay, so here we understand
now the electrons, now, realistically,
what is the amount of, what are the values
of the charges? That realistic or laboratory
values of charges are in the order of Pico
column and nanocoulomb. What does pico? Pico means? P or pico means tan to
the power negative 12th. Nano means then towards
the power negative nine. Again, what does this mean, e.g. it means 1/10 to the power nine, or it means a one multiplied
by minus z, minus z. So the electric
charges or electrons, or the cause, or the one which form is
the electric current. The electric current,
we denote it by I, which representing the
electric current and the electric charge
is denoted by Q. So Q, representing
how many charges, all you're representing
the electric current. The electric current is formed of a group
of electric chores, or the motion of a group
of electric charge. So that electrical charge, or the electrons to
be more specific, are the one which causes
the electric current. Okay? So what is an electric current? Electric current is
the rate of flow of electric charges
through a conductor. Okay? So electric current is the time rate of
change of the charge, and that's measured in amperes. And pears, or a is a unit of
measurement of the current. Current is denoted by all. Are you representing
the electric current, which is the rate of
flow of electric chores? This can be
represented like this, d Q over d t, the derivative of Q, or that electric
charges with respect to time rate of flow
of electric charge. So the electric current
is the result of the motion of electric
charges inside our system. Or an hour conductors. So remember the
previous figure of the, we had a pottery, we have wires and we
had the lamp also pulp. Now we had our
positive terminal of the battery and the negative
terminal of the battery. And then we said we have a very large here
amount of electrons, which I would like to
go to this poster. Or the positive of the battery. And the wires itself
has electrons. So what will happen
is that we will have a very high repulsion force between the negative terminal of the battery and these electrons, which will lead to motion
of these electrons through that wire going into
the positive terminal. You can see this motion
is what happens exactly. This motion form is
an electric current. The flow of electrons
or the rate of change of a charges
with respect to time, it Charles, or electric
charges with respect to time. Now, this flow of electric
charges through Paul, but e.g. will lead to consumption or using the electricity or absorption of
electrical power. Okay? Now it's the same figure here. You can see this figure we
have suppose that terminal, negative terminal ends,
electrons starts moving. Now something which is really, really important, what is
the direction of current? So you can see here that the electrons are
flowing like this, right, from the
negative to post them. However, a scientist,
all of the scientists, they agreed on one direction. They agreed on it selecting
the z direction as the one which is opposite to
the direction of electrons. So what does this mean? It means that the electrons are flowing like this from
negative to positive. Now was a scientist said that. Electric current will
be the opposite T2s, it will be like this. So if the electrons
are flowing like this, then the electric current is the one which is
obviously a toy. They call it the
conventional current. Conventional current. Okay? So if you look at this circuit, we have a Patrie, we have positive and
negative of the battery. Now, this ball step you
can see here as current, current going from
positive to negative. Remember, current
always coming from, suppose they've gone
to the negative. But what is the
direction of electrons? The electrons are
moving from negative, going like this to
the other side. Okay? You can see that the
direction of the electrons are opposite to the direction
of current. Why is this? Because the scientists
agreed on this. Okay, actually, what is the direction of the current
is similar to electrons, and that is what
happens in real life. However, they selected,
they agreed on selecting the
direction of current as the one which is
obviously totally, okay. So this is for
understanding in real life, it moves it from
positive to negative. The current flowing
from positive, from the high potential to the negative or
the low potential. Okay? It's called
conventional current. And it is measured in amperes. Okay, Here's another figure
to understand what tiny. So you can see is that we
have the post of the battery, which means have
high holes, e.g. hypostyle halls. And do we have here negative
or negative electrons? So the electrons would
like to go through that pottery like this and
go and fill this holds. Each electron we have here hi electrons and here
high balls, the balls. So all of these electrons
would like to fill this hole, would like to fill this one, this one, for this
one, you'd like. So that's the goal like this. So you can see that the flow of electrons leads to the
presence of current, the actual current, okay? The actual Can, the
radial current. However, when we are talking in our course or in anywhere, we say that that rail, that conventional current or the current which we are talking about is the one which is moving like this from
positive to negative. So this is the actual current. This is a conventional
current which we use in our circuit analysis
in our daily life, in every equation, okay, from positive to negative. Okay? Okay. Okay, so we said that
the electric current is the time rate of
change of charge. So we can say is that I
is equal to dq over d t, the derivative of Q, or the charges with
respect to time. Okay? So the current is
measured in amperes, and each one ampere is
one column bear second. So as you can see, current, which is one at Bear, means that we have one column
over 1 s. You can see here, Q is measured in column a and t is measured in second column of our second
column over seconds. So one ampere is one
column over one-seventh. You can remember that one column is very large amount
of electrons. So if you get back here, here, you can see that one column is equal to 6.24 multiplied
by ten to the power 18. Very large amount of electrons. Okay? So you can imagine that
one embedded GIFS, very large amount of currents. However, in real life, there are very large amount of very large values of current. You can find thousands
of ambient then thousand and bears in case of electrical
voltages and so on. So from this equation, this equation, if we
would like to get Q, or the quantity overcharges, or how many charges
than simply q will be the integration of
current with respect to time. Integration of current
with respect to time from any initial
time to any final time. Okay? Now, what are the different
types of electric currents? You will find that in real life, we have two main types. We have that direct
current or direct current, which is a DC current. It is a current that
remains constant with time. Now, what does this
constant mean? Constant, that means it has a unidirectional
one direction. So you can see the erected
current is one direction. It doesn't, it change its legs. But it can have a
valuable magnitude. Magnitude. So as you can see
here, we have the current with respect to time. You can see that the
current value, e.g. let's say this value
is two unpaired. So you can see this value
is constant to waste time. Okay? So this current, it has a one direction. You can see it as four-step
to impair all the time, which means it is
unidirectional, has one direction, and at the same time it
has a constant value. So it means that it
is a DC current. Another one, e.g. you can have
something like this. Like this. This one is also at DC current. Why? Because it has one direction. You can see e.g. this peak, e.g. let's say 1.1 and bear e.g. you can see the value
of the current itself, which changes 0-1 and then 1-0. So it is always positive. It means it is
unidirectional OTA, or it has one direction, which means it is also at DC. Dc current. The most
important thing about DC current is that it
is uni, directional. It has one by erection. If we look at another type, which is the
alternating current, it is a current that varies
sinusoidally with time. What does this mean? You can see it is forming
a sine wave like this. This one is the one which comes from electrical
generators. Ac, electrical generates,
it's called the alternating. What does alternating mean? It means that it is
switching directions, keeps changing its direction. So what does this mean? As you can see here, e.g. here in this figure, you can see this part
is positive, right? So it is one direction. However, after a certain time it starts switching direction. We have here negative, positive and the negative, then another time positive, then another time negative. So you can see it
is alternating, keeps changing its direction. Sometimes positive, sometimes
negative, and so on. So we call this one an
alternating current. So if you look at this figure, this figure will help
you understand the idea. You can see this
is a DC current. You can see that the
electrons here, again, this is a conventional, Conventional which we choose. The one which we use
in our analysis, not the actual current, but the conventional,
which we'll use as the current goes
from negative to positive to negative convention. So the electrons here assumption or we can say
halls or whatever it is, it moving like this, like this to the next. Okay? So you can see it
is one buyer exam from positive to negative, that electrons are
moving like this. However, if you look at AC, AC or alternating current, you can see sometimes
moving from here, moving, moving like this. And other times
moving like this. Okay? So sometimes it moves from
here, going into here. So sometimes this one is
positive and the negative, so it moves like this. And after a certain time, it changes its direction, it becomes negative bolster
the current moves like this. So you can see it's called
alternating and C keeps switching, switching
its direction. Sometimes from here to here, sometimes from here to here. That's why it's
called alternating, because it is changing
all the time. Okay? This one is a DC. You can see it's constant
to one direction from positive to
negative doesn't change. Okay? So in this course, we will discuss mainly the
DC current and then we will start adding more lessons
about the alternating current. Finally, we have something which is called system of units. So what does System
of Units mean? You can see that zeros
and international, international units
that everyone uses. Okay? So e.g. e.g. Zan lens, that international
standard or what all of the people agreed on is that meet the mental lens in meters. So if you look, there are some countries
which a measure is a lens of anything in feed, e.g. or inch or whatever, and others use meat. So the international standard, or what all
scientists and people over Walden standards agreed
on is that they use meter. So meter is called
International Unit. Okay, or standard
unit mass, e.g. for the mass, how many kilogram? Kilogram? Some countries use kilogram, other countries
use a pound, e.g. for the international
standards are using kilogram by him is
measured in seconds. Current image and bear
that temperature Kelvin. Okay? You will, as you know,
that some countries use Celsius as our countries
use Fahrenheit. However, the one which
is a standard values is that Colvin veil and so on. So this is a unit of measurement and this is a sample
that they use. Okay? So each, each quantity, lens, mass current,
charge, voltage, whatever. They have, one basic
unit and one sample, which is the
international standard. Okay? Now we have prefixes
here, SI prefixes. What does this mean? What does this mean? It means that it is an
abbreviation for large numbers. E.g. when we say
middling, mainly meet. Okay, We say mainly meet. What does mainly meter means? Let's say e.g. three millimeter. It means it is equal to three. Multiply it by ten to the power negative
three of the meat. Three multiplied by ten to
the power negative 3 m. So mainly here is an abbreviation for ten to the power negative. As you can see, Meli, e.g. abbreviation for ten to
the power negative three. Now, as an example, micro to use is done to
the power negative six. As you can see, it's
simple, Micro like this. Like this. And do we have here Nano, which will exhaust said before, nano n ten to the power
negative nine pico, ten to 12, femto is ten to
negative 15, and so on. You'll find here Kilo
ten to the power three, mega ten to the power
six, and so on. So you can see this
table help us, you know, the abbreviation of
lots of multipliers. So instead of saying, oh, point, let's say 0.003 e.g. which is 1234, Let's
say e.g. 512345. Okay, let's add
another one here. Zero here, like this. So we have 123456. So instead of typing
this larger number, we can say three multiplied by ten to the power negative six. You can see how many zeros
or how many decimals. 123456. So we have three multiplied by ten to
the power negative six. And instead of
typing this or this, we can say three. Mike, Okay? Whatever microphone, rod, micro column,
whatever it is. Okay. So I hope the idea of the System of Units
is clear for you. So in this lesson, we started discussing some of the basic concepts of
electric circuits, including electric
current, electric charges. We discussed system of
units in the next lesson, we will have some
examples on this. Then we are going to
discuss the voltage and the power energy and so on. Okay?
4. Solved Examples 1: Hey everyone. In this lesson we are going
to have some solvent, the examples on the electric charges
and electric current. So as a first example here, or how much charge is
represented by 4,600 electrons. So the question here is, would like to know how many columns that for cells and 600
electrons equal Q. So it is pretty straightforward, as you can see, electron, each one electron,
as we learned in the previous lesson,
is negative 1.602. Multiply it by ten to the
power negative 19 column. And you have to know that this negative sign,
what does this mean? It means this negative
sign because electrons, because electrons are
negatively charged. Okay? That's why we added here
is a negative sign. If we're talking about
winds up protons. So e.g. then in this case you are going
to say all stiff vein. Okay? So we know that
each one electron, one electron is equivalent to 1.602 multiplied by ten to
the negative 19 column. So how much is Josh? How many columns is
4,600 electrons? So simply we will take this
value number of electrons and multiply it by that column
to get the total charges. So as you know, is
at negative 1.60 to multiply by 1010 to
the power negative 19. Coulombs bear each
electron for one electron. So if we have 4,600, so we'll multiply this together. We will get negative 7.3, 6.9, then two
spawning 16 column. Okay? Now let's have on us or one, if the total, total charge
entering the terminal, given by q equals five t sine
four Pi t. Mainly colon, finds the current at
time equals 0.5 s. So first, what does
a terminal amine? So if you have e.g. a. Battery like this with the
positive and negative. Now this part, this wire
and this wire or this, to be more specific, this part and this part
are called terminals. Terminals of the battery. So if the total charge entering athermal,
what does this mean? E.g. if you know that we
have electrons here and this electron goes to this part, here, is all of these electrons are traveling and entering, that goes stiff terminal
entering that course. So if the total charge, the amount of Q, which is flowing
inside the wires, negative charges going into
the positive terminal is equal to five t sine four
Pi t, mainly column. What does this mean? It means it is T here
representing time. So e.g. q at time equal one. Okay? After time equal 1 s, then we'll substitute with
one in this equation. So it will be five multiplied by one sine four pi
multiplied by one. If e.g. our time
equal tos and we'll substitute with t
equal to n, so on. Okay? So this means that our
q is a changing with time. What do we need is that
we need the current. So if you remember that
we said before that the current is equal
to a DQ over DT, or the rate of change of the
charge with respect to time, or the derivative of Q
with respect to time. So it will be derivative of q, which is five t sine four Pi t. Here, five t sine four by T. Mentally calm. Remember mainly here, ten to
the power negative three. So the derivative of this one, we have five t sine four Pi t. Now we have to, if you don't know what they are doing here in the
derivative assembly, we would like d over d t
for two variables, x and y. So you can say X is this one, this one is x, and
this one is y. The derivative of
two multiplication. It will be equal to
derivative of the first multiplied bys
a second as it is, plus derivative of the second multiplied Paul
is that first one. So the derivative of a
forest which is five, t derivative of t is five. And y, which is the second one, sine four pi t, sine four Pi t. Plus derivative of the second
we have signed for Pi t. Derivative of this one is
four pi multiplied by cosine, four by T. So it will be, its derivative is
four pi multiplied by cosine four Pi t. You can
see cosine four Pi Ti. And Do we have here
four pi, okay? Then multiplied by x, which is the first
one, which is five. So this multiply it by five t. So we have
four pi multiplied by five t gives us two when t is Pi t When Tea Party
or to any party. Okay? Okay. All of this is
Mendeley and pair. And bear is a unit of
current and Middle East. And so we are talking
about mainly colon. Okay? So what I need this
as the current in general with respect to time. Now I would like the value of
current at time equals 0.5. So I will take 0.5 and substitute
it in T here and here, and here, like this. So by substituting
0.5 in this equation, you will get that z
value of current at time equal 0.5 is solid to 1.42. Mendeley and bear. Okay. Now let's have another one. Determines that totally
charge entering a terminal between time equal 1 s
and time equal to second. If the current passing
through the terminal is I equals three t
squared minus t. Okay? So what do we need here
is that we need the queue or amount of charges going through a terminal
of a pattern between time one and time equals
2 s. We have given, in this problem, we have I given the value of current
with respect to time. So if you remember a queue, which I have discussed before
in the previous lesson, you can get q or the amount of charges by using integration. You know that i equal to dQ over Q is the
integration of currents. So we can say it
is integration of current with respect to time from t naught from any initial time
to any final time. So the initial time here is 1 s. And our final time is 2 s. Because we need the amount of charges between these two times. And our current
itself is three t square minus t. This is
the equation of current. So you will have like this
integration 1-2, digression, 1-2 for the current as a function of time current
which is three t squared minus t. So the integration
of this one first, the integration of
three t squared. Integration in general,
if you don't know. Here again, just as
small reminder for this integration of x
is x to the power 2/2. Or in general, if
you would like to integrate x to the power n, it will be x to the
power n plus one. We add one to the power and
divide by the new power. So this one will be equal to this t square will be t
cubed divided by three. So three will go
with this minus t. It will be t squared
divided by likes us. So we have t cubed minus
t square divided by two, and the boundaries are 1-2. So what does this mean? It means that we will
substitute with two and this equation will
substitute with two, then minus the
substitution of one. So when we substitute with
two in this equation, we will have eight minus two k because we have two
to the power three, which is 8.2 to the power 2/2 gives us two minus
substitution of one. It will be one minus half. So N Z N D will
have 5.5 columns. Okay? So this is the amount
of charges entering terminal of a battery
between this two times. Okay? So in this lesson, we had some solved examples
on the current and charges.
5. Voltage, Energy, and Power: Hey everyone, In
this lesson we will discuss another term
in electric circuits, which is a voltage. So what does voltage mean? So in order to move
an electron inside a conductor in a
particular direction, it requires some work
or energy transfer. This work is
performed by an EMF, or external electromotive force, typically represented
by a battery. So what does this mean? So here, if you remember that we have here electrons, okay? We have these electrons, okay? So our wire, so we
would like to push these electrons toward
the positive terminal. Okay? So how can we do this? So in order to push
these electrons, we need an external force. This force is provided
by a battery. A battery provide is
the work required to push these electrons toward
the support staff time. Okay? Now this work or this EMF, or the external
electromotive force, is the force itself is represented by the
voltage of the battery. The voltage of the battery. As a voltage of the battery
provide is a force required. And the higher the voltage, the higher the force that will be provided
by the battery. Okay? So conventionally we can think about it like
this, let's say e.g. here we have the conventional current which removes it
from positive to negative. This is a conventional cat. So the higher the
voltage of the battery, the higher the force, which means we will push more electrons
inside that conduct. Okay? Okay, So here is the voltage
between two points, a and B. In an electric circuit is
the energy or the work needed to move a unit
of charge from a to b. So what does this mean? Let's say e.g. we have here our fat-free plus-minus, okay? And we have here
an element, e.g. resistance or an electric bulb or whatever it is,
any electrical load. Okay? So in order to move
electrons, let's say e.g. electrons from a to b, we need a force. This force is provided by what? This force, all
the work required to push these electrons through this element from a to b
is called the voltage. So the voltage is the force
that will be provided all the work done to
move at charge e.g. let's say e.g. we have
a positive charge, whatever it is, opposed to
push it through this element. Okay? So the work required
to move from a to B is representing
the voltage. So the voltage which are,
as you can see, VAB, which is voltage of a minus
voltage or be zoster members. Okay? Because we will need
it later in this course. So here we have a is a positive terminal and
b is a negative term, then a is supposed to
have tenement means that the current will go through a through and through this element and goes
out from B. Okay? So plus, minus or
the potential here, it's called sometimes a
potential difference or voltage or the front
is called the VAP. So the voltage or the
potential difference, is the energy required to move a unit of charge
through an element at, as measured in voltage. Okay? So here if we have amount of
charge one and charge e.g. you wanted to charge, you wanted to charge
means one column, one column, one column
of the charges, which is equivalent to a
large number of electrons. If I would like to push
this large number of electrons from a to P. This representing the
voltage, the work, or the energy required to push this amount of electrons
from a to P. Okay, Hobbes idea is clear. So from this definition, VAB, or the potential
difference, is equal to d Omega over d q
naught omega d W over d Q. Or we can say war over Josh. Okay? So as you can see, energy required, energy or work. You can see energy or work. So as you can see here, work to move a unit of charge. So the amount of energy required pair column divided by q. So you can see 1 v, what does evolve to mean? It means one joule per coulomb. Joel is also a newton
meter pair column. So 1 v means we need one joule for each
column of a charges. Okay? So let's say if this
element requires ten volt, it means that it needs
a ten joules amount of energy needed for each column, for each column,
or one column of electrons to move
it from a to B. Okay? Here's this figure will help you understand what is the
meaning of voltage. Here we are using
Zack conventional. Okay, so forget completely or
assume that, let's say e.g. here we have a
battery plus minus, okay, with a certain voltage V. So we will always sync for now and until the end
of this course, we will think about the current moving from positive
to negative. Like this. The flow of electrical charges
from positive to negative. So you can see this
is the electron flow. Conventional,
conventional current, not the real current convention. Okay? So moves it from
positive to negative. So this here, between this point and this point there is that
potential difference. So what does this mean? It means that this one e.g. have a zero volt
and this one e.g. has a two volt
potential difference with initial difference, which is the voltage
will be two minus zero, which will give us two volt. Okay? Usually is a negative z usually connected to the ground
which has a zero voltage. Okay? Now, in order to understand
what a voltage you mean, you can look at this figure. You can think about
what's the difference in voltage as a
difference in height. So as you can see,
this is the height of water from here to here, okay? And you can see this
is a flow of water. Now when this height
increases, the flow increase. Similar to here. If you
look at this figure, Let's say e.g. here. If you look at this figure here, we have this,
electrons are flowing. So higher voltage means the
higher difference in height, which means higher
flow of water, which means higher flow of
electrons or higher current. More energy that will be, more energy will be
provided for electrons, or more energy will
push these electrons. Okay? Similar to E. So how large the difference in height means a
larger voltage. You can see it's a
flaw of water is high. But when that
potential difference is low or the difference
in height is low, you can see that the
flow of water is small and the flow of
electrons is small. Okay, look, now wins. This one changes. Look at here and here. You will see that this
electrons will move like this, can seem much faster and
the bulb is much brighter. Why? Because more electrons
passing through. Okay, So this is
called the voltage. Now, when we have an
element Example here, plus minus nine volt and
the negative nine volt, this one is similar to this one. How was this simply VAB? We are looking for
the voltage V A B. So when we say VAB, it means VA minus VB. Vb. Voltage VAB means voltage
of a minus voltage of B. You can see here a has a host of sign and the
B has a negative sign. So you can see
here positive with a and the negative was, okay. So nine volt, which is VAB, it means that a is higher
than B by nine volt. Okay? Now let's look at this one. We have here also
VAB, VAB. Okay? So we have here the negative
sign selected for a. So we will say negative and
the positive sign with B. So it will be plus Vb, which is equal to negative nine. Here. If you take here negative
as a common factor, so you can say
negative VA minus VB. So negative VA minus VB
equal to negative nine volt, which means that it's
equal to negative VAP. So from this equation
you will find that VAB is equal to nine,
similar to this. So these two representations represent this as same voltage. Okay? Okay, so here you can also think about
it in a different way. Mine voltage here, which is the difference between
two voltages, VB. Here's a positive sign with b. And the negative sign
is a equal to 9 v, which is negative VAB, which is negative of mine. So VAB is equal to nine. Okay? So in the MD will have VAB. What does VAB mean? It means VA minus VB, okay? Vb is equal to VB minus v. So the first VBA means
B first minus second, a first minus second. So if you look at here, you will find that
VAB is equal to negative vb, okay? Okay. So what is energy and power? So you will find that we
have current and voltage, which are our basic variables inside our electric circuit. However, they are not
sufficient by themselves. We need more representation or more definitions that will help us understand
electric circuits, such as energy and power. So usually if you always
hear that we have a bulb or 400 watts or any
electrical device with a certain wattage. What does that mean? It means power. So power of the element of
power required by the element. So e.g. we have 100 watt and we have a 60-watt
and usually hundreds, what means more power? So it gives them more
light than the 61. Now, when we pay our bills, we don't pay in wattage. If you know any
electricity bill. It is measured in kilos. What our okay. So we have kilowatt, which are representing
the power. And our itself is our time, which means that we have
power multiplied by time, which means a kilowatt
hour, representing energy. So when we pay, we don't
pay for our water devices. We pay for the energy consumed. Energy consumed in one month, e.g. how many kilowatt-hour? Okay? Okay. So those power
and energy calculations are important in
circuit analysis. So here, what is power? Power is the rate at
which work is done. And it's measured in watts. So as power is equal to
work done over time, since it is a rate, remember rate at
which work is done. Okay? So we can say, since we
are talking about trade, we can say that power is
equal to d Omega over d t, d w, sorry, not omega
d W over d t here, W representing the
walk done. Okay? So the war codon with
respect to time, this is a general definition. This is if work is a constant value and with respect to
in a certain time. So finds that power
is measured in what? So one watt is equal to war, which is measured in
joules, divided by time, which is in second
one What representing one joule of work done in 1 s. So we have power, which is
work divided by time d w, d t, rate of exchange, or rate at which work is done. Now, if you remember
that we dw over d t, If we multiply here by
dq and divide by d q. If you look at this equation, we have dw over d t, dw over d t. You can multiply it
by a DQ and divide by dq as if you
didn't do anything. So when you do this,
you have dw over dq. Dq over d t here representing the war required or the
energy required pair column, which is what,
which is a voltage. And this representing
the rate or the flow of electric charges per unit
time, which is the current. In z and you will
find that power of any advice is equal to
voltage multiplied by current. Like this. Now, here we have
something to understand. Let's say I would like we have an element here and I would
like to know the power. Okay? So here the power is equal to voltage
multiplied by current. Okay? So the voltage here is a
polarity here is selected. I select the polarity
as I would like. I can make, let's say e.g. we have a resistor which we will discuss if I would like to
make it like this, okay? If I would like to make this one is positive and this
one is an active, as you would like. It's your choice. Unless it is defined
in size of problem, it is your choice to select whatever signs you would like. Let's say we select a
positive, negative. And the current is flowing like this going through the element. So the power will be equal to the voltage multiplied by the current entering
is a positive sign. So is the current entering? Yes. So it will be all. Now, if we look at this one, you will find that
the power is equal to voltage multiplied
by current, right? However, the word current, what current is the current
entering Zappos design. However, here the current is leaving the element.
What does this mean? It means this one is negative. So when the power is positive, it means that this element, such as a resistor or
an electrical load, it absorbed this power. If the sign is
negative like this, it means it is supplying power. Now, let's make
this point clear. So if you have e.g. a. Battery like this, and you have a resistor such as
a pulp or whatever. And we have this signs plus minus and this element
as this supply is 2 v, the current is flowing
like this, okay? Current going from the
battery through the loop. Okay? So let's say this
current is equal to one and bear for simplicity. Okay? Now, let's see what
is the power of this battery and what's the
power of this resistor? So if you look at the battery, power will be equal to voltage multiplied
by current, right? So what is the value
of the voltage and voltage is 2 v, okay? Now is the current. So we have here positive, negative, similar to
this case, balls, the negative ends are current, leaving is above step leaving as opposed to going
out from suppose them. So it means it is
supplying power. So it will be negative. So it will be multiplied
by a negative. How many amperes? One and bear. So to be negative two. Now, let's look at this resistor
similar to this element. You will find that power is equal to the voltage across it. The voltage across
it is two volt. And we will understand
later in the course, why is this multiplied
by current? So you can see the current
is entering, entering. So it will be plus one and bear. You can see entering
positive so it is consuming or absorbing power. So it will be almost the VI, so it will be two to one. So what does this mean? It means that the
battery negative two, what it means applying electrical
power and that resistor or the load is two
what it means it is consuming electrical power. Okay. Hope it's clear here, similar to what I
just said right now. Here, as you can see here, the positive and the negative current
entering is positive. It means it is consuming. So it will be positive sign
for all multiplied by three. Okay, Well bought this one. If you look at it, if you see this current is
moving like this, like this, like this, like this. So in the end, the current is entering the ball
step similar to here. So it will be four multiplied
by three here with a supplying power
and elements such as a battery current going
out of the post here, e.g. like this, like this, like this current
going out of support. So this element is
supplying electrical power, supplying electrical power,
so it will be negative. Okay? So here is the passive sign
convention is satisfied when the current enters through the positive terminal
of an element. So B will be plus V. So when's the current entering
the positive terminal? It means it is consuming. So it will be plus VI if it is entering the negative
terminal, similar to here. So it is all leaving from
the positive terminal. They are the same. It will be negative or
supplying electrical power. Now we have to know that
the law of conversation, conservation of energy must be obeyed in any
electric socket. For this reason, the
algebraic sum of power in an electric circuit at any
instant of time must be Z. So some measure of power at any instant is equal
to z as unexamined. You can see here we have, let's delete all of this first. So we have here an element like this supplying
electrical power, negative two all connected to another element like
this, like this. Plus minus. So this is a consuming similar
to this one, also 12 volt. So the summation of this 212
or -12 is equal to zero. Okay? So the total voltage
inside the circuit at any instant must be equal to z. Now what is energy? Energy is the
capacity to do work, and it's measured in joules
or what pair or what second. Okay? So remember that energy
in general is equal to power multiplied by time. This equation is used
when power is constant. When power is constant, similar to this equation here. Here, power equal to walk over time when work is
a constant value. Similar to here, this
equation, that derivative, if the work itself, is it changing with time. Remember this? So here we have energy equal
to power multiplied by time. So power is what, what n time is equal to second. So you can see what second. As you can see here, all Joules. Energy is about applied for
a certain amount of time. Here in general, you can say, is that energy or what is the integration of power with respect to time
b multiplied by t. Okay? So integration here,
b multiplied by T from T naught to any time t. And the power is
equal to voltage multiplied by current d t. Okay? This equation, this equation is used
as a general equation. If e.g. the voltage and the
current is constant values, then we say power is
win-win Zen we say energy is equal to power multiplied by time or
v multiplied by t. The integration is used when, when power or
voltage or current, any of these two is
a function in time. If it is constant as n, We don't need to do
any integrations. Now the final point here is
that you have to note that the electrical power
utility companies measure energy in what hours
or what are where? One watt hour equals
3,600 Joules. Okay? So as you can see here, we said joule is equal to 1 s, 1 s, 1 s. So let's say you would like
to convert seconds here, which is our time, into what? Our. So how can we do this? Assembly takes a
second and multiply it by 60 to convert
it to minutes, lauded by 60 to
convert it into hours. So 60 multiplied by 60 is 3,600. So what our is
equivalent to say 1,600. What second? And what second is
a joule similar to 1,000 x only Joule or so
thousand 600 watt second, which is equivalent to 1 h.
And this is a unit which we use in our measurement
of electricity. Okay?
6. Solved Examples 2: So now let's have some
examples on this. Owns the energy and power, voltage, energy and power to
understand these concepts. So we have an energy
source that forces or provides a constant current
of two and bear for 10 s. So to embed is our current value of the current is equal
to two and payers. The energy source, e.g. a. Battery gives a current of two amperes for time
equals to 10 s. So the time is equal to ten. Second. For flow through a
light, a light bulb, if 2.3 kilojoules as given off in the form of
light and heat, energy. Finds a voltage to drop
again across as open. So here we have 2.3 kilo joule is given off in
the form of light and heat. This is our light bulb, which gives us light
and heat energy. So this energy is
2.3 kilojoules. So we say that our energy
is equal to 2.3 kilo joule. We said before that kilo is equivalent to tan
to the power three. Okay? So we have here energy, we have currents, we have time, we need the voltage. Okay? So if you remember that
energy is equal to voltage, let's say power first
multiplied by time, right? Or V multiplied by time. So we have energy
to 0.3 kilo joule. And we have voltage is
the one which we need. And we have
guaranteed to impair. And we have time 10 s. So from here you can get the
amount of voltage, okay? So as you can see here, the voltage total charge. This is another way. Instead of, this is
a method which will exhaust, think about, okay? So you will find that here
from this equation. Let's see. At first, we have
voltage equal to energy, which is 2.3 multiplied
by ten to the power three divided by two multiplied
by two multiplied by ten. This will give us the voltage. Now if you look at
the voltage here, which is the final form
in this other method, you will find 2.3 multiplied
by ten to the power 3/2. Okay? Now, what does this method? It is the same idea. You can do it in another way. We know that the
voltage is equal to change in water or
energy bear column. So here we have energy which is 2.3 kilo joule and we need to Q, amount of charges is equal to the current multiplied by time. Okay? Now where did we get this? Remember that current
is equal to Q over t. So from here, Q is equal to
current multiplied by t, multiplied by t. So we
have two multiplied by ten gives us 20 columns. Using this one, we
will get the voltage. All of them will lead
to the same result. Okay? Now let's have another example. So finds a power delivered to an element at time equals
three milliseconds. So we need the power
at a certain instant. It was a current entering. It's a positive terminal, is equal five cosine
60 pi t and bear. And the voltages
in the first case, we have v equals three. In the second case we
have v equals three, d over d t. Okay? So first we need
here is the power. So the power simply equal to v multiplied by voltage
multiplied by the current. So the current here is
five cosine six t by t. This is the equation of current. Now what is the value
of the voltage? Voltage is in the first case, three is three times this value. In the second case it will be three multiplied by
derivative of this guy. So let's do this first, we will get the voltage. So in the first case we
have v equals three. So the voltage is
V equals three. I is three multiplied
by this column to give us 15 cosine 60 pi t. And the power equal to voltage
multiplied by current. So we have the voltage
which is 15 cosine 60 y t. And we have current which
is five cosine 60 Whitey. They're multiplication will
give us cosine square. Phi multiplied by
15 gives us 75. Then what is the next
step? Very easy things. The spine, which is
three millisecond, and substitute it
here like this. Okay, we substitute with time
equals three milliseconds. So we will have this
amount of power. Second equation is
that V equals three d over d t. So we'll get the
voltage same as before. Three, d over d t is the derivative of current
with respect to time. So we have here our current. So we have three first, we have three here. And the derivative of current
with respect to time. So we have here our current. This is a constant value, so we would keep it as it is. Okay? Then we need that d over
d t or the derivative of cosine 60 Paul t. So the derivative of
cosine is negative sine. We have negative sine t by t, sine t. Then multiply it by
the derivative of the angle, derivative of cities
to buy tea is sick, is t. So we have 60 pi here. So this multiplication
will give us negative 900 pi
sine two Pi t volt. This is our voltage. Now what is the value of power? Power will be this voltage
multiplied by the current. So we have this equation, and this equation of
current takes us one, the blood by this one you
will have this equation. Then what are we
going to do assembly, we are going to get the
power at three milliseconds. So we'll take this
three millisecond and substitute in t here. And in this t like this. So we will have in the end power equal to negative 6.3961. Okay? So what does this mean? It means that at this, this case, it is positive. It means that this element is
consuming electrical power. So here in this case, p equal to negative 6.396. It means that it is
supplying electrical power. Okay. Another example here. How much energy does 100 watt electric bulb consume in 2 h? Very easy example. This one is the
easiest one of them. So energy is equal to power
multiplied by time. Okay? So what does the value of power? Hundred watts multiplied
by that time? What does our time 2 h. So the energy consumed
200 kilo watt hour, kilowatt hour, no known
as there is no key here because it is only what? So it will be 200 watt
hour or we can say 0.22 kilo watt hour. Okay, ten to the power three, so 12.3, 0.2 kilowatt. So as you can see here, not the same answer, okay, I will tell you now why. Here, what are all the energy is equal to power
multiplied by time. So power here is 101. And time 2 h. If you multiply these two, you will get 200 watt hour
or 0.2 kilowatt hour. That is the first solution. Second solution, if you
would like it in joule. And instead of what our, I would like it in
what second or joule. So in this case,
you will convert that 2 h into seconds by multiplying this by 330,600
or 60 multiplied by 60. Okay? So you will have 7 ω, 20,000 Joule or 720 kilojoules. Okay? So this is similar to this one. No difference except
that this one is kilojoule and this
one is kilowatt-hour. Okay? As you would like, says same. The same once again, as you can see here is the same as what equal to power
multiplied by time 101 multiplied by 2200 watt
hour or 0.2 kilowatt hour. Okay? So in this lesson we
discussed the some examples on the energy,
power, and voltage.
7. Dependent and Independent Sources: Hi, and welcome everyone to this lesson in our course
for electric circuits. In this lesson, we will discuss different types of
elements in general. And we will discuss
the difference between an independent source
and dependent source. So first, what are the
different types of elements? So you will find that we have two main types of elements that are found
in electric socket. We have the passive elements, and we have the
active element. Okay? So what is the
difference between passive elements and
active elements? And active element is an element that can
generate energy. However, app as if
element is not. Okay. So an active element can be a
source of electrical power. However, a passive element
consumes electrical power. Examples of that
passive elements are resistors, capacitors,
and inductors. These elements will be discussed in details
inside our course, the resistors, the
capacitors and inductors. Then we have active elements
such as generators, batteries, and
operational amplifiers. Operational amplifiers will
be also discussed in detail. So these are active element. Act of parliament can means
that this element needs, can provide electrical power
or can provide energy, or it does not operate unless there is an
electrical source. E.g. the operational
amplifier needs a supply or a voltage source so that it starts to operate or do
the function required. As we will see
inside this course. The resistors, capacitors, and inductors does not
need any source. Use us to add it to
the circuit and they perform a certain function,
as we will see all. Now, let's discuss
the difference between independent
and dependent source. But before we do this, you have to know
that these elements will be discussed in
class. Don't worry. This is just an overview
about the difference between the active elements
and passive element. Okay? So first, what is an
independent source? So an ideal independent
source is an active element, Is that provides a
specified voltage or current that's
completely independent of us, our circuit elements. So it is an independent source. It means it provides voltage or provides current regardless
of the circuit elements. As an example, here we
have a voltage source. You can see this one. This is a voltage source, and this one is a
voltage source. Okay? Now, this voltage
source is called an independent source
when you see is a circle or you can
see this figure, it means that this
one is independent. What does this mean? As an example? If this one
has a voltage equal to, well volt, or this one has
a voltage equal to 10 v. It means that this source, such as a DC voltage
or a pottery, e.g. or this source. What does this mean? It means that these
two sources will provide that well
volt or the ten volt, regardless of the other
circuit elements. So it does not depend on
Zach currently flowing in the circuit or any other
voltages or any other elements, it provides us that, well volt or Gibbs
is at ten volt, regardless of the other
circuit elements. Okay? Now similar to the
current source, you can see this simple, this circle and
inside it, an arrow. It means this arrow gives
us a direction of current. So as you can see here, here, the arrow is going up. It means that the current is
coming out from this spot, like this coming from here. Then it goes through the
circle and it comes back here. Okay? So this is called an
independent source. As an example, it will
provide five and bears. Okay? Or equal five and bears, it means it will
provide a constant. The value of current. Independent tones are voltages or currents inside the circuit. Here, similar to this one. Here you can see that we have some positive and negative
terminals post have negative. It means that current
will go out from supposed current going
out from the post. Okay? Now what is dependent source? It means that our
dependent source, or a controlled source, is an active element, similar as independent source. However, this element
is the bend and What does this mean?
It means it depends on other elements of
our circuit, e.g. it depends on the voltage or
the current of the subject. As you can see here, e.g. you can see this as
a dependent voltage or a dependent current source. This one is this
volume and shape. This diamond-shaped
means it is dependent. Dependent. If it is
a diamond shape, if it is circle, then it is an independent
source. Okay? Now, what does this mean? As an example? This voltage dependence on other elements inside
our cells, e.g. you can see here, here four
types like this, okay? E.g. you can see
this circuit, e.g. let's say this one. Let's see this one first. Okay? We have a voltage source
here with a certain value, but you can see a circle, let's say it is 10
v. It means that it will give 10 v between
this point and this point, we have a ten volt. Regardless of other
elements inside the sack. This one is independent source. Okay? However, if you look
at this one, e.g. you can see this one is a dependent source.
Now, why is this? Because it has a diamond shape. Now also, why it is dependent? What, what element
does it depend on? It depends on the voltage. So as you can see here, you can see, is this
a current source? It's a value, this
current source, it's a value equal to b. Multiply it by v x. B is a certain constant,
let's say e.g. to a certain value. So it means that the
current is equal to two multiplied by v x. Okay? Now, whereas vx, vx is a voltage drop
across the resistor as the voltage are required
to pass an electron from here to here, okay? To pause this resistor. Ok. So in order to pass
this resistor, we have Vx, our voltage
required called the v x. Now, this x can change dependent
on the circuit itself. Okay? So as you can see, this
current source is equal to two vx as an example. Now, it says if this voltage
is equal to five volt, it means that the
current will be equal to two multiplied by
five, which is ten and Ben. So this current source will provide what will
provide ten and Ben. If this vx is equal to, let's say 1 v e.g. you will find that
the current will be equal to two multiplied
by one equal to omega. So as you can see that the
current source value itself, it's the current source value. It changes depending
on the value of v, x. So it is called a dependent
source in the Bendis. It depends on us, our element inside our circuit. So as you can see,
a dependent is an active element in
which is a source, a quantity, the value e.g. of current is controlled by
another voltage or current. Okay? Hopes ideas clear. Now we will find this one, e.g. you can see here one circle
and another one here. You can see this one is a
voltage source, 20 volt. Since it is our circle, it means it is
independent source. It does not depend on any
other element of this act. However, if you look
at this one, e.g. you will find that this
one is ten and pears, which is also independent because it is a circle and
it has a constant value. However, this to
our diamond shapes, diamond, diamond,
which means they are dependent source, e.g. the first one here
is equal to 0.8 VAP. It is a current source. This auto means this plus
minus means the voltage. So this arrow means it
is a current source. Its value depends
on the voltage, VAB, voltage between a and B. This one is point 5ix. This plus minus means it is a voltage source and its value
depends on the current IX. So here you will find
something interesting. Interesting here is
that you will find that we have four types
of dependent source. Okay? It is really easy. So you can see a voltage
controlled voltage source or current controlled
voltage source. Voltage controlled
current source or current controlled cancels. So let's understand what
does this even mean. So you can see for us
that we have number 1.2, voltage source and
voltage source. Like this one, this is
a voltage source plus minus a voltage source
like this. Okay? So the first one, N
something voltage source, something voltage source. Second, 13.4, you can
see current source, current source like this
diamond shape current source. And this one is also
a current source. Okay, now let's see
the first sentence. Sentence here. You will find, is it a voltage
controlled voltage source? So it is a voltage source. It's, a value is
controlled by a voltage, by another voltage
in the circuit. What does this mean? As an example, you can
see this one, e.g. let's say it's a value
v is equal to 0.5 VAB. You can see it is
a voltage source plus minus our voltage source. Controlled by a voltage, a voltage controlled
voltage source controlled by a voltage. You can see this voltage source of value is controlled
by another voltage. If this one is a
current controlled, it means that
instead of this one, instead of this,
we can say e.g. I. X. So it is a voltage
source plus minus. It's a value, is
controlled by account. It's a value, it's
controlled by a car. I hope it's clear. For the same idea for
the current sources. You can see a voltage
controlled current source. So it is a current source like this one controlled
by a voltage, voltage controlled
current source. It's a value controlled
by a voltage, a current controlled
current source. It means that like
this one, e.g. let's say this current
source is 0.8 RAX. So it is a current source. Control the y
account by account. Okay? Okay, so let's
have an example on the dependent and
independent sources to understand more
about this elements. So here you have a very
simple electric circuit. We have a voltage
source of 20 v. We have an element here
is that consumes 12 volt. We have another element here
that consumes eight volt. We have here a current source. Its value is 0.2 I. So first point, you can
see we have a circle here. It means that this element is
what is independent source. Okay? It's a value is constant, independent of any other
elements inside ourselves. Okay? If you look at this one, e.g. you can see it is 0.2. It is our volume and
shape which is dependent. Okay? And its value is
dependent on a current. What current? Current? Whereas I, I is here, I'm here. Okay? So its value is point to this. Oh, okay. Okay. Now we need to find the power supplied or
absorbed by each other. So let's just start. We have B1 first element here. Okay? So B one equal to. So we have the positive
terminal here, and then we have the
negative terminal here. And we have the current
flowing like this. I would from the positive
side equal to current, equal to five and bears
going out from suppose them. So since as I can, and if you remember sensors, that current is going
out from the positive, it means it is supplying
electrical power. Okay? So in this
case the power would be negative, negative what? V. So it will be equal to
negative as a voltage 20 volt. Multiply it by the value
of current, which is five. And there's so we
will have like this, we have that p1 equal 20 multiplied by negative five
equal to negative 101. It is supplying power. If it is consuming power
is in, it will be posted. Since it is supplying or going out from suppose
that terminal, it is supplying
electrical power. Okay? Now let's look at
the second element. This one, you can see
a positive, negative. So the power is equal to V, the voltage and the voltage across it across this
element, which is at 1 v. And the current
going through it, which is a five and back. Five and back. Okay? Now, what is the value of
what is the sign here? Is it bolsters non-negative. Now, you can see that
the current entering is a positive sign. So it means that this
element is consuming. So it will be posted. So the power consumed by two is equal to 12 0
multiplied by five. As you can see, 60 watt
which is absorbed by what? Absorbed power. Why? Because the current entering
is a positive sign. Here is the current
leaving the positive sign. So this one is entering, so it will be consuming
and this one is leaving, so it will be supply. Then we have the
third element here. Also again, you see the current entering
is a positive sign. So it means the power
will be equal to Austin. Since it is consuming electrical power, absorbing
electrical power, voltage multiplied
by the current, like this, eight
multiplied by six, which is a 48, what? Absorbing power. Then we have the final
element which is before. So let's delete
all of this first. So we have b for which is v
multiplied by i ends of sine. Now if we look at here, this element is a
current source. Current source. So what does the value of current sources is equal to 0.2, 0.2 multiplied by the current, whereas the current I is
equal to five and bears. So it will be
multiplied by five. So this is the value of
the current Going out. Okay? Value of current going out. So this will be equal
to the current, which is this 1.2
multiplied by phi. Okay? Now we need the voltage, okay? You will find that something here that the voltage
of this element. So let's say we have
here positive, negative. Now we will find that
the voltage across the current source here is
equal to eight volt two. Now, why is this? Because these two elements are
in parallel to each other. You can see this branch. And this branch is
parallel to each other. So the voltage here
will be equal to the voltage here will
be equal to eight volt. Okay? Now, the current, as
you can see here, leaving the positive sign, so it will be negative. So we will have the
power four equal to negative eight multiplied
by 0.8 multiplied by five. Okay? That is the
first solution. Second solution you can think
about is that Let's delete this, like this. Okay? You can think about here
20 volt and 12 volt. So the difference
between them is 8 v, which is here and here. Same ID. Okay? So you can see before
is equal to eight, which is a voltage multiplied by negative 0.2 multiplied by
five, which is negative 81. Okay? Don't worry, we will, we will discuss this again. You can see that yEd vote
because they are parallel. We will learn in this course about seniors and the
parallel circuit. So don't worry about anything. Okay? So here we have the
supplied power, we have absorbed the power. Now we need to do a
small check on this. So simply, we can
say is that in order to make sure that we are satisfying the law
of conversation, we need to conservation, we need to make sure that the supplied power equal
to is absorbing power. So you can see that
the supplied power, negative 100 plus negative
eight is negative 108. The consumed power
can see 60 plus 48, which is also 108. What? So the summation of these two, you can see some mention of
all powers is equal to zero. Okay? So the total supply
power equal to the total absorber power
supplied power BI is a 20 volt and the current
source is equal to the q power consumed by
these two elements. Okay? So in this lesson, we discussed the dependent and
independent sources, and then we had a
small example on this.
8. Cathode Ray Tube and Electricity Bills: Hey everyone. In this lesson we are going
to have some examples on the applications of
the basic concepts that we discussed in this
section of the course. So first, we have
application number one, which is called Zach
cathode ray tube, or the abbreviation is
CRT, which is this one. As you can see here,
this cathode ray tube is used in all the TVEs. Okay, So if you are from
the old degeneration that had TVs like this one
or a screen like this one. Bc screens like this one, or EPC monitored to
be more specific. You will find that this is a
very big, large background, or the back of this
monitor is very large because it is working on the principle of the cathode
ray tube, which is this one. So what is an important
application or one of the important
application of the motion of the electrons is found in both transmission and the
reception of TV signals. Ads or transmission
and TV camera Reduces, Reduces as seen from an optical image to
an electrical signal. So here we have that TV camera which takes an optical image. Then it converts the
optical image into electrical signal that is
sent to our homes, e.g. at the receiving
end at our home. The image is reconstructed by using a cathode ray
tube, which is this one. So how does that
cancelled rE two walks. First, we have here an
electro gun or electron gun. It is maintained at a high potential
fires and electron p. So here is this one has
a high potential, e.g. at 25 kilo volt, very high voltage that
fires and electron beams. You can see this, this
electrons, you can see this. All of these electrons are
fired by electron gun. It provides us a larger
beam of electrons. Okay? Now when this electrons
move like this and hits a fluorescent screen, we will have a bright spot. So as you can see here, we have when this electron
beam hits as a screen, we will have a dot like this. Then by doing this
several times, we will have adult here, another one here, another
one here like this. Then we can form any shape
we would like e.g. a. Circle, e.g. so when
we look at the screen, we will have a circle like this, which is format from
larger number of electrons when z hat
floors and screen. So when electrons hit
the fluorescent screen, we will have a bright spot. Now, how can we control if it is here or here, or here or here? In which point, while
using a deflection plates. This deflection plates or
horizontal deflection plates, which controls the horizontal
motion left and right, like this here or
here, or here or here. And we have a vertical
deflection plates which controls up and down. Is it here? Anywhere over here
and anywhere here. So we can select any
point on the screen, we will, as we would like. So here, the beam passes
through two sets of plates, which is a horizontal
and vertical plates that are used for vertical
and horizontal deflection. So that's a spot on the screen
where the beam strikes, can move right and
left and up and down, right and left using the horizontal deflection and up and down using the
vertical deflection. When the electron beam strikes
the fluorescent screen, it gives off light at this spot. Okay? So by using this idea, we can draw any figure
world like the beam can be made to paint a picture on the TV
screen like this one. Okay. Now let's have us
symbol example on this one to understand how
does an electron beam, or how can we deal
with an electron beam. So we have an electron
beam which is a source of electrons or the electron gun. It provides this electron
using a certain voltage. By applying a very high voltage, we can accelerate electrons
and provide an electron beam. Okay? So as you can see,
an electron beam in a TV picture tube occurrence at ten to the power 15
electrons per seconds. So we have number of
electrons is equal to. Ten to the power 15/s. As our design engineer
determines a voltage V naught, we need to find the
voltage required to accelerate the electron beam to achieve a power
of four watts. So we need the power of four 1s. So how can we do this
assembly? You can. What is the relation
between voltage and power? You know that the power is equal to voltage
multiplied by current. The voltage is equal to power, which is for what? Divided by the current. Power divided by current. Here we have for what? We need to find the
value of the current. So if you'll remember
that current is equal to dq over d t or delta
Q over delta t. So in 1 s, we will have how
many charges does a number of charges is equal
to number of electrons. Multiply it by the charge, the value of one electron, which is 1.602, and so on. And number of electrons
is ten to the power 15. From here we can
get the current. So let's these sees
us in more details. So the charge on an
electron is E equal to negative 1.6 multiplied by ten to the power
negative 19 column. This has a charge
of one electron. Now we have how many electrons? Ten to the power 15. So the total Q provided by this electron beam is equal to E multiplied by
number of electrons. And doing all that, the
current is equal to DQ over DT and the q is
equal to n. Okay? So it will be in Z. And you can see here
number or zero, charge one and charge one value 1.6 and number of electrons. So we will have the current
provided is negative 1.6 multiplied by
ten to the power negative 14 from the power. That power is equal to voltage
multiplied by current. Voltage would be equal to
a power of our current, which is 4/1, 0.6 multiplied by ten to
the bonding to four. So we would need 25 kilo volt
to achieve this function, or to achieve the fourth
power of the screen for what? The power of the screen itself. Okay. Now, another application
is the electricity bills. Okay, so usually if you e.g. this is a sheet which are
representing the Excel sheet, which are representing
the monthly consumption of a household appliances. So as you can see, we
have our water heater, we have a freezer, lighting, dishwasher, electron,
electric iron, and so on. Each one of these
devices consume a certain amount of
energy kilowatt hour, which we will find, which we find in
the electricity. So we have here kilowatt-hour, as we discussed before. So by adding all of this, we will have the total
consumption of a household. Okay? So the cost of electricity, the bands on the
amount of energy consumed in kilowatt hour. Okay. So we usually pay our
bill depending on how many how many dollars or how many pounds for
each one kilowatt-hour. Okay. So here, e.g. you
will find that the kilowatt hour in
order to get us how many kilowatt hour we get as
our wattage of any device. And multiplied it by
the number of hours, then divide by one selves. So as an example, if
we have a device which consumes 100 watts
lot such as e.g. a. Light bulb, e.g. this one is operating for 10 h. 10 h. So in this case, how many kilowatt hour
is our consumption? So first in order to, we need to multiply what
by hours to have 1 h. Okay? Then to convert what our Endo kilowatt hour
divide by 1,000, like here. Okay? So this is a general
formula to get how many kilowatt hour consumed
by any electrical device. So as an example, if our bulb of 100 watts operating for 1 h, it means it consumes hundred
what our as an energy. Or point to 1 kw hour. So now let's have an
example on an electricity. So a homeowner consumes 700
kilowatt hour in January, determines or electro
staple for some ounce using the following
residential rate schedule. So we have a base
monthly charge of at $12 and we have the first
hundred kilowatt hour per month at a sixteenths
and next 200 at Tencent and overthrow
hundred kilowatt hour at 6%. So what we need, we need
to find the electricity, but how can we do this? Forrest, our bell is equal to the price or the consumption of each one multiplied
by the equivalent price. So at the beginning
we have the pay is monthly charge, we have $112. Okay? This is the first one. Now we have our consumption
as 700 kilowatt hour. So the first hundred kilowatt
will be charged at 16 cent. So we will have plus
hundred multiplied by 16%, which is $0.16. Okay? So first we have 17
minus hundreds of 17. This is a total consumption. We will take from it the forest
100, as you can see here. So the remaining
will be 600. Okay? So the first 100 we charged
it at points because $10, then the next 200
kilowatt at Tencent. So next 200, so we'll say plus 200 kilowatt at
$0.10, which is 0.1. We will subtract from
600, 200 minus six. We will have the remaining 400. So the total 701st, 100.16 s or 200.1. So the total here,
300 kilowatt hour, subtracting it from 700, we will have 400
kilowatt-hours. Okay? Now, if we have more
than 300 kilowatt hour, after, after this one, anything greater than
surrounded kilowatt hour will be charged at 6%/kw hour. So we'll say plus. So we have a 400 which
is the remaining value. So it will be 400
multiplied by 6%, which is 0.06. Okay? So summation of all of this
will give us z electricity. So as you can see here, forest 1.16 multiplied
by 100, which is $16. Next 200 multiplied
by 0.1, which is $20. We have a 400
multiplied by 0.06, which will give us $24. Where here, where
we're here, okay. Then summation of all
of this will give us the total
electricity bill of 72. So we will find first
100, then next 200. So this total is 300. Okay? So if your electricity
bill is below 300, we are going to
use this formulas. If you are beyond
the city hundred, then you will start
getting the children there any kilowatt-hour remaining
greater than zero hundred, well, be charged at
6% as we did here. Okay? So let's say e.g. so this is a
solution or solution of this example. Let's say e.g. we would like to know the average cost for each
1 kw hour, the average. So we have the total energy consumed and the
total money spent. So if I would like to know what the average value the
average cost will be. That total money,
$72 divided by 700, which is the total kilowatt. Okay? This will give us
that the average cost of a kilowatt hour is 10.2 cents
per kilowatt hour. Okay. Let me see. 10.2 sons? Yeah. I think so. Because we have 72/700. Yeah, I think so. Yeah. Almost 0.102 and sync dollars
that so it will be Tencent, okay? Exactly, right, right. Okay. So this will give us the average cost for
our electricity, but, okay, so I hope this example
helps you understand more about what the basic concept
of electric circuits.
9. Introduction to Basic Laws: Hi and welcome everyone to this part of our calls
for electric circuits. In this part, we're
going to start talking about the basic laws. So if you remember in the previous part of
electric circuits course, we discussed the basic
concepts such as e.g. Zach, current, voltage, and
power in an electric circuit. Now in order to find
this variable z, any electric circuits,
we need some laws. We need some laws such as Zara Ohm's law and
the Kirchoff's law. Also, we need to
understand how can we combine resistors in
series or parallel. The voltage division,
guaranteed division, delta to y and y to
delta transformation. So that's what we would like
to get from this section. We would like to
learn all of these. Okay? So first we will start by talking about with
that resistance. Okay? So what does that resistance me? So does that resistance is a physical property or
the ability to resist current and it is represented Paul is a simple graph, okay? And it's the only it is ohms.
So what does this mean? Let's say if we have
a supply like this, Let's say we have
a voltage source. And this voltage source will
be connected like this. So there is a current which will flow through this circuit. Let's say the
current will be e.g. five and the value of the current that flows
through the circuit. Now, if we add an additional element knowing
as the resistance, okay, so if we add an
additional element such as a resistance like this, and it is drawn
like this resistor. This resistor or the resistance. You can see here, when
we add this element, which is having a resistance, are measured in ohms. This resistor, what does it do? It resistors the ability
of the current to flow in instead of having R5 and bear of current
flowing, we, e.g. have a current equal
to three and pair. And we will understand
how can we get the value of the current
based on the resistor? That resistor, its function
is to reduce the cost. Okay? So what does, as a
resistor looks like? It is something like this. You can see this is a conductor. The conductor which will
carry electric current. In a conductor. Any conductor in real life has a certain resistance, okay? So this cable has assaulting cross-sectional area
and certain lens, okay, similar to any
electrical wire. Okay? So this one is
called a resistor. It's an aim is resistor and its equivalent inside an electric
circuit is a resistance. So the material itself
is called a resistor. That representation
is called resistance. So we have a resistance R, which is represented
by this sample, this sample representing the resistance that
resistor there. Okay? Now you will find
that any material, any material, any cable, e.g. or any conductor, has
a certain resistivity. And it is measured in
how many ohms meter? That is its unit. Okay. What does this resistance,
resistivity mean? It is a property, property of the material itself to resist that electric current. Okay? So what is the
difference between resistance and resistivity? The difference is that the
resistance is the one which we use in our circuit analysis. Okay, So any material has a
certain resistivity. So e.g. is a good conductor. So e.g. the copper or
aluminum or silver, gold, copper, steel, seawater. All of these have
low resistivity, so they allow electric
current to flow. That's why we call
them good conductors. Other materials which are
electrical insulators, such as rubber or glass, oil, diamond, wood, paper. All of this are having
high resistivity, which means they resist
the electric current. So they have a certain value, rho, which is the resistivity. And the, when we start dealing with them in
electric circuits, we don't use that resistance. We don't use the resistivity. We use that resistance R. So we'll find that the
resistance in general, which are, which combines all of the properties
of the material, is debonding on that uniform
cross sectional area a and the length L. So
when I have a conductor, any wire in wine like this, it has a certain length and
it has a certain resistivity, which is a resistance
to electric current. And it has a certain
cross-sectional area. This cross-sectional area. When we combine all of
these three elements, area, lens, and the resistivity
of the material, we will get the resistance. What is the value of
the resistance is that we use is equal to rho, which is the resistivity, multiplied by the length of the conductor divided by area. So as you can see, as
resistivity increases, it means that we have more equivalent resistance is that we have in our
electric socket. If the length of the
conductor increases, lens increases, we will have
more resistance, right? However, higher cross
sectional area, larger cross-sectional area,
bigger cross-sectional area, means that we will
reduce our resistance. Okay? So why is this important? Because if you look
at any electric, keep, any electric
conductor or a cable, you will see that as the area
of this k will increase, it can carry more
electrical current. Now why is this? Because more area means
lower resistance, which means that we have
more current that will flow. Okay, That's why
it's a big cables, indicates that we have a
large amount of current. A small cables mean
we have a small cat. Now, here is a table that
shows you the values of the resistivity of
different materials, e.g. silver, copper,
aluminium, and so on. It's value, as you can see here. If you compare e.g. e.g. if
there's a couple ways, e.g. glass, you can see that
the resistivity of copper is 1.72 multiplied by ten to the power
negative ten, which means 1.7 2/10
to the power eight, which means a very,
very low resistivity. That's why copper is
used as a conductor. That conductors electric current or allows the flow
of electric current. If you look at
something like glass. Then to support, we're at 12
is ten is one end episode. 12 zeros, okay, 12 zeros, which means very
high resistivity. That's why it's a glass
is used as an insulator. Between these two materials. We have carbon,
germanium, and silicon. These three materials,
or don't have low resistivity and don't
have high resistivity. It is a medium resistivity. In this case, we
use them in type of electric circuits
called semiconductors. They are used as a
semiconductor, e.g. in electronic circuits. So in this lesson, we had an introduction
to that resistance, meaning of the resistance and resistivity of the material. In the next lesson, we
will talk about Ohm's Law, which will help us to
understand the relation between voltage,
current and resistance.
10. Ohm's Law and Conductance: Hey everyone. In
this lesson we will talk about the Ohm's Law. Ohm's law, what does it mean? It means that we have Ohm's Law states that
the voltage across the resistors is directly proportional to the current I flowing through that resistor. Okay? So if you have e.g. a. Supply like this, this one, e.g. let's say ten volt. And we have a resistance here. Let's say this
resistance is 1 ω. Okay? So we'll find that
the current flowing here, as we will see right now, is equal to ten and pairs. So you have to understand
that in order for the current to move
through a resistor, it will consume voltage, okay? It will consume certain voltage. Here. You can see that the current is entering here from the
positive direction, entering the positive terminal. So when's the current
enters as opposed to determine it will
consume some voltage. Now why is this in order
focus I can and to pass this resistance and continue
to the negative terminal. So if I would like to find the voltage across the resistor, what is this voltage? This voltage will
be equal to the current multiplied
by the resistance. So the current passing through the resistor,
which is e.g. ten amperes, multiplied by the value of the
resistor, which is one. Which means that we
will consume ten volt so that we can pass
through this resistance. Okay? So all of the 10 v coming
from the supply will be consumed inside this resistor
for the current to flow. Okay? So own defines a constant
of proportionality for a resistor to P is a resistance R. So what does this mean? You can see here
that the voltage is directly proportional
to the current. So V directly proportional
to the current. So this constant, so we can say V equal to a certain constant multiplied bys account
this constant two is our resistance r. Now, as you can see,
voltage equal to the current multiplied
by the resistance. Or from this
equation we can have that the resistance
is equal to V over I. As you can see, the
resistance which is on 1, ω equal to 1 v pair
one and bear, okay. Okay. So the resistance,
as we said before, it is the ability to
resist is a flow of electric current is
measured in how many ohms? So let's understand a very important concept
in electric circuits, which is the open and
the short circuit. So what does an
open circuit mean and what does a
short circuit me? I short circuit is a
circuit element with a resistance approaching
zero, almost zero. Okay? So if you look e.g. if we have this circuit,
this circuit e.g. let's say, let's say for
simplicity voltage source, a voltage source like this. Okay? So this box, let's say e.g. it is a voltage source. If we apply a short-circuit legs as a short circuit like this. Okay? Any short-circuit,
What does this mean? A short circuit means that the voltage here and the voltage here are
similar to each other. So you can see that here. This wire have the same, has the same voltage. So the voltage here,
let's say e.g. if this one is 1
v, as an example, this point is 1 v, then this point is also 1 v.
This one is also 1 v, 1 v. And since we have
also a short-circuit, or let's say a wire
with a zero resistance. So this point, 1 v, 1 v, 1 v, 1 v, and so on. So all of the voltage
is equal to each other. So we have, what does that
potential difference here? Difference between
here, which has 1 v, difference between here
which is the same voltage, 1 v difference v delta
v is equal to zero. Okay? Now why is this? Because, because we say
is that the wire itself, this solid line representing a wire with a zero resistance. So when we apply Ohm's Law, you can see that the voltage
which we need as equal to the current flowing
multiplied by the resistance. Since this wire is I have a
zero, has zero resistance. Or equal to zero. It means that our voltage
will be equal to zero, as we have seen here. Okay? So what we can conclude is that when we have a short circuit, we have a zero resistance. We say that all equal to
zero or zero resistance. And at the same time, the voltage will
be equal to zero. What, what voltage is
a difference between this point and this point? Potential difference equal to Z. From where we get
this from Ohm's Law. The same idea. If we
have an open circuit, which means that we have a very large resistance like this. So e.g. if we have a
supply like this, okay? Now, the supply,
any supply, e.g. ten volt, let's say ten volt. Supply would like to provide a current that will
flow like this and goes through this wire and the comeback to the
negative terminal. However, when we have
an open circuit, you can see this, this is a resin space between
these two terminals. Open circuit. They are not connected
to each other. So in this case we say when
we have an open circuit, it means that the
resistance is equal to infinity, very large resistance. Now as you can see that the
current or the electrons, let's say we'd like to
go move from positive, going through the air
gap to the negative. However, no current will pass zero current toolbox because
it is an open-circuit. How does the current will
go from here to here? It is an open circuit. So current equal to zero and the resistance
equal to infinity. So what is the value
of the voltage or what is the value
of voltage here? It will be equal to the supply. So you can see that this point
here is equal to ten volt. This point is equal to z. So we have a voltage equal to 10 v and resistance
equal to infinity. So let's say I would
like to get the current. So the current will be v over r. So ten divided by infinity, divided by infinity
is equal to z. Zero current will flow. This is using Ohm's law. By logic. Logic, you can see that we have all of this wire
is the same wire. Zero voltage, YouTube. No voltage is consumed because
it has a zero resistance. So we say that the voltage
drop is equal to zero. Here we have an open circuit. So open circuit means that no
current will flow by logic. So r equal to zero, or we say that the
resistance here is very, very large, approaching infinity because it is an open circuit. Now let's talk about
another thing which is a fixed it and the
variable resistance. So resistance itself
can be fixed it, e.g. like this one or this one. This is a fixed
value of resistance. As an example, let's
say if one of them is 25 kilo ohms, any of them. Okay. What does this mean? It means that when we
take this terminal and this terminal and
connect it to our supply, like this, plus minus
connected like this. And let's say this
one is at ten volt, then the current
will be equal to the supply divided
by the resistance, which is 25 kilo-ohm. This resistance is a constant, does not change because
it is a fixed resistor. So the material itself, among the material,
the component itself, it is called resistance. Okay? So let's just explain the
difference between this stream. So you don't get confused. Okay, so we have a resistor. Resistor is that element itself, k, element itself as a
component that we use. This component, it's
an aim, is a resistor. This component has a material, is made of a certain material. The material itself,
material itself has a certain resistivity Rho dependent changes from
one material to another. Now when we combine the
row with the area of the resistance resistor itself and the length of the resistor, the lens of the resistor. When we combine all of
these three elements, we have a final representation
which is resistance. Resistance is the
one which we put in our electric shock when
we do a circuit analysis. So again, resistor is the element or the
components that we use. This is called as a storage. This one is called the resistor. Each material has a
certain resistivity, which is the ability to prevent the flow of
electric current. When we take the resistivity of the material with
the area and length, we have a final resistance that we use according
to Ohm's Law, the electric circuit analysis. Second type is called
the variable resistance. So you can see is this one. And this one is called the
variable resistance. So e.g. if you look at here at this
point or using this one, if you rotate, rotate this part. You will change the
value of the resistance. So by rotating this screw, you will be able to change the resistance so
that resistance itself is not constant,
it is variable. When we have a representation
of the resistor. That is, the story itself
is represented like this. This one is a fix-it. Fix it like this one. If this resistor is valuable, valuable it changing,
we add this line. You can see this arrow
representing that, this one or this row, this row or this
arrow representing valuable resistor or
valuable resistance. Okay, for talking about the circuit analysis,
variable resistance. The first one is used for a variable resistors
such as this one. And there is another type of variable resistors are
called the potentiometer, which when we change
as this scroll, we will change the resistance z. I have the same function off-state difference
in construction. Finally, we will talk
about the linear, linear and the
nonlinear resistors. So you will find that
we have two types, linear resistance or a resistor and a nonlinear resistance. So what's the difference between them is a linear resistance is a linear resistors
that obeys Ohm's law. What does this mean? It means that there are, at any instant is equal to
voltage divided by current. So if we take the voltage
and divide by current, we will get the resistor. An example of this, this graph. At any point, you
can see this line, constant line, its slope
representing the resistance. So at any instant voltage
divided by current, we will get the resistance. And it has a constant, the resistance and the VI graph is a straight line
passing through the origin. You can see passing
through the origin point. A non-linear resistance on
the other side and z does not obey Ohm's law and its
resistance variables account. So you will find that in the nonlinear or does
not equal v over r. It is a variable resistance. It changes with the
value of the car. As an example, if you
look at this graph, you can see that here we
don't have a constant line. We have a variable
resistance at each point. We have here our slope. Here we have a slope,
another slope, as the slope of the line. It changes the slope at any point gives us
the resistance. So you can see that
the resistance itself is changing with time. It is not constant. Okay? Okay. Finally, we will talk
about the conductance. So what does the
conductance mean? Really, really easy. Conductance is the inverse
of the resistance. So it is the ability of an element to conduct
electric current. So that resistance is the
ability or the resistivity is the ability to resist
that electric car. Conductance is our ability
to conduct electric current. So it is a reciprocal of the
resistance R, denoted by G. And its unit is more and more, which is the inverse
of home more. You can see on MOHO in
the spelt backwards. And its symbol is the
inverse of the arm. Arm like this. It is the inverse or that
is pro con of their own. And its unit is Siemens. So we say mono, the unit of conducts
mono or Siemens, or some of them are correct. So we can say is that g is one over our reciprocal
of the resistance, or I over V. And as you can see, this has units, as you
can see here. Okay? So let's just understand again. So we have resistance that prevents the flow
of electric current. We have the inverse
of the resistance, which is one over G. G here, representing
that conductance. Conductance. So we have resistance and we
have conductance. Now with the same
idea, the same idea. We have for any material
we have resistivity. Resistivity. That resistivity is
the inverse of what? Of conductivity. Conductivity. Same, we have our resistance, which is a representation in
the electric circuits we use resistance or conductance is
the inverse of conductance. Resistivity is the
inverse of conductivity. So this is a property of the material itself to
conduct electric current. This one is a probate. Two of their electric material
to resist electric car. Resistance is a
representation of that. Resistance to electric current. G is a representation of their conductance or conducting
the electric current. As you can see, HG is equal
one Siemens or 1 mol, which is the inverse of on. And as you remember
that from Ohm's law, voltage equal to I multiplied by R or I equal to V over R. So one over r can
be represented as g. So we can say that the current is equal to
v multiplied by g. And since g is the inverse
of that from this equation, you can see G is
equal to I over V. So we can say 1/1 volt, as you can see, which is
opposite to the resistance, which is 1 v over one and bear. So let's delete all of this is so you can see that the power dissipated by a resistor can be also expressed in terms of G. So we have Zika as a
power is equal to I squared multiplied
by, all right? So we can say I squared over g, like this one, I squared over g. And instead of R1 over g, So it will be I squared over g. And we have y-squared off, okay? And power equals VI, VII. Second one is this
V squared over R. So we can replace one
over r minus g squared g. Okay? Well, all of this will
lead to the same solution. Okay? So what you will
understand is that g is one over r, Okay? So if you have a circuit and you would like to get
the conductance, then the conductance
will be one over r. Okay? So in this lesson, we discussed the some important concepts
about electric circuits.
11. Solved Examples 1: Hey everyone, In this
lesson we will have some solvent examples
on Ohm's law to understand how can
we apply this law. So we have an electric iron
that takes two and bear current of two and beer
at a voltage of 120 volt, we would like to
get the resistance. So from Ohm's law, you know that the
voltage is equal to current multiplied
by the resistance. So from here,
resistance is equal to the voltage divided by current. And we have the voltage
hundred and 20 volt. We have the current
equal to embarrass. Our resistance will
be equal to 60 ω. Let's have another one. In this example,
we have current. We would like to
get the current I, which is flowing from
the supply through the resistance five kilo-ohm, and it gets back to
the negative terminal. So we need for the current, we need that conductance if g, and then we need the Power BI. So as you remember, if you look at this circuit, you can see that this terminal is a boast of
terminal is connected to this terminal of
that resistance, and negative terminal
is connected to this terminal of the resistance. So you can see that
the voltage applied through the resistor
is a t volt. So you can see that
the current from Ohm's law will be
equal to the voltage applied divided by
five kilo ohms, right? So you can see that here, the current will be
the voltage divided by the resistance or resistance. It will be equal to
six milliampere. So this is a current
flowing from the supply through the resistor
and gets back. Okay. Now, what I would like to get a second requirement is
that I would like to get their conductance G. So if you remember from
the previous lesson, we said that the conductance
G is equal to what? Equal to one divided
by that resistor, or the reciprocal of the
resistor, one over R. And we have our resistor
is equal to five kilo off. So our gene will be 1/5 kiloohm, which will be 0.2 milli Siemens. Okay? Last requirement
is that we need to find the power p watt power, power consumed in
sign this resistor. So we have three options. We have, all of them
will give us the same as your nose at the power
from the previous lessons, we know that the
power is equal to I squared multiplied
by the resistance. So the square of the current multiplied by the resistance or power will be equal
to v squared over r, which is a voltage, a square applied through as a resistor, through the resistor
divided by five kilo-ohm, or power equal to V squared g, or power equal to
square of energy. You can see all of them will
give us the same answer. So as you can see,
voltage multiplied by current I squared
R or V squared g. All of them will
give us a power consumed over 180, mainly what? Now, let's have another one. If we have a voltage source with a value of 20 sine moiety. So what does this even mean? This value of this
voltage source is called the an AC supply. Okay? Ac supply. We will discuss this in, later inside our course
of electric sockets, will find a section about
the basics of AC circuits. So this is representing
an AC supply or a variable voltage source connected across our
five kilo-ohm resistor, find the current
through the resistor. From Ohm's law, we know
that the current to voltage divided by
resistance voltage, which is two when t sine
pi t ands are resistor, which is five
kilo-ohm like this. So it will be for signed
by t, mentally impaired. Then we need to find the power dissipated or power consumed. So that's our power is equal to the square of the current, power equal to v
multiplied by I, or power equal to v squared
multiplied by the resistance. All of them will give
us the same vein. So voltage multiplied
by current. So you can see current
and voltage. This one. So for matplot by 20 gives us 18 signed by t multiplied
by sine pi t gives us sine square pi t. All of this mainly because we have here
mainly and bear, okay. This is a power consumed. We call this type of power. This type of power is called
the instantaneous power. Okay? So if you look at this circuit, Let's just illustrate this. We will discuss, of course, all of this in the AC power
analysis inside our course. So we have this source, it looks like the
census as a sine wave, it will be something like this. Okay? And the maximum value of 20, so this is a voltage
with respect to time. It is a changing with time. The current is the same
direction like this. Current. It will be like this too, but with a value of four. When you multiply these
two waves together, you will have the final wave, which is a power. The power, which is the
sign, but it's squared. So it will be
something like this, I think, if I remember
correctly, okay? So anyway, this
will give us what the instantaneous power is, the power at any given instant. So in this lesson, we had some solvent examples
on the Ohm's law.
12. Branch, Nodes, Loops, Series and Parallel Connection: Hi and welcome everyone to this lesson in our course
for electric sockets. In this lesson, we
will talk about the branch nodes and loops. Also, we will talk about with the Kirchhoff's Voltage Law, kirchhoff's Current
Law and more. Okay? So for us, we would like to understand what
is the meaning of branch node and loops,
branch node anaerobes. So we have this circuit, okay? The first one at branch, what does a branch amine? It represents a single element, such as a voltage
source, current source, a resistor, a capacitor,
and inductor, whatever it is, each element
representing a branch. So if we look at this
electric circuit, we have this voltage source. So this part alone
representing one branch is S1. Another branch, this one, another branch, branch
and the Swan a branch. Why each of these branch, because all of these rubber presenting what
representing elements, each element to
representing one branch. So how many branches
we have here? We have five branches. Remember, five prime. Okay? So what does the node mean? The node means is
that a point of connection between
two or more branches, between two or more plants. So we have this branch
and this branch. So you can see z are
connected together using a one point, a. This point is called node, that node 0 for that circuit, this one is anode, is called our point
of connection. Remember it is really, really important because
we will use this in our analysis or the
methods of analysis. Okay? So it is important to
understand this concept. So that node is a point of connection between
two or more branches. Here you can see we have all of this representing one node connecting between this branch, this branch, this branch. Okay? Why all of this is known? Because you will find
that here we don't have any elements between them. You can see all of
this one big line. So we consider all
of this as one node. And this one is another
note here is this one. See why? Because we don't
have any elements here. And it connected between the
front, the front branch. So you can see this one and this one and
this one and this one, all of them consider
the one node. Now why NYU one node? Because we don't have
any elements here. Okay? All of this one line, this 11 line here, we don't have
accepted this point, so this one is another node. You can see in this circuit
we have three nodes. Now, what does that mean? Allo is any clause
the ball in a socket. So you can see this
squared, this one. When we have
something like this, you can see one big loop. This one is considered the loop. This one is considered
as another loop. This one is also considered
as another loop. And sometimes we
have super loop. What does this open-loop mean? Like this? We can say that these
two R1 elements, so we can say this big loop, like this, or from this square. So we can see all
of this big loop. Okay? So you can see any clause the boss in a socket is
considered as a loop. Okay? If you look at any electric
circuit here, e.g. you can see we have
the main lobes. The main lobes,
you can see 12.3. Okay, So we have how
many loops that we have? Three loops. How
many nodes here? How many nodes, how many nodes? 123. So we have three nodes. And as this equation shows you the relation between branch
loops and nodes, okay? Minus one. So you can see here, three plus three minus
one gives us five, right? If you look at this circuit, we have how many
branch 51,234.5. Okay, so I hope this concepts
are clear for you now. Now we have also in
electric circuits as series and
parallel connections. So what does this mean for us if two or more elements
are in series, what does this mean? It means you have
a single node and consequently carry
the same current. So you can see here, we have a ten volt. We have 5 ω, 2 ω, 3 ω. And to ambient. Now if you look at this
supply and this one, they are cheering
OneNote, right? It is only one common. So you can see for the five ohm, it has this node
here, this node here. For the ten volt. It has this node here. This node here. Now you can see
45.10 z are sharing one single node, a single note. Therefore, you will find that they carry the same current. How will this, if you
look at this circuit, if we have a current
coming out this supply, where does this current will go? It will go like this, shows a 5 ω, same current. So same current going so 5 ω. So it means that the ten volt, see here is with the 5 ω. Okay? Okay. Now, what does a parallel
connection mean? A power connection, it means
that these elements have the same two nodes and consequently have the
same voltage across them. So as an example, you can see is that Zhi Chu, arm to arm here. And the three ohm
and Joe and payers, all of them are parallel. Now why is this? Because if you look at 2 ω, 3 ω and two again, if you look at where's
the first one, this node. And for the 2 ω, this
node, this node, for the 3 ω, this
one and this one, and this one and this one. So if you look at
that three branches, you can see that the first
node is same in all of them. All of them have the same node. In the second denote all of them also have the same second mode. So this branches have the
same first and second nodes. What does this
mean? It means that these branches are
parallel to each other. And what we can learn is that
z have the same voltage. So let's say e.g.
if I connect here, a voltage source of
five volt, like this. Since this one and this node, and this node is similar to
this node and this node, all of this is one big node. And all of this one big node, it means that five
volt battery to two unpaired parallel
to three to 2 ω. So what I can learn is that you can see is a
potential difference between this point and
this point is five volt. So it means that here difference between this point and this 0.5 volt between this
point and this point, also 5 v between this point
and this 0.5 volt and so on. So it means that z
have the same voltage. So when, when the
elements are in series, they have the same current. Same if they are in parallel, such as here, it means they have the same voltage across them. Okay? So in this lesson
we talked about with ZAP branch nodes and loops. And also we talked about the series and
parallel connection.
13. Kirchhoff's Laws KVL and KCL: Hey everyone. In this lesson we will talk
about the Kirchhoff's law, Zach KVL and KCL is a Kirchhoff's voltage law and the Kirchhoff's current law. They are really, are really important in the
electric analysis. And we use them to build a sum circuit theorems in electric circuits yourself slow are really, really important. First one, we call it
Z factors of saccharin to law and the
Kirchhoff's voltage law, or this one is
abbreviated as KCL. This one is abbreviated as KVL
for us to one which is KV, which is a KCL. First law is a KCL. This law is based on the law
of conservation of charge, which means that the
algebraic sum of the charges within a
system can not change. Or we can say in Amazon method, in another easier method is that the algebraic sum
of the currents entering anode or a closed
boundary is equal to zero. So the summation of all of the currents entering
anode equal to z, the sum of the currents, all we can say is that the
sum of the currents entering anode is equal to the sum of the currents leaving the node. All of this means is
the same thing that KCL or the Kirchhoff's
current law. So here we can see
that that total current entering
equal to the total current leaving all
the summation of all of the currents
is equal to z. Okay? So let's understand this idea. So understanding the
previous lesson about the branch nodes and loops. So the node itself
is this point. Any node in an electric socket. If you look at here, you can see that the
current entering, you can see current
entering means it is coming toward us this point. Entering this point, there is another current
leaving such as I5. You can see coming
out of Zeno's i4, entering, I3,
entering i2, leaving. So let's store to buy. The easier method is the easier method to
understand the KCL. First, you can see
that using this law, it means that the summation
of all currents entering anode is equal to summation of all
currents leaving a node. Okay? So let's try equally. So what are the
currents entering? You can see that all
you want entering one i4 entering or for entering. Only three entering. Entering. Equal to what? Equal to
the total current leaving, or E5 plus i2, five plus. Okay? So this is that Casey low. Or you can see that summation
of currents equal to zero. How can I apply this one simply, you can say equal to zero. Okay? So let's assume that
any current entering, let's make it positive. And any current leaving, Let's make it negative
to apply this one. So what are the currents
entering i1, i4, i3. So the rpoS, the IL-1
are you for and I3, which accounts are
leaving I5 and I2. So we will assign them
a negative value. So we say negative
I2, negative Wi-Fi. So you can see this one
is similar to this one. Okay? If you take this
to the other side, you will have negative
two and negative five. So this is called
the Zak ECL low. Now why do we use this? Because we would like to
analyze our electric circuit. So in order to identify is that currents entering anode
and the living anode. It will help us to get
us our values required. When we go to some
solved examples, you will understand
what I mean exactly. So you can see here I1, I3, I4, positive values
and negative two, negative five, negative
y to negate the Wi-Fi. For this one, you can
see some mention of current entering equal
to summation of current, as you can see here. So let's have another
example, e.g. if you look at this circuit, we have current sources. This can solve produce
current i1 and I2 by exists. And I is three, likes us. So if you remember from
the previous lesson, you will see that all of
this is one node, right? This one. So if I apply KCL, let's say is a total current entering equal to the
total current leaving. So I can say entering
i1, entering I3, and anyone living, Let's
make it a negative sign, negative I2 and negative I
taught all of this equal to z. So we have an equation that representing the relation
between these currents. So you can see I total plus i2. Since it's the negative signs, we can take them
to the other side. So it'll be i2, i2, i2 equal I1 plus I three. Or you can add this, take this one to the other side, so it will be I taught and
equal to I1 plus I3 minus two. Now, let's talk about the
Kirchoff's voltage law. Kirchoff's Voltage Law,
similar to the KCL. But then instead of
dealing with currents, we deal with voltages. In this law, it says that the algebraic sum of all
voltages around that clause, the boss or a loop
is equal to zero. Or the summation of the voltage drop equal to
summation of the voltage rises. So if we apply KVL
summation of all of the voltage in one
lobe equal to zero. So if you look at this circuit, we can get this or apply
this gives you a loan. What is the benefit of this law? And we tell us to
get the relation between the voltages
inside a circle. So you have two options. Let's first start with this one. I don't use this one. I don't use this one. Usually because sometimes
in electric circuits, we have elements that provide electrical power or absorbed
by electrical power. So it's hard to apply this one. However, you will
find we will apply another message that will
make it much easier. For us. You can see some mention
of the voltage drop equal to summation of
the voltage rises. So what causes a voltage drop? Resistance? Any resistance
cause voltage drop. So I say V2 plus V3. And this element,
which I don't know, let's say it makes a voltage drop equal to
summation of the voltage rises. What makes this a
voltage rise is sources. So we say v1 plus v four. Okay? Now remember, remember, this is one message saying
voltage is dropped. Some mention of the
voltage drop equal to summation of
the voltage rises. And sometimes in
electric circuits, it may be confusing. Sometimes you don't know if the voltage of the source itself is dropping voltage or
all supplying voltage. There are some electric
circuits which you cannot identify this. So what I'm going to
do in the next method, or the general methods that I am going to use in
all of the course. Here. We are applying. We have this big loop, right? This is a big loop. Now you have two options. Concentrators me please. You have two options, either to have a clockwise
loop or anticlockwise. They will give you
the same answer. But I usually use at clockwise or the general case or
use clockwise loop. So let's understand
how can I apply this. So I will apply a clock
wise loop like this. Okay? So how I'm going
to apply KVL loop? So first I will go like this. I'm going clockwise, okay? So I'm going to lie
exists facing V1 to V1. So I am going like this. I met V1. Okay. Which assign data did I see? That I met? The first signs that I
have seen is negative. Okay, so when I made
a clockwise loop, so I go here and the
meat than negative one. So I would say here negative V1. Then I will continue like this. And the phase plus forest V2. So I will say plus two. Then I will continue like
this and the mean plus V3. So I say plus three. Okay? Then I continue like this. And the meat negative
V4, negative V4. Then I will continue like
this and the mean plus V5, so I say plus v phi equal to z. So this is a summation of
all Walter's equal to zero. Okay? So you can see here,
if you look at this equation and this one, you can see that we have v2, v3 and v5, V2, V3, V5 equal to V1, V4, V1, V4, or summation of the
drop equal to the supply. Okay, So this is a
much easier method, even if you don't
know if this one is a source or a supply, you can simply
apply this message. Okay? Okay. So here we can see the summation of all voltages using the slope, you will get this equation. This one is similar to this one. So in the next lesson, we will have some examples or the Kirchhoff's voltage law and Kirchhoff's current
law to understand how can we apply these laws?
14. Solved Examples 2: Hey everyone, In this
lesson we will have some examples on
Zach, KVL and KCL. For the circuit shown, find the voltages V1 and
V. So we have a supply, we have our resistance 2 ω, and we have a resistance 3 ω. So what I'm going to do first step is that
you will apply KVL. We have learned so far about Ohm's law and KVL, KCL, right? So if I need voltages, I will apply caveat. Okay? So how can I do this? All you will assume
clockwise current, y-axis. Okay? Assuming that the current
will flow like this. So I assume a clockwise loop. Okay? Okay, Nice. So how can I write the equation assembly as we'll
learn it clockwise, right? So we go like this clockwise. We emit first negative two, negative two, and
then I go like this. And the mean plus V1 plus V1. Then Angola exist and
the meat negative v2, negative V2 equal to what? Equal to z. So we'll find the likes us. So what I did is that I wrote
it in the form of what? V1 and V2. Okay. That is the first point. Second point is
that how you will write it in the form of current. So you will have
here negative two and then the current flowing like this
through the resistor. So we say I to I plus
two and go like this. The current flowing through
the 3 ω, so it will be three. So plus three equal to zero. This equation is
similar to this one. So you can see current
going like this. So the sign will be
in the entering port. So it is entering like
this, so it will be I. So V1 will be two
multiplied by the current. Here you can see that
current going like this. So that science should be plus, minus the voltage,
let's say v three. So v3 should be equal to what? Equal to current multiplied
by the resistor or three. However, you can see we had here opposite direction,
negative, negative direction. So V2 will be negative three I1. So V2 is negative 3.1, so it will be plus. Anyway, you will see like this. So we have V1 equal to I, we have V2 equal
to negative three. You can see negative
V2 equals three. So V2 equal to negative three. Okay? So by applying KVL negative 20 plus V1 minus V2 equal to zero. And substituting this value, we will have this
equation. This equation. So the current will
be four and pairs. So if I would like
to get V1 and V2, I will take this value
and substitute it here. Like this. Okay? So again, normally we have a current coming like
this. Let's say current. So this current entering
the resistor whose sine, sine will be voltage
drop will be plus, minus, like this, plus minus the voltage drop.
What is the value? It's the value will be the 2
ω multiplied by the current, so it will be too high. So you can see
plus minus and V1, the same sign, plus minus V0. V1 will be equal to two y. Same current going like
this through the three ohm. So going like this. So it will be plus minus the voltage drop should be three multiplied by its occurrence where you are. However, you can see that V2 is opposite to the
original sign. So V2 will be negative three. Okay? Let's have another example. Same idea here, finds a voltage V naught and the
current in the circuit. So what are we going to do? You are going to apply again. The KVL likes us. In the same direction
of the current. So first, if you go like
this, go like this. Facing first. Here we have a current,
direct current, current. So this current
multiplied by 6 ω. So we can say six
and negative 12.4. And going like this clockwise, plus two V naught
plus two v naught. Going along this negative
four to zero, right? If we apply KVL. So you can see that here, the low been talking
about will be like this negative 12th for
I to V naught -4.6. Oh, okay. Now this is the first equation. We need another equation
because we have two variables. We need V naught and we need, so we have V naught and I. So what is the relation between
V node and the current? So you can see current
flowing like this, flowing like this,
like this, like this. So when the current
goes through six ohm, it will cause a voltage
drop of loss minus. So entering point is a positive
of the voltage to drop. So what is the voltage drop? Plus minus it will be six. Okay? However, you can see V naught
is always a sign that I am. I have done. Okay? So it means that V naught
will be negative six. So V naught equal
to negative six over the toes or the
direction of the current. Okay? So what will happen? We'll take this one
substituted here. We will get the current
and the forms the current, we get the voltage. Okay? Now let's have another one. We have this circuit. We have a current
source. Current source. Don't worry about this shape. We will learn about it in that dependent and
independent sources. And we have a
resistor 4 ω, okay? So we need what we need, I naught and V naught. So you can see that I
naught its direction going downwards like this. So it will cause a
volt drop plus, minus. So voltage drop
will be for nought. The direction going like this. And you can see V node has the same sign as
the voltage drop. So V naught will be
equal to four I naught. That is the first equation. Second equation. How can I get this from KCL? You can see that
this big node here, this big node here, has a current
entering and leaving. So you can see is that
for a 0.5 or inode entering three and bear
entering and leaving. So what I can say is that some mention of the
current entering, which is 0.5 plus three and bear all of this entering equal to
current leaving, which is node three and beer. So 0.5 I plus three
equal to I naught. From here, we can get all node and substitute here
to get V naught. So you can see here is
the equation of the KCL 0.5 I naught plus three
equal to I naught. So the classics I'm bear and by substituting
in this equation, we get the value of the voltage. So in this lesson, we had some solvate the
examples on the KVL and KCL.
15. Voltage Division, Current Division, Analogy between Resistance and Conductance: Hey everyone. In this lesson, we will talk
about series resistors and voltage division and also the butter resistors and
the current division. So that's a start. So if we have a voltage
source like this, and do we have two
resistors in series? You can see is the
cheering OneNote. And you can see that the
current flowing through R1, similar to the current
flowing through R2. So both of them are in series. Now the question is, I would like to
take R1 and R2 and add just one
equivalent resistance. What will be the value
of this resistor? What are we going to do? First, we will find that the
equivalent resistance of any resistors
connected in series is a summation of the
individual resistance. So what does this mean? It means R equivalent
is equal to R1 plus R2. Now let's understand
where did we get this? So we have the current
flowing through R1. So V1 is equal to i, R1, and v2 is equal to i2, r2. Now, let's start
by applying KVL. So we have our KVL in the
clockwise direction like this. You will find that if we apply this gives me
a lot like this. You will find that we
first have negative V. And the goal x is plus V1
plus V2 equal to zero. So we will see that
the summation of the voltage is equal
to V1 plus V2. So the supply voltage
is divided into a voltage across R1 and
the voltage across R2, V equal to V1 plus V2. And we know that V1
is I R1 and V2 is i2. So from here we have all
between two brackets, R1 plus or chew. From here, the total
current flowing shows a series resistor circuit. It will be the supply divided
by the total resistance. Also the voltage. Supply voltage will be the current multiplied by the
equivalent resistance. Okay? So what we can
learn from this is that the equivalent
resistance of a series circuit is equal to the summation of
the two resistors or more. Okay? So as you can see here, v, The Fourth Circuit
is equal to the current multiplied by
the total resistance, which is R1 plus R2, which is similar
to this circuit, which is all multiplied
by R equivalent. So from here we can see that
R equivalent is R1 plus R2. Now, so if you have
a circuit e.g. not just R1 and R2 and R3, R4, R5, whatever it is, the equivalent resistance in a series circuit will be the summation of
all of the resistors. Okay? So if we have a
group of resistors, they're summation is the
equivalent resistance. So we can remove all
of these resistors and add just one resistance. Now, what if I would like
to get V0, V1, and V2? So you can see that v1
is simply equal to V0. V1 is equal to current
multiplied by R1. And we know that the
current itself is equal to so supply divided by
the total resistance, R1 plus R2 multiplied by R1. Okay? So you can see that
v1 is equal to the supply voltage multiplied by R1 divided by R1 plus R2. Okay? Same idea. You will find that v2 is equal to R2 multiplied by the voltage divided by the
summation of the resistors, that what we call
voltage division. Okay? Now, if you would like
to remember this much, in a much easier way, simply if I would like
the voltage V1, okay? So I say that v1 is equal
to the supply voltage, supply voltage applied,
which is V, Okay? Multiply it by a voltage
across the resistor required. So I need the voltage V1, which is the voltage across R1. So I say R1 divided by
the total resistors. Okay? So it will be R1 plus R2. Okay, now let's say you
have another resistor, R3, then it will be R1 divided
by R1 plus R2 plus R3. So summation of all resistors
inside the circuit. You can see that the
voltage of any resistor, let's say v n, it will be the resistor R n, which is the same
resistor at which we measure our voltage
multiplied by Czar supply divided by the total resistors
inside our cell. Now why is this?
Because you will see that V divided by R1 plus R2 plus until our n. This
gives us the total current. When we take the current and the multiplied
by the resistor, we get the voltage. Okay? Now let's talk about with
ZAP parallel resistors. So we said that if we have
two resistors are parallel, if they have node number one, node a and B are the
same nodes of R1 and R2. So if we have a supply V
connected to R1 and R2, so the voltage here
is equal to V, which is similar to the supply. And the voltage across
R2 is also V K. Why? Because they are
parallel to each other, R1 to R2 parallel to the supply. So all of them have
the same voltage. Now, the current coming out of the supply will be
divided to two carbons. You can see that the
current flowing like this, it has two ways. Part of it will go through R2 and part of the current
will go through R1. Then y exist, they exist, then they will be collected
again and get back, choose a negative of the supply. Okay? So first, we have
said that here that the voltage of the
supply is similar to the voltage across resistor
R1 and the resistor R2. So what is the voltage of
this resistor is I1, R1, and this one is i2, r2, which is equal to zero supply because all of
them are parallel. Now, you will find that
if I would like to take these two resistors and replace
them with one resistor. What is the value
of this resistor? You will find that the
equivalent resistance of two parallel
resistor is equal to the product of these
resistors divided by summation. Okay, so let's prove this. So we have this V equal I1, R1 equal to i2. Our job. Now we can, we can see that
here from this equation that I1 is equal to the voltage across it divided
by resistor R1. And i2 is a voltage which
is V divided by R2. Now can we apply KCL? If we apply KCL at this
node here, node a, you will find that
the current entering, you can see that all of
this is node a, okay? All of this. You can see that
current entering, which is equal to the
total current leaving. So I equal to I1 plus I2. I1 is equal to V over R1
and R2 equals V over R2. Okay? And the current itself, the current of any electric
circuit is equal to the voltage divided by the
equivalent of this parallel. So it will be V
over R equivalent. So you can see that the
current itself is the voltage divided by the equivalent
resistance of this part, the i1 and i2, V over R1 over R2. So you can see that this, we have a V as a common factor. So we can take V as
a common factor. So to be one over
R1 plus one over R2 equal to V over R equivalent, which is the current. Now what you can note from
here is that one over R equivalent is equal to one
over R1 plus one over R2. So the equivalent
of the resistor, one over R, equivalent one
over R1 plus one over R2. So from here you can get that all equivalent of a two
parallel resistors, remember to butter resistors
is equal to R1, R2, their product divided
by their submission. Okay? So the equivalent of
two power resistor is R1, R2 product divided by summation. Well, what if we have
more than two resistors? In this case, one over R
equivalent will be one over R1 plus one over R2
plus one over R3, plus one over R4 and so on. This role is zoster for
two parallel resistors. If we extend this
to a general case, you can see that for N
resistors in parallel, which will be one over R
equivalent one over R1 plus one over R2
until one over n. Okay. Now, what if we have all of the resistors
equal to each other? Okay? So what are we going to do? So if these resistors
are equal to each other, you will find that R equivalent. So we'll find that
R equivalent is one resistor divided by the
total number of resistors. So let's understand this. If you look at this equation
here, let's say e.g. we have three resistors, okay? And all of them are
equal to each ours. So we have one over R equivalent
is equal to one over R1, which is R, okay? Plus one over R2
plus one over R3. If all of the resistors
are equal to each other, R1 equals R2 equal R3 equal r. So we have one
over R1 plus one over r plus one over R. So you can see this will
be quite three over r. One over r equivalent. Okay? So from this equation
you can see that R equivalent is equal
to r over three. So you can see that in general, we have three resistors, so we divide it by three. So if we have n no resistors, then will be divided by n. You can see that N resistors, we will divide by n. So what we can learn
from this week alone as a battering combination or parallel formation of resistors lead to smaller resistors. So you can see that R equivalent
is always smaller than the resistance of the
smallest resistor in the parallel combination. Okay? So if we have e.g. one arm and we have here 10 ω, then all are equivalent. By using this role, you will find that it is less
than the lowest resistance. Okay? So in the end, the parallel formation leads
it to smaller resistors. Okay? Now let's talk about with
AZEK current division. Okay? So here you can see that
the current V is equal to current multiplied
by R equivalent, okay? And the oral equivalent
is a product divided by the summation. Okay? Now, what if I would like to get the current through R1 and R2? So you can see this voltage
is the voltage here. And I need I1. I1 will be equal to
voltage divided by, divided by the resistor R1. Okay? So if we take this
voltage which is equal to multiplied by
z equivalent R1, R2 over R1 plus
R2 divided by R1. You can see that R1
way go with this R1. So we will have R1 equal to
R2 divided by submission, R2 divided by summation. And for the second resistor I2, you will find that
also the same idea, V over R2, the voltage here
divided by that resistor. So if you divide this by R2, you will be remote. So we will have i
R1 over R1 plus R2. Okay? Now, if you would like
to remember this, if you'd like to remember this, this is really, really easy. Let's say I would
like the current I2. Okay? So the current i2 will be equal to i2 will be equal
to total current. Total current is
the supply current multiplied by the
other resistor. So I would like the current R2. I will use the other resistor, which is R1, divided by the total resistance.
As you can see here. Same idea. If I need I1, it will be total current multiplied by
the other resistor. I am talking about IE1. So I will use the
other resistor, R2 divided by R1 plus R2. Okay? So we will have
this final equation. Now, what is the analogy between resistance
and the conductance? So we said that resistance is
inverse of the conductance. So if we have a two
parallel resistors or a general several
parallel resistors, one over R equivalent
is one over R1 plus one over R2
plus, and so on. And then we said that G or the
conductance is one over R. So we can say is that one over R equivalent is g equivalent. One over R1 is G1, G2, and so on. Okay, now same idea. If we have a series circuit, we have our equivalent equals
R1 plus R2 plus, and so on. So our equivalent
will be one over J equivalent R1 would be
one over g1 and so on. Same idea for Zack
currently division, you can see all u1 is
equal to R2 over R1 plus R2 and id equals
R1 over R1 plus R2. Now, you can see that
here it is confusing. It is the opposite. Instead of R, we use g.
And then instead of R2, we use g one. And then instead of R1 use J2, everything is reverse it. Okay? So this is just to help you see the difference between using resistance and the conductance. So usually of course, in all of our problems, so we use the
resistance equations, the equations of the resistance. Finally, let's talk about the open circuit
and short circuit. Now here, if you look at them shore to sell it first, okay, So we have a resistor
to assure the circuit. I showed the circuit. So what does a
short-circuit mean? It has a zero resistance. So we have R1 with a certain
value and parallel to it, R2 with a zero resistance, okay? Now we have a current
I1 and the current i2. So biologic, biologic, the current it chooses
the lowest resistance. Suppose most of the current goes to the smallest resistor. So you can see we
have zero resistor and we have a larger resistor. So biologic that, that
total current will go through that short
circuit back to the supply. Okay? Now, why equations simply, it is really, really easy. Let's say I need i2. I2 is equal to supply
multiplied by, if you remember, the resistor R1 divided by total
resistor R1 plus R2. And we know that R2
is equal to zero. So E multiplied by R1
over R1 gives us one. So it will be equal to I. The current i2 will be equal
to the supply current. You use e.g. one, okay? Or E1 equal to total
current multiplies? Boy, I need the current
flowing through R1. So it will be the
other resistor, R2 by the total resistors. Okay? Now we know that R2
is equal to zero. So this part will
be equal to zero, so that current
will be equal to z. So what we can learn is
that all of the current, if we have a short circuit, remember this, it's
really, really important. If you have a short circuit
barrel toward resistor, we can remove this resistor. It does not exist at all, as if it does not exist. So as if our circuit will
be something like this, okay, we don't have
this resistance. Okay? Now what about a short circuit? So we talked about
short circuit. Let's talk about
the open circuit. For the open circuit, as you can see here, are two open circuit
and we learned before that open circuit means
infinite resistance. So how does the
current will behave? So let's just store, to
buy the current here. A biologic, as we said before, that zero current
to flow is through an open circuit
because the current will not pass through
this air gap and go here. It can to do this. So the current here
should be zero. How can I prove this? Simply, i2 is equal to the total current multiplied
by another resistor, R1, divided by the summation of the two resistors,
R1 plus R2. Now, r2 itself is
equal to infinity, and anything that word by
infinity gives us zero. So i2 will be equal
to z, okay? Okay. What about I1? I1 equal to total current multiplied by the
other resistor R2, divide by the total
resistor R1 plus R2. Now, in this case, you will find that our
two approaches infinity. Okay? So we have here infinity and
we have here also infinity. So what we can do in this case, simply we say by using the limit tending to infinity when we have two
parameters or like this, it will be equal to
multiplying by one. Okay? It will be lead to the
end to one from the limit, when are two terms
to be infinity? So the total current I1 will be similar to the supply curve. Okay? So in this lesson, we talked about with parallel resistors, voltage division, the open
circuit and short circuit. And in the next lesson
where you will have some solved with examples
on czar resistance. And how can we combine them?
16. Solved Examples 3: Hey everyone, In this
lesson we are going to have some solvent examples on zeros and the
parallel resistors. Okay, So let's just start
by having this socket here. And I would like to know what is the equivalent resistance. I would like to replace all of these resistors
with one resistor. So don't worry about anything
users to go step-by-step. Okay? So if we look at this circuit, you will find that
the closest thing that you can see is that
you can see that 1, ω and the five ohm
are in series, right? They are series with each other. So we can say is that this part, it's equivalent,
is 6 ω like this. Okay? Now what about this one? We have the two on my exerts. Okay? And you will find
that in this part, you will see that this node, first one and second node, you will see that the six ohm is parallel to the three own. So we have zero
equivalent like this. Let's say R equivalent one, which is six ohm parallel
to the three ohm. And then we would have, there are 4 ω and the 8 ω. Okay? So let's start step-by-step. So we will start with this one. You can see we have six ohm
parallel to the three on. Their equivalent will be
zero product divided by zero summation product
divided by summation. Sign means that
six ohm parallel, these two parallel lines
means the parallel, okay? So this simple means parallel. Okay? So six ohm parallel to the 30 means product divided
by summation gives us 2 ω. Okay? So the equivalent of
this part is 2 ω. So how can I add
this resistance? Okay, So if I would like to remove this and add one
equivalent resistance, it is really, really easy. All you have to do
is that you will remove any resistance
such as six ohm. I will remove this as if it
does not exist like this. The boats are 2 ω instead
of the three own like this. So we will have the two
ohm series with the two. Okay? You can see that 1.5 are
series with each other, so they're equivalent
is 6 ω zero summation. Okay? So you will find that our
circuit will be four on it. As you can see in
this equivalent is 6 ω ends at 2 ω series
with another 2 ω. Okay? So from this socket you
can see that the 2 ω, 2 ω, so zero summation
will be 4 ω. Okay? So we have this part, they're equivalent is 4 ω. So what I'm going to do, I'm going to change
one of the resistors, make it 4 ω, and the other
as if it does not exist, as if it is a short
circuit like this. Okay? So in the parallel we
cancel the other resistor. In series, we add
a short circuit. Okay? Okay. So now we have here four ohm, which is equivalent
of this part. Of course is a forearm
is parallel with the six ohm, right? You can see that the forearm is now parallel with the 6 ω. So there equivalent is their product divided
by zero summation. So it will give us 2.4 ω. So what I'm going to do simply, I will do like this, make this one on
open-circuit cancel any of these two resistors and the
boat instead of six ohm, we will make this
12.4 like this. So you have 4 ω two point 4.8. So if you look at this circuit, what is the equivalent
resistance? It will be four ohm series with the 2.4 series with a tone, which means it is zero
summation like this. Now let's have another one. If you have this circuit
and we would like to get the equivalent
resistance R a P, which is a resistance
between this point and the other point between a and P. So therefore we have
here as ploy like this voltages also e.g. and I would like to get the equivalent
resistance of this part. Okay? Okay. So don't worry about this. It is really, really easy
to apply what we learned. So what you can see
here is that we have 1 ω series with a five on. So they're equivalent will
be one plus y, which is 6 ω. And that makes all other one as a short circuit like this. Like this. Then if we
look at this part, it seems complex, but
it's really, really easy. If you look at the
three and the six ohm, you can see z have
the same initial node and same final note. So you can see this is a six
ohm and this is a three ohm. So what does this mean? It means that six ohm is
parallel to the three own. And if you look at this
one, this one here, you can see that four
ohm and the 12 on, you can see the initial
point and the final node. So from here we have four
on parallel to the 12 volt. Okay? So what you can see, three parallel to the
six ohm gives us 2 ω. And the 12 ohm parallel to
the four ohm gives us 3 ω. So how can I draw this simply? You will delete any of these resistors and the ions
are vanished to the drawing. You can see that
also this part is series with each other
as we said before. So you can see one
series was five, gives us six ohm. Now we have one arm here, like this, and we have
a short circuit here. Short circuit here. Now you have this two are
parallel to each other. So we will delete any of this
as if it does not exist. Delete this one at all. And then instead of
3 ω, we have what? We have to remove
this and add two. So we have this part 2 ω and we deleted the
original resistor here. For the second part four ohm parallel to
that world for all, we can remove this one
at all as if it does not exist and replace the
4 ω with the three. So you will see 3 ω and
we deleted this branch. Okay? Now what does an extra step
you can see is that the three on parallel to what? Battery? To the six ohm. Same initial node,
same final node. So three power 26 is zero. Multiplication divided
by z are some mission. So what we can do
is simply that we can sense as the R in parallel, we can delete one
of them like this. Why exists? Delete them? And
then instead of 3 ω, we will add two arms like this. So you would have ten. We have one arm, j2 ohms. And the tool from this drawing, you can see that one arm
is series with that 2 ω. One series was 2 ω. Zero summation will give us 3 ω. So they're equivalent. We will make this one a
short circuit like this. Remove this, make sure circuit
and it makes us one 3 ω. Okay? So we will have, then, we will have something
like this, then ohms. We have 2 ω and we have the three own lives
this by exist. So you can see that the arm
is parallel to the 3 ω, same initial, same fine. You can see the same drawing
here. As you can see. To own para to the three own. So again, three parallel to six, this gives us 2 ω series
was the 1 ω gives us 3 ω. So there's also these brands. This part becomes what? 3 ω, as you can see here. And two parents to the
three gives us 1.2. So we can remove one of them and they make
the other 11.2 ω. The equivalent will
be 10 ω plus 1.2 ω. The equivalent
resistor will be 11.2. Okay? So in this lesson, we
discussed the ZAB, some solvent examples
on the resistors.
17. Delta Wye and Wye Delta Transformations: Hey everyone, In this lesson, we are going to
start talking about with that delta Y connection. So you'll find that there are some situations which you will
find in circuit analysis. Windsor resistors are nicer
in parallel nor in series. So as an example, you will see this tool
transformation that star, you can see is Y or
a star connection. And you can see delta
or Pi connection. So this one, delta or pi
are similar to each other. And the star, star or y
are similar to each us. Sometimes we call it
that T formation. You can see that if you look at these two circuits or
this for circuits, you will find that
the resistors R1, R2, R3 are not in
series or pattern. Also are a or B or C are
not series or parallel. Now how is this? If you look at R1, e.g. you can see that if you
look at R1 like this, current flowing like this, okay? Okay. Now R one is R1, R2 or not. It is not serious. Y, z have the same initial node. However, the current
flowing here is not the same current
flowing through R2. So they are not in series. The second question, all the parallel nodes,
they are not parallel. Why? Because they don't have the same initial node
and the same final node. So it means that R1 and R2
are not parallel and series. So in this case, it is difficult to analyze circuits in this way because
they are not serious. Or however, this formation or the y formation or
z or t formation or the star connection can be transform it to
delta connection. And this connection
that will help us simplify our electric socket. So we can it change it from
this foreign materials, this one or from this
one to this one. This will help us to simplify
our electric sockets. Okay? So let's start by learning about that Delta
towards transformation. So let's say I have delta a, c, b, or a, b, c, whatever it is, you can see this triangle, this one representing
a delta connection. Now what I would
like, I would like to convert this into Y connection. So how can I do this simply, I take from each point you
can see a being unseen. Extend our resistor. So I drew first
resistor like this, then second resistor like this, then the resistor like this, and all of them are
connected to one point, which is then neutral point. We say it n or the
neutral point. Then what we are
going to do when we get the value of R1, R2, R3, we can just delete the delta formation and we will have only
our Y connection. Okay? So how come I get R1, R2, R3 simply will find
using this equations. These equations, you
can see R1, R2, R3, R1. Remember how can I get the y formation from
delta formation? So in order to get
this branch or one, you can see that we have beside it the nearest
to two resistors, R or C and a or b. So we say that R1 is equal to c multiplied by RB or
rp multiplied by RC. You can see RP multiplied by RC divided by the summation
of the three resistors, or a or B or C. Okay? Now, let's say I would like R2. R2 will be equal to, like this below the summation
of the three resistors, or a, or B or C, or a, B or C. And above
here you can see R2. What is the true
resistors it beside it, the two resistors,
R or C and RA. So we say RA, RC, as you can see here,
lost one, e.g. if I need all stream, then it will be RA or B, or a or B divided
by the summation. This, how can you transfer or transform Delta connection or a delta formation
into a y formation? Soul finds that each resistor in the Y network,
which is this one, Y network, is the product of the two resistors in the two
adjacent delta branches. So you can see adjacent, you can see this one and
this one for R2, RC and RE For all three or a or p, divided by the summation of
the three resistor, or a, or B or C, As you can see here. Okay? Now what if I would like to
convert from y to Delta? So we have R1, R2, R3. All of them are
connected to form, it forms three points, a and B and C. So in order to draw delta, we draw a resistor between a and the B resistor between B and C, resistor between a and D, C. Okay? So we have Delta and y. So if I have y like this, you can see why you like this. Like this. This is why you can see why if I would like
to draw delta, then I connect a resistor
between H22 points like this, like this. Okay? So you will have this part which representing a DLT, okay? Okay, now I would like
to get this delta r, C, RA and RB. So how can I do this? First, you will see
that we have our a e.g. whereas RA, RA, RA, RA will be equal
to the product of the two resistors divided by
is perpendicular resistance. Okay, so what does
this even mean? You can see or a, or B or C. You can see. We have three resistors for R1, R2, R3, which one is
perpendicular to RA? You can see that R1 is
perpendicular to RA. So we divide by R1. So you can see divided
by R1 for P e.g. whereas be this one, be the perpendicular is R2. You can see R2 forming a perpendicular or
90 degrees was at. Okay? So it will be divided by R2. Rc is a perpendicular
to it, is R3. Okay? So that is a first word, second part, what
are you going to do? You are going to
multiply each resistor. Pi is the second one. You can see that this term, this term, and this term are
all similar to each other. So what does it represent? R1 multiplied by R2, R2 multiplied by all three, and all three multiplied by R1. That's it. Really, really
easy. So our a e.g. it will be R1, R2, R2, R3, R3, R1, okay? Divided by the forces to one, or we can say is a
perpendicular one, the one which is very, very far from RA or
perpendicular to r. Okay? So we say that each resistor
in the Delta network is the sum of all
possible products of why resistors
taken two at a time. You can see R1, R2, R2, R3, R3, R1. This is all of the
possible product of two resistors divided by
the opposite. Why resistor? You can see this one
is opposite to RA. Or for C are three
is opposite to it. For ROP, R2 is opposite to it. Okay? Now, what if Delta
and y or balance it? What does that balance it mean? Balanced it means is they
have the same resistance. Okay? So e.g. if four is
a Y connection, if we have a y formation
and it is balancing, it means that R1, R2, R3 are equal to each of us. Can see R1, R2, R3 equal to one value. And if that delta is balancing, it means that all ARP or C
are equal to each of us. Okay? Now in this case, you will find that our y, the y formation, Is equal to each other delta
divided by three. Or each Delta is equal
three times the resistor. Now where did we get this? Okay, let's get back to
any of these values. So let's look at this one, e.g. you can see RA is
A-delta, right? R1r2, all our star connections, star connection or Y connection. Now what are we going to do? Simply assembly. You can see that all
of the resistors are equal to each other
and R1 equal to RY, R to R Y, and also equal to our y. Okay? So if I substitute it here, we have R1, R2, which means RY, RY. So it will be our y
square plus R2, R3, R2 multiplied by all three
is also R y squared plus R3. R1 is also R y squared
divided by R1, which is why you can see
this part will be three, or y squared divided by R1. So it will be three or y. You can see that when
there, That's three. Resistors are balanced or
the system is balanced, it delta or star. You will find that
the delta value of Delta is equal to
three times RY. As you can see here. This case, when z or
policy or Windsor, all of the resistors are
equal to each other. So in the next lesson, we will have some
solved examples to understand why is delta and wide transformations
important in the simplification
of the resistor or in the pure
resistive circuit.
18. Solved Examples 4: So let's have some
solvent examples on that Y delta transformation. So as you can see
here in this example, we would like to convert Delta network here
you can see here APC forming A-delta similar to adult or like this triangle, or a or B or C with
each values showing. Now what I would like to do
is I would like to convert this network into an
equivalent why network? So how can I do this? Okay, let's first
delete that delta here. So I have Delta with
a story points, a, B, and C. What are we going to do first? You are gone. We need y from delta. So I will extend resistor exists and extend
another resistor like this, and extend another
resistor like this. All of them are
combined to one point, which is the neutral point. Okay? Okay. So let's say e.g. this one
is R one, Let's say e.g. this one is two, and this one is our string. So you will find that
R1 is equal to R1. You can see the product of
the adjacent resistors. You can see R1
besides RA and RB. So it will be all a are B divided by the summation
of the series resistance R a plus R being plus
RC for R2 e.g. or to or to a product of
the adjacent resistors, or C or B. So we can say or
b or c divided by the summation or a
plus r p plus or C, or three equal to product of the adjacent
resistors, or a, or C, or a or C divided by the summation
of this V reserves stores. So you can see here e.g. you can see R1, R2, R3, their product
divided by summation. However, you will
find that here. Let's draw the final one. You can see that
here. This one is R1. Here I am writing it as R2. So we can say that this
one is instead of auto, let's make it R1, R2, R3, R2 is R3, okay? Exists. And they make R1 or
like this, okay? So we will have this one too. It doesn't matter, just zoster
the name of the resistors. So you can see that R1 is RB, RC divided by summation or B
or C divided by summation, or to, or to RCRA. Rcra, the war, the
Poisson mission, or three or a or p divides
the boys summation. Now if you look at here R1 or C multiplied by
a or B, R1 or CRP. R2 is our c multiplied
by RA or RCRA, or three, or ARB
or ERP, and so on. Okay? So now after finding
the three resistors, we built their
values on the graph. Then what are we going
to do for the delta? We will remove the
Delta complete. So you will have this
one is invisible line. It does not exist. We have only the y for. Now. Let's have another example
to understand this id. So we would like to get
the equivalent resistance between a and B and use it to find the
value of that color. So what does this mean? You can see between a and B, we have this big resistors,
okay? This part. So what I would like to do is that I will have those apply law exists under 20 v coming out
from that current equal. And we would like
to replace all of these resistors with
just one resistor. Like this are equivalent. Okay? So the current will be the voltage divided by the
equivalent resistance. So first we need to get the equivalent
resistance of this part. Now, let's look at
this circuit to understand whereas y and
delta connection, okay? Now if you look
here, if we have a current coming out
of the supply chain. So current, it will
be divided like this one going here
as our goal in here. Okay? And if you look at here, you will find that
what does it do? It will go here and here. And then it will go
like this or like this. Okay, you don't turn also the
direction of the current. Now why is this? Because you don't know if this resistors are
in series or parallel, they are not serious
and do not fire. Why? Because if you look at here, you can see as this one. Form is a connection, a connection or a
delta connection. You can see that this
point is one node. So you can put it like this. And you have resistors
are ten exists. And then you have point n, okay? And then the phi form exist. And then point C, Then
we have 12.5 lines. This 12.5 connected
to the same pointee. So if you look at
this part alone, you will see it is
delta connection. Same idea. You will find that this part, this point connection
is also another debt. Okay? So you can see we
have two delta here. And how many star? If you look at here,
you can see that this part forms or store. You can see we have one
resistor, two resistors, three resistors connected to one point, which is a neutral. So we have here one star to
another store is this one, this one connected to this
one connected to this one. So we have the second star. Now, if you look also
again, if you look again, you will find that this
board form is another dealt. Now why is this?
Because you can see we have ten connected to
Thursday, connected to ten. All of them form is
also delta connection. So you can see we have
three delta and two stars. So how can we deal with
something like this? You need to do a trial and
then what do I mean by this? You need to transform
any delta or any star to the other time and see if you can
simplify the socket. You can see in this circuit
we have two networks, this one and this one, and we have three delta 12.3. So what are we going to do? You will have many,
many solutions. As an example, all of them
will lead to the same answer. Okay? E.g. I. Will take that wide
network of 510.20 is this Y network and convert it
into the delta connection. Okay? Okay. So we have three resistors
connected to one point. How does the delta
will look like? Between each point, we
will add a resistor. So we will have one
like this, says, one exists and a resistor
between a and b like this. So here, if you look at
here we have R1, R2, R3, this resistors, this one is R1, this is R2, and R3 is 55 ω. So let's look at
our circuit first. Okay? So here, if you look
at this circuit, we have RA, RP and the RC. So let's say e.g. let's say e.g. R1. Okay? The problem here
of this example is that I didn't show our
AARP and our scene. So let's say that
this one is our a and this one is RP
and this one is RC. Okay? So in order to convert the star connection to
delta, what do we do? Somebody? We have our a, which
is the first one. It will be the product of all of these
resistors, each pair. So five multiplied by ten, then multiplied by 2020
multiplied by five. As you can see here. This we are going to
do for each resistors, you can see this
multiplication 750. So we use the same value
in the other resistors. Okay? Now, the second part here is
that we would like our a. So we have RA, what is the
nearest to two resistors? That five ohm and 21, which is a false cyst or the way resist to the one which
is away is that 10 ω. So we use the 10 ω for RAM. So you can see RA
divided by 10 ω. So we get the first
value for RB. You can see it's NURS to
two resistors at 10.5. So what is the
resistivity is very, very far from it. That when T, So we use
that when T1 for RC. What is that perpendicular
order for this to resistor? You can see the nearest resistor is ten and the 20 ends up false. This one is 5 ω. So we divide by 5 ω. So when you calculate
all of this, you have 35, 17 point 5.7. So our a would be five, RP which is 17.5, and RC which is 70. Okay? So we have this
delta between a c b, c, a c b. Okay, okay. That well, for 0.5 ω
is the same as it is, 15 ω the same as it is. Okay? So 2 ω as this. Then after we add the Delta, we delete this one as
if it does not exist. So you can see we will
have here an air gap. So you can see delta and the
store is completely removed. Okay? Okay. Now, what
does an extra step? Now, as you can see, as you can see in the circuit, you can see that the source you own is parallel to
the seventh genome. So 70 as parallel to 13, 70 parallel to search. Same initial node,
same final note. You will find also that 17.5 and 12.5 or
parallel to each other. You can see same initial
node, same final point. You will also find is that
the 5.15 are parallel. You can see same initial point, same final 0.15 to 35. Okay? So that's 31. Parallel
to each other will give us this one. What
is the next step? We will remove this resistors. So 70 parallel to this one, butter to this one gives us 21. So e.g. I. Will make this 121 and
delete this one complete. So we'll have this
branch 21, 12 0 point. Why parallel to the 17.57, 0.2. So I will delete this
one completely as an example and changes
this one to 7.292. So you can say 7.29
to 15 uncertified. We will delete this one and instead of 15 we will add 10.5. So we'll have 7.29
to 10.5 and 21. Okay? Of course, as you can see that this resistor and this one, our series with each other and the cirrus compliant nation
is better to that one-to-one. So you can see that
7.2 and 10.5 or cirrus and their combination
is parallel to that when T1. So we will have our
equivalent 9.6 632. So the equivalent
current will be voltage divided
by this resistor. Okay? Okay. So that is the first solution. We convert this star into what? Into delta. Okay? Can we have another one? Yes, you can take any Delta, any store and transform and see if you can
simplify the circuit. As an example, we will take
the delta which is formed of 105.12 0.5. Okay? Then 5.12, 0.5 is this one. You can see this
delta n. I would like to transform this into a star. So what are you going to do? First we have this point. All of this is OneNote. So I can put like this resistor, a resistor coming
from this point. The resistor coming
from this point, and the neutral point. So we have three resistors here representing a Y connection. Okay? So what are
you going to do? Let's draw this first figure. Okay? So you can see that we have this resistor and
these two resistors. Okay? So the first one is R a D. So here is the D representing
the middle of phi. So let's make this one D. Okay? So first resistor is what
I would like to get, is R a D. If you look at this resistor, what is the two resistors? The beside it? Two resistors is that
then on and 12.5. So it will be ten multiplied by 12.5 divided by summation, then multiplied by 12 E15
divided by summation for all c, d or c, d, this resistor. You can see what are
the two resistors that beside it to one point, 5.5. So it will be 12.5 multiplied by five
divided by summation. As you can see, same
summation of course. For our NAD, NAD, the two resistors,
the beside it, 10 ω and 5 ω, then multiply Y5, the
wider poisson mesh. As you can see here.
Then you will find that the first resistor is 4.5 545. Second resistor to 0.273, sources to 1.8 182. Now as you can see, we
found the three resistors. So we will delete this one
as if it does not exist. We will delete this one as if it does not exist and
delete this one. So you can see 20 series
with the resistor here, 20 series with the
resistor here. And you can see this resistor
in series with the 15, this resistor series with 50. Then we have the final
resistor connected to a, final resistor connected to a, and we have a sorority or, okay? Okay. So what you can see
that this branch, these two resistors are
in series with each ours. These two resistors are
in series with each ours. That combine nation
of this one and the combination of this one
are parallel to each other. You can see same initial node, same following or not. Okay? So you can see that this combination parallel
to this one, this combination is
1.8 182 plus 20, which is this first part. And second to combine nation to 0.273 plus 152.273 plus 15. So product divided by summation, we will give us this equivalent
resistance of this part. So what we can do is that you can just add one resistor like this of 9.642 of course, and leads a sport like this. Delete this part like this. And you can see
that this resistor, we'll be series with this one ends at formation
will be parallel. So you can see 4.454
series was 9.6. So you can see 9.642
series was 4.46. Okay. And this formation is
battery to the sergio. So this branch aspire
to the salty own. So it will be thirsty
multiplied by Z series, salty multiple advisors series divided by the summation
of all resistors. So we will have the same
equivalent resistor as what we get from
the first solution. So our current will
be the same value of. In this lesson, we had another example on the
delta Y transformation. I hope you are understanding
the importance of Delta Y transformation
and why do we use them in electric circuits.
19. Application on Basic Laws with a Solved Example: Hey everyone, In this lesson, we are going to have an
application on the resistors. So you have to
understand that that y is a resistor is important, or why is the
resistor important? That resistor are used to
model devices that convert the electrical energy into heat energy or any
other form of energy. So as an example, we can use the resistor to represent these are
conducting wires, Zao, y or z itself, the conductors that will
carry electrical power. We can use it to represent
as a light bulbs. We can represent
the electric heater is using the resistors, ovens, and loud speakers. Okay? All of this can be
represented by the resistors. Also, e.g. if we have
an electric motor, let's say we have
an electric motor. This electric motor
can be represented by a resistor and another
element called inductance. We can represent it by that
resistor and inductance. The inductance will be
discussed in the course. Okay? So we use the resistor
along with another element called the inductance to
represent any electrical. Okay, so now you
can see also that when we look at our home, in our home, you will find
that inside the wall itself, we have a socket. The socket is a place
in which I will add, I will plug in like this and connect to this
one to any electrical load, let's say e.g. a. Poll. Okay. So this bulb will take from
the outlet itself or the socket itself will take the two terminals and
connect it to our poll, okay, in order to produce
electrical power or provide electrical
power to this ball. And this ball will give
us heat and light. So we can rubber listen
to this ball but by a resistor like this. Okay? Okay. Now, you have to
understand that usually, usually or in our home, that power block itself is we connect our
loads in parallel. Why in part because
we would like all of them to have
the same voltage. So if you look at
the outlet e.g. you'll find that the
voltage is 110 volt, e.g. in my country, 220 volt. Okay? So this is a
potential difference here. Now we connect to all of the loads in our
house in battery. Why in order for all of them
to have the same voltage, which is 220 volt. Okay? Now sometimes in some loads
you will find that we have electric bulbs that will
be connected in series. So summation of all of the voltage across
this bulbs will be the 220 volt or the
supply voltage. So this is a general
case in our house. We connect to all of the
electrical loads in parallel. And sometimes we have a long wire which
have several bulbs. So this bulbs will be
connected in series. Okay? Okay, so what we learned from
this is that we can take this ball and the
represented by a resistor. Okay, Let's have an example. So let's say we have
a battery, mine volt, and it is connected to
it to parallel loads. So you can see two
bulbs in series, and this bulbs are parallel
to another one of 21. Okay? So you can see this
poll consumes a 515, what? This ball, but consumes
10-watt bulb consumes 21. Okay, now what I
would like to get, I would like for us to get the total current
supplied by the battery. Secondary comment is that I need the current through each to pop. So I need the current
through this branch, the current through this branch. Then I would need to find the czar resistive representation
of each of these bulbs. So I would like that
resistance here of this bulb. The resistance of this bulb and the resistance of this part. So what we can do is that we can represent our circuit like this. Each tuple can be
replaced with a resistor, okay, to Paul as a resistor. Let's say e.g. this one is R1, R2, and see, okay. So the first requirement
is that we need the total current
supplied by the battery. So you have to understand that. We have here this
representing our loot. All of this load has
a certain power. This is a power consumed, right? So where did this come from? It will come from the battery. So the power is one whole
supply, electrical power. Voltage source is the one which will supply
electrical power. So what we can see
that according to the law of
conservation of power, we know that the power supplied must be equal
to the power consumed. Power and supplied by this voltage source
will be equal to summation of all
of these powers. So you can see that the power
supplied by the battery is equal to the total power
absorbed by supports. The power or equal to
15 plus ten plus 20. So this power is the
power coming out from the voltage source
going into this loads. Okay? Okay, so how does
this will help us? If you remember that the power supplied by a battery or consumed by a load is
equal to the voltage. It's a voltage multiplied by the current coming out from it. So that current coming out from the battery
will be equal to a power which is 45/9 volts. Okay? So now we obtain
the total current, which is five mLs. Okay? Okay. Now what is the next step? The next step is that
I will need that current through each bulb. So how can I do this? Simply? If you think about it, you will know that
the voltage here, which is equal to 9 v, and the voltage here
is equal to 9 v. So if you look at this
branch here as this one, you can see Power when to what? The voltage 9 v. So I can get the current. So the current will be
equal to same as here. Power divided by voltage. The power two n divided
bys of mine volt. Okay? So you can see here, let's go to the
other side. Here. You can see here is
a power which is 20 watt divided by the
voltage which is nine volt. So this will give
us 2.222 and bear. So we know that the current
year is 2.222 and the current here is unknown and the current coming from the
supply is five and bears. We have Karen to supply it
and to current going out. So if we apply KCL here, you can see that
the five and bear, which is incoming current or current going into that node, is equal to the two currents coming out of the anode, okay? So it will be 2.22 plus
R into the current will be five -2.22 by applying KCL
at node, let's say node a. Okay, so what is the next step? Now we have all of our accounts. I would like to get the
resistor R1, R2, and R3. So we have current and we have, we have all of the
currents and we have all of the 15 water, all of the power. So if you remember that the
power of each one is equal to I squared multiplied
by our resistor, equal to power divided
by square root of. So simply if I would
like R1, let's say R1, it will be when t What
divided by 2.22 square? If I need two, it will be 15 What
divided by 2.778 squared? If I need all three, it will be 10-watt
divided by 2.778 square. So we will have for your
final resistors R1, R2, and r three. So in this lesson,
we talked about as simple application on
using the resistor. We can use a resistor to
model our electrical devices.
20. Methods of Analysis and Nodal Analysis with No Voltage Source: Hi, and welcome everyone to this part of our course
for electric sockets. In this part, we will talk
about methods of analysis. So in the previous
part of the course, we discussed the
fundamental laws of circuit theory, such as e.g. Ohm's law and cycle shifts
low or Zach, KVL and KCL. Now we would like to use these laws or the
KVL and KCL laws in order to develop two powerful techniques
for circuit analysis. What are these techniques? We have the first one
which is nodal analysis, which is based on Zach ECL or Zach
Kirchhoff's current law. And then we have the second
one which is a mesh analysis, which is based on
Kirchhoff's voltage law. That two techniques
are so important that this part will be regarded as the most important
part in the course. Now why is this? Because as you will see
that we are going to use mesh analysis and nodal analysis or load
in electric circuits. Okay, it is a very,
very important circuit analysis
methods that we use. So by using dimensional analysis
and two nodal analysis, we can analyze any
linear circuit. And what I'm implying linear, it's consisting of linear
components such as already linear electric
circuit elements, such as e.g. the resistors, inductors,
and capacitors. So we will use the mesh analysis and nodal analysis to have some simultaneous
equations that will be solved to obtain
the required values of current or voltage. So we will start
in this lesson by talking about the
nodal analysis. So we have two types
of nodal analysis. We have no data analysis
with no voltage source, and the nodal analysis
with a voltage source. In this lesson, we
will start with nodal analysis with
no voltage source. Okay? So the nodal analysis is
used for analyzing circuits using the node voltages
as a circuit variables. So by choosing the anode
voltage instead of the element voltage as
a surrogate variables. It is convenient and will reduce the number of equations
required to solve it. So how can we apply
another Ana's first tool? Select the anode,
a reference node. And we will assign
voltage V1, V2, V2 to the remaining nodes
in the circuit itself. And the voltage
will be represented with respect to the
reference node. Then we'll start
applying KCL to each of the n minus one
non-reference nodes. And we will use Ohm's law to
express the branch currents. Then we want to start
solving these equations. Okay? I know that you don't
understand anything till now, but don't worry, you
don't want to start. When we start applying
this nodal analysis, you will understand everything. The first step you are going
to do is that you select and reference voltage or a reference
node inside the circuit. So the reference node
inside the circuit, you will find e.g. in the electrical
circuits in general. In the examples which we do, you'll find that
as these samples, you can see the samples. What does this mean? They mean reference voltage
or ground when the voltage. So what does this even mean? It means that these
voltages are equal to zero. So this point in the electric
circuit is equal to zero. So if you look at any
electric circuit, e.g. this one, you can
see that we have this as ground the
sample like this one. It means that this
point, the node voltage, node voltage here,
the node voltage here is equal to zero. Why? Because it's connected
to the ground. So that is a first step. You will usually see in
any electric circuits that we have a point at which
we will put the ground. Okay? Okay, then what
is the next step? Next step is that each node
inside that circuit itself, we will start giving
it a voltage. So if you look at this circuit, this is the original
circuit here we have the reference value
equal to zero. Now, how many nodes in
this electric circuits? You can see we have
the first one, first node here, and
second node here, and the third node here. So we have three nodes here. We have this node, this
node and this one. This one is zero, which is the reference. Now we have certain volts. You can see node number
one, node number two. So we'll say is that this one we will assign into voltage called V1 and assign this
one as a voltage V2. So we can see we have this
node here is a voltage V1, and this node here is a volt V2. So what does this voltage,
this voltage, e.g. if it is two volt, it means that this
point with respect to the ground has a potential
difference of two volt. So this point with respect to the zero reference is
equal to two volt. This point, let's say
V2 equals three volt. It means that this point
with respect to the ground is 3 v more than the
ground applies three volt. Okay? So again, first step we have the reference node
here, which is zero. Then we assign for each other nodes inside
the circuit H naught here, we will put it a number, e.g. V1, V2. Is that as a second step? Then what is the next step? We will start applying KCL to each node
inside that socket. Forest node here,
which is this one, we will start applying KCL. So as you can see
that we said that KCL say is that the
current entering, all current entering equal to the all of the
currently living. Now as you can see that here, you can see I1, I2, and I3 discounts
were not visible. If you get back here. You can see this is
our original circuit. And what we do, what we do is that we
assume that we have a current I1 coming out here, or E1, and the current
i2 and current I3. This is an assumption. You can add any direction, e.g. instead of saying that I1
coming out from this node, you can simply say Taiwan coming like this
as you would like. Okay, in the end, when you get these values, you will understand
if it is positive, it means that this
direction is correct. If it is negative, it means that this
direction is false. So it doesn't matter what
direction you are selecting. Here we can see that
current entering or anyone entering to node n1. And what are the currents
leaving, leaving i2, i2 and i1, i2, i2. And one. For node number two here, you can see that i2 entering, i2 entering and i3 leaving. So we will have this i2
plus i2 equals three. Okay? Now this is a key
Cl, this is a KCL. Now we have this I2 and I2, I1 capital I, capital I, capital I, capital E. We have this value and this
value given in our problem. Now what about I1, I2, and I3? We will get them
using Ohm's Law. So you can see that e.g. I1, I1 coming from this one going through that
resistor to the ground. So we have plus minus because the current
entering from here. So I1 will be equal to V0, V1, V1 minus zero divided by R1. Difference in voltage divided
by the resistor. For i2. It will be a2 will be going like this,
entering it from here. So it will be plus minus. So it will be this voltage minus this voltage divided
by the resistor V, V1 minus V2 divided by R2. What about I3? I3
entering like this, so it will be plus, minus i3 will be equal to V2
minus zero divided by R3, V2 minus zero divided by R3. So as you can see here, the first equation,
second equation. So the equation sign we will start substituting
in this equation. So we will have this final form. Okay? That is a method that people use or the methods
that they apply KCL, then they apply Ohm's law. What I do is that there is a very simple methods
which are use. So what does this
method assembly? If you look at
this circuit here, let's say I would like to get, so say we have equation
one and equation two. I would like this equation. How can I get this? Let's start with the first node. The first node here, this one, V1, or node number one. What I'm going to do
is that I will assume, you will assume that all
of the currents entering. Okay. All of the currents entering or not the currents entering, all of the currents leaving. I will say that current
coming out of V1, current coming out of V1. V1 current coming out from V1. Okay, then I would get
each of these current. So I will say is that all of these currents will
be equal to z. Okay? So summation of all
cannons equal to z. Here, I'm assuming
all coming out. So let's just start
with this first one. You can see this current
coming out however, I E1 is entering. So it is opposite
to that direction. So I say negative I1. Second the current
year coming out, it will be V1 minus
zero divided by R1. So we say plus V1 minus
zero divided by R1. Guarantee from here leaving. So it will be plus V1
minus V2 divided by R2. Then currently leaving,
you can see currently coming out in the
same direction of i2. So it will be plus i2. So if you look to this
equation and this one, you will find that z are
similar to each other. So if you take this
one to the other side, you will have all U1
equal to all of this. So you can see all U1
equal to all of this. Now, let's type the
second equation. I am working with what? With node number two, this one. So I will say all currents
coming out, I assumption. So the first, the
current coming out, you can see currently
coming out with however i2 opposite to it. So I say negative I2. Now we have a current
coming out from here, so it will be V2 minus
V1 divided by R2. So it will be plus V2
minus V1 divided by R2. Okay? Then guarantee coming
out it will be V2 minus zero divided by R3. So to be V2 divided
by R3 equal to z. Now, if you look at this
equation and this one, you will find that they
are similar to each ours. Okay? So how is this? If you take this one to the other side and this
one to the other side, you will find that
v2 divided by R3, which is this one, equal to i2 plus minus, minus V2 minus V1 plus V1
minus V2 divided by R2. So if you look at this equation, you will find that
similar to this one. So what did I do
instead of doing KCL, saying which accounts
are entering and which occurrence all leaving,
then start applying. Ohm's law. I did all
of this in one step. I take any node here, then I assume all
currents coming out. Then I get the value of
each current similar to V2. I say all the
currents coming out, then get the equation
number two and so on. Okay? So final solution, what
are you going to do? You will see that we
have as three equations. Remember this two equations that we obtain, these two equations. And we have this values
of currents that we substitute it to
get this equations. Now it's the same
idea you can simply say instead of V1 over R1, you can say G1, which is a conductance. And instead of resistance, you can say conductance. And the one over R2 is
G2 and one over R3 G3. Then you can substitute
in this equation, are replaced each
one over R2 by G2, one over R3, G3, one over R1, G1. To get this equation, okay? It doesn't really matter. If you use g or use
resistor, it is the same. Then you will use this
one to form a matrix. Form a matrix like this. Okay? Why we will form a
matrix in order to use a method called
zach Kramer resumes, traumas method, which
is used to solve several equations by
using matrices. Okay? So how did we form this matrix? First, what are the
variables here? Our variables are
v1 and v2, okay? All E1 and E2, all of this are constant. So what I'm going to
do is that I will make V1 and V2 in one side. And the n equal to
something here. Okay? So let's say e.g. you can see that
we have I1 and I2. So if we take this one
to the other side, it will be I1 minus I2. So it will be equal to G V1, V1 plus V2, V1 minus V2, V2. Okay? So let's talk about V1. So we have V1, V1 has G1 and G2. So we say V1 plus V2 plus second
variable, which is V2. You can see V2 has negative. G2, can see G2 and the negative. So it will be negative g two. Okay? Now for the second
equation, same idea. You can see i2 here. So we'll keep it as it is. And take this part
to the other side. So it will be negative, negative V2, V1 minus V2, V2. So for V1 we have
negative V2, V1, V2, v1. For V2 we have
negative, negative, so it will be plus V2 here. So negative negative
plus g2, g3 here. So it will be G3
plus G2, like this. So you can see we
have V1, V1, V2, V2 equal to a certain value, equal to another value. Now, if we put this
in a form of matrix, you can see I1 minus I2, I2, I1 minus I2, I2, V1 and V2, V1 and V2, V1. You can see the first
column will be V1 plus V2 negative G2, g1, g2, negative G2, second column, negative G2 and G3 plus G2, G2, G2 plus J3. Now why are we going
to do this to use grammar method to get V1 and V2. Now zeros, notice this
is not the only way. The second way is
that you can get v1 as a function of t equal
to something, give V2. Then use this equation and
substitute here to get V2, then get back and they get V1
by reducing the equations. Anyway, I'm going to show you the Cramer method because you can use this when we have
three equations or more. So we have our equation here. How can we solve
with this assembly? If you remember, here
we have x and y, which is V1 and V2, V1, V2. And this is the first column. This one representing here a, B, C, and D. Kayla exists V1. And let's say x and y. X
and y equal to e and f. So this matrix representing
this one, okay? So if I would like X, which is V1, V1, then what I'm going
to do is that forest, you will get the determinant a. What is the determinant a
is a coefficient matrix. What matrix? A, b, c, d, This matrix, you will get the
determinant of a, B, C, D, its value. If you don't know about
determinant or metrics, you can go to back to our mask clauses in
order to understand it. Then we have here the first
matrix here, first matrix. You can see that we
have a, B, C, D. Now I would like X, X representing v1 or the
forest to column, this column. So what I'm going to
do is that I will take this colon and
substitute it here. So it will be this
colon is E F, E, F, first column, and the
second column as this B, D. Okay, you can see E, F, BD. Now, same idea. If I would like v2. If I would like v2,
if I would like v2, then what I'm going to do the same determinant a
is this determinant. And what about the first one? Why is the second variable? So I will take the second column and replace this
one with this one. So E F will be the
second column. So I put here e, f. And the first colon
as it is, a scene. A scene. So again, if I would like e.g. apply this, V1 will be equal to the determinant
of this matrix. And G1 plus G2 minus
g2 negative j2, g2, G3 determinant
of this matrix. And what is here? Here we will add v1 is
the first variable. So first variable means
first to call them. So we'll take this one and add it to the first two columns. So I say i1 minus i2, i2. Second the colon as
this negative G2, G2 plus j. Okay? So this is called
zach Kramer methods. This is used to
help us to solve, to solve two equations
or even three equations. So if you have three equations
like this, A1, A2, A3, B1, B2, B3, C1, C2, C3, D1, D2, D3. Okay? So we have z m matrix, original matrix, this column, this column, and
this column, A1, A2, A3, B1, B2, B3, C1, C2, C3. This is a determinant D, or the similar to here,
the coefficient matrix. So we get the determinant of the coefficient
matrix as normal. If I would like the first
two variable X, we need x. What I'm going to do
is that I will take this all on the boat it here. And instead of A1, A2, A3. So you can see D1, D2, D3, D1, D2, D3, and the rest as it is. If I would like e.g. Y, which is a second variable. So I will take this variables and the boat it in
the second column, b0, b1, b2, b3. So you can see V1, V2, V3, d1, d2, d3, and the other columns, as it is, if I
would like season, I will replace the last column. As you can see, this is
called is a chromosome. And so those are Cramer's
rule for three variables. Okay? So in this example
or in this lesson, we talked about nodal
analysis without, with a voltage source. And we talked about grammars
and Mason's that is used to solve two equations or more. Okay? So anyway, you have to
understand that this method, this method, which is
the Cramer method, is used in, in general, it is not related
to nodal analysis or mesh analysis or
any other analysis. It is used to solve the
two equations or more. If you have other methods, you can apply any method in order to solve with
these equations.
21. Solved Example 1: It's two. Now let's have an
example, a solvent, the example on the nodal
analysis with no voltage source, you can see that
this circuit only consisting of current sources. And I would like to get the
node voltages in the circuit. So how can I get
this anode voltage? As you can see here,
we have first step. We have the reference
or the grounds. So this point grounded, it means it is a zero voltage. So we obtain all of the node voltages with
respect to this ground. We have the first node
here, this node here. Let's say V1 and the
second node, v two. So what I'm going to do, I need to KCL. You can start applying KCL, then apply Ohm's law, then combine them together. But I told you I use another method which is
very, very simple here. So what I'm going to do simply, we started with the
first node here. I assume that all of the
currents coming out, all currents coming out. So current coming out here, out here, and current
coming out here. Okay? So the first account
you can see currently coming out is always the
tools of five and Bayer. So it will be negative five and bear second the
current coming out. Okay, So it will be here plus, minus, plus, minus
current entering. It will be using Ohm's law. This current will be
V1 minus V2 divided for V1 minus V2 divided by four. And last the current
coming out here, it will be V1 -0/2. So it will be V1 -0/2. All of this equal to zero. That is the first equation. Second equation is
that if you look at this node here as this one, we assume again, since we are talking about the
second denote here, we assume all currents leaving. So currently leaving, current leaving, leaving and leaving. So this guarantee leaving is
always a to ten and bear, so it will be negative ten
plus this current leaving. So it will be V2 -0/6. So it will be V2 divided by
six plus guarantee leaving, so it will be V2
minus V1 divided by four and V2 minus
V1 divided by four. Okay? Then the last, the
current here coming out in the same direction
of the 5M Bayer. So it will be plus five and bear all of
this equal to what? Equal to z. Okay? Okay. So what next? So we have now two equations. Equation one, equation two. These two equations has
two variables, v1 and v2. So what you can do
is that you can pay, you can get v1 as a function of V2 or V2 as a function of v1. So you take one equation
both to the other side, so that we can have
V1 equals something. If V2 or V2 equals something V1, then you take this reduced
equation, V1, e.g. and substitute it in the
other equation to get V2. Okay, Let's see. Again here. You can
see, let's do this. You can see here negative five. Okay, so let's take this one to the other side
will be negative five. So we have negative
five and V1 minus V2 over V1 minus V2 over four, and V1 over V2, V1 over two. Okay? Second equation
here you can see. Let suppose this one
to the other side, we have five and then five plus V2 -0/6 takes
us one to the other side, would be negative ten. Take this one to the other side. It will be plus V2
minus V1 over four. If you look at this equation, you can see negative
ten, V2 over six. You can see V2 minus V4
over V1 over for V2 minus V1 over 4.5 equal to z. So this equation,
these two equations obtained directly by the
method I told you appellant. Okay? So here you can see that normal methods that people
use is that they say, what are the currents entering? What are the currents leaving? Was they apply KCL first, then each current, we
will apply Ohm's law. Then we will obtain
the equation. Then the second DMSO, see what currents entering, what currency leaving, and assume currents,
then substitute. And the same, same thing. However, the easiest domain, so as I told you is that e.g. node one, then I exist node alone and assume all
currents coming out. And they get this, all of
these currents equal to zero. If I'm talking about v2, all currents coming out obtain this guarantees
equal to zero. You will get the same equations much easier and those are wet. And without thinking, okay, this is really, really easy. So when we have
these two equations, we said that we can
solve with them. So we can simplify this. We will have this equation, simplifies this and they
have this equation. Then you can get v1 and v2. Okay? So if you would like
to understand what I'm talking about,
the substitution. So as an example, we have here three V1
minus V2 equal to n. So if we rewrite this equation, you can say is that V2
equal to three, V1 -20. Alright? V2 equals three
from this equation. So we have V2, something V1. So what I'm going to do, I will take this V2
and substituted here. So we have negative
three V1 plus five. What is the value of V2? V3, v1 -23, V1 -20 equal to six. Okay? So you can see we
have one big equation with only V1, okay? So if we continue, we have negative three v0, v1 plus five multiplied by 315, V1 minus hundred equal to six. So negative three v
1.15 V1 is 12 V1. And take this to the
other side will be 160. Okay? So V1 will be equal to
hundred and 60/200 and 6/12. If you divide this by four
by four and this by four, you will get 14/3,
similar as here. Then after getting V1, you substitute in this
equation to get v. Okay? So this is a method
of substitution. The second demonstrated
is that you add, you form a matrix
using this one. For my matrix and solve them using their
chromosome muscle. We will have an
example on this in order to understand this ID. Okay? So as you can see here, is that after getting this voltage is what is the
requirement in the problem. You can see we need the
node voltages V1 and V2, so we obtain the mode
voltage V1 and V2. Now, as you can see that when
we assume currents, e.g. assume i2 like this,
ice-free, like this. If you have, if you have the
current as opposed step, it means that this
direction is correct. If e.g. i2, as you can see, we assume that the
current I2 going, go from V1 to V2 y-axis. So when the current
becomes a negative, it means that the
correct answer is that I2 is flowing like this in
the opposite direction. So you can see that i2 negative
means that the current flows in the direction
opposite to the one assumed. Okay? So in this lesson
we had a solvent. The example on the
nodal analysis.
22. Nodal Analysis with a Voltage Source: Hey everyone. In this lesson, we will talk about the nodal analysis with
presence of voltage sources. So we discussed
before that if we would like to do
the nodal analysis, we simply apply KCL and
then we apply Ohm's law. Or by some method which
I explained is that by assuming all of
their current going out with and then
obtaining the equation. Okay? So in the previous lesson when we discussed the
czar nodal analysis, we didn't have any current
source, a voltage source. So now what if we have
a voltage source? Okay? So if you look at here e.g. if we would like to do
and nodal analysis, okay, so we have this
node V1, node one. Node one here has a count, let's say going out like this. And I currently coming from
the voltage source, right? So if we apply nodal analysis, we will simply say V1 minus V3. I exist difference in voltage
divided by that resistor. Plus for this point, V1 minus V2 divided by the two, V1 minus V2 divided bys to own, then plus the current
coming out from here. Okay, so the current
coming out for you, how can I get it? Okay? It will be V1 minus zero. So it will be V1 minus zero divided by the
resistance, right? This resistor, so we
have a voltage source. What is the resistor here? I don't know. Okay, equal to zero. So the problem here is that when we have a voltage source, I cannot get the nodal analysis or I cannot apply
nodal analysis. So in this case, what
I can do assembly is that we have two cases here. When we have voltage source, we have this case and this one, we will understand what's
the difference between them. So first one, if the
voltage source is connected between
the reference node and the non reference node, we simply set the voltage ends. The ads are non-reference
anode equal to the voltage of
the voltage source. Example in this figure, V1 equals ten volt. Okay? What does this mean? If you look at here, if the voltage source
is connected between a reference node and
the non-reference node, then between reference
and the non-reference. So whereas the reference
node, this one. Okay? Whereas the non-reference
V1 or V3 over V2, all of this are
non-reference nodes. Why? Because the reference
node is the one which has a zero voltage. Now, if it's connected between voltage RL and red non-reference
and our reference, then the voltage itself will
be equal to the supply. So in this case, V1
will be equal to 10 v, okay, so we already
know no value of V1. Now, if you would like to
understand this logic, we have here a supply, right? This supply, you can see that the potential difference between this point and this
point is ten volt. So it is plus, minus ten volt, right? So the difference between this point and this
point is ten volt. So it will be V1 minus
the voltage here, which is zero, equal
to the ten volt. So what does this mean? It means that V1 is
equal to 10 v by logic. So if it is e.g. the opposite, ten volt, e.g. ten volt about negative
plus like this. Okay? So what does it mean? It means that I exist. It means that here we have
both negative then volt. So it means that the
difference between this point and this point
is equal to ten volt. So ten volt equal to difference between this
point and this point. This point is equal to z minus
this point, which is v1. So v1 will be equal
to negative 10 v. Or you can think
about in another way that plus minus negative. What does this mean? It means that this point is higher than this
point by ten volt. This point is zeros and therefore this point
is negative ten. Okay? Okay? Now the second case
is that if you have a voltage source between
two non-reference nodes. So in this case, if
the voltage source dependent or independent
doesn't matter, connected between two
non-reference nodes. The two non-reference
nodes will form a generalized node
or super node. So what are we going
to do in this case, we apply KCL and KVL to
find that node voltages. So this node, which is
called the supernode. Now why supernode? Because it's connecting between two non-reference nodes and between them is a
voltage source. Okay? So it's warmer by enclosing dependent or independent
voltage source between two non-reference nodes and any elements connected
in parallel with it. Okay? So how can I benefit
from this supernode simply you will find that
we will apply KVL and KCL. So you will find that forest. If we apply KCL, we assume currents, which
assumes that current. We assume first that this
supernode is one big note. All of this is one node. And we look at, we apply KCL at the supernode. What does this mean? It means that we see all of the currents entering and the currents
leaving from this supernode. So you can see this super nodes
connected to this branch, this branch, this
branch, and this one. So we look at it
as one node, okay? So we assume here some currents. We assumed I1, I2, I3, I4, as you would like. Then we apply KCL
to this big note. So you can see that we
assumed i1 entering. So I1 will be equal to the currents entering
is I1 and I4, and the currents
leaving i3 and I2. So you can see i1, i4, i2, i3 currents entering equal to the current leaving I1 itself is equal to what
coming from here to here. So it will be V0, V1
minus V2 divided by two. And the current I4, R4 coming, entering here, coming
from V1, going into v3. So it will be V1 minus
V3 divided by four. And we have current i2, i2 coming out from here. So it will be V2 -0/8 and
I3 coming out from here. So it will be V3 and -0/6. You can see that as if we
combined two KCL in one. So instead of doing
to understand the idea and instead
of doing KCL for V2, only seeing the currents
entering and leaving. And then doing another case, L4, V3, this one and this
one and this one, what we do is that we
can combine this two KCL together into one big KCL
into a KCL to this big node. And instead of V2 and V3 only. Okay? So we have now with the first
equation coming from here, we will get another equation
from KVL. How was this? You will see that we have this voltage source
included in this loop. You can see that V2 is the voltage between this
point and this point. V2 and V3 is a voltage between
this point and the ground. So if we apply KVL here, you will see that,
let's say e.g. clockwise, like this one. You can see that as
we learned before, as a KVL going like this, negative v2, negative V2, then going like this plus five, then going like this, plus V3 equal to z. So what we can get from here is that if we take this
to the other side, here, you can see five, okay? So you can see five equals to take this one
to the other side, V2 minus V3, which
is the equation. This is using the KVL. Now I do, I usually
don't do a KVL. It is really, really
easy without KVL. How, if you look at here, you can see that
we have a source. We have a point V2 and V3. If we look at this source, it means that plus,
minus five volt. So what does this mean? It means that the
difference between this voltage and
this one is 5 v. So it is means v2
minus V3 equal to 5 v. Or you can see that
the positive with V2, it means that V2 is higher
than v3 by five volt. So we can say V2 is equal
to V3 plus five volt. This one is similar to this one, similar to this one. Okay? So in this lesson
we talked about the nodal analysis when
we have a voltage source. In the next lesson, we will start taking
an example on this.
23. Solved Examples 2: Hey everyone, In this
lesson we will have some soul with
examples about that. A nodal analysis with
a voltage source. So in this example you can
see we have a current source, current source, and
a voltage source. And you can see we need
the node voltages, node voltages in this circuit. You can see v1, which are representing
this node. And this node is V two. Now we would like
to get v1 and v2. Okay? So the first step, first step, if you look at this circuit, the easiest equation, okay? So you can see that first, first we have two
variables, two variables, or two unknowns,
which is V1 and V2, we would like this values. So in order to get
them, we need what? We need two equations. So if we have two variables,
we need two equation. If we have three variables
and we need three equations, four variables, four
equations, and so on. So we need two equations. The easiest equation,
the easiest one, if you look at the
source to vault, you can see plus -2 v. So it means the one which
add a positive side, which is V2, is higher than V0, V1 by two volt. So you can see a plus-minus
the difference between this point and this 0.2 volt. This point is V2. And this point is V1 equal to two volt or V2 higher
than V1 by two volt. Okay, what did we get this from this first equation here from KVL or by looking at the supply,
really, really easy. Okay? Second equation can be
obtained from the supernode. Now, where do we
have a supernode? Because we have a
voltage source here. So we can combine them all
together as one big note. So you can see here, you can see applying KCL
to the super-node. You can see that we assume
that this is a big node. And we assume some tenants, we have a current source
coming like this. And we assume the
current going like this. And we assumed another current going like this and
another current Alexis, It is all assumptions. So we have current coming
like this, two unpaired. We have current coming
out or E1 coming out, i2 and the seven and bear. Okay, now let's apply KCL. You can see that
current entering to impair equal to
the total current leaving that too and pair
will be equal to I1 plus I2 plus the seven and bear current entering equal
to the current leaving. So I1, if you look at here, is the difference
between this point, V1 minus zero
divided by the two. V1 minus zero
divided bys a tool. And V2 minus zero divided by the four ohm
gives us I2. Okay? So from here we will
have an equation and other equation which is
second equation here. One from TCL and one from the KVL or by looking
adjusts to the supply. So you can see by applying
KVL here and this loop, we will get that V2 is
equal to 2 v to plus V1. So by solving these
two equations, we will get V1 and V two. Okay? Okay. Now, again for Zack ECL, you can see applying KCL Ohm's
law to get this equation. Now, the easiest
method is that I look at this one as a big note. And I assume all currents
going out like this. Okay? Then, alright, the equation. So first the current going
out is negative two. So negative two. Then second the
current going out. So it will be V0, V1 divided
by two on like this. Then this current going out, V2 divided by four. And this kinda going out
with seven and bear. You can see this
will give us five. So it will be V1 over V2 plus V2 over four plus five equals to z. Okay? So we have this equation. So if you multiply
this equation by. For you will have four
multiplied by V1 over V2 to V1 plus V2 equal to negative 20. Okay? Now, if you take this
one to the other side, eight -28 is negative
20 equal to V1 plus V2. You can see this
equation similar to the one obtained here. You can see that when we
apply that super-node here, you can see that
then all is useless. It doesn't do anything for us. Okay? Okay. Now someone will ask me why. It doesn't matter. If you look at here, at this point, this is
a supernode, right? So if we say e.g. if we add this current, you can see that
this is a big node. So we say that we have a
current coming out from it, which is the same
current entering. So as if, if this current is I1 coming out is the same
current i1 entering. So if you add it here
to this equation, so we say that here current entering equal
to current leaving. So which one is leaving? All E1, let's say
RAX, RAX sleeping. So it will be plus I x. Then what current is entering
again is I x plus I x. So as if you didn't do anything, this will go with this. Okay? So again, the resistor here has a current guarantee coming
out from the supernode, which is the same
current entering is a supernode from
the other side. So this resistor as
if it doesn't exist, This gallant will
cancel each other, as you can see here. Okay? So let's have another example
to understand this idea. We have here in this
example we have two voltage sources and would like to apply
nodal analysis. So we have V1, V2, V3, and V4. So we have how many variables? We have four variables,
all four unknowns. Which means we need four
equations involving V1, V2, and V3 in order to
solve this problem. So let's just store
to buy is a KCL. So let's start by KCL. So we have currents. We assume this one will be supernode and this
one will be a supernode. Why? Because as you can
see, it is between two non, non-reference modes. And this is between two
non-reference mode. So this is a super node, and this one is a supernode, as you can see here. Then you start assuming
cans or E1, E2, E3, or E5, I4, I1, as you would like, in any direction,
it doesn't matter. In the end, it will
give the same answer. This is just an assumption. You can assume I1 going
like this or you can assume i1 entering
as you would like. Okay. Okay, so we have here as a supernode less
supplies, a KCL. So at supernode, one
to this supernode, you can see current entering, current entering and current
leaving current levy. So I1 plus I2 equal
to I3 plus ten. As you can see
here, then I1 will be v1 minus v4 divided by three, V1 minus V2 4/3 i2
will be V1 over V2 at 10:00 A.M. bear
entering i3 will be v3 minus v2 divided by six. So this will give us
this final equation. Now, if we would like to
never assume any currents, if you don't want to
assume any currents, I can get it like this. Assume all currents
coming out like this. So we will have v1 for us to current v1 minus v4 divided by three plus V1 over two plus
it will be negative ten. Negative ten because it is opposite to that current source. Current allowing our V2 minus V3 divided by six equal to z. So this will give you
the same equation here. Okay? So that's the first
case L second the KCL at this node here, we will say all of the
currents leaving except I1. I1 will be equal to I4, I5, I3. As you can see, one, what's the value of R1? It will coming from V1, so it will be v1 minus v4
divided by three, is three. V3 minus V2 divided by six. Or E4, I4 coming from before. So it will be V4 -0/1 in
five coming out from here. So it will be V3
divided by four. So you will, when
simplifying this, we will get this equation. So we have the first equation, we have the second equation. Now how can I get more
equations by applying KVL? So we have here V1, V2, and V1, V2, and V3, and V4. So if you look at here, we can apply KVL to get the first equation and apply another KVL here to
get second equation. So you can see that for loop
number one, this lobe here, you will find that
V1 minus V2 equal to n. Or simply if you
look at this one, you can see that this point is higher than
zero point by two. Or this point minus this
point that gives us 20. So V1, this point, V1 minus this point V2 equal
to 20 without any given. Here. If you look at this one, apply KVL to get an equation. If you look at here,
you can see V3, this point higher
than this point, you can see plus this point higher than this
point by three vx. So V3 higher than
before by three vx. And what is the
value of v x itself? If you look at
VAX, look at here, vx is the potential
difference between this point and this point. So vx is a potential
difference between this point which is V1 minus, minus this point which is V4. So you can take this equation
and substitute it here. So we will have V3 minus
V4 equal to three Vx, which is v1 minus v4. So we will get Texas one
to the other side, 31. Negative three is
negative three V4. Take it to the other side. It will be plus, sorry, V4. So it will be plus
V4 equal to zero. And v3 will be as it is. Will find this equation
came from here. You can see here three V1 here saying negative
V3, negative to V4. Okay? So why is this? Because if you look at here, you can see three V1, three V1. So when we take it
to the other side, it will be negative 3D
view on this one should be negative 31, negative three V1. And you can see if
it's three V1 plus V3, negative three plus two
is four, negative two v4. So if you take e.g. and negative, if you take e.g. a negative as a common factor, it will give us
the same equation. Okay, so it is correct,
nothing changing. Okay? So you can see vx is difference
in between v1 minus v4. The loop number two,
which is this loop. You can see V3 minus V4, V3 minus V4, three vx. By applying the slope is three minus V4 equal to three vx, vx. So if you take this one to the other side and this
one to the outside, you will have the same equation. Or by applying that gives
me negative three plus v, v x plus V4 equal to Z, which is the same equation. So in the end, you can see
there are different muscles. All of them will lead
to the same solution. Nothing changed. All of them will lead
to the same answer. You can see we had, in the previous here, we had this equation number one. We had this equation number two. And we had here this
equation number three. And this equation number four. We have four equations
with four variables. Now, I would like to
reduce this equations. So I will use this one, which is V1 equal to 20 plus V2. Or you can say V2 is equal
to V1 -20, whatever it is. Then you take this equation and substitute in number four. In number one, and
substitute it in number two, we will have only
three equations. So as you can see, V2 e.g. equal to V1 -20. Okay? So when we take this
equation and substitute it in number one, number three, and number four, you will
have only three equations, which is this one, this one, this one with three variables, v1, V3, and V4. Now what I'm going to do, I'm yoga and I'm going to use the Chrome or Chrome or
message will be like this. We will form, we have three
variables or three unknowns. V1, v3, v4, as you can
see, equal to 084840. And then we will
have three columns, 366, negative one, negative one, negative five, negative two, negative two, negative 16. Okay? So we have this matrix. Now, if you have
a MATLAB program, if you will know about MATLAB, you can use assembly
solver this, or by using the Cramer method. How can I get this? We need forest to
get the three delta. So all the four delta
we have delta itself. We have delta one delta
three, delta four. Delta is the coefficient
matrix determinant, the determinant of the
coefficient matrix. So this is a coefficient matrix. So determinant of a
is the same as it is. We get its determinant, it will give us negative 80. Then we get the
determinant for V1. Okay? So how can I do this? Simply takes a Skolem and substitute it in instead
of the forest to column. So it will be 08040
and the rest as it is 0840 and the rest as it is. Then if I would
like delta three, which are representing V3, we will simply take this
column and substitute it here. So it will be this as it is, this one as it is, and this one will be 084. Then the last one before, we will take this column
and substitute it here and the first
two column as it is. So we will have all of our
Delta now in order to get V1, V2, V3, v1, v3, v4. It will be delta y over delta, delta, delta, delta four
over delta, and so on. So we get all of the
voltages v1, v3, and v4. And finally to get v2, it will be V1 -20. So it takes us one
subtracted when t will get the voltage. So in this lesson,
we discussed or had some solvent examples
on the modal analysis. And how can we apply this
to the electric circuit?
24. Mesh Analysis with No Current Source: Hi, and welcome everyone to another lesson in our course
for electric circuits. In this lesson, we will discuss another
method of analysis, which is called
the mesh analysis. So the mesh analysis provides another general procedure
for analyzing the circuits. In this case, we use
something which is called the mesh currents as the
circuit variables. Using the mesh
currents instead of the element currents
as circuit variables. It is convenient and will
reduce the number of equations that must be
solved simultaneously. And what is exactly
a mesh is a loop that does not contain any
other loops within it. Okay, so let's
understand this ID. So if you look at this circuit, this one, if you remember, we discussed in the
previous section of our course the concept of loop. We've talked about
examining golf loop, right? We said that a loop is any clause the palace
inside the circuit. So if you look at this one, we have this part. This one is
considered as a loop. So this part is first loop. And if you look at
this part here, you will find that we
have another loop. So what do we do exactly
in the mesh analysis? We assume that in each of these lobes we have a
current that flows. So e.g. we assumed, we assumed either Karen's, we assume currents in clockwise direction or anticlockwise direction,
as you would like. Usually use what you
will find that we usually assume all of the currents in the
clockwise direction, okay? For Loop Number one, we assume that there is a
current i1 flowing here. And the for loop number two, we assumed a current
I2 flowing here. Okay? Then, now what are
we going to do? We are going to
apply KVL here and another KVL here to
obtain current I1 and i2. And from these currents, we can get any thing
we would like. Okay? Okay, So if you remember
in the nodal analysis, nodal analysis we used, we applied all we
assumed each node, node number one,
node number two. And so on, each one, what did we do? We apply KCL, KVL, and KCL here, and so on. Here in the mesh
analysis, we have loops. So we have log number one, loop one, we have loop two. And what are we going
to do in each one we apply KVL, KVL. So what you will learn is that nodal analysis
is based on KCL. We do KCL several times. That mesh analysis is
based on Zach KVL. We do give you several times. Okay. Let's delete this. Uh-huh. Okay. So that in the first case, we will discuss mesh analysis
without any current source. If you remember, in
the nodal analysis, we discussed nodal analysis
without any voltage source. And then we discussed
the nodal analysis with a voltage source in which
we had supernode, right? In this case, we will discuss mesh analysis without
a current source. Then we will talk about mesh analysis always
a current source. Okay? So in the first case
you can see all we have here is voltage sources. So what are we going to do? We are going to apply
KVL in each loop. Okay? So the first step to
apply mesh analysis, we said number one, we assume a current
in each loop, you can see i1 and i2 assign mesh currents
to the n measures. What does that mean? Means loop. Okay? Then apply KVL to
each of the n mesh. So we will apply KVL
here and KVL here. And then you will have
several equations which you will solve to get i1, i2 until I N. Okay? So first we will apply
KVL in this loop. So how can I do this? Exhaust really already
concentrated with me in order to understand how can I
apply mesh analysis. Okay? So similar to any normal KVL. So you can see, I will go like this in the clockwise direction. So I go like this. Negative V1 exists. Then I will go like this. All you want flowing through R1, so it will be plus one or one. Then it will flow
like this, okay? Then we have all
three, remember, all three, so we say plus R3. Multiply it by you what? The current flowing through it. Okay? So if you look
carefully here, look carefully here that
we have one like this. This is one. E1 and i2 is flowing
in this loop. So we have I2 flowing like this. So I ask you now, I'm doing KVL in this direction. So what is the
current flowing in this downward through R3? The current flowing
is I1 minus I2. Now why is this? Because I1 in the
same direction of the KVL and i2 is always
a two KVL by action. And all of this
will be equal to Z, which is this first equation. Okay? So let's delete all of this. Then we will apply mesh
analysis to the second row. So you can see we have
this clockwise to i2, i2 flowing through R2. So i2, r2 then
flowing like this, meeting a forest is
a positive sign. So it will be plus V2
going down like this. And the 2 h three plus
R3 multiplied by what? You can see, all three, what is the direction of
the loop is clockwise. So the current I am
talking about flowing like this, i2, okay? But you see that I1 is always it to us,
RUN going downward. So it will be I2, which is flowing in the
same direction minus I1 equal to zero. Okay? So you can see i2, r2, V2, and also i2 minus
i1 equals to z. Okay? So the most important thing here is that when I'm using I1, I1 like this, I
always say to us, so it will be I1 minus I2. When we're using i2, it will be i2 minus i1. Okay? Okay. Now what are we going to do? We have all of the values
except the i1 and i2. I1 and i2 are the unknowns
in this equation. So we have two equations
with two unknowns. Solving this
simultaneous equation, we can get I1 and I2. Or by applying zach
Kramer method, we will put them in the
form of a matrix i1, i2, V1 minus V2, which is the value
of the voltage. Both this to the other side and the boards is
towards the other side. You will have V1 minus V2. And the boat this in
the form of AX plus BY equals to e and c x
plus d y equal to f, as we did before. E and F is this values, and x is i1 and i2 a
and the b coefficient. So if you remember the same
idea which we have done in the Cramer method
in the nodal analysis. Then what are we going to do? After getting I1 and I2? We obtained I1 and I2. Now e.g. I. Need current I1. Current I1 is different
from this one. R is small, this one is capital. So let's say, I
would like all U1. How can I get it simply, you can see I1 is the current
flowing in this loop. Are you unlike this one small. So you can see that I1 is in the same direction
of I1 capital. So from here we can
get I1 is equal to I1. Let's look at this current i2, i2 flowing like this. I too small. So i2 small is similar
to i2 capital like this. Okay? Now, the last one which
will help you understand, you can see I3 flowing
downward what I3 capital. You can see that this current, it will be equal to we
have i2 flowing like this. And we have I1
flowing like this. So I3 is in the same
direction of what? Of I1. I3 will be equal to I1 minus I2. Why? Because I want in the same direction or
E2 is opposite to us. So they're subtraction will
give us three required. Okay, so in the next
lesson we will have some solvent examples on the mesh analysis without
any current source.
25. Solved Examples 3: Hey everyone, In this
lesson we would like to get or have some
solvent examples, owns a mesh analysis without, with a current source. So you can see in this
figure we have how many we would like to
get as a branch currents or E1 or E2 and E3 how by
using the mesh analysis. So you can see we have
assumed current I1 in this loop and the
current i2 in this loop. Okay? So what are we going to do? We are going to apply KVL in H0. Okay? So let's start with our u1. So our loop like this, i1 flowing lungs, this
meeting negative 15, negative 15 plus following
glycolysis through 5 ω. So it will be plus five
multiplied by I1 small. Okay? Forget all of these
currents we are talking about only the
mesh currents, okay? Okay, then flowing
lines us through 10 ω, so it will be plus. Then you can see current flowing downward,
i2 flowing upwards. So it will be all y1 minus y2. And then flowing like this, we will meet all stiff ten
plus ten equal to zero. Okay? So this is the
first equation here. Negative 15 plus five, I1 plus I1 minus I2
plus ten equal to z. By simplifying this, we will
have the first equation. Okay, so let's
delete all of this. Then we are going to apply KVL
again for the second loop. So if you look at the
second lobe here, I2. So if we look at I2 like this, starting like this, I
two multiplied by six. So six I2, remember
I, too small. This one is not important for us who are talking
about this one. Okay. Then I to flow is exists
through the forearm. So it will be plus for i2
flowing legs as negative ten. Negative ten, then flowing
like this rose at 10 ω. So it will be plus then E2. E1 is always it to us. We are moving like this
item opposite to us. Negative one equal to
06 i2 for i2 plus ten, i2 minus i1 minus
ten equal to zero. So now we have two equations here by solving this equation. So we will get the value
of all U1 equal to one and pair i2 will be also
equal to one and bear, okay. So if we look at our E1, E1 is similar to how
you want it small. So I1 required,
which is a branch, the current I1 capital will be equal to I1
equal one, and pair. I2 is in the same
direction of i2 small. So i2 capital will be
equal to i2 small, equal to one and bear. Okay. What about I3? I3 flowing like this? One in the same direction, I two opposite to it. So i3 will be equal to
I1 small minus i2 small, which is equal to z. Okay? Okay, so this is
a forest example. Second example, we
have this system. We would like to get
the current node, this current inside
this circuit, we have how many loops
you can see 12.3. So in each of these slopes, we have a certain mesh current. So in order to get the current, I know do we need how many, how many KVL we need? Three KVL. We have i1, i2, i3, which is three unknowns, means we need three. Okay? So let's
start by this one. So we have like this I1 moving like this
clockwise, negative 24. Then flowing like this. Suppose that 10 ω, so it will
be plus ten on flowing like this are U1 minus
I2, I1 minus I2. Again, why? Because i2 lie exists. And i1, which is a loop we are talking
about, is like this. So ten on I1 minus I2. Then flowing like this, you can see I3 always it to us. So toward the blas, U1 minus I3, I1 minus I2, I3 equals to z. Lying this first equation. Then second equation, i2, i2 like this. So it will be 24. O2. Going live exists. We have the four
ohm plus for E2, which is a loop we
are talking about, and I3 opposite to us. So it will be minus three. Then going like this
through the 10 ω plus ten multiplied also I2. You can see in this
resistor we have our one opposite to
us, i2 like this, and I one is opposite to it, so it will be negative I1
equal to zero. Like this. As you can see. Last equation is three
plus four I naught, okay, remember this is a
voltage source. Plus four. I know it's in bone
like this. We have 12. So it will be plus 12 is three, which is our loop. Minus I1, I1 always
it to us minus y1. Then going like this, we are flowing like this. So it will be plus
four or a three. Then we have through
this resistor, we have i2 opposite to us, so it will be minus
I2 equal to zero. Like this same equation. However, you can
see our unknowns, I1, I2, I3, I1, i2, i3. But you can see we
have our node here, so we need to remove
this I naught and the, make it i1 and i2 and I3. So I node itself, you can see I node like this. We have all E1 in
the same direction, i2 all visit to it. So what does this mean? It means I naught will be equal to one which is in
the same direction, or U1 minus one which
is opposite to it, i2. So I naught, we will take
this one, n substituted here. As you can see, we have
now three equations. We have 12.3. So we have i1, i2, i3 equal to something i1, i2 iso equal to something i1, i2 equals equal to something. So how can I solve
these three equations? You have several methods. One of them is using
the Cramer method. So we will put them in the form of a matrix as we
learned before. Like this Zach coefficient
matrix equal to something. Then we will get as S1 is that determinant of this
one, which is delta. Delta is a determinant
of this part. Then we will get delta one, which is taking this column
and replace as a forest one. So it will be 1200 and the
others will be the same. Two will get us delta one. In order to get delta two, we will take this colon and
replaced as a second one. In order to get delta three
will take this column and replace them by getting the determinant of
the three deltas, we will get delta, delta one, delta two,
and delta three. Then i1, i2, i3. It
will be like this. I1, i2, i3 equal to
delta one over delta, delta two over delta, delta three words that
we have our currents, I1 itself, RE1 capital
will be the same value. Sorry, we need I-naught, okay? We need, I know we said
that I naught is equal to I1 minus I2, I1 minus I2. So we will take this
value and subtract it from this way, like this. So we'll get our current
equal to 1.5 and bear. So in this lesson we
had some soul with examples on the mesh analysis.
26. Mesh Analysis with a Current Source: Hey everyone, In
this lesson we will talk about with
the mesh analysis, but with the presence
of a current source. So as you can see
in this figure, we have this loop and this one. But as you can see, the second loop contains
a current source. So how can I deal with
something like this? You'll find is that mesh
analysis much easier in this case as it will reduce
number of variables. Now, as you can see, when a current source exists
only in one mesh. So as you can see in
this loop, this one, there is only one current source or a current source
existing in just one mesh. What does It, does it mean? It means that this cancels
is not between two measures. So as you can see, five and bear only existing in this loop. So what does this mean? It means that currently itself will be equal
to the current source, however, with a negative sign. Okay, so what does this mean? Okay, so as you can see, if we look at this loop, this loop contains
a current, i2. I2 is the current
flowing in this loop. So if you look at it here, we have i2 flowing like this. This i2 is the current which is flowing through 3M pair as Rosa 3 ω and at same time
flowing here, right? So poly logic i2 flowing here. However, the current is force
it to be five and payers. So what does this mean? We have a current five Umberto, going like this, and the
current assumed like this. So what does this
means? Means i2 is equal to five and bear
but with a negative sign. Okay? So as you can see, i2
equals negative five. Now why is this? Because you can see that i2
is the current flowing here. And we have five
and bear which is a current flowing also serosa 3 ω. So biologic or E2 is equal
to negative five years. Okay? So how can I apply
KVL simply like this? So first step, we have
the first equation I to equal negative five
and bear first equation. Second equation
came from this KVL, which is like this,
negative ten. Then I1 multiplied by four. Then we have i1
and i2 like this. So it will be 6 ω multiplied by I1 minus I2 equal to zero. And we already obtained the i2. So you can see that
the mesh analysis is much easier here. We didn't have to
do any KVL here. So in the end we will
have equal to zero. So the current will be equal
to negative two and bears. So in the second case, when we have a
mesh analysis with a current source about
between two measures. So when a current source
existing between two measures, in this case, we will create
a super mesh between them. Similar to what? Similar to the supernode. So if you remember when we had a voltage source in
the nodal analysis, we all form it as super
node here in the same idea, when we have a current
source between two meshes, we will form a supernode. So as you can see
that here, e.g. if you look at this figure, we have six and bear
flowing like this. So the current flowing here, we already know that this current is equal to six and bear poly logic for this branch is false it to B6 and
Beta equal to the supply. And from there, what we learned from the
mission analysis, that current or one like this, and I2 like this. So the current flowing here will be equal
to j is only e.g. equal to i2 minus i1, right? Because i2 is in the
same direction of this current source and
I1 is opposite to it. So we already know that I
is equal to six ampere. So we have a very
easy equation is that six ampere equal to I2 minus I1. So this is considered
as equation number one. Equation number one. Okay? So as you can see here, Here's this one similar to
by applying KCL or pi logic. You can see that from this, we have six and Beta
equal to i2 minus i1. From here currently exists i2 in the same direction
but opposite to it. So the result on the current, which is six ampere, will be i2 minus i1. That is the first equation. Second one can be obtained
by using a supermesh. So now we can not do a KVL here, and we can not do
a KVL here. Why? Because we already obtained an equation here
for the current. So what are we going to do? We are going to combine these
two loops as one big loop. So as you can see,
this will be our new. As if this branch does
not exist like this. And each one with its own cans. So what do I mean by this? Let's start by doing. This is called, this big lobe
is called the supermesh. So when we apply KVL to the
outer loop or the big loop, you can see like this,
negative 2020 volt. Then going like this, we have current flowing
through six ohm. What is the current or E1? So it will be plus six I1. Then flowing like this. What is the current
flowing shows that 10 ω is i2 plus i2. Then why is this four ohm
plus four equal to zero? As you can see here. Okay? So what did we do
exactly as if we combine these two loops
into one big loop, or we combine the two
KVL into one big caveat. So remember that TV is applied to any loop,
any closed loop. So this one is a loop. This one is a loop and also
the outer one is a loop. So since we have an
element between them, so we form a supermesh combining these two
loops together. And of course,
each element has a current flowing through
it, or E1, e.g. here, and here, I2. So as if we combine these
two key, we're now finally, when you have these
two equations, you will get value of I1 and I2. So let's go on to
have some solvent, the examples on the mesh
analysis whose account source, to understand the idea.
27. Solved Example 4: Hey everyone, In this lesson, we will like to have
a solvent example on dimensional analysis
with a current source. So as you can see here, we need the current I1, I2, I3, and I4 is this
loop currents. Okay? Okay. So if we look at this circuit, we have how many loops? 123.44 lobes with current I1, I2, I3, and I4. Now, if you look at this loop, does not contain
any voltage source, any current source, this loop does not contain
any current source. So we can apply
here KVL normally. Okay? Okay. Now what about this
loop? This loop, e.g. this loop and this one consisting of all contains a
current source between them. So you can see that we have this current source between
this slope and this tool. So what are we going to do? We are going to get an
equation from here. You can see that three
naught equal to, which is the current
going downward, equal to i2 minus i3 minus I3. And if we look at this one, at this current source, this current source five and bear equal to Zach guarantee
needs a direction I2, clockwise in the same direction, or E2 minus one which is
opposite to it or you want. So you can see that
we obtained equation one and equation two. How did we obtain them? We obtain them using their current source
between the two loops. Okay, So we have
already two equations. Rubber presenting
this current source, a current like this, and another one like this. It's a complaint. Or their summation
gives us three. I naught and z are
some measuring gives us five and bear here. So let's first see this too. You can see that
i2, i2 minus i3. So from this equation, or i2 minus i3 equal to negative three
I4, write this equation. Now, where did we get this one? You can see i2 minus i3, similar to here, equal
to three or inode. Now i-node itself, you can see I naught flowing like this, and i4 flowing like this. So I naught equal
to negative I4. So you can see is
that we can take this negative I4 and
substituted here. It will be negative
three or a four. As you can see, this
equation representing the equation of the due to the presence of
this current source. The second one, which
is the five and bear, which is i2 minus i1. You can see i2 minus i1
gives us the five and bear. So we already have two
equations representing what, representing the presence
of this current sources. Okay? So if you remember, we have how many variables
or four variables? So we need four equations. Four equations. I already obtained the two
from this one and this one. Okay? Now let's continue. Then we have another
equation coming from, coming from this loop. If we apply KVL here, you can see plus ten volt. Then going in like this, we have IL-4 and
IL-13 going down old. So it will be eight on
multiplied by I4 minus I3. Okay? Then going like this, two multiplied by I4
to I4 equal to Z. So which one of these here? This equation, where here's
this equation to I4, eight by four minus I3 plus ten. Okay, so we have this equation. So as you can see, we have how many equations we obtained. We obtained the one. We obtained another
one here, two. And the third one here. One here. This one, as you can see, this one
similar to this one. Okay, nothing can
change it here. So we have now three equations and we need one final equation. Now, where do we are going to get this from the supermesh? Now, as you can see, between this one and this one, there is a current
source, right? So we have to apply a KVL
big like this supermesh. However, if you look at
this, this two loops, you can see there is
another current source, which means that we
have to combine these two together like this. So by combining this
and combining this, we will have a big mission, this big mesh, the outer loop. Now, why did we
take this big one? Because we have here
a current source. So as if this branch does
not exist and we have here another cancels as if this
branch does not exist. That's why supermesh
will be this outer loop. So let's just start writing it. So as you can see, that's
a start from here. You can see here all you
want going like this. So it will be to u1. Then moving downward here for all I3 flowing through four ohm. So it will be plus
or R is three. Then going here through this, it all we have current
flowing like this, which is i3 minus i4, which is the current
opposite to it. Then we go like this
and six, so on. So it will be plus six
multiplied by what current i2, i2 is the one flowing
here, equal to zero. So if we look at here, we
have 21438/3 minus i4, i2 equal to z. So now we have how
many equations? So we have four equations. So how can we solve with them? We can simply reduce
this equations. How can we do this simply? We have I2 equal I1 plus five. So I can take this and
substitute it here. Take this one and
substitute it here. Here. We don't have
i2, we don't have i2. We will have here n equation
of I2 and I1, I3, I4. And do we have here
another equation so we can take this one, we can take out two
and substitute here. So we will have I1, I3, I4, I1, I3, I4, and we have I4. And then we take these three equations after
substituting with this one, we will have a metrics
using Chrome or method. We can get I3, I4, I1, then we will get I2 or e.g. you can simply keep
substituting in the equations to
get I1, I2, and I3. So Paul is solving these four equations in
any method you will like. You will get I1, I2, I3, and I4 with these values. Now as you can see, when IE1 is equal
to negative 7.5, what does this mean? If you look at here, one is the one which is
flowing through the 2 ω. So I1, I assume that
R1 is like this. Movings rules at 2 ω
in this direction. But since it is negative, it means that the
correct answer is that the current 7.5 and bear is moving opposite to
the one which I assumed. So the current here, 7.5 ampere is flowing
in this direction, opposite to that original
proposed one. Okay? So in this lesson we
had a solvent example, another Soviet example
on the mesh analysis. Don't worry, we are going to
apply nodal analysis, again. Mesh analysis, again in many,
many circuits theorems. We are going to apply
this in superposition in North and serum in seven in serum in the next
section of the course. Also, we are going to apply
this to the AC circuits. So we will have
many, many examples. You will understand
how can we apply mesh and node analysis
in different circuits.
28. Nodal vs Mesh Analysis: Hey everyone, In this lesson
we would like to give zoster a small comparison between nodal analysis and
mesh analysis. When should I use nodal analysis and when
should I use machine hours? Usually it doesn't matter if you use mesh analysis
or modal analysis. It will give you
the same answer. Okay? However, there are some cases
in which nodal analysis may be easier or mesh
analysis may be easier. Okay? So e.g. if we have a
network consisting of several series
connected elements or voltages sources or super meshes are more suitable
for mesh analysis. If we have a network with
parallel connected elements, current sources, or supernodes, they are more suitable
for nodal analysis. Also, if you have a circuit
with, for your nodes, then measures the nodes in the circuit are much
smaller than the measures, then we are going to
use nodal analysis. However, if we have a
few are mentioned Senza, nodes in the circuit, then it is better to use a mesh analysis depends
on the circuit itself. So the key is to select
the muscles that will lead to smaller number of equations that smaller the equations
that are less than Alice's and much easier to
get the solution. Node voltages are required, then we will apply
nodal analysis. If the branch or the mesh currents are the
one which requires it, of course it's better to
apply the mesh analysis. Now also one
important note about, note about the mesh analysis
and the modal analysis. We have some circuits
called Zap learner circuit. To the circuit, you can see two, d, x and y. There is also an
unplanned or circuit which is in the three D. You can see x and y
ends at 3D circuit. So we have a planner circuit, which is a 2D circuit, and the other one which is a 3D. You have to understand that
for non planar circuits, for non-blank are circuits
that nodal analysis is the only options because mesh analysis only applies
to plan on network. So mesh analysis, we can
use it in this network. However, we cannot use
it in a non planners or mesh analysis is used
for two D networks. For 3D networks, we can
not use mesh analysis. We can only apply nodal analysis for the
non planner network. Okay? This is really, really important in real life. Also will find that
know that analysis is easy to program in computer. So what we can learn
from this lesson, we learned is that
mesh analysis cannot be used for non-polar NOR
circuits or networks. However, nodal analysis
can be used for both plan on networks and
the non planar network. Also finds that when we
have an electric circuit, if the number of nodes is
less than number of measures, then we use nodal analysis. If there are a
number of measures, is less than number of
nodes in the circuit, then we use nodal analysis. So fewer nodes in the
circuit less than missional. And then Senza measures, we use nodal analysis. Measures lower than nodes is n, we use nodal analysis. Okay?
29. Application Transistor Circuit with a Solved Example: Hey everyone, In this
lesson we will have an application on the mesh
analysis and nodal analysis. So the application is simply here we have a DC transistor. So what does this
circuit elements do? We have an element in electric circuit called
the transistors? They are used in amplification or switching of electric
circuits. So e.g. as you can see in this figure, we have a BJT or known
as the Bipolar Junction, or abbreviated as BGT. Use the for amplification and switching of electric
circuits, e.g. you will find that BJT, e.g. it is used in power electronic
circuits as a switch. So what does this mean? It means e.g. in instead of having a manual
switch like this. For the circuit, circuit, open circuit like this. And when we close the switch, it will be a closed circuit. This thing cannot
be done manually. Okay? Why? Because the power
electronics circuits are really, really fast. So I need a very
fast switching of electric circuits to
achieve a certain outcome. So we cannot do this
using manually switches. We have to use
transistors or e.g. we need to use something like a BJT or something like that. Most fits or diodes, all of these circuit elements are used in the power
electronic circuit. So if you join my own course
for power electronics, you will learn about this. Okay? So anyway, these types of transistors are used to
switch electric sockets. That is the most
important function that are used in
power electronics. Okay, in order to turn off
and on an electric shock. So what you can see, e.g. for zippy, It's consisting of three points or three terminals. To be more specific, you can
say three terminals, 12.3. These terminals are first
one is called the Bayes. Second one, collector
and emitter, base collector and emitter. Each one of these has
assaulting current, base collector and
emitter transistor. We find that the current
of the image over here is equal to the summation of oil-based multiplied
by a collector. And also the collector
current is equal to the base current multiplied
by a certain constant beat. Okay? Okay. Second, the things that we will notice here is that there is a potential
difference between B and D called VBE. And we have a
potential difference between C and D called VCE. And we have also a potential difference between
C and the B called Vbc. Okay? Or VCB to be, to be more specific, VCB coming from C to P. Okay, so we have
three voltages. Okay? So z is sports have a potential
difference between them. Now, how can I analyze
a circuit like this? We can analyze it using
the mesh analysis. Okay? So as you can see, if
you look at the circuit, we have this voltage source. We have a resistor, and we have our transistor
here consisting of current I, base, collector, and emitter. You can see that there is a potential difference here, VBE, which is the voltage
between base and emitter. Here we have the
metal and we have a potential difference between collector and emitter
called the VCE, or in this figure, V output. Then we have 100 ω and
another voltage is 6 v. Now what we would
like to get from this problem is that
we need to find the current IEP or a collector
V output in this circuit. Given that p.sit take 115 and
VBE equal 0.7 voltage 0.7. So how can we analyze
a transistor? We apply KVL, knowing
as that input KVL or input mesh loop and
output give you out. Okay? So let's start by the input KVL. You can see KVL like this,
clockwise, negative four. And current. What is the current
flowing here? So it will be plus p
multiplied by 20 kilo ohm. Then like this, plus
VB equal to zero. And VB0 is given
as 0.7 v 0.7 volt. So from here we can get our API, as you can see, equal
to 165 micro and bear. Okay. Now the second part is
that we will need IP, we need ICIC assembly. What equal to or
equal electoral? As we seen in the
previous slide, beta multiplied by 0. Ebay's sorting gain
multiplied by obeys. So to me means a 50 multiplied
by or a base like this. So it will be equal to
8.25, mainly compare. Okay? Last requirement is V output. So how can I get the
output voltage or VCE by doing an hour
would give you out. So I will give you
negative V output. And the current flowing
here is opposite to IC, so it will be negative i c multiplied by the
resistor hundred. Then long lines plus
six volt equal to zero. And the IC is equal to
this way, like this. Negative V output, negative 100 IC plus six equal to zero. And we have IC. So you
get V out as 5.175 volt. Okay? So as you can see here from this example is that our
voltage like this KVL. So we have, let's
say any current, let's say the current
I multiplied by 100. And this I equal
to negative y, z. So I say negative I c multiplied by the
resistor, hundred ohms. Okay, So in this lesson we had
a soul with the example on an application which is a transistor for more
APA was at transistors. And how can we use them in power electronics circuits
to form a rectifiers, AC shoppers, DC
shoppers, inverters. You need to go to our course
for power electronics.
30. Introduction to Circuit Theorems: Hi, and welcome
everyone to our section for circuit theorems
in electric circuits. So in this section we will
discuss the different types of circuit theorems that we can
use to analyze our circuit. So in the previous section, we discussed Kirchoff's
laws such as the KVL, KCL, in order to obtain the voltage and current
in our seconds. And in addition
to learning about the mesh and the nodal analysis to obtain the voltage
and the current. Now, however, these
methods can be useful in symbols circuits and
simple and easy circuits. We can use the KVL, KCL mesh and the model answers. But what if we have complex
and the large circuits? And the large and
complex circuits, we need to use other types of theorems which we are going
to discuss in this section, such as the
superposition theorem, source transformation
that seven and serum, the Norton's theorem in order
to analyze our circuits. Okay, So in this section, we will discuss
the superposition, the source transformation
seven and and Norton's here. This CRMs are will
help us to analyse our electric circuits easier. And in order to simplify
large circuits.
31. Superposition Theorem: So in the first lesson, we will discuss
superpositions here. So what does the
superposition theorem or what does it mean? So if we have a
circuit like this one, circuit like this one has two or more independent
source, okay? We have two or more than
two independent source, independent, such
as the circuit. We have six volt
independent source, and we have three and
bear independent source. So in order to find the
z-value, for example, for the voltage here
at this point between the resistance four ohm load to find the voltage
or the current here, or in any part of the circuit. One of the methods
which we have used is that the nodal analysis or mesh analysis by applying
give you all here and another KVL here,
or nodal analysis. We can obtain the voltage or the current we need
inside this circuit. However, there is
another method. Another way is to deter
mine or determine the contribution of each independent source
to the variables. So we need to find the contribution of each
independent source, independent, not
independent, but independent variable
and then adds them up. So what does this mean? As an example, we need
the voltage here, right? So the voltage here, we can say that the voltage
V is equal to summation of two voltages, V1 plus v2. Okay? Now what does this mean? V1 and V2. V1 is the contribution of the six volt source and V2 is the contribution of
the three and bear source. So by adding this contributions
of these two sources, we will get the voltage needed. For example, if we
need the current, then the current will be
our u will be equal to o, u one plus n2. Which means the contribution of the first voltage source and the contribution
of the three and bear. Okay. So what if we have three
sources and it will be V1, V2, V3, i1, i2, i3, and so on. So what we are going to do in the superposition
theorem is that we will take the
contribution of each source. We will understand
how can we do this? The superposition
principle states that the voltage needed, such as v here, or does the current
needed throw an element current
through here or here, or whatever it is
inside our circuit. In a linear circuit, linear circuit means it's
consisting of linear elements, such as the resistance
inductance capacitor. It states that the
voltage or current is equal to the algebraic sum. Algebraic sum, which is
like this plus sum of the voltages or the current
through the element due to each independent
source acting alone. Okay? So such as V1, V2, V3, and so on. Now, the principle of
superposition helps us to analyse a linear circuit with more than one independent source by getting is a contribution
of each independent source. Now, let's have some notes about the superposition theorem
before we understand the steps. In order to find the
contribution of each source, then we use or consider one
independent source at a time. Why, while others are
independent source, all the other independent
sources or turn it off. So as an example, if I need a contribution of the six volt, then we deactivate
this free and bear. Okay. So we are using, we are analyzing our circuit
one source at a time, one independent source
at times the six volt. And we deactivate the
three and then we will, if we would like to find the contribution of
the three and bear, then we deactivate
the 6-volt and analyze our circuit with
the three and bear only. Okay? So we use one source at a time. For example, if we have a
three sources, for example, then we will deactivate
the other sources. And the key one source. Now the question is, how can we deactivate assaults? Now, the deactivation of the voltage source
is by making it, the voltage is 0 or making
it a short circuit. So as an example, if we are using the three
embarrassing weekend, we will deactivate the
six volt by making this a short circuit as
if it does not exist. Okay? And if you are dealing
with a three and bear, we can make the current
source is 0 and Ben, or an open circuit like this, remove this by making
this point open circuit. So again, that voltage source is deactivated by making it a short-circuit or making
it so voltage 0, it is the same principle. And each current source we may want to make this
current equal 0. So by making it an open circuit, this helps us to have a very simple and more
manageable circuit. Now, remember that the dependent
sources are left intact. We don't do anything to
that dependent sources. Why? Because they are controlled by the
circuit variables. We don't deactivate them. We can deactivate
them because they are dependent on other elements
inside our circuit. So what we deactivate with the, activate only the
independent sources. Independent sources. Now is analyzing circuits
using superposition is that it can involve the more work, sometimes the mesh doing the mesh and the nodal
analysis is easier. And other times a
superposition that makes the circuit more
easier to analyze. Now what are the steps of applying the
superposition theorem? Now the first step is
that we the activate or turn of all independent
source except to one source. So we have in the circuits
3M bear and six volt. So if we are considering
the six volt, then we deactivate the three and bear and keep only the six volt. Then we will find that require the voltage
or current due to this one source activated using the techniques such as MS, mesh analysis, nodal analysis, KVL, KCL, and so on. Okay? So we obtain the voltage here due to the contribution
of this one. Only. Then we do the reverse. We activate this one, end. Keep only the three and bear and the fight
against the voltage. Then after getting the
two voltages or currents, we add them algebraically
in order to find all the contribution of the independent sources
and define the finally, the value of the voltage. So let's go and have an example on the
superposition theorem.
32. Example 1 on Superposition Theorem: So example one, example one. Here we have the same
circuit which we have showing in the
previous lesson. We have use the
superposition theorem to find this voltage
inside our circuit. So we have two sources here. We have the six volt and
we have the 3M payer. So we will start to by
deactivating, for example, six volt and defines a
contribution of three and pair. Then deactivate
the three ambient and find contribution
of the six volt. So first sensor, we have two sources here
in this problem, we have six volt. We have this free and bear. It means that the
voltage needed, this voltage will
be the summation of two voltages, V1 and V2. Where V1 is the contribution
of the six volt supply. And V2 is the contribution of
the three and bear supply. Okay? So we will start by
getting the contribution. Getting the contribution of V1, obtain V1 or the contribution
of the six-volt supply. So in this case we are
considering this one only. So what we are going
to do is that we will deactivate the
three and bear. We said before, how can we
deactivate the three and bear by making it 0 and bear or
making it an open circuit. So it will be like this. So as you can see,
six volt, eight ohm, four ohm as it is, and the activation of
the street and bear making it an open
circuit like this. Then we need to find V1. So as you can see,
we have six volt, we have eight on, we have the fore arm. So we can get V1, V1 by KVL
or the voltage division, similar to each other. So by doing KVL, we have negative six volts, as we learned before. Negative 60 volt. And eight are u1 plus
v1, like this. Okay? So we have y are blind KVL, we have negative six volt. We can write it like this. Negative six volt by
exist negative six volt. Then we have 84 in series, which gives us 12 on. So it will be I1
multiplied by 12 volume. So it will be plus 121, Okay, equal to 0. So as you can see, this equation
is similar to this one. So it will give us, our y1 would be equal
to 0.5 and bear, which is a current
flowing here, are y1. Y1, v1 is equal to one
multiplied bys of four. So two will be like this. V1 for which is resistance and
multiplied by the current. And the current is 0.5. So it will give us two volt. This is v one. We would like to get
the contribution of the source or the
current source. So here we will deactivate the six volt and then
making it a 0 volt. So what does a 0 volt mean? It means short circuit. So we'll make this one as
short circuit like this. We have this circuit, we
have a three and bear. We have eight ohm and four ohm. So here you will find that
the current three and bear going gear to
the arm and forearm. Now, let's see as
the voltage here, V2 is equal to I3. Multiply advisor for all right, So what does the value of i3
from the current division, or is three is equal to the total current 3M bear
multiplied by the resistance, eight ohm divided by the total eight ohm divided
by the total four plus eight. So we will have a two. And this is from what, from current division, which we have discussed in
the previous lessons. So the voltage V2 is equal
to I3 multiplied by four, is three multiplied by four, which is eight volts. Okay? So now we have V2
and V1 equal to two volt. So here we will have
the total voltage required in this problem
will be equal to V1 plus V2 contribution of the first source plus contribution of the second
source gives us ten volt. So this value. If you do, for example, a mesh analysis like
this or nodal analysis, you will get the same
value of voltage. But as you can see, we have
simplified the circuit. When we do the superposition, you make the circuit
much easier to analyze.
33. Example 2 on Superposition Theorem: Now let's have another example. So use the superposition
theorem in order to find the current I
naught in this circuit. So as you can see,
the circuit has how many independent
and dependent sources we have in the circuit. We have two independent
sources is a four amp here. And the 20 volt
independent sources and one dependent source. So as you remember, the
superposition, deactivate what? Deactivate the independent
sources. Only. This one. We don't do anything to it because it is a
dependent source. We are concerned only with
the dependent sources. Okay? So let's start. So we have here in the circuit
and the dependent source, so we will leave it as it is. Then we have i naught, which is equal to
two currents are in a dash and the
i-node double dash. This is due to the
contribution of the first source
and second source. So as you can see,
is I naught dash is the contribution of the four
and Bayer source, this one. And I know double dash
is the contribution of the 20 volt source, this one. Okay? So we need to get
the contribution of four and pear and contribution
of the 20 volt alone. So we will start with
the dash four and bear. So in this case, we
deactivate the 20 volt, we make it 0 voltage. What does this mean?
It means that we will make it a short
circuit like this. Okay? So let's see. We have our
circuit, the same circuit. However, that 20 volt becomes a short-circuit as
you can see here. Okay? Now, what we would like to have, we would like to have all
this current flowing here. So we can obtain this current by doing mesh analysis like this, give you all here, KVL here
and another caveat like this. Okay? One important to note is
that as you can see that this source is five I naught. I naught is the current we need. So when we activate this supply, we have our dash. So this one will be five dash. Okay? Now let's start. So we have here in this
circuit we have three lobes, I1, I2, and I3. I1. As you can see,
the current I1, like this, is in the same
direction of the forearm bear. So in this loop we
have all U1 equal to four and bear all U1
equal four and bear. Now in the second loop, loop number two, this
loop, this loop, you will find that here we have negative negative
year to two elements. As you can see here, we
have the three ohm. Okay? We have to, we have one
arm and volume data. So let's delete all of
this first and type it. Okay? So let's see i2 here
to moving like this. So the first is apply, this one is a current
source voltage. Source voltage is also
not a current source. So we have negative five, negative five I naught dash. Then we move like this. We have two currents. We have I3 here. We have i2 minus i3
multiplied by one. So we have plus resistance
one ohm multiplied by I2 minus I3 flowing through
here to minus three. Then we have here a
non-zero current, I2 going here and
one going here. So we have plus three
multiplied by r2, i2 minus U1 equal to 0. So we have negative five, I know dash plus one, i2 minus i3 plus
three, i2 minus i1. Now let's see what we have here. We have negative five, negative five, I know dash. And we have i2, i2, so it will be plus
for i2, right? So we have plus for i2 and minus three and
minus three i1. However, we have another
element here that two ohm here. Don't forget about the two ohms. So we can say plus two i2. To Err into, okay, so we have one I2, I3, i2, i2 plus C plus
one gives us six i2. So we have here
six i2 and we have negative i3, negative ice ring. And we have one negative
31, negative 31. So all of this equal to 0
and we have I1 equal four, I'm bear, so we can
substitute this one here. Okay? So we have i2, i3, or dash. Now for the third loop, loop number three here, you can see that we have, if we do like this, Let's start from here. For example, we have four
ice-free for all three. Then we go like this. Then we will meet five ohm. So five ohms. And we have here I1. So it will be plus five
multiplied by three minus one. Then plus, we have here 1010, i3 minus i2, all
three minus two. Okay? Um, don't
get step-by-step. Okay? Then we have here plus
five I naught dash plus five or no dash gives us 0. Okay? So we have four I3, I3, I1, I3. So we have ten I3. So as you can see,
them ice cream. Then we have negative 51, negative five by one. Then we have negative
I2, negative two. Then we have plus
five dash, dash. Okay? So as you can see, we
have this two equations. Okay? So let's delete all of this. So we have these two equations and I1 equals four and bear. So we substitute
with foreign bear here and for Amber here. Okay? So when we substitute here, so we will have
three unknowns, I1, I2, I3, and I know dash. Okay? So as you can see here, I know dash is the current
flowing here, right? So I naught dash is equal to one minus I3 are one minus
I3 forms a mesh analysis. So are you one? Our dash here will
be one minus I3. Okay? This equation is similar
to this equation. Okay? So we have a relation
between dash and all three. So we can say from here, I3 is equal to four
minus I naught dash. Then we take this equation
of I3 and substituted here, I3 here, and I3 here. So we, in the end we will have
i2 and the dash, two dash. Then we solve these
two equations to obtain the value
of R naught dash. So by solving these
two equations with all of these values, we will have no dash
equals 52 over 17 and bad. Now, the second one, which is 20 volt, we need the contribution
of the 20 volt. So we will activate the
four unpaired by making it an open circuit like this. So four and bear became
an open circuit. So we have three own
51220 volt and forearm. So here I naught double dash. So I know double dash, double dash, double
dash, like this. Now what are we going to do? We need I know dash so we can use again the mesh analysis. Okay? Now, for loop number four, here, we can say is that here. If we go like this, like this, so we have to multiply by I4
and three multiplies by I4. So this tool will
give us five I4. And do we have here 10? So it will give us 646 I4. And that we have here like this, negative five I naught w dash, negative five I NOW dash. And we have our E5 going like this in the
opposite direction. So it will be negative
one multiplied by five, negative one multiplied by phi. The second loop, this is
similar as we did before. In this lobe we have four
exists for I5 and five I5. So we will have mine I5. And do we have
here is a one ohm, so it will be ten i5. So we have ten I5, okay? And we have negative 20
and the slope negative 20. Then we have like this plus
five dash, double dash. Then we have one ohm. I4 is opposite to us, so it will be negative I4. Now, I'll double dash is
equal to negative I5. I5 is like this. So all in all double
dash is obviously to I5. So I5 is equal to negative
i-node double dash. Okay? So in this three equations, we have I4, I5, I know w dash I4, I5, no double dash, I5
and R double dash. So we can, by using
these three equations, we can get the value of
r dash by solving them. So we will find that R double dash will be equal
to negative 60 over 17 and bear now as a
final value of current. So we have our inaudible dusk is a contribution of four and bear. And I know double
dash contribution of the 20 volt there some mission will give
us the total current. The total current I
naught will be this plus this one as negative eight
over 17, or negative 0.4706. And bear. Of course,
as you can see, instead of doing this, we could just do
mesh analysis, okay? We could just do it
from the beginning. Mesh analysis here
and here and here. Since we have done analysis
almost a three times in each of these
superposition problem. However, in this problem we request is that you have to do it using the superposition. But as you can see
here in this example. In this example you can
see we did a lot of work. However, if we exhausted
it, a mesh analysis, we could just get this value of current faster than
we did right now. Okay? That's why superposition. Sometimes it can be useful. Sometimes it can give
us more work to do. That's why is that
usage or selection of the theorem is important or very important and helpful in making that circuit much
easier or much more difficult.
34. Source Transformation Theorem: Hi and welcome everyone to another lesson in our course
for electric circuits. In this lesson, we will discuss another theorem inside
that, electric circuits. So which is a source
transformation theorem. Okay? So we would like
to understand what does a source
transformation mean. So first, as you remember in the previous section
of the course, we learned that the series
parallel combination and Y delta transformation
help us simplify circuits. Okay, So we use a series
parallel connections or the Caesars Palace
combination of the resistance in the future inductance and capacitance and
the y delta transformation to simplify the electric circuit or simplify the resistance, the inductance, or a capacitance
of an electric circuit. So here we are dealing with is the passive elements or the resistance inductance
and the capacitance. However, we would like to
understand how can we simplify an electric circuit by
changing our source? So the source of transformation
theorem deals with it changing there source itself from voltage to current assault, from voltage source to
current source and so on. So the source transformation is another tool for simplifying
the electric circuits. Now, how does it work? So the transformation
is the process of replacing our voltage source VS. This H&M is VS in series
with a resistor R by a current source IS better with a resistor R or vice verse. Okay? So we are changing our source from our voltage
source to our current source, or from a current source
to voltage source. So as you can see the circuit, we have our voltage source,
ideal voltage source. So it has a
resistance in series. And we have a non ideal
current source which is IIS, but with our
parallel resistance. Okay? So remember that
in Zao, non-ideal, non-ideal, non ideal sources. In the voltage we had, we had a resistance in series, and for the current, we had our resistance
in parallel. So if the source is ideal, then R will be equal to 0. And in that current source, R will be equal to infinity. Or an open circuit. This is in the case of the ideal voltage source and
the ideal current source. So what does social
transformation do as our social transformation? A change is a circuit, or the changes that circuit
from the voltage waveform to the current form or from the
current form to voltage for. So the question is,
how can we do this? So as you can see, simply the relation is very, very easy. You simply have
the voltage source equal to the current source
multiplied by the resistance. And the current
source is equal to the voltage source divided
by the resistance. Okay? So assembly, if we
have this circuit, V source and a resistance
in series, ok. So in both cases are here in the voltage source is similar to R in the current source. Okay? So this, our, uh, similar to this are, okay. Now second step is
that, for example, if I would like
to change it from the voltage source
to current source, first I will add the
current source like this. The current source. Now what is our value of
current source IS now what is, IS-IS assembly equal to the supply VS divided
by the resistance here. Oh, okay, So this is a
value of the current. So the current year
I S is equal to v. S is a supply of the voltage source divided
by the resistance. Now, let's say we would
like to do the reverse. We would like to change it from the current source
to voltage source. Assembly will take
this resistance and the voltage in
series with the source. The value of the source
is equal to the current multiplied by the resistance
IS multiplied by R. Okay? So these two circuits are
equivalent to each of us. Okay? So this is what's called
the source transformation, serum conversion from
voltage source to current source or from current
source to voltage source. To simplify our circuit. Now we have to understand
that this method or this source
transformation theorem is suitable for the dependent sources
and independent sources. Okay? So in this case we have
a dependent source. So we can use the
same methods, or S, or the current is
equal to V S over R. And V S is equal to, IS multiplied by r. Both of them are
similar to each other. Okay? So the voltage source
transformation, or the source
transformation theorem is available for or can be used in the conversion of dependent sources and
independent sources. Now we have to understand
that here that the r of the current source is directed towards the positive terminal
of the volt source. So as you can see
that, as you can see current going like this, right? Okay. So in this case we will have a voltage source pointing the
same direction plus minus. So the positive
terminal pointing to the current here is
the same direction here. Okay? So here we have a
current going like this. So this source should give
current in the same direction. Okay? You can see it's
a positive here, pointing to the same
direction of the RAS. Second is that the
source transformation is not possible when r equals 0. This is the case was an
ideal voltage source and current source
with r equal infinity. These two cannot be used.
In these two cases. We can not use source
transformation when r in this case of the current source is
equal to infinity. Or IS, is an ideal source. Or here r equals 0 when the
voltage source is ideal. In these two cases, we can not use source
transformation. Salsa transformation is
used only for practical, non-ideal voltage source
or current source. Okay? None idea. So the R here should have a value and odd here
should have a right? It cannot be here infinity or
cannot be here equal to 0. So let's start by
having an example on source transformation to
understand how can we use it.
35. Example 1 on Source Transformation Theorem: So in this lesson we
are going to have an example on the sol
transformation serum. So here use us also
transformation theorem to find v naught in the circuit. So as you can see
in this circuit, we have applied to all volt. We have a three arm to arm
three and bear four ohm. And we have here
at home and would like to find this voltage. Okay? So what we need is that
use source transformation, not KVL, KCL or
any other theorem. We need to use source
transformation. So here we have this role, converting the voltage
to current and the equations that we have here, our first step, as you can see, we have this part and
we have this part. Okay? So we have
three ohm series with a 12 volt and we have four ohm series parallel
to x3 and bear. So what we would like to
do is that we would like to convert using as a
source transformation, this current source, and
this voltage source. So how can we do this first? Let's take this part, for example, this one. So we have a three arm, then we have a 12-volt. So in this part we will have three Ohm battery to
a current source. The current source is
the same direction of the plus terminal. So it will be like this. Okay? So this part, we have a three arm parallel
to a current source. The value of the current
source is equal, IS, is equal to 0 supply
divided by resistance. So the supply is 12 a
volt divided by three. So 12 with whether by three
ohm gives us four there. Okay? So this
circuit representing the equivalent of that 12-volt
series with a three ohm. Then we started drawing
our eight ohm like this, which is a voltage V naught. And we have to form, exists as it is. Then you will see we have a
three and bear and four ohm. So we can do like this. We have resistance. The current source parallel to for almost become resistance, which is for all series
with a voltage source. Okay? So this voltage source is that bold step is the
same direction here. So it will be like this plus, minus because it is in the same direction of
the three unpaired. Now, what is the
value of the supply? Supply VS is equal to the current multiplied
by resistance. So 3M bear multiplied by the
four ohm gives us a 12 volt. 12 volt. Okay? So this is our circuit. So let's see. So we have that 12-volt
pointing downward, similar to the 3M pair series
with the same resistance, four ohm, four ohm and four ohm. Then the two ohm
will be as it is, the eight ohm as it is, then Czar for three
and bear will be converted to as a 12
volt and threonine, we convert 234 unpaired
for an pair and three ohm. And as you can see, the
current pointing upward, similar tools, the current going out of this supply
is at one volt. Okay? So now we converted this to
supplies into the circuit. So let's delete all of this. Okay? Now, what is the next step? We would like to simplify
our circuit more. Okay? So as you can see, we have the four ohm
series with that too. Okay? So this combination can be equal to
six, so on. Okay? So we have a 12-volt
series with a six so on. So we can convert this, this part into a current
source once more. So we have here a
current source like this pointing downwards
because the current, suppose the V here,
so it will be pointing downwards like this. Okay? And we'll have parallel to it a resistance of
four plus torches, the six ohm like this. Okay? So the current value
of current is equal to the voltage divided by the resistance 12 with
the wider by six, it gives us two and bear. We have a two amp arrow
pointing downward and six ohms. So let's see is a circuit. So we have the two and
bear pointing downwards. And six on eight
omega is this rearm. And for Ambien, now what
does an extra step? You can see that here in this
example we have the four unpaired pointing upward
providing current abort. And we have a two and pair
providing current downward. So this TO supplies are
opposite to each ours. Okay? So they're summation. We have four ampere upward and we have two
unpaired downward. So four minus two gives us two and pair pointing
upward. Okay? So it will be like this. So it will be two
amp pointing upward. That is a forest topology. It almost be as it is. Then we have here as three arm
battery to six ohm, right? So as you can see, the
first to note here and here as the first mode
here, similar to here. So six ohm is parallel
to the screen. Now what is their equivalent? We have the equivalent of z. M is equal to six, so multiply by three, as we learned before, divided by the summation. Six multiplied by three, divided by six to plus three. So we have 18 divided by nine, which will give us two. So the toe arm is
the equivalent of the three ohm and six ohm. So as you can see,
we have two oh, this representing
the equivalent of this two circuits to resistance. So now we simplify using
the source transformation. We simplified our circuit
from this larger circuit, not very large, but large with respected to
the final form here. Okay? So what does the next step? The next step is
that you can find the voltage here by
doing two methods. The first method is that
you can convert that to an bear and two ohm and
two like this. Okay? Eight ohm, keep it as it is. Then you will find here
two ampere and to own. You can convert it to
a resistance to own. And a current supply
can be converted into a voltage source
plus minus I exist. And the value of the voltage is equal to two multiplied by two, which is four volts. Four volts. So we need the voltage across
the eight or so. The voltage here of the
eight ohm is equal to the total voltage multiplied
by that resistance, eight, which we
would like to get divided by the summation
which is ten. Okay? So we have, as you can see, 32 divided by 10.23 volts. Okay? So this is a method we can use. We can convert this, this is a ploy into a voltage source series
with a resistance. So we can use voltage division
to get us the voltage. Another method is using
Zack current division. As you can see, we
have a supply to unfair by exist
going here and here. So the voltage here
will be equal to eight ohms multiplied
by current. The voltage can be equal to the eight ohm multiplied by the current
flowing through it, V naught eight
multiplied by current. Like this. So it will be equal to eight, multiply it by the current. So what is the value of current? Value of that current
is equal to that two and multiply it by with
the current division. The other resistance sensor
we need here is a current. We need the other resistance
divided by total resistance. So resistance is two divided by the
summation, which is ten. So we'll find that eight
multiplied by 21616 multiplied by 232
divided by 10.23 volts. Okay? So this is the second solution. Is there another one? There is another one. The third one is that the
voltage here is equal to the voltage or the voltage
across all of this. So we can say is that V
naught is equal to the current multiplied by the
equivalent resistance, which is two multiplied by eight divided bys of
summation, which is done. So it will give
us also 3.2 volt. So as you can see,
there are lots of solutions for this problem. Okay? So let's see them again. You can see that when we use the current division to
get the current here, it will be equal to the
total current to embed multiplied by other resistance
divided by the total. The other resistance,
which is two, all we need is a current flowing here divided by the
total resistance. So it will give us 0.4. And
so the voltage will be equal to current flowing here
multiplied by the resistance. So it will be eight multiplied
by 0.4 gives us a 3.2, which we have obtained
several times. The second method is that
since they are in parallel, so they have the same
voltage V naught. The voltage V naught
is equal to the total current in the circuit multiplied by the
equivalent resistance, which is a parallel tutors or to own eight parallel
to the two ohm. So eight parallel
to the two ohms. The equivalent is there
multiplication divided by summation multiplication
eight multiplied by two and summation eight
plus two, which is ten. So it will give us also 3.2. So as you can see, there are several solutions to
the same example. Okay? So this was the first example on the source transformation.
36. Example 2 on Source Transformation Theorem: Now let's have another example. So example two in
source transformation. Okay? So use a source
transformation theorem to find that V x
in this circuit, which is a voltage here, or the voltage V, VX is a
voltage across the two ohm. So let's just start. Again. We have, in this circuit, we have a 6-volt
series, was too long, and we have four ohm battery
to a current source, not a voltage source, it is a current source. Remember this is
a current source. The first step is that
we can convert this, this six volt series with
a two ohm similar to here, into our current source
parallel to our resistance. So what is the value
of the current? Current is equal to the voltage divided
by this resistance. So six divided by
two gives us three. And Bayer. And direction of the
current similar to the direction of the positive
terminal which is abort. Like this. Barrier to resistance
which is due on. Okay? So we have converted this from voltage source to
current source, okay? Like this. Now, then we will have 18 volt here
in this terminal. 18 volt plus, minus 18
volt, like this. Okay? We have this source, which is a current source
barrel to a resistance, okay? Karen, source barrel
toward resistance. We can convert into a voltage source in
series with a resistance. So the voltage
source, so we will have the plus terminal here and the negative
terminal here plus terminal of the voltage source, similar to the same
direction of current. So it will be like this. Plus minus as the
value of the voltage. What is the value of voltage? Is the current multiplied
by resistance. So format the blood by 0.25, it will give us and vx. Okay. Series was what? Series with the same resistance, which is a four on
my exists for all. Okay? So let's see
the result like this. So as you can see three
and bear parallel to the two or three and Bear
power to the two ohm. That two ohm is a disease. This one, it will be here. And this point to
all of the VX, okay? Then we have 18 volt v x
for all 18 volt vx forearm. So let's delete all
of this like this. So is there any problem
or any problem if we take the two ohm barrel to Vx at
two parallel to the 2-ohm, knows there is no problem. Because the voltage vx
is the voltage between this point and this point, okay? Plus, minus v x. Okay? So remember, between
this point and this point. So it is okay to take the
equivalent of these two. So the equivalent of the two ohm and two ohm is equal to what? To multiply it by
two divided by 0. Summation gives us
one arm, right? So the two ohm parallel to
the term gives us one ohm. So the equivalent of
this too is like this. The three and bear. Okay? And parallel to one on. Okay? Now we can use source
transformation to convert this into a voltage source like this. So how is that voltage will be, voltage source will be equal to three and bear
multiplied bys of 10. So it will give us three volt. And the resistance will
be the same one arm. Okay? So remember, the spot exists. Okay? So as you can see, the equivalent of
three and bear to omentum is the three
volt series was one ohm. So as you can see, like this, delete all of this. The three volt series
with the one on the plus, minus vx is up between
here and here, between this point
and this point. Now, as you can
see, an extra step. So as you can see, we need vx. Okay? So we can apply KVL in this very large loop,
like this here. Current here, this current. The current is equal to I. So if we do the loop, we will have negative
three volt exists, negative three volts. And we go like this, one on, multiplied by the current and for all
multiplied by current. So it will give us five ohm,
the blood by his account. So we have five. Then
we continue to exist. We have plus v x plus v x. Then we go like this. We have plus 18, plus 18. Okay? So we applied KVL
in such large loop, in the outer loop. Now, what if we apply
KVL in this loop? And this is small loop? We will have negative
Vx exist, negative vx. And for i, four multiplied
by I plus four, I plus V x plus V x plus 18 volt plus eating equal to 0. So we'll have the second
equation we have. So as you can see,
we have in this, from this equation you can see
negative vx goes with v x. So we will have the for I equal, for I equal to
negative 18. Okay? So the current will
be equal to negative 18 over four, right? Or negative nine over two, which is negative 4.5 and bear, which is the current
here, as you can see, another method instead
of doing in this loop, we can do in this small loop. So we have plus V x plus V x plus V x negative three volt, negative three
volt and 11. Okay? So as you can see, we have this equation and this equation with
two variables, which is all a vx. And are you vx? So by solving these two, we can get the value of v
x and value of current. Here, as you can see, we obtained the v
x as a function of I is n. We took this one
and substitute it here. Okay? So we got finally we got current equal negative 4.5. Okay? Now from this we can find vx as the voltage
vx, the voltage v x. We can take this current
and substitute it here. Vx is equal to three minus I. So it will be like this. Three minus I, which
is negative 4.5, gives us 7.5 volts. Okay? So again, you can, where did we get this equation? Where the equation
from this one, vx plus or minus three equals 0. So that vx is equal to
three minus three minus I. So here you've got vx from here, from this equation, which is
similar to this equation, or you can just substitute
in this equation. All of them are similar to HR. Okay? So this was another
example on the soul. So transformation.
37. Thevenin Theorem: Hi and welcome everyone to our lesson in our course
for electric circuits. In this lesson we are
going to discuss is another theorem in
electric circuits, which is 77 and Xian. So in the previous lessons, we discussed this
superposition theorem, that source
transformation theorem, which can help us
simplify the circuit. Now another important
theorem which you will hear a lot is called z 70. So what does the seven and
serum or why do we use 70? So let's look at this circuit. So in this circuit we have
a resistance of four ohms. We have a current source
of three and bear. We have resistance eight ohm
and the two ohm resistor. In this circuit, for example, we need the voltage across here between these two terminals or across the 0
resistance it, okay? We need V naught across
the resistor eight. Okay? So as you can see, what if, what if we need, for example, the voltage
here across that resistor. When this resistor R. This
is a resistor called, or I would like to find V
naught when r equal to eight on another time when
r equal to two ohm, another time when r equal
to 150, as an example. Okay? So as you can see, we have three different
values, 8215. So if I would like to find
the voltage in this case, or in this case, or in this case. Then each time we need to start analyzing
all of our circuit. So in each time
we need to do KVL or KCL in each of these values. However, there is another
method which is called 707. And what do we do? We replace the constant
part is A-bar, which does not change, which is this part
in our circuit. This part, two ohms
three and Bayer forearm. This is our part
which is constant, which does not change
in our circuit. So we can replace all of
this with one voltage source in series with a
resistance. Okay? This resistance is called the R7 and the voltage
is called the V7. And then we connect
our load, which is, for example, eight ohm, eight ohm or two ohm, 15 ohms. So we have a very simple circuit which we can get the
voltage we need. Okay? So the seven and serum helps
us to replace large part of our circuit into
a voltage source in series with a resistor. And instead of analyzing
our circuit every time we change the resistance. Okay, so let's start
by learning more. So first you will find
that in practice we have a particular element in a
circuit which is a variable, likes a resistor here is, this is our variable element. It, it changes. It can be eight
ohm, it can be six, or it can be seven on this
usually called the loop. Okay? So for example, in our home, the outlet that our left, which should we take
from electricity. Okay, you will find in
the wall, for example, we have something which
is called the outlet, which we take from
it electricity. Okay. So this outlet is connected to a load such as a refrigerator, I'm afraid at TV. Any fluids. Okay. So this load is variable. And each time we're
connected this load we have a different voltage
and the current. Okay? So instead of analyzing
our circuit every time, okay, we just use the theorem to find
the required values. So as an example is a
household outlet terminal may be connected to
different appliances, leading to a variable load times a variable element it is, it changed it for examples
or resistance from eight to six to 15 to
whatever its value. Or for example, it changing
is a load from Frederick two, TV2 mobile or anything. The entire circuit must be
analyze it all over again. We need to do KVL again, KCL again every time. So to avoid these
problems as 70 theorem provides a technique in
which is a fixed part, which is constant and does not the change is replaced by
an equivalent circuit. A very simple
equivalent circuit. So if we have a linear
two-terminal circuit, this circuit contains
a resistance. In that resistance. And voltage sources, current
sources, whatever it is. So we take this
large circuit and we convert it into a
small circuit like this. So as our large circuit
consisting of lots of elements, we can convert it
into one source, one voltage source
in series with one resistance,
which is called ZAB, V7 and seven and voltage
and the resistance, okay, R7 and the V7. And then after replacing this with one small
equivalent circuit, we can connect it
to our load to find the voltage or current or
whatever we would like to have. The severance
theorem says is that our linear two terminal
circuit can be replaced by an
equivalent circuit consisting of one
voltage source, V7 in series with a
resistor R seven, where V7 and is the open circuit voltage
at the terminals. That open-circuit voltage
here at this terminals. And zap R7 and R7
is the input or the equivalent resistance at the terminal ends of
Windsor independent, okay, remember independent
sources or turn it off. Okay? So here we
have our circuit. So what does the V7 and V7 in is the
open-circuit voltage. So what does this mean? So if you have a
very low circuit, the equivalent here is
a two terminals here. The voltage between them is the open circuit voltage
or the voltage we need. Okay? So when we make this
two terminals opened, we removed as Zach connected
load or remove this load, we have an open-circuit voltage here between these
two terminals. The voltage here we
get is called the V7, which we will use in
our equivalent circuit. The second part
is called the R7. R input. The R7 is the resistance
which we have, or the equivalent resistance
when we look at the circuit. Okay, So when we have
an open circuit here, and at the same time we clause or we turn off all the
independent sources, all the independent
sources or equal to 0, similar to the superposition. If you remember, then we find
the equivalent resistance. So we will have a resistance
R in the R input, or the R7 is a seven and
resistance used here. Okay? This is in what, in the case
of independent sources only. So if we have a circuit with
only independent sources, then we turn off all the
independent sources and look at our circuit and find
the equivalent resistance. Okay, Don't worry, everything will be clear when
we have an example. Now. So when we don't have
any dependent sources, when we have no
dependent sources, only independent sources. In this case, we turn off
all the independent sources, we replace them by 0 sources. Then we look at the
circuit and we will have the input resistance
is called the R7. Okay? Now second
case is that when we have a dependent source. So if our circuit has a dependent source is
the first step is again, turn off all the
independent sources we turned off by making them 0. Similar as the
superposition theorem. In the superposition theorem, as we remember that depend on sources are not to be
turned off because they are controlled by the circuit
variable dependent on the voltage or a current
inside our circuit. So in this case, what we do is that at the two terminals here, we connect a voltage source, or we connect a current source. Okay? So we add a voltage source or
do we add a current source? Okay? Then the R7 will be equal
to the voltage source which we added divided by the
current coming out of it. Or in the case of Zara, adding a current source, then it will be R7. And OB is a voltage between, across this source divided
by that current source. Okay? So someone will ask me, what is the value of the V naught or what is
the value of i-node? You can select any value, any value for the
voltage source, any value for the
current source. And then in case of
the voltage source, we find that current from
the circuit analysis. And hero, if we select this or I naught or
the current value, then we find the voltage
from the circuit analysis. So for simplicity, we
use when we select the voltage source equal to one volt or the current
source equal to one m bear. Or you can use any other values. Or you can, for example, select is a voltage
as two volts, three volts, 100
volt, whatever it is. All of this is acceptable. But for simplicity, I'll
usually use the voltage as equal one volt and the
current equal one and bear. Okay. So sometimes you will find
that when we get the R7 and sometimes you can find that it has a
negative sign, okay? Or the R7 is a negative
value. What does this mean? It means that here at
these two terminals, we are not connecting a load, we are connecting supply. Okay? So in this case is a
negative resistance means that the circuit
is supplying power. It is not connected to a loop. Okay? Now, after we get V7 and all
seven at the two terminals, this representing the
equivalent of our circuit. At the two terminals we
connect our load RL. Then, for example, if
we need Zachariah, dozens or current will
be V divided by R seven and plus RL V7 over R seven
plus R L from the Ohm's law. If we need the voltage here, then it will be as
a supply voltage, V supply or visa, and then multiplied by
the resistance over the summation or L divided
bys or some measure. Or it can be equal to IL, load current multiplied by RL or is a load
resistance like this. So both of them are correct. Okay. So let's go on to have some examples on czar
seven and serum to understand how can we analyze our circuit or simplify our
circuit using civilian.
38. Example 1 on Thevenin Theorem: So in the first example
on seven and serum finds us seven and equivalent
circuit of the circuit a, b. Okay, so we have here our
Lord circuit, as you can see, between two terminals, a and B, we have a loop, which is our n. This load is variable. It may change at six ohm, 16 and served six ohms. Okay? So what we need is
that we need to find the equivalent circuit of the circuit AABA to the
left of the terminals. Okay? So we need to find the equivalent of
this larger circuit. Okay? We need to change this
constant apart into V7 and R7. And then we would like to
find the current through RL flowing through R
L Windsor resistance is 616 and served six. So what we need is
to change this part into seven and equivalent
circuit of V7 and R7. So first step, let's
us see the circuit. So the circuit here consisting
of independent source, independent voltage source,
independent current source. So here we are going to find
that R7 and by looking at the circuit and
V7 and hours of V open circuit by using KVL, KCL. So let's start. So we
find the all seven n by turning off the
supply, as we said before, since we have only
independent sources, then we turn off this supply of the supply in order
to find the R7. So if we don't as this supply, it will be a short
circuit like this. So let's delete
this, all of this. So it will be a short
circuit like this. Okay? Then we have the forearm. My exists. Okay? Then we have at
12 ohm exists 12. Then we have the two and pair. When 2M bear is turn it off, it means it is an open circuit, so it will be an open circuit. Here. Then we have the one ohm, one ohm like this. And here a and the B. Okay. As you can see,
it's a one on here. We don't do, we remove
the load completely. So this as if it does not exist. Okay? So what we need is we need to find r seven. So what does R7? R7 is the equivalent resistance when we look at this circuit. So how can you find this as if you are looking at the
circuit? What does this mean? As if we have a current going like this, I
would like this. Okay? So this current will go, we have one ohm. So we say one arm
going like this, then the current would
be divided through that arm and forearm. So it will be plus, since it is divided here as
ends at 120 m is parallel to the floor or the 12
parallel to the forearm. So it will be 12 volt
multiplied by four divided by 120 plus
four, which is 16. Okay? So this will be equal to R7. Okay? So here if you look at the
circuit like this from here, this perspective, you will find one arm series with the
equivalent of four and the 12. Okay? So four multiplied by 12 is 4848 divided by
16, I think three. This one is three,
one plus three. So R7 and will be
equal to four or. So. Let's delete all of this
and see again like this. So as you can see,
we turn it off 32 volt by making it
a short-circuit. We turned it off to embed by
making it an open circuit. Then we find the R7. And R7 is one ohm series with the equivalent of four
ohm and 12 VO, like this. So as you can see, for
parallel to 12 plus one gives us a four ohm
as we found before. So this is our survey. Now the next step
is that we need V7. So V7 and is the open circuit voltage
between here and here. So as if this does not exist. So we have the same
circuit but V7. And as you can see here. So how can we get V7 and the voltage here across
these two points? Now, something which is really, really important is that if
you look at this circuit, these two terminals are
now open circuit, right? Open circuit. So is there any
current can boss here, throws a one-arm, knows the
current year is equal to 0. Following this was a one ohm, it is 0. Why? Because here the one arm
is an open circuit here, between here and here
is an open circuit. No current will flow here, 0 current flowing here. Okay? So in this case we can remove the one as if it does not exist. So we can delete all of this. And we can say is that, that V is the voltage between
this point and this point, or this point and
this point here. Since two is parallel
to that world. Now what does an extra step? The next step is
that we can apply KVL here and give you L here. So the KVL here will give us i2 will be negative
two and there are A2 will be equal to
negative 02:00 AM bear. Okay? And I1. So from KVL like this, we have negative two, negative two plus we have I1 for I1 for our one. Then we have that will warm. So it will be plus
12 i1 minus i2, or one minus I2 equal to 0. Okay? So as you can see here, i1 flowing like this, and I2 flowing zone
access from this loop. So we will have 12
ohms multiplied by I1 minus I2, I1 minus I2. Okay? So negative 32. We have for I1 and I2 I1, so it will be 16 I1, I1. Then we have negative 12 I2, negative 12 or two equal to 0. And the i2 is equal
to negative two. So we will have negative
32 plus 16 or one. Negative 12 multiplied
by negative two gives us plus 24 equal to 0. So here we have 2432. So two summation will
be negative eight, takes us one to the other side. So we'll have 16
I1 equal to eight. So R1 will be equal to, okay? So we have I2 and I1
from the mesh analysis. So the V7 M, V is equal to, you can see here V7 and we said it is between here and here. So we can say it is V7
and here plus minus. So the current flowing
here multiplied bys at 12. So that, well, we're all multiplied by the
current flowing here. The current flowing
here is I1 minus I2, I1, which is 0.5 minus I2. So it will be 0.5 minus i2. I2 is negative two, so it will be plus two. So it will give us
2.5 multiplied by 12. So it will be 2.5 multiplied
by 12 will be 24682, I think, 30 volt, I think. Okay, So let's see
again what we did. Delete all of this excess. So we have here, as we did exactly where the two
loops, the first loop, which is this 1
second loop here, I2 is equal to
negative two and bail. So by substituting this here, we obtain the I1 equal half
as we did before, then V7, and it will be 12 on the
blood by I1 minus I2, which is in the end
30 volt as we did. So we have V7 and 30
volt and R 74 ohm. Okay, So let's go. So we have V7 and 30
volt and all 74 ohm. So we will make our
circuit like this. So we have the 30 volt. And for all series with
the load which we need. We said before that
we need to find the currently here in three
different guests. So the current is equal to V7, which is a 30 volt. And all seven in which
is a four ohm and R L, which we are going to change. So we will have like this I l will be equal to V7
N divided by 4771, which is a four ohm plus z load, which we are connecting. So we will have this
three cases like this. When r l equals 61 or a liquid
cysteine or RL equals six, we Assembly makes us
1616 and served six. Then we'll have i l equal
to three and bear 1.5.75. Okay? So this was an
example on how can we use seven and serum
to simplify our sucked. Now as you can see that if we do the mesh analysis
and nodal analysis, normally, you will find that
each time in order to get, for example, the current
when RL equals six. Then why do we need to do mesh analysis all
of our circuit. And then F18 at 16s
and we need to do mesh analysis again at
Sussex mesh analysis again. So instead of doing this
using is a seven and serum, we did mesh analysis
only one time. To simplify our circuit. Then we got any values
we need easily. Okay.
39. Example 2 on Thevenin Theorem: Now let's have another
example on 70. So in this simple example, we need this seven and
equivalent circuit of the circuit a and B. So we need seven in-between
these two terminals. So we need to replace
all of this to one voltage source in
series with our 17. Okay? So as you can see in
this circuit, we have, remember we have an
independent source. And in this case we have
a dependent source. So it means that when
we get our seven, we need to add here, our supply. Is there a voltage source
or a current source? So let's start. We will say for simplicity, we add here voltages
source of one volt. As we said before, why? Because we have a
dependent source. Okay? Now, the first step is that we deactivate all of our
independent sources, which is a five and Ben. So by activating this, it will be an open circuit here. As you can see here, it disappeared because it
is an open circuit. Now we have the fore
arm to arm to vx, six ohms, two ohms, and all of our circuit. And we connected here
is our voltage source. So in order to find the R7 and we need to find
the current year, then all seven and
B is a voltage divided by current.
So let's start. So we have here are
three lobes, I1, I2, and I3 using mesh analysis. Okay? So it will find that by applying mesh analysis to
loop number one. So as you can see, we
have this loop I exist. So we have here our u1 to
multiplied by I1 minus I2. So we have i2 like this. So it will be too I1 minus I2. Then we have here in this loop, negative two vx, negative
two dx equal to 0. Okay? So from this equation, we can find that V x is
equal to I1 minus I2. Second equation, the
second equation. So we have here all i2, i2. We can go like this. So we have negative v x. And the second row, we
can say negative v x or we can say for i2. So we have here for i2. Then we go like this. As you can see, we have to
multiply by i2 minus i1, two multiplied by I2 minus I1. Then we have here two currents, I2 minus I3 multiplied by six. So six ohms, i2
minus i3 equal to 0. Okay? So this is the second
row, the third loop. Here. We have in this slope, we have exist plus one. Okay? So we have here plus one. Then we go like this. We have I3 minus I2
multiplied by six ohm, ice cream minus I two
multiplied by six ohms. Then we have here two multiplied
by two multiplied by I3. So as you can see, we
have in this equation, I2, I1, I3, I2, I1, I3. Okay? And do we have in
this equation that V x is equal to what is equal to negative I2
multiplied by so for all, so vx is equal to
negative four i2. Okay? Why? Because
I do like this. And VX is equal to
the current flowing in this direction going
into the plus terminal. So it will be negative
I2 multiplied by four, negative I2 multiplied by four. So we can take V, x and substitute it here
in this equation. So when we substitute
this one here, we will have i1 equals
negative three i2. So we have one
equation, two equation, three equations with three
variables, I1, I2, I3. So we can find the value
of the three current. Okay? Now what does important for
us that we need I naught. And what is the value of final? You'll see ice Rayleigh exists. This is always three. So I naught will be
equal to negative three. Okay? So from this equation, we have I3 equals
negative one over six. So I know it will be one over six since it
is negative i3. Then what does R7 and
R7 and will be equal to the voltage divided
by the current. The voltage is one volt divided by the current,
which is one over six. So we'll have six
ohm resistance. Now, second part is that we need to find the voltage V seven. Between these two terminals, V or V open circuit. So how can we find this? Again? Since it is an open circuit, then this resistance
as if it does not exist because there is
no current flowing here. So this point here, and here we have v sub n. Okay? So as you can see here, we have five and
bear vx six ohms to two VX and VY survey or V open circuit between
this and this V7 it. So again, we're going to have
three loops, i1, i2, i3. The by doing this
I mesh analysis. Using mesh analysis, again, we have I1, i1 equals
five and bear. Are you on equal five and bear. In the second loop, which is this one. This one I screen. We have by exist, we have to own multiplied
by i3 minus i2, i3 minus I2 multiplied by two. And we have here
negative two vx, negative two vx, vx. So from this equation
is i3 minus I2. Okay? The second lobe here, we have i2 to own multiplied by I2 minus I3 to own what
allowed by i2 minus i3. And do we have here in this one, we don't have any
current except by two, so it will be six i2, i2. And here we have negative
vx, negative vx, which is similar to for i2, i2 minus i1, for i2 minus i1. Okay? And we know that V x is equal to I1 minus I2 multiplied
by four ohm, I1 minus I2 multiplied
by forearm from where? From here. Vx is a current flowing here
multiplied by four ohm. So the current flowing here is y1 minus y2 multiplied by four. So it will be for all
multiplied by I1 minus I2 gives us vx. Okay? So we can take this one
and substitute it here. So we will have a relation
between I1 and I2 and I3. In addition to this equation, in addition to I1
equal five or ambient. From this equations, we can get the value of current i1, i2, i3. And the voltage V open
circuit is simply the current flowing here multiplied
by six so on. Since this part and
this point is V7 and so V7 and will be equal
to the current flowing here, which is a2 multiplied by six. So it will be I2
multiplied by 61. And from this equations, I2 will be equal
to ten over three. So the V7 or the V open
circuit is 20 volt. So by taking these two values, we have this circuit
that when two volt and six ohm
Sierras, who is it? Ok, so this is a V7
and equivalent circuit of this large circuit. Okay? So this was another example
on seven and serum. I hope it was helpful for you.
40. Norton Theorem: Hi and welcome everyone to another lesson in our course
for electric circuits. In this lesson, we
are going to discuss another serum in
electric circuits, which is a normal tone theorem. Okay? So in the last lesson we discuss
this as seven and serum. Now I would like to discuss
is an all-trans here. So what is the difference
between seven and CRM? And the Norton theorem. So remember that in
seven and serum, we took as circuits
are linear to turn our circuit and
converted it into a V7. V7. And if you remember, V7 in series with R7. Okay? So we simplified
our circuit and V7 and R7 now in the
Norton Theorem, and instead of having V7 and R7 and we will change our
circuit and convert it into I Norton better
to our north on. Okay. So instead of having a V7 and all seven and similar tools
of source of transformation. As you know that we can have this one equivalent
to this one when, when R7 and similar
to or Norton or seven m is equal to our
Norton and I north on I, n is equal to V over R seven. Okay? So this circuit which is
using source transformation, we can change this I
V supply and series resistor into a current
source to a resistor. Okay? So it's similar
to each other. So C is seven and serum uses a voltage source and
series with a resistance. For the Norton Theorem, it uses a current source
and better to resist. The Norton theorem states that linear two-terminal circuit can be replaced by an
equivalent circuit consisting of a current source IN parallel to our resistor R N I N is our Norton and
our n is our Norton. Okay? So how can we get INI n
can be obtained by using a V7 and over r seven
or V7 and over R n. Or by applying a short circuit and the findings are current. So this current will
be the Norton current, similar to the open
circuit voltage, which are representing
V7 N and R N, which is the input or the equivalent Systems
ADS terminals when all of the independent
sources alternative, similar to seven and serum. If we have only
independent sources, then we activate all of them and look at our circuit
and defined or input. Okay? And at the same time, if we have dependent sources, then we're going to add
here a voltage source or a current source of
one volt or one and bear. So here as you can see, an altered serum and seven and serum or seven and
north on transformation. As you can see, this one
is equivalent to this one. This is our Norton's theorem. This is a seven and serum. So as you can see,
V7 is equal to i, n multiplied by n, or n is equal to V
divided by R seven. And from source transformation. And our N is similar to
R seven, as you can see. So the first step is
that we need to find, in order to get the
equivalent Norton circuit, we need number one is the
short-circuit current I in, the Norton current IN, okay? And i n is also equal to
V sub n divided by R7. So he will get the open circuit
voltage and divide it by R7 and we can get the
Norton current, okay? Or you can just obtain it by applying a short circuit
and the circuit analysis, we can find the
Norton current for R7 and as you remember for
R7 and R7 and or are, they are similar to each
other when we look at our circuit and find
the input resistance. Okay, so let's have
some examples. Owns an OT on serum
to understand it.
41. Example 1 on Norton Theorem: So in our first example, find the Norton
equivalent circuit or this circuit a, b, this circuit. So we need to find
the Norton equivalent of this logic circuit. We need to convert
this into a current source battery resistor R n. Okay? Or in like this, and this is a Norton current. So this is what we'd like to do. So we need IN and we need our n. Okay? So let's start. The first step is that we
need the Norton resistance. Okay? So as you remember, is
an auto resistor is similar to the seven resistor. So how can we get this? We can get this by deactivating
all of our supplies. Okay? So we have here is at 12 volt, so we can deactivate that well Volta by making a short circuit. And we can deactivate that too. I'm bear market making this
an open circuit like this. Okay? So as you can see, we have here an open circuit which
are representing the place of the two
unpaired and short circuit here instead of the 12-volt. Okay? Now we need to find the R N
or the Norton resistance, okay, is the equivalent
of this resistance. How can we do this? So, since we need between a and B, okay? So you can think about this how as if there
is a supply here, our current to go into here. This current when it goes
here, what will happen? It will be divided to five ohm exists and they
get back to that p.sit. So we have here all E2, okay? And the first one, I1, i1. We'll go here through eight ohm, then serosa forearm, then
through the eight Ohm, then it will be
combined at this node. Okay? So we have here I1, one
and this is our total. Okay? So if you think about this, you can see that the eye
is divided to i1 and i2. So from here we
can know which one is parallel and
which one is series. So as you can see,
it's a five ohm here. This five ohm is parallel to eight ohm plus four
plus eight on. So we can say is
that R naught is equal to the five ohm battery to the eight plus four
plus eight because this one and this
one and this one, all of them are in series. So eight plus 816
plus four equals 20. So it will be equal to five. The blood by 20 divided by 25. Okay? So it will give us four ohm. Okay, let's see if we
are correct like this. So as you can see,
it's all Norton is equal tos of five ohm. Battery is a series combination. So it'll give us five to 20, which is XeF4 on. Okay, So this is the first step. We obtain the Norton resistor, which is similar to our 17. Okay? Second step is a two-way needs
a short circuit current. So how can we do
this or the Norton current by doing a
short circuit here. So by doing a short
circuit here IN, okay, we can get the
Norton current like this. So as you can see from a
to B, a short circuit. So we have here, okay? Now let's look at our circuit. So we have here the two
unpaired is that well vault 885 ands are
short circuit. Now look at this circuit. So we have here a
short circuit here. Okay? Short-circuit barrel
to follow you home. Okay? So as you can see, the
initial node here, similar to the initial
and the final nodes here. So it's a five ohm is
battery to a short circuit. So what does this mean? It means that we can
cancel the five ohm. Why? Because there is a
short circuit here. So no current will pass
through the five ohm. So we can draw our
circuit like this. So we have to embed 12 or
488, as you can see here. And here at this two terminals, we canceled the five ohm because it is parallel
to a short circuit. Now as you can see
from this circuit, we have two loops, are I1 and I2 from this one are u1 is equal to two and so on. One equal to two, I'm there. Okay? Now, what about i2? I2, by doing hierarchy exists. So we have eight
multiplied by I2, eight. O two minus 12. Here we have the four ohms
always say plus four on. Okay? I2 minus 12, minus one. Okay? And do we have here
another eight? So we can say eight plus
eight, which is 16. So we can say this 16 i2. Okay? So we have here
eighth series with aid. So you can say 16. And i2 minus 12 was a supply. And for I2 minus I1, I1 is equal to two
and bear here. Okay? So we have 16 i2, i2 plus for i2 to minus 12 equal to 0, of course, minus 12,
which is this one. Okay? Minus ru for I1, which is minus eight. So as you can see
from this equation, negative 20, and here
we have to enter i2. So this will go to
the other side. We will have I2 equal
to one and bear. Okay. So the Norton current is equal to i2 since the z
have the same direction. So the Norton current
will be one unpaired. So let's see if we made
czar right calculations. Okay, so as you can see here, we ignore the five ohm because
it has been short-circuit, as we said before here. And applying mesh analysis, we have R1 equals to m pair and the second loop here of the i2. So we finally get I2
equal one I'm pair, which is similar to
the Norton current. So we have here one and
pair is an autumn current, and we have the four ohm, which is a resistor. So this is a Norton
equivalent circuits. Okay? You can also change
this into IV7, like this plus minus
V7 and or seven. Or seven and is similar
to our north on. So it will be for all and V7 and is equal to a for all
multiplied by one ampere. So it will give us a four volt. Okay? So this is an
equivalent circuit. So you can change
the seven n into Norton's Norton equivalent
circuit or from north on to seven and
equivalent circuit. Now let's have another solution, another solution
for this problem. And instead of getting
Norton current, we can get V7. Okay? So how can we do this by getting the open circuit voltage
V, open circuit, and by getting V7
and divide it by ordinal tone or R7 and we
can get the Norton current. I will just an order
to show you is that Z are similar
to each. Awesome. So as you can see
that the current I3 is equal to two and there are three equal to two and bear. Okay, and second loop, we have eight ohm. We have five ohm, eight ohms, so eight plus 816
and five-year 21 I4. So we can say 21 for here. Negative 12. Going like this. We have the four ohm. So the for all metabolized
by I4 minus I3. So plus four multiplied by four, minus three equal to 0. And the i3 is equal
to two and bear. So we can say that 21 I4 plus
four I4 gives us 25 four. And we have negative 12. Negative for I3. I3 is equal to two and bear. So negative two multiplied
by four gives us negative eight equal to 0. So our E4 is equal to
negative 20 divided by two, the other side, 20 over 25. Okay? So we can get V7 and here V7 n is equal to the current
flowing is a five ohm, which is I4, multiplied
bys a five ohm. So V7 is equal to I4, the wind over 25. Multiply it by that resistor, which is five or 255, gives us 20 over five, which is four volts. Okay? So we have here
is a four volt V7 and similar to what we
just did that here, if you remember here, similar to the V7 and which is also obtained from this V7. And we can get that
all Norton is equal to four volts divided
by the R seven and, or, or nor too much was four. So it will give us one and bear. Okay, It's similar as here. Now let's see the steps. Okay, let's delete all
of this, like this. Okay? So as you can see, we can
find IN from V7 of our R7. So we obtain the V7 and from the open circuit voltage by using the mesh analysis
ICU equal to m pair. And from second row
we have i4 equal 0.8, which is 0.8 is 20 over 25. Then the V open
circuit is the V7 and, or the voltage between these
two points here and here, which is a current
here, flowing here, which is I4 motto, blood buys a five ohm. So we have V7 and
equal four volt, then Norton current
will be one and bear. So both of them will lead to the same solution
as you can see. Okay.
42. Example 2 on Norton Theorem: Now let's have another example
on Zhan nor tones here. So using the Knowlton serum, find our n and I n
of the circuit a, b. Okay? So as you can see
in this circuit, we have a dependent source. In this case, when
we get the RN or the Norton resistor or the R7 and we need to add
here a supply. Okay? So let's start. So the first step
is that we add, for example, a supply, which is for example, one volt. One volt as a supply. As you can see, V supply, the one volt plus minus, okay? Now, one important thing
which you have to remember is that when we obtain the
R7 and or our Norton, we need to deactivate
all independent sources. So the only independent sources
we have is at ten volt. So we activate this by making it a short circuit like this. As you can see here. Okay? So now we have the supply
five on two IX and current. Ix is the current flowing
through the forearm. Now something which
is really important, as you can see that I xx is the current flowing
through the four. All right? However, the fore arm is
parallel to assure the circuit. So what does this mean when
this part is parallel to a four or a four ohm resistor
parallel to a short set. It means that the
current flowing here through the
resistor is equal to 0. Or we can cancel the four
ohm as if it does not exist. So as if we have only
a short circuit here. So since I x is equal to
0, what does this mean? It means this supply
will be equal to 0. So that 0 current supply. What does this mean? It means that this one
will be an open set. So our circuit, as you can see, will be simplified like this, is a short circuit. Here is a short text of the ten volt and the four
ohm is completely removed. Also, the Ixx will be canceled because the
current is equal to 0. So it will be like this. This will be canceled
at the open circuit. So we will have five ohm
series with the V supply. Okay? So as you can see, V naught
is equal to one volt. Okay? So what do we need here? We need the resistance. So our Norton will be
equal to the voltage, which is a one volt divided by I-naught,
which is a current. Okay? Or are not on
its own directory, okay, It is the same. So anyway, I naught is equal
to one divided by five. So this is one divided by five. So one divided by one over
five gives us a five ohm. Which is similar as if you
look at between a and P, you'll find that
as one resistance only, which is the five. Okay? So it is the same ID
as you can see it. Okay. Now, what do we need next? We need the short-circuit
current or the Norton current, like this by making a
short circuit here. So I, Norton is a short
circuit current here. So how can we get this
using different methods? Okay, you can do as
a mesh analysis. You can do so nodal
analysis, whatever it is. But you can see here
in this circuit, we have a very simple solution. Okay? So if you look
at this circuit, we can say that the
Norton current here, our Norton from KCL, it is equal to the current
year plus the current year. So we can say is that I
know tone is equal to x to x plus the
current flowing here, which is, for example, or u1 plus i1. Okay? So now what we need is
that we need to get the oil one and do
we need to get Ixx? So if you look at this circuit, you will find this
point and this point, this point and this point
are similar to each other. So as you can see that ten volt. And the second part here, you can see that the ten volt is parallel to the fore arm. So from here, what
does this mean? It means that the voltage
between these two points is similar to the voltage
across the resistor. The voltage here is ten volts, so the voltage across the
resistor is ten volt. So I x is equal to ten volt, which is the voltage between
this point and this 0.10 volt divided by the
four on like this. So this is a value of I x. Okay? Now, so the supply here, this current will be equal to two multiplied by this value. Now what do we need
next is our E1. Now, if you look at
the same idea here, the first to terminal here, here, which is forest
terminal of the five ohm, is similar to the
plus of the supply. And second, the point
here of the supply is similar to the second here of the five senses are parallel
to each other, okay? Or they are parallel
to each other. So I can see this point here. This is all, all of this is
one big note like this. Okay? So what does this mean? It means that the five ohm
is parallel to the ten volt. So again, i1 is equal to the
ten volt divided by five. Okay? So we have IN, or the Norton current will be equal to two multiplied by x. So it will give us two
multiplied by ten over four gives us ten over
two, which is five. Plus ten over five
is equal to two. So it will give us
seven and bear. So I hope I am correct. Let's see. Okay. So here as you can see, the first step is that
you will find them. To get IN. We need to do a short-circuit, as you can see here. From this figure,
you will find that the four ohm is better
to the ten volt and the parallel to the phi form for all the ten
volt and wife home. All of them are embarrassed. So I X would be equal to the voltage divided
by the resistor. And by applying
KCL at this point, we can get this equation. Then on 15 is a
current flowing here. Okay? Funds as a short
circuit current, which is a Norton current
equals seven and pair. Okay? So now you have the final circuit which is the Norton current 77
and pair like this, parallel to the R7. Okay, or seven
hours or R naught, which is five ohm, which is why my exists. So this is the Norton
equivalent circuits. Okay?
43. Maximum Power Transfer: Hi and welcome everyone to another lesson in our course
for electric circuits. In this lesson, we will discuss the maximum power transfer. So this theorem discusses
that you would like to, when we would like to transfer the maximum
power to our load. So let's say for example, we have a large circuit,
two terminal circuit, linear circuit, which it has an equivalent
circuit, V7 and R7. Okay? So this is the equivalent of
a very large circuit, okay? And it's, the terminals are
connected to a variable load. Okay? So the RL changes. It is not constant, it is a variable load. Okay, so our LEA
can be any value. So according to the value of RL, current, absorb it,
it changes. Right? I it changes when our LEA changes because the
current is equal to V Thevenin divided by
the total resistance, or seven n plus l. Okay? So here when we change as RL, the load current changes, which means that the
power also changes. As a power of the
absorbed by solute is I squared multiplied by RL. So when RL it changes, the current also changes. Okay? So if we look at this equation, we can say that the power, That's a power absorbed
Budweiser load is equal to V over R seven plus R L square, which are representing here. Is this representing
the current, okay? So the power is a square of the current absorbed by selfhood multiplied by RL or the
resistance of the load itself. Okay? So as you can see
here is that when our L increases, for example, you will find that this
parameter increases, which means the power
should increase. However, you will find
that R L also here. So when RL increases, the value of the current
starts decreasing. So find two parameters or two parts of this equation which are contradicting
with each other. They are not by directly
proportional to each other. You will find that when RL increases, the
current increases. You will find that
the, when RL or the resistance increases,
the current decreases. And the same time the
resistance itself increases. So the total power
it changes according to the product of
these two parts. So we'll find that
the relation between the power and the variation
of R L in this circuit, you will find it will be some centralized at the beginning, starting from 0 resistance. When we start increasing, the resistance of
power increases, reaching a peak values, then it starts decreasing again. Okay? So here you will find there
is a certain resistance here, which we have the maximum
power P max, which we need. Okay? So the maximum power transfer means we would like
to find this value, is the value of resistance, which gives the maximum power. Okay, so what is the
value of resistance here? What is the value here that
gives us the maximum power? You will find that
from circuit analysis. And from the probe which you are going to take is that
the value of R L that produces the
maximum power is seven. Okay? So when r is equal to seven and we have the maximum power
transfer, the twos's loop. Okay? So this is a maximum
power transfer city. Okay? So it says that when we, when the load resistance is
equal to R7 and resistance, then we will have the maximum
power transfer to the load. So let's delete all of this. So first, why do we need
the maximum power transfer? Because in many
practical situations, a circuit is designed to provide power to a load, such as here. Okay, So this is our circuit. Is that for whites
power to our load, which is our L. There are
applications in areas such as communications where
it is desirable to maximize the power
delivered to a load. So we need to maximize
or old provides a maximum power to our load. Okay? So in order to find to absorb the maximum
power from the circuit, we need to make the resistance or the load resistance
equal to seven, or resistance or
R L equal to R7. And Sousa, the maximum power
is transferred to our load. You'll find here is this. Here again, the relation
between the power ends or L from 0 to a very large
value to infinity, for example, you will find that at the beginning increases, then it starts to decrease
when we change our L. Okay? And you'll find that
the maximum power occurs when our l equals 078. Okay? So the power absorbed
by the load is equal to the current square multiplied by the resistance of the root. And the current is V7 and divide the boys are total resistance. Okay? So as you can see from
this equation and when we substitute with our L equal to R7, okay? So when you take R, L equal to R7 and or
the maximum power, you will find that here are
seven and plus R7 is to R7. And so we have V7 n squared over two or seven and square. So the square here is
divided to V7 and here, so it becomes a V7 square. And the two here is
devoted to this part, or seven Ambrose or L
or R Thevenin plus R7, which is to R7 and all
squared multiplied by RL, which is now R7. Okay? So we'll find that here, R7. Let's type it here. V squared divided by
two r squared becomes a four or seven and square
multiplied by R seven. Okay? So in the end we will have, if we delete this was this, you will find that the maximum
power will be equal to V sub n square divided
by four our 17. So let's delete all of this. This is the maximum power V7 square divided by
four R7 it again, how did we get the
maximum power by replacing the RL with our seven? Okay, when we substitute here, we will get this equation. Now, how can we get
the maximum power? Or what? What is the proof
that identifies that? Are l When l becomes R7 and we will get
the maximum power. Okay? So if we draw the relation
between power and RL again, which is the previous figure, you will find that here
this is a curve like this. Okay? So we need this point. Okay? So we need this point. Now, if you look carefully in
this point, at this point, the line passing through here, the tangent line passing
through here, has a slope. The slope of this
line is equal to 0. Okay? So we have the equation here, b and r l, b and r
l is this equation. Okay? So at a certain point when we have the maximum
power at this point, you will find that
the tangent line, its slope is equal to 0. Okay? So if you remember, if you remember
from mathematics, you will find that the slope
of the line is equal to 0. What does this mean?
If we need this point, it means that the derivative dy, the derivative of DY over
DX or db over the DRL. Our L is a variation of the
power with respect to RL, which is our x, is equal to 0. Okay? So in order to find this point at which
we have this slope, then we derivative or
gets a derivative of the y axis with
respect to x axis, or d v over d DRL, or DY over DX. Okay? So that's what we did to
get the maximum power. We differentiates our power, which is our y-axis
with respect to RL, which is our x-axis, and equate it with 0. So we get the derivative
of dv over a, D or L, Okay? So it gets a derivative of this equation with
respect to r l. Okay? So how can we do this? It is really easy. So first we can say is that
here we have V7 and a square. So we can say V square,
multiply it by. We have here or L two
will be like this. And do we have here a square
of r seven and plus RL or seven plus r all squared. Okay? So we have our L divided
by R seven and RL squared, or seven and RL squared, okay? And V7 a square will be outside. So V7 a squared does, is not affected by the differentiation because
our variable here is RL. So V7 a square will be, we'll leave it as it is, as you can see here outside. Now, we have this part. Okay? We need to differentiate
this equation, or L over R seven and
plus RL squared with respect to r l and
equate it with Z. So how can we
differentiate a fraction? For us? It will be square of this denominator is the
square of the denominator, which is the square
of this part, which is the square of
r seven and plus r, all squared is R7 plus
RL to the power four. Okay? And then the
differentiation of R L, which is one multiplied
by the denominator. So it will be one multiplied
by R seven and plus RL squared minus xy differentiation
of the bottom here, which is as its
differential will be to have an embolus or L. It will be to R7 and plus RL multiplied
bys and numerator, which is our n. Okay? So again, square of the bottom, then differentiation of the top multiplied bys and
bottom as it is, minus is a differentiation of the bottom multiplied
bys top as it is. Okay? So as you can see, this is our equation. Equate it to 0. Okay? So let's delete this. Okay, So we have
here V7 a square multiplied bys this law
should bracket equal to 0. So we can say, we can
cancel this part. So we'll have our seven and plus RL whole square minus two RL, R seven plus R L over R squared. If you simplify this equation, you will get all
seven plus or minus two RL divided by R
seven and plus RL all q. Okay? So when we say is
this equal to 0, we can take this
to the other side, which means it will
be multiplied by 0. So we will have our seven
and plus RL minus two RL. So 0 will be equal
to R7 and plus or minus two RL is this part. Okay? So we'll find that our
seven will be equal to RL. So then we take this one and
substitute here we will, so we will get the
maximum power. So here we understand
now is that when we make our resistive load
equal to the resistance, it will give us
the maximum power. So let's go on to have
an example on this.
44. Example on Maximum Power Transfer: So let's have an example. Find the value of R, L, which will help us to get the maximum power transfer
in this electric circuit. Funds that we have here
at 12-volt, six ohms, 12 ohms, three ohms
to unfair to all. And as these two
terminal we have RL. We need to find the
value of R L that will give us the maximum
power transfer. Okay? So as we know that
in order to get the maximum power
transfer in our circuit, we need to make our
L equal to R seven. Okay? So how can we get R7? And first, we need to get our input between a and b
between these two terminals. Why the activating all
independent sources? So when we deactivate all
the independent sources, this will be a short circuit and this one will
be an open circuit. So we will have like
this short-circuit here, six ohm, short-circuit
here, six ohms. 12 arm, as it is three arm. This will be an open
circuit since we, the activities are two
unpaired open circuit here. Then the two ohm becomes two ohm here and two terminals here. Now we would like to get R7. So what is the
value of R7 and R7? And if we look at this as if we have a current
flowing here, so that to own a series
with a three ohm. So we have five ohm plus the current flowing here
will be divided shows at 12 ohm and six ohm. So the six ohm is
parallel to the 12 ohm. So it will be 66 multiplied
by 12 divided by summation, which is 18. Okay? So this will give us, as I think, four ohm. So five plus four
gives us a line on. Okay, So this is our R7. And so as you can see here, are seven is a two plus three, which is a serious component, plus the parallel component, six ohm and 12 volt. Okay? So we will have online on, so this is the value
of resistance that will give us the
maximum power transfer. Now, if we need to
find also the value of the power itself is a
maximum power transfer. Then we can simply
get the visa and then V7, how can we get it? It is open circuit
voltage between these two terminals,
okay? Like this. So the open-circuit voltage
between these two terminals then only you will
apply to mesh analysis. So it's a forest, a mesh
analysis or first loop, you will find here I2. And do we have
here two unpaired. So i2 flowing like this. So I2 is equal to
negative two and band. So i to negative two. Okay? And I1. In this loop, we
can find like this, we have negative 12
volt, negative 12. Then we go like this. We have I1 multiplied
by six ohms. So we say plus 61. Then we go like this. So we have here two currents. So it will be plus 12. Here we have I1 minus I2. One minus two equal to 0. Okay? And do we have i2 is
equal to negative two. So this part will be plus two since we have negative
and then negative. So we will have negative 126
i1 and told E1 is 181801. And do we have here plus
12 and the plus two plus 24 equal to 0. So we have negative 1224 is 12, negative 12 plus 12. So go and get it
to the other side. So we will have all U1 equal to negative 12 over 18. Okay? So we have I2 and I1. So let's delete all of
this, as you can see here. So we have i2 is equal to
negative two and bear and I1. After simplification, it will
be negative two over three, which is from this law. And we take this current
and substitute it here. Now, if we would like to get
V7 and how can we do this? We can do a very large
loop here, like this. Okay? So we have this large
loop, negative 12, okay? Then six, I1 plus I1 here. Then we go like this. We have the three arm and I2. So it will be three I2. Then we go like this. We have the two ohm. Is there any current
passing through the 2-ohm? Know, so it will be 0 because no current will pass here
because it is an open circuit. Then plus v is happening
in the slope plus V7. So we have I1 and I2, so we can get V7
and equal 22 volt. So the maximum power
transfer will be V7 and square over four R7. And as we got before,
so we will have, since our L will be
R7 equal mine arm, it is a maximum power transfer. So substitute this one
here and V sub n is 22. So we'll have the
maximum power transfer to our circuit is 13.441. Okay? So this was an example, owns a maximum power transfer.
45. Introduction to Operational Amplifiers: Hi and welcome everyone to this section in our course
for electric circuits. In this section we are going to discuss their
operational amplifiers. So in the previous
section of this course, we discussed that different
electric circuit theorems. Now we would like to discuss an important element or an important device in
our electric circuits. And it is used in
several circuits, such as in power
electronics and so on. So what we are going to
discuss is called the operational
amplifiers or op amp. And this is how it looks like. Okay? Symbol chip with several pens. Okay? So what is an
operational amplifier? Operational amplifier is
electronic device that is designed to reform some
mathematical operations. So it can do, for example, the addition of signals, subtraction, multiplication,
division, and so on. Okay, So how does it do this? It does this when we connect external components
to its Ben's, for example, resistors
and capacitors. Now first we discuss
the resistors. Now what is the capacitor? The capacitor will be discussed in the next section
of this course. It is one of the
passive elements inside our electric circuits. So the op amp is an
electronic device consisting of a complex
arrangement of resistors, transistors,
capacitors, and lights. Where are all of
these are inside it? When system of resistors
and capacitors, transistors and bite that capacitors and inductors will be discussed in
the next section. Transistors and diodes will be discussed in our course
for power electronics. Okay? But you have first, you need to know first the
basics of electric circuits in order to go to our advanced set goals
of power electronics. The op amp is an active
circuit element. So what does an act of
circuit elements mean? Active. The active circuit
element means that it is operating or do its function or the mathematical
operations when it is connected to supply. That passive elements such
as resistors, inductors, and capacitors, all of this, they don't need any supply, just add it to the circuit. So the active element, that means that it
will aid needs supply. In order to do the functions. The passive element does
not need any supply. You can connect it
directly to the circuit. Okay? So what are the
functions of the op amp? It can be used in
addition, subtraction, multiplication, division, differentiation,
and integration. By connecting it to
our supply and adding several elements such as
resistors and capacitors. That additional amplifier
can be also used in making or the design of the voltage or the current controlled
current source. Okay? Remember that dependent source, which we have
discussed it before, we can make this using the op amp or the
operational amplifier. So first, let's discuss what are the components of the operational
amplifier or the pens. You will find here several pins, 1234 and on the
other side, 1234. Okay? So this operational amplifier
has a total of eight pins. As you can see here. It has a total of
eight bands, 12345678. Okay? So the op amps are
commercially available in the integrated circuit
packages and several forms. An integrated circuit
means the IC. Okay? So find this in several
electronic circuits because it is very important in
the electronics industry. So that typical one is it Ben, dual in-line package or DIP? It consisting of eight Ben. Okay, you can see here 44. Now, first, we will find that Ben number eight
here in our course. It is not connected,
it is unused, and you'll find that
Ben number also 1515. It is not also important
in our course. For now we will discuss
the pin numbers 23476. Okay? So the five important is 23476. You will find that
two inverting input, three non-inverting input. You will find that
number six is called the output of this
operational amplifier. V plus and v minus our CSA supply connected to
the operational amplifier. So as you can see,
what components, first, the inverting
input, pin number two. So when we connect
something here, it's called the inverting input. What does this mean? It means it inverters as a
sign of the input. If the input is positive, then the output
will be negative. If the input is negative, then the output
will be positive. That's why it's
called the inverting. It inverts the signal from positive to negative or
from negative to positive, as you will learn
in the section. We will learn in more details. Now is on non-inverting.
Here it is. Non-inverting means
that when we add also, the output will be posted. If the input is negative, then the output will
be negative as it is, no change in sign. The inverting involved as the input non-inverting does
not invert the input signal. So we'll find that here. Number two and number three are the two inverting input,
non-inverting input. And you'll find that
this representing that simple of the op-amp
or operational amplifier, which will use, we will use or add to our electric socket. Okay? So we'll find that
Penn number two and number three is the
inverting and non-inverting. Inverting sense it is
inverting all find the hidden negative sign
because it inverts, we multiply it by negative sign. And the non-inverting
means it is positive, it is as it is. Okay? So that's what it's called. This is a negative inverting. The plus is the non-inverting. Okay? Then we have the output out of this
operational amplifier. So the input here,
number two is the input, number three is
input and output of this device or the operational
amplifier is number six, which is the output. Now we will find that
number four and number 45, which is pin number seven, and number four for representing they're
all steps of the supply, negative supply. Okay? So for example, if we
have here a DC supply, for example, like this, okay? Plus and minus, this is
a DC supply is ends up. Bolster will be connected here, and the negative will
be connected here. 105 or not, we are
not using them, okay, So we can assume that the
Donald does not exist. So the plus connected
to the positive here and negative connected
to the negative here. So let's see here,
bolster supply, negative of the supply. So again, most of minus
supply terminals here, the negative and the plus ends out two inputs and one output. Now, an input applied to the non-inverting terminal,
which is this one. Remember, non-inverting
will appear with the same polarity
ads out here. While an input applied to
the inverting terminal here, it will appear inverted
at the output. Again, here if we have a plus, plus, if we apply this up, all stiff to the non-inverting sends out would be blessed. If we apply a plus
to the inverting, then it will be negative. So here, non-inverting
preservers or keep the same polarity and the non-inverting or the inverting reverses polarity
or a change as oblige. Now, barring an op amp. You will find that here when we look carefully about powering, which means we are
connecting it to a supply or operating gate. So again, we said
before that the op amp is an active
circuit element, which means it needs a supply. So finds out the supply here, as you can see, we can
represent it like this. Here. The positive terminal is connected to
the positive here. As you can see, the
negative terminal here. This negative is connected
tos and negative here, which is number four, like this. So this tool, you have
to know that here, this is a ground point, ground means of 0 voltage. And you can see that here, this battery plus minus, which means that the voltage here is greater than the
voltage here by VCC. So this will be, the
voltage here will be VCC or bolster value. Okay? So this will be the
positive terminal. Now what about this one? You can see that
here this one is 0. And this one is what you
will find that here. Plus. Monuments, visa, see,
what does this mean? It means the voltage
of this point is higher than the voltage
of this point by Vcc, or Vcc is equal to this voltage minus
this voltage, 0 minus vx. Vx will be negative. We see what is vx? Vx is this point, which means that this
value is negative Vc. So we'll find that the negative
voltage is connected to four and the bolster
volts connected to seven. Okay? Now we will find that here
since we have two inputs, it means there are two currents, are U1 and I2. And do we have here a supply, blood supply, and
the negative solar, which means that we have currents guaranteed
in both here, in bold here, and one output
which is coming out of here. So we have four currents, or E1 or E2, either positive or
negative going through. They're going inside the
op amp and the one output. So we'll find that from KCL that the output current is equal
to summation of all of these guns equal to I1 plus I2 plus I
posted plus i negative. Okay? Now what is the
equivalent circuit? Now will have to know that
the operational amplifier, in order to analyze the circuits which contains
the operational amplifier, we need to find an
equivalent circuits. So we have this
symbol negative plus. Now what is inside the op amp? I would like to know what is
the equivalent of this part, is the equivalent of
this part in order to help us to analyze our
electric circuits. So we'll find that
here we have 23. Welcome to the negative terminal and the positive terminal. If we go inside the op amp, we can represent it by
v1 and v2 between them, inside them, there is a
resistance between V1 and V2. This voltage and this voltage. What is the resistance here? It's called the R
input one and input resistance seven and
equivalent circuits. Okay? Seven and equivalent resistance, or R7 and gold are input when we look at the circuit from here. So we'll have our
inputs which are representing the
equivalent circuit. Now we will find that here. Since we have this
Zappos tech terminal, this is a negative terminal. So we have a voltage difference
between these two, Vd, which is the difference
voltage or the differential, differential voltage
between this and this, which is V2 minus V1. Okay? Vd here plus, minus, plus is supply positive
caramel, like here. This point. And then negative is negative. So V d is equal to V2 minus V1. Okay? Now for the output, when we look at it, we will have also a
seven output resistance, which is called our output. So both of this and
this are 27 and resistors when we look from here and the other one when we
look from the other side, in a way you don't know, you don't need to know these two values and you will know why in the next lecture. Here we are representing
a non-ideal. Remember, mom, ideal op-amp
or operational amplifier. So as you can see
that the output here, V out, what is the
value of V output? You will find that here we have, again, this is called, again, which is a value multiplied by the differential voltage or the difference in voltage
between these two terminals. So when we multiply
this by again, and you will find
that here minus the voltage drop
across our output, we will have the output voltage. So the output section
consisting of a voltage controlled
source, as you can see, voltage control
because we have again, multiplied by the
difference between these two voltages in series
with a resistance R out. So it's evidence
that from the figure that the input resistance R i is this resistance is a seven
and equivalent resistance seen as the input terminals. When we look at the
circuit like this, we will have an input
equivalent resistance or output resistance, or output is seven and
equivalent ads out. When we look at the
differential input voltage, VD is V2 minus V1. Now, as you can see here, we have V1 and V2. Again, V1 is the voltage here, and V2 is the voltage here at the negative terminal and
the positive terminal. So V1 is the voltage between the inverting terminal
and the ground. So V1 is the voltage between
this point and the ground, okay, so as we can see here, so the voltage
between this point. And the ground is a potential difference
between these two. And V2 is the
potential difference between this point and this one. Okay? Okay. So here's the OPM
sensors are differences between the two inputs and
ontologies by a guinea. So it Vd is a differential
between these two, is multiplied by again a. It will give us the output. However, here we
will have our out. The V output here should be
equal to this value minus, or I would multiply it
by the current I out. The current flowing here. Okay? So it'd be, should be
this one minus this one. However, in the ideal op amp, we can neglect as this part. So we will have V out
is equal to a V D. Now, where did we get that the ideal is equal
to this value will find now for us to all
find that a is called the open-loop voltage
gain because it is again at the op amp. Because again of the OPM without any external feedback from
the hour to the input. So it is again,
without any feedback, feedback, what does this mean? From our two input from
here to here, like this. So there is no feedback, the feedback or
feedback resistance without any feedback
from here to here. Okay? Now what are the optimum
values for the op amp? Okay? So we have here a non-ideal
op amp and ideal opamp. So in the non ideal op amp, you will find that
that typical range for xenon ideal value are
funds that are input. For example, our input is from ten to the power
five to ten to the power 17, and its ideal value is infinity. This is the ideal value. And xenon idle we use
this typical range, and in the idle we use this one. Now is the output resistance, or output is between
ten to a 100 ohms. And ideally it should
be equal to 0. Supply voltage, it can
be from five to 24 volt. And the open loop gain
is very large value from ten to the power five
to ten to the power eight. Now, what are the regions of operation of an ideal op-amp? So the non ideal op amp has
three modes of operation. Okay? So first you will
find that here. You notice that the
output is equal to here. If you go back here, you will find that the V output, ideally, considering that
are out, is equal to 0. You will find that V out
here is equal to a VD. This is an ideal value without considering
the voltage drop here. So V out is equal to a V d. Now, what is the
maximum value of V? The maximum value is what? The maximum value is, of course, equal to V supply, V supply, which is VCC. So find that here
during examples of saturation or the
negative saturation, which means that when we
have very large, again, multiplied by V D, When this multiplication
is very large, it's a maximum value in the
positive region will be VCC, and in the negative region
will be negative VCC. So this is the maximum. You can not increase the output voltage
more than the supply. Okay? So here we have,
starting from here, as you can see, when the differential difference
multiplied bys again, AVD at the beginning, when VD is small, you will find that
the output will start increasing
like this, linearly, like this, until reaching a certain value when
it's equal to VCC, after which we cannot
have more than we see, this is a maximum value. So again, a V d, a is a constant value. Vd is a difference between as negative as a positive
and negative terminals. As VD increases, the output increase until
reaching the maximum value, which is a supply away. Okay? And if we increase
VD more than this, we will still have the
saturated value, which is Vcc. This is the same case
in South Boston, boston region and in
the negative region. So again, as you can see, most of saturation, we, our goal VC is the maximum value in
the positive direction, negative saturation
in maximum value in the negative direction. And then between them, you will find here that
V out is equal to a V D, which is between Vcc
and the negative Vc. Okay? So in this lesson, we discussed what
is the meaning of m and the different pens
insights that op amp. And we'll discuss
that the op amp consisting of two
terminals, a plus, minus and in consisting
of an input resistance, R i and R output. And we have again multiplied by V d to get the
output voltage. Now, in the next lesson, we will have an example
on the non ideal op amp. And from this example, you will understand that it is difficult to deal with
mom ideal op amp. And from which we will
use the ideal op amp. You will find that the
ideal op amp gives us very, very values, close, very close
values to xenon idea OPM. You will find that the
non-ideal is very, very close to the ideal op-amp. Okay? That's why in the
rest of this course, you will find that we use the ideal Op Amp
instead of Zama item. So firstly, let's have an
example and then we will understand the ideal op-amp.
46. Example on Non Ideal Op Amp: So let's have an example
on the non ideal op amps. So as you can see,
we have this op amp with the negative terminal, positive terminal that out and connected to
several elements. You can see at 20
kilo-ohm bit connecting between the output
and the input. And often V supply ten kilo. So find the 741 op amp. Each of them has its own number. And each of this number
representing something about this op-amp has an
open-loop voltage gain of two multiplied by
ten to the power five. Again, what is the
open loop? Voltage? Is a. Okay? As we said before, that
the output is equal to a multiplied bys that
differential input. The input resistance
of two mega ohm, which is the input
resistance when we look at the op amp from here. And out resistance of 50 ohms, the output resistance here, when we look at it from here, seven in both seven and output, the OPM is used in the circuit, finds a clause, finds a close the loop
gain V out over V S. What does this mean? We need the relation between the output voltage
and input voltage. And determines that current I, this current when
the supply voltage is equal to two volts. When the supply is
equal to two volts. So let's start by first
representing all of our inputs. Funds. That's again, or when low voltage again
to Motorola buttons. So R5 is 20 thousand, which is a homologue gain. Second one is input
resistance Domingo, which is our output
resistance of 50 ohm, which is the R output. And the supply voltage
is equal to two. Okay? Now in the second step
is that we are going to remember our circuit. So remember this is
a non ideal op amp. So the non ideal op amp,
as you can see, V1, V2, or inward or
outward AVD and V-out. Now how can we substitute with all of this in the circuit? Okay? How are we going to
do this assembly is this triangle will be
replaced by the string. So how can we do this? You can see V1 and V2 is the negative terminal
and the positive terminal. Between these two terminals, we have a resistance or input. I'm going to do like this. Okay? Then we have at the output here, this out is what is
connected to our output, AVD, or output connected
to the ground. So this is our ground. So we have here our ground, here, the spot and
here like this. Plus minus a V d. Then connected to our output and connected to the out point
and then cancel the spot. So again, you will
find between 12 are inputs connected to the ground and the between the output or our AVD
connected to the ground. Here, you will find Lexis. Okay? So firstly between
V1 and V2 are in, but you can see our input
between V1 and V2 are input. And between the output
and the ground, you will find that AVD are out, AVD or out, as you can see here. Look at this port one
and the ground exists. One connected to
the ground here, so sparked the ground. And second part,
resistance, AVD, ground resistance, AVD
then ground. Okay. So now we replaced
this is simple with the equivalent
circuit, this circuit. Now what is an extra step? The next step is that we need, remember what we need. We need the current
I and the winning the relation between
V out over V supply. Okay? So first is account, then V out over V supply. So the current here, this current is what we need. And do we need? So first,
let's get the car. Okay, is this the
most assemble sink? So how can we do this
simply by using KCL? You can see that here we have a current here flowing here. We'll go through here and another one going
through here like this. Then this current, which is i, you will find this
point K is a current. The same current will go here. So I will go through here. Okay. So let's do this. First. We will say that this
one is called V1. And here we have V out. So by applying KCL
here and KCL here, where you can get the
value of current. So let's just start at
node number one here. You will find that here. The three counts, 123. So the input current is equal
to two outgoing currents. The input current equal
to two outgoing currents. The first account,
which is in bold here. What is the value of
this current is V supply minus V1 divided by
the ten kilo-ohm. V supply minus V1 divided
by the ten kilo-ohm. Now, the current I would
go in for a stone, V1 minus 00 divided by two omega V1 minus 0, which is as if it
does not exist, divided bys at two mega ohm
plus secondary current to going V1 minus V out
divided by the 20 kilo-ohm, V1 minus V out
divided by 20 kilo. Okay? So now from this equation, Let's delete all of this. You will find that
we have V supply, which is given in the problem, and we have V1 and V out. Okay? So from this equation, if we simplify this, we will get that V1 is equal to two V S plus
V out over three. Now let's do another KCL
here at this terminal. So we have this current
is equal to this current. This current is equal
to this current. We can say like this, at node 0, you will find
that the current going here, what is its value is equal to V1 minus v 0 over 20 kilo ohm. V1 minus V output divided by the 20 kilo-ohm equal
to the same current. This current, current here, similar to this one. This one is equal to what? Is equal to this voltage. What is the value
of this voltage? First, you will find that
here this is equal to 0. And plus minus means that this voltage is greater than
this voltage by this value, which means that this
one is a reading. Okay? So we will have the
current flowing here is equal to v 0 minus
AVD over 50 ohm, V 0 minus AVD over the 50 or so. And we have the gain given as 200 thousand and
V d as given as, as a differential is
equal to negative Vg one. Where did we get this? Remember that the
differential V d is equal to V2 minus V1. And V2 is connected to the
ground, so it is equal to 0. So let's get back. You can see that here, here. Here, here V2 and
this one is v1. V2 is connected to the ground. Now one important note
here is that the a, which is again to multiply
by ten to the power five. So it means that it
is 200 thousand. So there is another
0 here and 0 here. You will find that here. I wrote it correctly, as you can see, 200 thousand. So this is correct. V d will be equal
to negative v1. Vd is equal to
negative V1. Okay? So from this equation
we will have V1 minus V 0 is
equal to this value. Okay? So let's delete all of this. Now, as you can see, as you can see, we
have this equation. V1 is equal to V S
plus V out over three. And we have V1 minus
V 0 is equal to 400 V out plus 200 thousand V1. Okay? Now, V1, we can take V1
and substitute it here. Like this. This way, we will have VSV out, what V out would be
our V1, V1 here also. So in the end, you will have water, you
will have one equation. If you use this
one and this one, you will have one
equation which gives us a relation between V out and V supply will find
that V out over V supply is equal
to negative 1.9969. Now you will notice
something which is really important that the output is equal to the supply multiplied
by negative two. So V out. Is equal to negative two, almost negative two multiplied
by V supply funds that the output is equal to the
inversion inverted value, inverted value of V supply
multiplied by again, okay? So the output is multiplied by two and it has an inverted sign. Now we will find that this, this is similar as
what we discussed. You will find that
here is the supply is connected to what is connected
to negative terminal. Okay? That's why, how, what is the
inversion of the supply? Because it's connected to
the negative terminal. That's why here
you will find that the V output is equal to
negative supply itself. Now second thing is that
we need the current. So what is the value of current? So the current can be
obtained by V1 minus V out divided by 20 kilo ohm. You can see that when
v sub y is equal to two volt v sublime is given. So the V out is
equal to negative 3.994999 from this equation. So we have the value of V out, V out negative four, and V1. What is the value of
V1? You can go here. So V out is equal to negative four and V
supply is equal to two. So from here you can get V1. So we wouldn't
have V out and V1. So we can get the current, which is V1 minus V
out over 20 kilo. Okay? Finds that the current
is equal to V1 minus V out over 20 keto. So what do you learn
from this lesson? You will learn from
this lesson is that working with
non ideal op amp, such as in this
example, is dy dx, which means it is
very difficult to, because we are dealing
with large numbers to omega 20 kilo-ohm and so on. So what is the solution? So the solution is
that instead of using non ideal op amp, we will use an ideal op-amp. So instead of having a
gain of 200 thousand, we will assume it is infinity. Instead of having our input. We will say that this input,
this input resistance, is equal to infinity, which means this one is an open circuit as if
it does not exist. And the output resistance
or output is equal to 0. So we can see that this
one is equal to 0, which means it is a
short circuit like this. Okay? So from this, what we will have, you will find that when
we are dealing with ideal op amp is that we
analyze our circuit, you will find that V out over V S will be equal
to negative two. And the current will be
point to Mendeley and bear. This is in what, in
the ideal opamp. So where did we get this tool? You will find this in
the next lesson when we discuss the analysis
of the eyes are open, you will find that the
two values are very, very close to each other here. So it is much better to
use the ideal op-amp, not Zama, ideal op-amp. To make the analysis
very easy or much easier than the Nine Manager obey.
47. Construction of Ideal Operational Amplifiers: Hi and welcome everyone to our lesson in this section
for operational amplifiers. In this lesson we are going
to discuss the idea of m. So in the previous lesson, we discussed the
non ideal op amp and we said before it has
these values for the gain, for the resistance
for the supply. Now, in the ideal op amp, what we have here is that
the gain will be infinity. That resistance, input
resistance will be infinity. The output resistance
will be equal to 0. So let's just start out
resistance equal to 0. It means that this one will
be a short circuit like this. And the input resistance
are all becoming infinity. Infinity means that
this one will be an open circuit as
if does not exist. So we will have v1 and v2 by exists between them
and open circuit. Okay? Now, this part, you will find that it
will be a when vD. So a is again in this
case will be infinity. And the VD video, which is a differential voltage, or the differential difference
between these two voltage, V2 minus V1 plus minus vD. So we'll find that
in the aisles. And so we have
here open circuit. And each of these voltages are, is a difference between
it and the ground. So in this case, you will
find that V d is equal to 0, the differential
voltage equal to 0, and V1 will be equal to v2. Okay? So in this case you will
find VDO equal to 0. So finds that a
multiplied by reading a multiplied by the SVD is
infinity multiplied by 0, which is, of course, this multiplication
is undefined. Okay? So we cannot get the value of the V output using this part. Okay? So we need to do another
circuit analysis in order to get V out. So let's again identify
what we just said. The ideal op amp is an amplifier with an
infinite overlooked gain, infinite input resistance,
and 0 output resistance. So finds that in this case, since we have here open circuit, then the current coming here, or the current here, I one or i2 will be equal to 0. Why is the R equal to 0? Because this part will be
an open circuit infinity, which means this part
will be an open circuit. No current will pass here. So finds that I1 is
equal to i2, equal to 0. As you can see, I1 is the
current input to v1 and i2 is the current input to the second determinant as
opposed to a terminal. And you'll find that the
differential equal to 0 and V1 here is a difference
between it and the ground. V2 is the difference
between it and the ground. V2 will be equal to V1
in the ideal op-amp, and the current
will be equal to 0. Here. Again, that
differential voltage, V2 minus V1 is equal
to 00 ideal opamp. So V1 to V2. So what we learn
from this forcing is that this voltage
in xylitol OPM, this voltage and
this voltage here at the terminals is
equal to Chaucer, V1 equal to V2. And the current here, these two currents
are equal to 0. So let's just start by having an example owns
the ideal op amp.
48. Example on Ideal Operational Amplifiers: So the same example as before. Remember this example
which we have used in Zara non-ideal OPM, the case when we had V supply, then kilo-ohm, 20 kilo-ohm 741. And we needed this current. We need is the V out endings
are non ideal op amp. We did large circuit analysis. If you remember, let's get
back here. This example. We did several analysis, as you can remember
from here, like this. And we obtain the
finally is that voltage. Here's the issue
is negative 1.999 and the current is 0.910
milliampere. Okay? This is non ideal Op-amp. Now remember these values,
negative 1.99.19 milliampere. Okay? Why should remember
because we are going to use them or compare between
them and the ideal case. So if we get back to
the ideal op-amp, here, we need the
current and the voltage. So what are we
going to do simply? We will start by
doing like this. So here what we learned
in the ideal opamp is that the voltage here is
equal to the voltage here. So V2 and V1. So V0, V1 equal to V2. Now, V2, as you can see here, is connected to the ground. So what is the voltage of V2? V2 is equal to 0. So we can say equal to 0, okay? So V1 to V2 equal to 0. Second thing is, which we
know is that the current here and the current going inside this
op-amp is equal to 01, equal to 0, and the i2
is equal to 0. Okay? So what do we learn from this
is as a current going from this supply flowing
through this resistance, like this, is equal to
the current flowing here. So let's say if this
one is equal to, is I capital than this I capital is current
flowing through here. I can written like this, or a capital which is
similar to why is this? Because in this case
in this node here, the current flowing inside
the slope m is equal to 0. So the current here
flowing through the ten kiloohm is
equal to the current flowing through the
20 kiloohm because no current will pass
inside that op-amp. Okay? So from this
what we can learn, how can we get the current I? So assembly is a current i, since we know this
voltage is equal to 0, equal to this voltage. So V supply is
equal to two volts. So from Casey, from the
voltage alone, okay? Or the Ohm's law, you will
find that the voltage here, which is VS minus the
voltage here, which is 0, divided by the ten kilo
ohm gives us the current, which is similar to, I, will find that this current will be equal to
the current flowing here. I is equal to V supply minus 0, V supply minus 0 over
the thing kiloohm. So V supply, which
is two volt minus 0 divided bys at ten kilo ohm
gives us 0.2 milliampere. This current is equal to
the current flowing here, which is o, because the
current flowing here is 0. So zr as if they are in series. So if you remember from the
non ideal op amp example, this value was in the
non-ideal case 1, mine, mine, mine
mainly and bear. So as you can see, this value
and this value are very, very close to each us. Almost identical. So in this case, you will find that
in our example we can use the ideal op-amp, which is much easier instead of the non ideal op-amp,
will find that here. If we use our voltage, let's delete all of this. So we know that this
one is 0. Okay? So the current, so this current multiplied by 20 kiloohm
gives us negative V out. Okay? So how did we do this? Simply is a current
flowing here. The current flowing here
is equal to 0 minus V out, 0 minus V out divided
by the 20 kilo-ohm. Okay? So we will have
negative V output, negative V out will be equal to 20 K multiplied by the current. So from here we know the value of current is 0.2 milliampere. So we can get the V output,
as you can see here, 0 minus V out here equal to current multiplied
by 20 ketone. So we will have v-out
equals negative four volts. This value, again, is very, very close to an ideal case. First of all, you
will get the gain. Gain is V out over V supply. So V out is negative four
and V supply is two volt. So when we divide these
two values y each other, we get negative two. If you remember in
the ideal case, this value is again was
negative 1.99. Okay? So this value is very
close to the negative 1.999 in the non ideal case. So what we learn from this, we learn that we can
use the ideal op amp, which is very easy in analysis, converted to the
non ideal op amp in the circuit analysis. So we can assume in our circuit, this op-amp is ideal. And instead of
Marmite approximation is very acceptable
and provide very, very small error. Okay?
49. Construction of Inverting Operational Amplifiers: Now let's discuss
another type of op amps, which is the
inverting amplifier. So what is the benefit of
the inverting amplifier? It's assembly inverters, the voltage and the
multiplies it by again. Okay? So if we have a V input
like this and the V output, so the output voltage will be V input or negative V input, V out will be negative V
multiplied by a certain gain. Okay? So we inverted the input and multiply it by a certain gain. So this is what does an
inverting amplifier. So first thing you
will notice in this circuit for the
inverting amplifier is that since we are
talking about inverting, it means that we are
going to connect our supply to where? To the negative terminal. So it's a supply connected
to the negative terminal. So it's called the
inverting amplifier. If we connect this supply
to the positive terminal, it will be non-inverting
amplifier, which we will discuss
in the next lessons. So first inverting connecting
to the negative ten. So let's start. So in this circuit than
non-inverting input is grounded, you will see that here is our non-inverting because we
are dealing with inverting. So the supply is connected
to the inverting part. For the non-inverting
connected to the ground. Okay? Vi is connected to the
inverting input through R1, resistance, R1 connected
with resistance R1. And the feedback resistor, RF is connected between the
inverting input and output. You will see that in this
circuit you will find that the output is connected
to the input using an RF. What does RF is the
feedback resistance. So the composition
of this circuit is first supply connected to
the negative terminal, Zappos, the terminal ground. Then we have two resistors. One connecting in-between. The negative terminal
and supply are one. And the one which is our
feedback connecting between the output and then negative
feedback or the input. Okay? So let's just
start by learning what is the relation between V out and V input in this circuit. So first, we will apply KCL. Remember that the
current flowing here is equal to what? 0 and bear. Because we said before that
no current will pass here. And at same time,
this voltage is equal to this voltage
in the ideal op amp. So V1 is equal to V2
and V2 is grounded. So V1 will be equal to 0 volt. And the current flowing here. And similar tools or
current flowing here. Okay? So from KCL, from KCL at this
node, you will find that V, This current is equal to
V input minus 0 over R1, V0 minus V1, which
is 0 divided by R1. And the current flowing here, the current flowing
here is equal to V1 minus V out over RF, V1 minus V out over
RF or V1, which is 0. Okay? So we'll find that
negative v1 over v0, input over R1 is equal to
negative V out over RF. And V1 is equal to V2 equals 0. Okay? So from this equation, when this one equals
0, and this one is n equals 0 in the ideal op amp. So we will have V input of our R1 equals negative
V out over RF. As you can see here
from this equation, we can say is that V out is equal to negative
R F over R1 VM. So what we did here is that the output is equal
to negative V input. That inverted voltage
of the input, reverse polarity or reverse the sign multiplied by
a third thing gain. This gain is dependent
on RF over R1. So as you can see here
from this circuit, we have V out negative
or F over R1 VM. So we inverted the input, multiply it by again. So an inverting amplifier
reverses polarity of the input signal and amplifies
it with a certain gain. This gain is equal to
our feedback over r one. And this game again is
V out over V in both, which is negative R, F over R, one.
50. Example 1 on Inverting Operational Amplifiers: So let's have an example
on inverting amplifier. So if we have in this circuit, V input is equal to 0.5 volt, we have ten kilowatt here, which is that resistance
connecting between the supply and the negative
terminal, which is R1. And you can see
that this circuit is an inverting amplifier. Okay? Why? Because the second positive
is connected to the ground, supplies connecting to the negative terminal
with a resistance. And then we have a
feedback resistance, 25 kilo ohms. Okay? So what we need here is
that we need to find V out and the current flowing through the
ten kilo ohm resistor. Okay, so it's very easy example on the inverting amplifier. So remember that V out in
the inverting amplifier is equal to negative
R F over R1 V. So this is what we
are going to do. V-out is simply equal to. Here was a first requirement is output voltage then
the current, okay? So the output voltage, V out is equal to negative
or F over R1 V input. So what does the input voltage? Input voltage is 0.5. What does our f, which
is a feedback resistor, which is here, 25 kilo-ohm. And what does R1? R1 is a ten kilowatt
resistor like this. So V out is equal to negative
25 over ten multiplied by 0.525 over ten is 2.5
and V input is 0.5 volt. Okay? So this is the output
voltage here you can see inverted and multiplied
by a certain gain. Now we need the current
is the current here, which is flowing through
the ten kilo ohm. Symbol is this current
is equal to here. What is the voltage here? Here is equal to 0,
equal to this voltage. Okay? So that current will be equal to the current flowing gifts
for the ten kilo-ohm is V input minus 0
over ten kilowatt. V input minus 0 over
010 kilowatt. Okay? So V input is 0.5
divided by 10 thousand. That gives us the
value of the current. As you can see here. Vm
what minus 0 over R1. So 0.5 minus 0 over ten kilo ohm gives us 50 micro admire, the current flowing here. Now at the same time
you can get it with another method assembling
the current flowing here is similar to the current flowing through
that 25 kilowatt. So we can say is that
he is as voltage is 0. So we can say is a current also is equal to
0 minus V out and 0 minus V out divided
by resistance, 25 kilo ohm, like this. So in this case, you will find that here negative V out is 1.25 divided bys at 25 kilo ohm gives us
the 50 micro and bear. So this was a very
simple example on the inverting amplifier.
51. Example 2 on Inverting Operational Amplifiers: Now let's have another example on the inverting amplifier. So in this circuit, we would like to get the
output voltage in this. Okay? So as you can see
in this circuit, we have how many supply? We have six volt, 20 kilo ohms and a
feedback resistor. And in that terminal B, it is connected to
the ground with Amazon Supply two volt. Okay? So can we use directly
V out is equal to negative feedback over R
multiplied by the input. No, Why? Because we have here a two volt. Okay, it's not connected
to the ground. So first, how can we get V out? Very easy. So first is the current
flowing here is equal to 00 current or equal to 0. The current going
into the op-amp. Second thing, which
is a and the b. You know that this voltage is equal to this voltage
in the op amp. Now what is the voltage of B
and B is equal to two volt. This point is two volt, okay? So a is equal to two volts. Okay? So we have this point. So we can say is that
the current flowing here is equal to the
current flowing here. So from KCL, six volt minus two volt divided by the 20 kiloohm
gives us this voltage, two volt minus V
out over 40 kilo. So again, the first, the current six volt minus two volt divided
by the 20 kilo-ohm. Okay? 20 kilo-ohm. Like this, equal to the current
flowing here, which is two volts minus
V out over 40 kilo-ohm. Like this. So we can take the
skill with this one. So we will have six minus two, which is four over 20 equal to two minus V out
divided by four. So from this equation, you will get V out with,
as you would like. Okay, very simple, KCL. Let's discuss what we
exhausted it here. First, we apply KCL
at this node a. So the current flowing here is equal to the current
flowing here plus this one. This one is equal to 0. So the current flowing here is equal to the current
flowing here. So V a minus V out, V a minus V out divided
by 40 k is equal to six volt or minus a over 2686
minus VA over 20 kilo ohm. So from this you will find
V out is equal to three va minus 12 and the
VA is equal to VB, equal to two volt. Okay? So in this case, we'll find that the output voltage is
equal to negative 601. Okay? So this was another example
on the inverting amplifier.
52. Construction of Non Inverting Operational Amplifiers: Hi and welcome everyone to our lesson in our course
for electric circuits. In this lesson, we
are going to discuss another type of
operational amplifiers, which is a non-inverting
amplifier. So in the previous lesson, we discussed the
inverting amplifier, which is we have an op-amp
and that ends the inverting. We connected our supply. If you remember, the supply is connected to the
negative terminal. Okay? That's why it was
an inverting amplifier. In this case, of the
non-inverting amplifier, is our supply is connected to the positive
terminal of the op-amp. So as you can see
here in this circuit, we have a short circuit here. We don't have any supply in the negative terminal,
only one supply. And suppose the
similar as before, we have our feedback which
contains consists a connected between the output voltage
and the negative terminal, the output end negative
term. Remember this. Now what are we going to do? We would like to get the relation of the
non-inverting amplifier, the relation between
V out and V. Okay? So first, as we remember
that the voltage of the positive terminal and
the negative terminal in the ideal op amp are
equal to each other. So if we say this
is V1, this is V2. V1 equal to V2 equal to V two. Now what is the value of V2? V2 is a value is v
in V m, like this. So what we need is the relation between V out and
this input voltage. So first we have here ground
voltage here is equal to 0. This is V out. So there's a voltage is V out. So as you can see that
the current flowing here is equal to the
current flowing here. Because the current
year is equal to 0. Any current going into
the op-amp is equal to 0. So I1 is equal to i2 from KCL. So I1 equal to two. So what does the value of I1? I1 is 0 minus V over R1. 0 minus V input over
r one equal to i2. The current i2 is
V M minus V out, V input minus V out divided
by the resistance R F. Okay? So we have, here are a few
simplify this equation. You will find that V out
is equal to V input one plus our feedback
over R one. Okay? So as you can see here, z have the same sign. If this is a positive lens,
it will you post them. It's because it
is non-inverting. And one plus R, F
over R1 is the gain. Okay? So let's delete all of this. So as you can see, I1 equals I2. So here, 0 minus V1 over R1 equal to V1 minus
V out over RF. And you'll notice
that V1 is equal to V2 equal V input here. So from this equation, we have this equation. Then it's simplified
as V output is equal to one plus RF over
R1 multiplied by u0 v0. This is our gain and
this is our input. So this is our gain. A non-inverting amplifier
is an op amp circuit designed to provide our
balls the voltage again, all stuff because it
is the same sign, non-inverting same
sign and gain, since we are multiplying by
one plus R F over R one. So as you can see
from this equation, is that V out is
equal to one plus RF over R1 multiplied
by VM, right? So you're not something which is really
important if that, if R F is equal to 0, you will find that V out
is equal to V input. This part will be 0, so it will be V out one
multiplied by V input. So the output voltage will
be similar to input voltage. Or if R1 equal to infinity,
very large value. So this part will be equal to 0. Anything divided by
infinity gives us 0. So V out will be V input also. So in this case
you will find that the voltage are
equal to each other. So do we use this? Yes, we use this in
our electric circuit. In this case, the op
amp is called a voltage follower or a unity
gain amplifier, because the output
follows the input. V-out is similar to VM. Now what is the
function of this? You will find here. This is in, as an example, when the feedback is equal to
0 and R1 equal to infinity. As you can see, this is called a voltage follower or a
unity gain amplifier. You will find that
RFA here equal to 0 and R1 is equal to infinity. Okay? Now, such a circuit has a
very high input impedance. That's why we use
this circuit as an intermediate stage or
a buffer amplifier in order to isolate one circuit
from another like this. So for example, if we
have a circuit here and another circuit and
we would like to isolate between
the stool circuit. So we can isolate
it using a buffer. I'm going to fire. Or a unity gain when
non-inverting becomes unity gain has a
unity gain when our F is equal to 0 and the
R1 equals infinity. Okay? So you'll find that V
input is equal to V out. But this part, when we added
this buffer amplifier, it isolated between
these two circuits because it has a very
high input impedance. Or input, or input is
equal to infinity. Okay? If you remember from
the ideal op amp. So this infinity help us, us to isolate between
these two circuits.
53. Example on Non Inverting Operational Amplifiers: So now let's have an example on the non-inverting amplifier. So find the value of Z V
output in this circuit. In this circuit, as you can see, we have six volt, four kilo ohm. We have a four volt
and ten kilo-ohm. Here is four volt is connected
to the positive terminal. And six volt with a four kilo-ohm connected
to the negative terminal. Okay? And we would like to
find V out. Okay? So the question is, is this circuit
inverting amplifier or a non-inverting amplifier. You will find that
this circuit is a mixture between these two, or a mixing between
these two circuits. It is an inverting and non-inverting amplifier
at the same time. Because this part representing
an inverting amplifier, this part representing
non-inverting amplifier. So how can we solve
a circuit like this? So the first method
is that we can use the superposition we
have discussed before. So we have two supplies, six volt, another
supply, four volt. So by using superposition, we can get the effect of six volt and the
effect of four volt. And we sum these two
voltages to get that V out. So by using superposition, that output voltage will
be V out with one plus V, where V output one is due
to the six volt source. And we are able to do
those are four volt input. Okay? So first, let's
say we need V out one. So I would voltage one is due to the sixth of also
due to this stores, due to this source. So in order to get this effect, we deactivate the supply by
making it a short circuit. So when this one becomes
a short circuit, you will find that we
will have six volt, four kilo-ohm than ten kiloohm. So it is connected to
the negative terminal. So this is what is this circuit? The circuit is an
inverting amplifier. So this is our feedback. This is R1 a V output, in this case, V out is
equal to negative V input. Multiply it by R, F over R or the
inverting amplifier. So it will be negative. We input six volt R, F over R, then
divide it by four, then divided by four,
as you can see, then divided by four
multiplied by six volts, that gives us negative 15 volt. So this is what the hour
due to the effect of the negative terminal
or the supply six volt. Now the same idea we are going
to do for the four volt. We will deactivate the six volt, making it a short circuit. So we will have a
non-inverting amplifier. So the output will be one plus RF divided
by R1 plus one divide multiplied bys is a four volt. So this is n
non-inverting amplifier. This is an inverting amplifier. Inverting because it's connected to the negative terminal, non-inverting because
it's connected to the positive term. So we will have 14 volt. So now the V output will be the summation of these
two voltages like this. So V out will be equal
to negative one volt. So this was the first method. In order to solve this example. The second method
is that we can, we can say is that this
point and this point, these two nodes are
equal to each other. So this B is equal
to four volts, and this one is equal
to four volts. Okay? So by applying KCL at a, we have this current here
equal to this current here. So we can say is that six
volt minus four divided by four kilo ohm is equal to VA, which is four volts minus V out, divided by the ten kilo ohm. Very easy by using
KCL at this point a. So we'll find that
six so minus VA, which is four here, divided by the four kilo-ohm. Here you can take this was this. So we will have 410, same idea, VA, which is four
volts minus V out. Okay, so let's
delete all of this. You will find that V equals VB. This point equal tos is
point equal to four volts. So we have finally V-out
equal to negative one volt. So as you can see, by applying KCL or by applying
the superposition, both of these methods
provides the same solution.
54. Construction of Summing Operational Amplifiers: Now let's discuss
another circuit in operational
amplifiers or op amps, which is called the
summing amplifiers, or sometimes known as
that summer circuit. So as a summing amplifier
is simply what does it do? It sums or adds different
signals more than one voltage. If you look at this circuit, this is summing amplifier. You can see that here we
have our feedback as before. The positive terminal is
connected to the ground. And here, this part is a
resistance with our supply. Okay? So if you neglect
as a store bought this and look at R1 with a
supply, you will have what? You will have an
inverting amplifier. So it is the same idea. Instead of having just
one inverting amplifier, one input, we have
several inputs here. So as you can see, V1, V2, V3. If you cancel v2 and v3
as f x0 don't exist, you will have an
inverting amplifier. So the summing amplifier is an inverting amplifier
with multiple inputs. So our summing amplifier is an op amp circuits that combines several inputs and produces an hour that is a weighted
sum of its inputs. Okay? Why waited? Because it is multiplied
by a certain thing, again, dependent on
the resistance here. All of these resistors. So let's see how
can we get this. It's very easy. You can see is that from KCL
at this point, at this 0, you will find that
the current I1 plus I2 plus I3 is equal to o. Okay? So again, V1 as supply V2, V3, each of these
produces i1, i2, i3. Observe some mission will
give us the current I. So we can say equal to
I1 plus I2 plus three. What is the value of current? I can be obtained from here, from 0 minus V out
divided by RF. So 0 minus V out divided by RF equal to the current I1
is V1 minus V2 over V1, V1 minus 0 over R1 plus R2. The current I2 is V2 minus 0, since the auditor
same node here. So V2 minus 0 divided by R2, V2 minus 0 divided by
R two plus V3 minus 0, again divided by R3. Will find that here
from this equation, V1 over R1, V2 over V1, V2, V3 over R3. So V out will be negative r. F takes this RF to
the other side, multiply it by all of this. So our F over R1, V1 plus V2 plus V3, V3. So finds that here
it is as if it is a non-inverting one
to inverting 1233, inverting amplifier, inverting amplifier
connected together. Okay? So as you can see here
again, like this. So all you would come
to I1 plus I2 plus I3. And each of these is the difference in the voltage
divided by that resistor. So we'll have and v is
equal to this point, is equal to this
point equal to 0. I'm sorry for repeating is the same idea because
it is very important. Some people would like to hear the explanation
more than one time. So here you will find that the V out negative R F over R1, V1 plus RF over onto
V2 plus RF over A3 V3. Okay? So here is our final circuit. Okay? So let's have
an example on this.
55. Example on Summing Operational Amplifiers: In this circuit, we would like, or in this example we would
like to get that V out. And the current IR would
in this op amp circuit. Having the circuits at two volt, we have one volt, 2.5 kilo ohms, five kilo-ohm. Both of these supplies are connected to the
negative terminal, connected to negative terminal. And we have ten kilo ohm, which is our feedback. This is can be considered as R1. This one is R2. And we have V1, the V2. Now what we need to do is
that we would like to get v, our zone of current IL. So let's start
with two by V out. So as you can see
this two supplies, supplies are connected to the negative terminal
with their own resistor. And the post f ten minutes
connected to the ground. So this is what, this is a summing amplifier
with two inputs. So the output voltage V output, be equal to the
first, the supply. First we will type
negative feedback, which is ten kiloohm divided
by first one, which is five, multiplied bys a two volt
plus the second one, which is one volt, multiplied by our
feedback divided by 2.5. So this will give us
the output voltage. So let's see. So as you can see,
that V-out equal to negative auto feedback over R1 multiplied by V1
plus our feedback r2, v2, as you can see here, on feedback ten kilo-ohm, ten, then R1, R2, R1 is five kilo ohm. R2 is 2.5 kilo-ohm, v1 and v2, two
volt and one volt, two volt and one volt. So we will have
negative eight volt. So this is the output voltage. Now what we need is the
current output from KCL here, that the output current
is equal to two currents. Let's say for
example, I x and our UI I X plus IY from KCL. The current flowing here is this voltage minus this voltage
divided by ten kilo-ohm. Voltage a is equal to
voltage v equal to 0. So this point is 0, and this point is V out. So I X equals V out minus
0 over ten kiloohm. So it will be V out
over ten kilo-ohm, ten kiloohm plus uy between
this point and this point. So V out minus 0 divided
by two kilo-ohm, V out minus 0 divided
by two kilo-ohm. And V out is equal
to negative eight. Negative eight. So we will have negative eight
over ten is negative 0.8. Negative eight over
two is negative four. So it will give us negative 4.8. And remember we have here
Gilo sent to the power three. So it will give us
Millie and bear. Okay, negative 4.8
million numbers. So let's see if I am correct
or I made a mistake. Okay, negative 4.8 milli
and beer as I said. So as you can see, the current I is the summation of
the two currents. And each of them has
V out negative eight. So V equals V, V equals 0. So the output current is
equal to plus second, first, 1 V out minus 0
over ten kilo-ohm, V out minus 0 over
two kilo ohms. So it will give us negative
4.8 mainly and bigger. So this was another example
on the summing amplifier.
56. Construction of Difference Operational Amplifiers: Hey everyone. In this lesson we will discuss another type of
operational amplifiers, which is difference amplifier. Okay? So what does a difference
amplifier mean? So simply the
difference amplifier provide this an output voltage, which is equal to the difference between two input voltages. So we have here two input
voltage, V2 and V1. Okay? So the difference between them, each of them of course,
multiplied by asserting gain. The difference between
them multiplied by a certain gain gives
us the output. So what does this
difference amplifier do? So as you can see, that
circuit composition consisting of four resistors, R1, R2, R3, R4. For the negative terminal, we have the feedback
resistor R2. And we have a negative
inverting part, which is v1 series with R1. And we have a
resistor V2 with R3, and we have our four
connected to ground. So this circuit representing
the difference amplifier. So let's start. So the difference amplifier, or sometimes known as
differential amplifiers, it is used to amplify the difference between
two input signals. So we have two input signals, a V0, V1, and V2. So the difference amplifiers
take the difference between these two
signals and amplify it. Let's just start. How can we get this? Okay? So as you can see, we have
here v1, v2, and V-out. We need the relation
between V out and V1, V2. So we have VA and VB, again, z are equal
to each other. Okay? So remember this
second part is that we know that from
KCL at this point, we know that the
current going here is equal to the current I
would going from here. Because the current
year is equal to 0. The current going here is
equal to V1 minus V over R1, V1 minus V2 over R1. This current is equal to the
current flowing through R2. So the current flowing
through R2 is V a minus V 0 divided by R2, VA minus V 0 divided by R2. So we have V1, V and V out. Okay? So from this
equation we can say that V out is equal
to this equation. Now, if we apply
KCL at this point, we know that the current flowing here is equal to the
current flowing here, because the current
here is equal to 0. So you will find that V2
minus Vb divided by R3. V2 minus VB divided by R3
gives us this current, which is equal to this
current flowing through R4, which is VB minus 0 divided by odd for VB
minus 0 divided by R4. So from this equation
we have that VB is equal to R4 over
R3 plus R4 V2. So now you remember that, okay, In the
operational amplifiers are ideal operational amplifier. We know that VA is equal to VB, VA is equal to VB. So what we can do is that we
can take this VB equation and substitute it
here like this. So V equals VB. So V out will be R2 over R1 plus one
multiplied by R4 over R3 plus R4 multiplied by
V2 minus R2 over R1 V one. We have this equation. So we have the V out as
a function of V2 and V1. We can simplify this
equation like this. You will find that in the end, in the difference amplifier, the output voltage is equal to R2 multiplied by one plus R1 over R2 over R1 plus R3 over R4 V2 minus R2 over R1 V0, V1. Okay? So as you can see, it
is difference between two voltages multiplied
by a certain gain. Okay? So now we will have to know something which is
really important is that since the difference
amplifier must reject a signal common
to the two inputs. What does this mean? It means that that input
or the output here should be equal to 0
when V1 equal to V2, because there is no difference. So the output voltage
should be equal to 0. This one should be equal
to 0 when V1 equal to V2, when V1 is equal to V2. So if we look at this equation, we have 0 equals to
this larger part. Okay? Let's say it's this
one, x, for example, x V2 minus R2 over R1. We have V1 equal to V2. So we can say V2.
So as an example, we are getting the values
of the resistance. So when V0, V1 equal to V2, and in this case, the V out should be equal to 0. Okay? So we'll find that v2
will be cancelled. So we'll find that R2 over R1, this part is equal to
this large aboard. Okay? So if you simplify this, you will get this relation, R1 over R2 equal
to R3 over R four. So here when you substitute with this
inside this equation, you will find that V out will be equal to R2
over R1 V2 minus V1. So in order to win V1
equal V2 and V1 equal V2, this out will be equal to 0. This is what we need. That wins the voltage are common or z are
equal to each other, is the output should
be equal to 0. And so to satisfy this equation, we should have this condition. So when we substitute
in this here, R3 over R4 is equal
to R1 over R2. R1 over R2 is similar
to R3 over R4. So finds that this part
is equal to this box. You can cancel this with this. So you have R2 over R1
V2 minus R2 over R1 V1. So you take R2 over R1
as a common factor, like here, V two minus V one. Okay? Okay. So now we have R2
over R1 V2 minus V1. Now, if R2 equal to R1, R3, R4, you will find that the difference amplifiers
becomes S subtract. Okay? So as you can see, if R1 and
R2 are equal to each other, this one is equal to this one, which is similar as
if R3 equal R4, okay? You will find that V out will
be equal to V2 minus V1, which is subtraction
or subtract. So what is the difference? What is a change between the difference amplifier and subtract the difference
amplifier is the general case. You will find here,
again multiplied by V2 and other again
multiplied by V1. However, in a subtractor is a subtraction of the
voltages V2 minus V1, as you can see,
without any gains.
57. Example on Difference Operational Amplifiers: So now let's have an example
on the difference amplifier. So in this example, we need to design a difference amplifier or a
difference Op Amp circuit with inputs V1 and V2. V1 and v2, such that the output voltage is equal
to negative five, V1 plus V2. So as we remember that
the output voltage is equal to R2 plus R1 over R2, R1 multiplied by this
minus R2 over R1 V1. So this is the general equation
of the difference op amp. And you can see this equation. We have V out equals two. We need V2 and V1. So V2 has three. So three, V2 minus
V1, minus five V1. So as you can see that
our voltage V2, V1. So if you compare
this two equation, you will find that this
part is equal to three. And this part, which
is R2 over R1, is equal to five. Okay? So we want to start R2 over
R1 equal five, like this. So R2 over R1 equal five. So our two equal five R1. Okay? So, and in the second part, which is this one, this part, you will
find that here. If we look again, my exists, you will find that this part is equal to three. So as you can see, the
spot which is this part, is equal to 31 plus R1
over R2, R1 over R2. R1 over R2 is, R1 over R2 is the
inverse of this file, so it will be one over five. Okay? So we can say it's, the
sport is one over five. And we have here
R2 divided by R1. R2 over R1 is five. So as you can see
here, five, okay? So as you can see here, one plus one over five
is six over five. And this five will, will
take it to the other side, will be three over 53 over five. Okay? So we'll find that in the end, we have the equation R3, R4. We can get a relation from this. R3 over R4. R3 will be equal to R4 from where by
simplifying this equation. So now we have R2 equals five, R1 and R3 R4. Okay? So what are we going to do? We will assume values
as an example, we can say is that this R1
is equal to ten kilo ohm. R2 will be 50 kiloohm. R3, for example, 20 kilo ohms, R4 will be 20 kilo-ohm
as an example. So as you can see, if we
choose R1 to be ten kilo ohm, R2 will be 50 kilo,
as you can see here. If we choose all three
to be 20 kilo ohm, then R4 will be similar to it. Okay? So this is design. What does our design mean? It means you can
choose any values. So as an example, in instead of choosing, for example, R1 equals ten, we can choose five kilo-ohm
to when tequila on 30 kilo, any value would like. But the most important thing
is that this equation, and this equation
should be satisfied.
58. Cascaded Operational Amplifiers: Now let's discuss another
type of op-amp circuits, which is called Zach
cascaded Op Amps circuits. Now what does that
cascaded op-amp circuit? It means that we have, our voltage is exposed
to different stages. As you can see
here. For example, a three-stage means it's
consisting of three stages, 123. So we have here is our
input voltage, V1. This voltage is exposed
into three stages. Stage one, stage
two, stage three. Stage one, for example. It has again of A1. So the voltage is
multiplied by this gain. So the output voltage is A1 V1. V2 is equal to A1 V1, which is out of the first stage, then the output of
the first stage is an input for the second stage. So as you can see, V2 multiplied by this gain gives us V3, A2 V2 gives us V3. And this is an input
for this stage. So it will give us V out is three V3 funds that V1 is
exposed to is again A1, A2, A3 then gives
us the final out. So this is what's called the cascaded
operational amplifiers. So to be more
specific, for example, this stage could be a
inverting amplifier. This stage could be
non-inverting amplifier. This stage could be a buffer circuit or a
voltage follower circuit. So all of these stages will give us our final
output to reach a certain value would like,
I guess get conducted. A connection is a head to tail arrangement of two
or more op amp circuits, such as at the output of one, the input of the
next output of one out of the first stage is
the input of the next stage. When Op Amp circuit
are cascaded, each circuit in the string
is called the S stage. Okay, So this one is a stage. Stage, stage. You will find that the gains, for example, if this one gives, again A1 game in one gain, gain is three, so the output
will be V1 multiplied by A1, A2, A3, like this. Okay? So this is a total
gain of this circuit.
59. Example on Cascaded Operational Amplifiers: So now let's have an example on the cascaded op-amp circuits. So find the V output and
I output in this circuit. So as you can see, is this
act cascaded op-amp circuit. Yes. Why? Because as you can see here, we have one op amp
and another opamp. The first op amp here, this one consisting of this
one which is our feedback. This is our feedback. Because connected between
output and input. As negative terminal nodes, the input but the
negative terminal. Okay? So this is our feedback. And as you can see, the
negative terminal also has a resistor which is
connected to the ground. And our input, our input, which is 20 millivolt, is going into the
positive terminal. So as you can see
from this circuit, this circuit alone is one stage representing one representing
a non inverting amplifier. Non-inverting amplifier. Why? Because that input
is supplied to the balls. And we have a feedback
with a resistance R. Okay? So this is our first stage, which are representing a
non-inverting amplifier. Now, the output of
this stage is Va. This is what, this is the
output of the first stage, which is input to
the second stage. The second stage is also what is also a non-inverting amplifier. Because the VA is connected
to the positive terminal. And do we have here our
feedback and the resistance are connected to the ground to give us as a final
output to this. Representing the second stage. We will start to by
getting that v out here, then the V out here. So as you can see,
it's a first amplifier which is non-inverting is one plus our
feedback over R1. The feedback resistor
is that, well, we kilowatt connecting between the output and the
negative terminal. And the second resistor, which is R1, is a
three kilo-ohm. This one multiplied
by the input, which is that when
team may live old. So it will give us
finally 100 millivolts. This voltage is V a va
equal 200 millivolt. This is the input to the positive terminal of the second op amp or
the second stage. So it is another
non-inverting amplifier. So the output will be
one plus our feedback, or feedback, which is the
ten kilo ohm divided by R1, which is a resistor at
the negative terminal, multiplied by v, which is the input here to
the second stage, which is really a 100 millivolt. So you will get finally
is that the V out or the output voltage is
equal to 050 millivolts. So as you can see,
is two stages. First one, then second one. But to use 350 millivolts. Now, is there another solution? Yes, V output is equal to A1, A2 multiplied by the V input. So the input voltage
is 20 millivolt is multiplied by zack gain
of the forest amplifier, multiplied by the gain
of the second amplifier. The gain of the first
amplifier is one plus RF. Or what are the gain of second amplifying is
one plus ten over four. Multiplication of them
multiplied by the input will give us 050 millivolts. So it is the same idea. Now, the next question is, we need to find
that current I out. Okay? So you have two solutions here. We can say is that
the current I is equal to this point,
this voltage, the Albert minus VB
divided by ten kilo ohm. So we can say Albert is
equal to V out minus VB divided by the
ten kilo-ohm V-out. So 150 volt millivolt. And the V input Va, Vb, Vb here. This voltage is
equal to Va. Va. Vb, Va is equal to 100 millivolt. So this one is 100 millivolt. Okay? So this is a first solution. Second solution is
that we can say that output equal to
the voltage here, V out minus 0 divided by
the total resistance. So we can say V out
minus 0 divided by ten plus 41414 kilo ohm. So as you can see, as this
will give us the same idea. Why? Because as you can see that
the output current year, current flowing here, similar to the
current flowing here. So this terminal current
year is equal to 0. So I output I out. Okay? So we can say
voltage from here to here. So 150 divided by 14
kilowatts or total resistors, or V out minus Vb multiplied
by divided by ten kilo-ohm. Or another solution, Vb minus 0 divided by
the four kilo-ohm. All of this, all of these
solutions will give you, all of these answers will
give you the same solution. Okay? So as you can see out here, which shows as a
solution V out minus VB divided by that then Kylo. Oh, okay.
60. Digital to Analog Converter: Hi and welcome everyone to our lesson in our course
for electric circuits. In this lesson, we will discuss an application on
operational amplifiers. So in this example we will discuss the application which is that digital to
analog converter. Okay? If you would like to convert digital signal into
an analog signal, this can be done using
operational amplifier. So how can we do this? How can we convert a digital
input into an analog output? As an example, in this lesson, we will discuss for
digital input or fed for bet digital
to analog converter. The 4-bit is similar to five
to six, whatever it is. Okay, it's the same idea. So what are we going to do
or what we would like to do? That they just are in both. Usually when we have
a digital input, it can be 0 or 10
volts or one volt, okay, which is called the
binary, binary, binary system. The 0 or one. So if we have a digital
input, for example, for zeros or four ones or
anything between them. What does this mean?
As an example? 0000. What does this mean? It means that the first value, or the least significant bit, or the one which
owns a right side. And this one we multiply
by two to the power z. Second one to multiply it
by one to the power 112. The power two here, two to the power three. Okay? So this is equivalent
to 22 to the power n is 12 to the power one is 2012 to the power 12 to four. Here, two to the
power three is eight. This is in the case of the
four bit digital signal. Okay? So this is equivalent to
what, what its a value. 0 multiplied by one is 0
plus 0 multiplied by two is 0 plus 0 multiplied by four is 0 plus 0 multiplied
by eight is also 0. So this gives us
an analog value of 0 volt or 0 as a number. And I look out, okay, now what if it's, for example, 0101 as an example. So this will be 0
multiplied by eight, okay? Plus one multiplied by four, plus 0 multiplied by 2M
plus one multiplied by one. Okay? So we'll find that here. This will be equal to 0. This will be equal to 0. So we will have four
plus one, which is five. So this binary input, or digital input 0101 is
equivalent to five volts. Okay? So this is what
we would like to do. We would like to convert
the input signal, which is binary like this, into analog out as
a voltage five. So for example, if we have 0101, okay, I would like the output
to be five volt, okay? Like this. So how can we do this using
an operational amplifier? So we will see how can we do this for us as a digital
to analog converter could transform is that
digital signals into analog for typical example, is four bit digital to analog
converter, such as here, four digits or four
binary digits, or four bit input converted
into an analog signal. Symbol realization is a
binary weighted ladder. Okay, so what does this mean? What do we mean by
binary weighted ladder? Here, you can see
that here this is multiplied by two
to the power 0. This two to the power one, to two to the power two to
the power three, and so on. So this is called
weighted louder. Each signal, each
signal here, okay? Signal here is
multiplied by its. Respective weight,
in which is a bits, are weights according to the magnitude of
their place value. By it's sending the
value of R F over R n so that each lesser pet has half the weight of
the anarchist higher. So as you can see, for
example, for example, you can see that here to
the power three is eight. This one is for, this one is two,
and finally one. So you can see that it
half of eight is four. However, for is to
half of two is one. So as you can see,
each lesser pet has half the weight
of Xenakis to hire. Okay? So what does this mean? It means that this one
is the highest value, for example, 8421. Okay? So the first one here, first, voltage V1
is multiplied by 8. Second 1 is multiplied by four. So the one multiplied by two
force one multiplied by one. So as you can see,
each lesson pet, which means we are going to, Darwin, has half the
weight of the next trial. It becomes 44, becomes
a 22 becomes a one. So how can we do this? Each of this one is
multiplied by R F over R n, which is gain of them. Inverting amplifier is the
ratio of feedback is constant. The RN is resistance in
series with this signals. Okay? Anyway, you will see now
what I mean exactly. So as you can see here, we have this one
which is summing, summing amplifier, but it is an inverting amplifier
summing, inverting amplifier. You can see that the
positive terminal grounded, we have our feedback
which connects between the output enzyme
negative terminal. And we have here V1, V2, V3, V4, which are representing
the digital input. Okay? This one is V0, V1, V2, V3, and V4. Each of these resistors
have R1, R2, R3, R4. Okay? So if you remember that the output of the
summing amplifier, negative V out or V out is
equal to negative all of this. You will know that
the output is V0, V1 multiplied by the feedback divided by R1 or feedback R1, V1, and our feedback R2, V2 and all feedback or
three V3 and so on. So here, if we would like, is this one to be eight
multiplied by eight, this multiplied by four, this multiplied by 21. How can you do this assembly? The eight is the ratio
between our F over R1 for is the ratio
between RF and R2. Two is the ratio
between RF and R3. One is the ratio
between RF and R4. You can see that here, V1 representing their
most significant bit. And this view for representing
the least significant bit, LSB and MSP in that
electric circuits. So by choosing the R1, R2, R3, R4 ends up or feedback. We can do this function. We can convert 0101 or whatever it is between 010101
or whatever it's order. We can convert this binary
signal by using this widths into an analog output which
representing the real value. Okay. So now let's go and have an example on the
digital to analog converter in order to
understand how it works.
61. Example on Digital to Analog Converter: So in this example, we have in this circuit, let r, f or the feedback
equal to ten kilo ohm. Here we have on R1 is
equal to ten kilo ohm, R2, 20 kiloohm, R3, 40 kilo-ohm, R4, 80 kilo ohm. Then o tensor analog output
for the binary inputs 000, which means what does this mean? It means that V1, V2, V3, and V4, and so on. So we need all of the possible, possible values for
different binary inputs. Okay? So first, what are we
going to do assembly, we are going to
write our equation. We know is that here
we have V out is equal to negative R
feedback over R1, V1 plus negative r to
r feedback over r2, v2 minus our feedback over R3, V3 minus R4, R3, audio feedback over
R4 and V4 and so on as a summing
inverting amplifier. So the outer feedback
is 80 kilo ohm and R1, R1 is ten kiloohm. So here we will have like this. So V1 as multiplied
by our feedback, the feedback, whereas our
feedback ten kilo ohm. So we have this one is done. And R1 ten. So ten divided by ten gives us one v1 plus outer feedback. Or to feed back over
to our feedback, ten kilo ohm divided by two, which gives us half, and so on. So as you can see, one, it's a half is 0.5, it's half is 0.25. It's how value of
0.125 and so on. Okay? So this representing
negative V out, so V out is negative. All of this. What
are we going to do? We are going to do four different values for
the different binary inputs. As an example, as
you can see here, if we have binary
input, we have V1, V2, V3, V4, so zeros, zeros 00. So it means that
this value is 0000. So V out is, of course will be
equal to 0. And so on. You will do the same idea. So as an example, this 1110011001100 here it will be V1 will be one plus one, which is 0.5 multiplied
by one plus zeros, zeros. This one would be 0 and
this one will be 0. So it will give us 1.5 volts. Okay? So as you can see, this value representing what
OT V output of this example.
62. Instrumentation Amplifiers: Now in this lesson we are going to discuss another amplifier, which is the
instrumentation amplifiers. Okay? So what does an
instrumentation amplifier? This is a symbol of this amplifier,
instrumentation amplifier. Similar tools or
normal amplifier, but with a resistance
between them. This whole resistance is
called the gain resistor, which we can control. So if you look at
this amplifier, it consisting of 123 amplifiers. The forest amplifier
has an input V1. Second amplifier
has an input V2, and the output is coming
from the amplifier. Okay? So here what we can
notice from this circuit, if you would like to know that mathematical
proof for this one, you can send me a message
and I will give you the mathematical proof for the equations which show
we will have, okay? So anyway, you will find in this between negative terminal, negative and the
negative terminal, we have a resistance
between them, which has a variable value. We can change it by
changing this resistance. We can change the gain
of this amplifier. So what is the benefit
of this amplifier? What does it do? We have two inputs, V1 and V2, and we have one out. Okay? So simply, one of the most
useful op-amp circuits for precision measurements
and control is the instrumentation
amplifier. It is called an
instrumentation because it is wide spread use in
measurement systems. It is used in isolation amplifiers as
thermo couple amplifiers, data acquisition
systems and so on. So what does this
amplifier to do? This instrumentation
amplifier is an extension of the
difference amplifier. And that it amplifies the difference between
two input signals. If you will remember that
difference amplifier or differential amplifiers, we said that the output
is equal to V2 minus V1. Okay? And each of these
is multiplied by a certain gain dependent on
the elements in the circuit. If you'll remember, we
had R1, R2, R3, R4, different resistors
which affect us z again, it increases or
provides to the hour, the difference
between two signals amplifies the difference
between two signals. Now what is the difference
in between difference or differential amplifier and the instrumentation
amplifier? The difference amplifiers
again is controlled by that resistance inside the
circuit R1, R2, R3, R4. In Xi'an instrumentation
amplifier, we controlled again by using variable resistance, RG
external resistance. So the gain of this amplifier, this system, this
amplifier system, is not dependent on the value
of our sides a circuit, but we can change it using
an external resistor, RG. Rg. By changing this, we can change the gain of
this amplifier. Will see is that the
output voltage of this amplifier as equal to a certain gain
multiplied by V2 minus V1. So the difference between two signals is unbelief
it by a certain gain, a v. This gain is equal to
one plus two r over r g. You will see that
here by changing this resistance as
external resistor, we can change again
as we would like. Okay, this is one
of the advantages of using instrumentation
amplifier. Now, you will have to know that the instrumentation amplifier rejecters the common voltage. So if V2 is equal to V1, then they cancel each other. However, if there is
a small difference between these two
signals, it amplifies it. So it amplifies that
small-signal voltages. So as you can see, if we
have a very small signal in V1 and V2, very small signal. By using that differential, by using instrumentation
amplifier, we can amplify a signal,
making it bigger. So we amplify is a
differential signal. The signal which is different, or the difference
between two signals. Difference between two signal is amplified and a common signal. If V1 equals V2, it will be rejected. It does not pass. Okay?
63. Example on Instrumentation Amplifiers: So now let's have an example on the instrumentation amplifier
in order to understand it. So we have this circuit of the anisole mentation amplifier. We have the resistance
are equal ten kilo ohm. We have the voltage V1 equal to 2.011 volt and V2 is
equal to 2.017 volt. You can see that the
difference between two signals is very small. And we would like to use the
instrumentation amplifier to amplify this small
change between them. The difference between
them now is the resistance RG is adjusted to 500 ohms. Again, resistor, I
was asked to 500. So what do we need? We need number one, the voltage again. Number two, the output voltage. So let's start. So if we remember that
the voltage again from the previous lesson, AV, or the voltage gain is equal
to one plus two r over r g. So that resistor is
equal to ten Kilo ohm. And R gene, or this resistor is equal to
what is equal to 500 ohms. So one plus two multiplied by 10 thousand divided by 500
gives us four hundred. Four hundred is again
without any unit. Okay? Now what is the value
of the output voltage, the output voltage assembly that gain multiplied
by the difference between the two signals. So as you can see,
the output voltage is again multiplied by V2 minus V1. So we will have in the
end 246 Mendeley volt. Okay? So as you can see, that the difference
between these two signals is very small value
by multiplying this, by inserting gain one
plus two out of our RG, we obtained a large voltage. So we can maximize or, and amplify that very small difference
between two signals. If we would like to
see something inside the signals such as
harmonics or anything. By amplifying it, we can amplify it using the
instrumentation amplifier. Very small difference
which can be amplified to an output voltage.
64. Introduction to Capacitors: Hi and welcome everyone to this part in our course
for electric circuits. And this part we are
going to discuss Zach, capacitors and inductors. So in the previous sections or the previous parts of our
course for electric circuits, we discussed the
resistive circuits. We discuss the different
circuit theorems that op amps and everything
about resistive circuits. Now we need to add two new and important passive
linear circuit elements, which is that capacitor
and the inductor. Okay? So what is the
difference between the resistors and
capacitors and inductors. Resistors as they dissipate
electrical energy, or they consume
electrical energy. However, the capacitors and inductors does not dissipate
electrical energy. However, they store this energy which can be retrieved later. Okay? So resistors dissipate or consume electrical energy. The capacitors and
inductors are used to store electrical energy in the form of electric field
and magnetic field. That's why is that capacitors and inductors are called storage elements or electric storage elements because they
store electrical energy. So in this section, we are going to discuss
that capacitors. Then how to combine
capacitors in series, in parallel, similar to the resistance resistors
in series and parallel. And we're going to do
the same for inductors. We will first start
discussing the inductors. Then we are going to combine
them in series and parallel. After this, we are
going to discuss some applications
for the capacitors. Capacitors that can be
used with op amps or the operational amplifiers
in order to form a two new applications
or two new op amps, the Integrator Op Amp and
a differentiator of n. So now we will start
with the capacitors. So we need to understand
what is a capacitor? What does this mean? What does its composition
and how it works? First, you will find
that the capacitor is a passive element designed to store energy in
its electric field. So it stores the
electrical energy in the form of electric field. And there's a passive
elements similar to the inductor and similar to the resistor. Passive elements means
it does not need an electrical source in
order to start working. Unlike op amp or bocce
needs our supply. It needs a supply in order
to do it's a function. So that's why it's called
the active element. However, capacitors,
inductors and resistors are called
the passive elements. So the capacitors are
used in electronics, communications, computers,
and power systems. Some of these applications
is a tuning circuits of that radio receivers use the dynamic memory
in computer systems. Now, what is the composition
of the capacitors? There are different
capacitors in our real life, one of which, which
is commonly used and you'll find it a
lot, is called disaster. To parallel plate. That two parallel plates. You will find that the
capacitor is consisting of two conducting plates to conducting parallel
plates separated by an insulating material
or dielectric material. So if you look at this figure, this figure representing what is this representing
a capacitor. This part is a capacitor. Okay, so what is this
capacity of this capacitor, as you can see here,
consisting of two plates. One plate here, as you can see, this plate, this plate,
this rectangular shape. And there's another
one on the other side. Okay. And between them
there is a material this material here as this
material between them, this material is called
an insulating material, which insulate k between this plate and the
one which is behind. As you can see here. You can see here we
have two plates, one another blade here
and between them here. Here. Insulating material which
insulates between these two, it prevents the contact
between these two plates. Okay? So this material or this insulating metric
can be plastic, can be air, for example,
has many compositions. So two conducting
plates between them, an insulating material or
a dielectric material. Ok. So in many practical
applications, the plates, these plates, these plates, these plates can be made
from aluminum foil, whiles or dielectric material, or the insulating
material can be air or ceramic or paper or mega, any of these, what
is its function? It insulates between
this plate and display. It prevents a direct
contact between them. Okay. So it is insulating between insulates between the
two metal plates. Now what happens here? When source, so we have this capacitor and it
has two terminals. One terminal connecting to one plate and also terminal
connecting to other plate, as you can see here, one here, and another one here. Okay? Now when a voltage
source like this one plus minus connected
to these two plates, the source deposits apples
if queue or a boast of a charge on one plate and the negative charge on
the other plate. That's why it's a
capacitor is set to store the electrical charges. Okay, So what, how this happens? Okay, So as you can see
here in this figure, we have a boast of supply here. Both the voltage and
the negative supply, or ZAP positive terminal of the supply and the negative
terminal of the supply. Now, this one has
a high voltage, this one has a lower
voltage, okay? Or we can say most of the charges and the
negative charge here. Okay? So now what happens here? This wire, This wires, it contains electrons, electrons, negative
electrons. Okay? So what happens here is that
you will find that here, these electrons here,
electrons here. Okay? This electrons which are at the beginning here for
the negative terminal here. You can see this is a negative term and this
electron has a negative charge. So what happens is that there is a repulsion force between
these two electrons. Between the electrons and the negative terminal
of the supply, there is a repulsion force. So what happened
to this electron? It tries to go away
from this supply. Okay? So, uh, tries to go
away from this supply. So what will happen
is this electron. So we'll start accumulating
on this plate. One of this plate, this plate will become
negative charge, okay? You have to know that
the charges are not the shortest inside the wires only, but the charges inside
the plate itself. This plate has also positive and negative
charges and metal which consisting of positive
and negative sources. So negative source
are accumulating here from the wires and
from the plate itself. And the boast of a
charge on this plate are going toward the supply. So in the end you will find that this plate will become
negative charge. It contains a large amount of negative charges or negative q. Now what happened
to the other side? To the other side, we have here electrons and we have
here also electrons. So these electrons,
all of these electrons are attracted by the positive
terminal of the supplies. They are going toward the positive terminal
of the supply. Any post of sorts are going
toward this, this plate here. Then negative with the electrons are having repulsion force. So z accumulate here and
negative charge because there is a repulsion force
between the terminal of the supply and this
negative electrons. Similar here as
opposed to the charges or repulsion force
in-between that most of the charges and the
Zappos to which calls itself the positive
terminal and the most of which are
accumulating on this plate. And the negative charge
on the plates are going into war this the supply. So in the end you'll find here both of the charges
and negative charges. So the question is, why does this is opposed to charge it and this
negative charge? We say something is always the recharges or bolster we charge it or
negatively charged, depending on the
number of electrons, combine the two as
opposed of the charges. Okay? So here in this part, in this plate and this plate, you will find that the number of poles there were charges are greater than the negative
electrons on this plate. So that's why we say it
is a post of a charge. For this plate, the
negative with chores are much Ahmad Zen supposed
to be charges. That's why it's called
a negative charge. Okay? So in reality,
similar as that atom, the atom itself is neutral. It doesn't, it is not
that both of which are Miser our positive
nor negative chores. So when the, when we
remove electrons from it, it becomes both the ventures. Or if we add electrons
and become negative and stores same idea
for this plates. So anyway, you will find that these plates will accumulate
electrical charge. So since there is an insulating
material between them, then there is a magnetic field that between these two plates. And magnetic field from Ball
State which was going into negative source electric field,
magnetic electric field. Okay? So this power, or this energy is stored in the form
of electric field. So the amount of charge
stored representing by Q is directly proportional
to the applied volt. So the higher the
voltage applied a more charges will be
accumulated on the plates. Will find that the Q is directly proportional
with velocity. Voice that note
velocity and 0 voltage. So as voltage increases, Zach, you increase the amount of
charge accumulating increase. Now, if you would
like to change or replace this constant
of proportionality, we can say that q
will be equal to ascertain constant c
multiplied by voltage. This constant is known as the
capacitance of this plates. Okay? Or the capacitance
of this capacitor. Okay? So as you can see here, c is called the constant of proportionality is known as the capacitance
of the capacitor. Now, the capacitor itself
or the capacitance, is measured in far-out. So when we say it's a
capacitance, how many far-out? One for odd one microfarad,
one millifarad. So it's a far-out ear, similar as resistance
measured in ohms. However, capacitance
measured in far-out. Why it's called the follow-up
because it is the owner of the English physicist,
Michael Faraday. Faraday. Remember this name.
Faraday is very, very important scientists in
the field of electricity. For are they added a very important law
which is called the induced induced EMF or
electromotive force induced EMF. Or how can we
generate electricity from as a variation
in magnetic field. So the induced EMF is a very important law proposed or designed
by Michael Faraday. Okay? This induced EMF is very important in the
electric generators. We can now generate
electricity without, with knowing the induced EMF. So thanks to Faraday, we had this law of
electromotive force, which helped us
understand how can we generate electric energy. Okay? So remember his name because
it's really important. Now, the capacitance, what is
the capacitance from here? From this equation, the
capacitance is equal to Q over V, right? Zach quantity of charge
divided by the voltage. So as you can see, ratio of
the charge on one plate, q is one plate. Okay, So when we given zack EOQ, is that representing of one
plate not the two plates. One plate only. It is a queue of wire. Capacitor is charged on one
plate of the capacitor to the voltage difference or the potential difference
between these two plates, or the voltage applied
between these two plates. The capacitance
measure the fraud. Okay? Now, we need to understand more about
the capacitance, okay? Now we have to understand
that the capacitance, which is the ratio
between Q over V, it does not depend on the queue
and does not depend on v. So whatever the supply is, a capacitor does not change, or whatever the Q accumulated, the capacitor does not change. So what does that
capacitance depends on? How can we get the capacitance? The capacitance, it depends on the physical dimensions
of the capacitor. Okay? So what does this mean? You will find now, for example, in this parallel plate capacitor is the capacitance is equal to epsilon multiplied by
area, multiplied by d. Okay? So the capacitance of a
parallel plate is equal to. Epsilon multiplied by area, multiplied by d. What is d? D is a sickness of the
dielectric material, or the sickness, or the distance in between
the two parallel plates. So the distance here between
these two parallel plates, or the thickness of the
insulating material is called d. Okay. Now, the second
part is the area. What does area? Area is the surface area of
one of the metal plates. So as you can see, this plate, this plate, for example, or displayed as consisting
of a lens and the width. So the length is the
distance from here to here. Okay? And the width is the
distance from here to here. So the area of this
plate is equal to l multiplied by W
or the rectangle, or the area of the rectangle. Okay? So this is area. Now as a
final property is called the epsilon or the
dielectric permittivity. Permittivity, permittivity of the dielectric material epsilon. Okay? It is, the permittivity
is a property which is related more to zap polarization of the
electric polarization. So what does this mean? It means that the
more permittivity is a more polarization
of these two plates. So more poll severe,
more negative here. Promisee polarization
or permittivity, which means that how
much it I will allow the electrical field to
boss from here to here. Electric field going from the positive to the
negative rituals. Okay? So anyway is the permittivity
is a very large appealed related to understanding the definition of
electric field, which will need its own
course in its own, okay? So we should not discuss
the permittivity anyway. You should just know
that the permittivity help us enzyme polarization
of the capacity. Okay, that's all what
you need to know. Now. Mod permittivity. The larger the area,
the more capacitance. Higher the distance here, molars apes the
distance increases in the capacitance
will decrease. If this distance is very small, then the capacitance
will increase. Okay? Now you will find that the area, the surface area of each plate, D is the distance
between the two plates. Epsilon is permittivity of the dielectric material between the plates, which are changed. Of course, the
permittivity changes depending on the
material is a plastic, is it air, whatever, each of these materials
has its own permittivity. Now, we will find that as
a surface area increases, that capacitance
increase, area increases, capacitance increases
as the spacing between the two
plates spacing d. The smaller the spacing, the higher the capacitance. If the permittivity
of the material, the higher is the
permittivity is a greater the capacitance
epsilon increases, capacitance increase. Okay? Now we will find that the
capacitors have values in that beaker far-out,
two microfarad range. Okay? So capacitors, usually,
usually they are not measured in 11 for odd or
too far out or higher. This values are
very, very large. Our usual, you
will find that one microfarad to microflora
to 0.1 microphone route. Sometimes you will
find it pico far out. Okay. So usually one followed until forever or not, the
common capacitor, common capacitors
are in microfarads in big difference because
it is a suitable value, one for our two,
for our ten follow. This is a very large
amount of capacitors. Now, one note so that you
can know is that there are some capacitors which
have this larger value. It can be one for all, too far out, 1000 for odd. Now, this capacitors are
not the normal capacitors. They are not the capacitors, which I don't usually use. These capacitors are
called supercapacitors. The capacitor rule is
a very large value. One for auto, for our
tenfold hundredfold are called super capacitors, which are used in
certain applications in in electrical power systems. Okay. Now, if you look here, we have two samples. This is the capacitance, two parallel plates
between them as small gap. Okay? So this symbol, what
does this represent? This representing our capacity. So we can, when we draw
an electric circuit, we add the sample in this form, two parallel lines between them. Again. This line, what
does this represent? This means that this
capacitor is variable. It is not a constant
capacitance. Okay? Now, there are two
types of capacitors. There is a fixed
value capacitors and there is a variable
value capacity, fixed value. It is a fixed value
and does not change. So as you can see, we have here, as you can see, three types. All of these are capacitors. One of them, the first one
is polyester capacitors, ceramic capacitor, and
electrolytic capacitor. Okay. So if you opened any electric circuits or any electronic circuits
in your own home, for example, you
will find this one. You will find this one and
this one and this one. You will usually
see is this one, this one in a cylindrical form. And sometimes when
zeta is a problem with your own
electronic circuit, it is usually this one
becoming defect, okay? You will find it is
about to explode. Okay? I'll find it is larger
than its own normal value. So this part is called
Zach capacitor. It is very, very important
in electric circuits. Now, another time you can see the variable value capacitors, this one and this one. Tremor and film, film
and trim capacitors. This tool types of capacitors
provide variable value. It means that when you rotate
for examining this one, when you rotate
this screw, okay. When you rotate
this one like this, rotating it, this part, you will find that you
can change the value of the capacitor by
rotating this screw. Okay, that's all. That's why it's
called the variable does not fix it like this one. This one, you add it to
the electric circuit, it gives you a
certain microfarad. However, this one, when
you rotate the screw, you can change this
amount of microfarads.
65. Equations of a Capacitor: Hi everyone. In this
lesson we are going to discuss the different
equations of capacitors. So first, we would like
to get the relation between the current and
voltage inside the capacitor. So as you can see here, we have Zach you
or the amount of charge equal to the capacitance
multiplied by voltage. Now remember that the current, that the current is equal to, current equal to d Q over d t. Or the current is
equal to rate of change of the charge
with respect to time. So as you can see
here, we have Q. So we can Q equals CV, so we can substitute
with this here. So this will be
equal to d over d t, the rate of change, or the derivative of Q, what is Q equals CV, c multiplied by V. So capacitance is a constant. It does not change with time. So we will take this
outside that derivative. So it will be C multiplied
by the derivative of the voltage with
respect to Phi. Okay? So the current of the capacitor
is equal to C DV over DT, or the current is equal
to the rate of change of the velocity of the voltage
with respect to time. The voltage with
respect to time. So as you can see here, this current is equal
to C DV over DT. So this is a relation
between current and voltage. Now we would like to get the
voltage or current relation. So we need to find what
is the value of voltage, the equation of voltage with
respect to the current. So as you can see here
from this equation, we can say is that takes
us one to the other side. So it will have dv over d t is equal to one over C
multiplied by the current. Okay? So we took this capacitance, so sales of science
or become division. Now we have dv over d
t equal one over c I. Now what all would like
to get is a voltage. So we will integrate this side and we will
integrate this site. Okay? How with respect to what
with respect to two times. So integration with
respect to d t, integration with respect to d t. So you can see that the integration of the
voltage dv is equal to voltage v equal to
the integration of one over c multiplied by d t. So it means that the
voltage is equal to the integration of current
with respect to two pi. Okay? Some larger ears, that current is equal to the derivative. So the voltage is equal to the integration of the current. From what time? From negative infinity. So anytime we would like to get, now, of course there is no time equal to
negative infinity. The lost to time is equal to 0. So t equals 0. So instead of negative infinity, we can say from 0 to any time t. Okay? Now, let's say instead of, we need to find the voltage starting from time
equals t naught. We need to find the
voltage at time t naught, starting from time t node at
which we start to charging. So we can say integration
of t naught to t one over c I d t plus the
initial voltage. So what does this equation mean? So here we have the capacity. We connected it to
a certain supply, connecting it to a supply. Now, this capacitor, if we start the charging at time
equal t naught, okay? Till time t anytime. Okay, so we started
charging at T naught to T. So what is the
value of the voltage? So I can see that the
voltage here is equal to the integration of one
over c I j dot d t. That, that duration of the current
with respect to time from the starting of a
charging till anytime t. Okay? Now, however, this representing a charging of the voltage
V2 in this time range. However, when we start a
charging this capacitor, it could have some
initial voltage. Can have some initial
voltage V naught. So we need to, so we need
to add this voltage plus the initial voltage at time equal t node adds a
starting, starts charging. Okay, so the voltage here across the
capacitor is equal to the initial voltage
at the time of a charging plus a
charging the period. Okay? So as you can see here, that voltage will be
equal to one over RC integration from t naught to anytime t d t plus the voltage
at t naught y voltage. Because the voltage
at t naught is the initial voltage at
which we started charging. So as I've all the
cholesterol can have a certain voltage when we
start the charging it. So we need to sell. So the total voltage will
be the initial voltage plus the voltage due to the
charging of this capacitor. Okay? And of course is the voltage at time t naught can be
obtained from the Q, Q T naught at time
equal t nodes. The amount of charge at time t naught divided by
the capacitance. Now, we have now the
equation of the current, equation of the voltage. Now we need to find
the power inside the capacitor and
energy of the capacity. The power, as you remember, is equal to the current
multiplied by the voltage, or power equals v
multiplied by i. Now, the voltage,
what is the value of the voltage v, which
will be as it is? And what is the value of
current is C dv over d t. So c d v over t. So it's a power is equal to
cv d v over d t. Now what does this
power representing? This is a Power which is
stored by the capacitor, or the power delivered
the tools or capacity not the stored energy is stored
the one power delivered. One important question is, this power is not
consumed power. It is the power which will be stored inside the capacitor. Okay? Now, the energy, as we
remember, what is energy? Energy assembly is
the integration of the power with respect
to time, as we remember. So we can say is that, that energy stored inside
the capacitor is equal to the integration of the power with respect to time. Now, integration from any time
to final time t. So again, the lowest time is 0. So we can say from 0 to time t. Now what is the value of power? Power is equal to CV DV over DT. So it will be c v d v over d t, since c is a constant, so we bought it outside
the integration. We have integration
from anytime to t v d v over r d d t dot d
t. So as you can see, d t can be canceled with
d t. So we'll have v dv. So you can see, see
integration v dv. So the integration,
as you can see here, a sweatshirt from integration
with respect to two time. Since we have d t
integration with respect to two voltage
with respect to dv. Remember this is
really important. Why are we changing? Because the integration
itself a change it from DT to dV Now, because we cancel
this with this. So we have only dv, Okay? Now we have C v dv. Now we are integrating with
respect to two was a voltage. So the integration of v is
equal to v squared over two. So if you don't know, integration of x dx is equal to, we have here to the
power one. Okay? So the first step is that we
increase the power by one. So it will be one plus one. So ultimately come to then divide it by the new
power, which is two. So x to the power one, integration of x is
x squared over two. Similarly here, x to the
power five, for example, dx is equal to first, increase this power by one, so it becomes a six and divide by the new
power, which is six. So this is the integration
if you don't remember it. So we have here half c
v square from, okay, you need to add this limits from vetoes at negative infinity
and V at time equal t. Since we are integrating
from here to here. So this integration will
be, after integrating, we need to substitute
with V as a function of t minus v at
negative infinity. Okay? So as you can see here, here we, this should be
written like this. Half c v final, which is V at anytime
t all squared minus half C V at negative
infinity. All the squared. Okay? Now v at negative infinity, or at a very small time, or a time equals 0, if this capacitor is
ionic charge, it. This carbon. So answer, so the energy
stored is equal to 0. Why? Because the voltage is 0, it is on a charge at a time equal negative infinity
or at time equals 0, to be more specific. Okay? So we will have only half c v as a function
of time square. So you will see is
that the energy stored inside a
capacitor, if it is not, if it is uncharge it, then it will be half c v square. Now we'll remember that this
equation is really important when you hear about
with Zach capacitor, what is the energy stored
inside a capacitor? You will always hear this
equation, half CV square. And I just told the
half CV square. It is commonly used in
electrical power systems. Now, we can write this
equation in another form, how you can see that
Q is equal to C V. Okay? So Q is equal to c
multiplied by v. So you can do like this. You can replace the voltage. Voltage here will be
equal to Q over C. Okay? So we have here half c v square. So we need V squared. V squared is equal to q
square divided by c square. So here, if you substitute with this here in this
equation, so we'll have w, or the energy store of c
multiplied by V squared. V squared is q squared
over C squared, q squared over C squared. So you can take the C
was one of these powers. So we'll have half q squared
over C, which is this one. Q squared divided by two
is half q squared over C. Okay? So this is another formula
for the energy stored. Okay? Now, the energy
stored inside a capacitor, it is a stored in what form? It is stored in the
form of electric field. You will have to remember this. How can we store electrical
energy in the capacitor? We store it in the form of electric field in the inductor. When you will learn about it, you will find that
it's stored in the form of magnetic field. So why does an
electric field exists? Because we have here
negative charges and here bolster the charges
on each of these splits. So an electric field is
formed between them. So that's why it's
called that the energy stored is in the form
of an electric field. Okay? Now, how does this
energy stored? We connect it to our supply. For example, any
supply like this. This capacitor
will be charged as those positive and negative
charges completely Jordan. Then what does an extra step, the next step is that to
you disconnect as a supply. Okay? So when you were to
disconnect is as supply, you will find that
we still have here both the visuals and
the negative source. And this is an open circuit. So this is Charles's will not be dissipated because there is
no pass or no load connected. So when you would like
to take the energy stored from this capacitor
is start connecting it, for example, to a
resistance like this. So it will start discharging and the providing
power to a resistor. Now as you can see, this energy can be retrieved the sensor, an ideal capacitor,
cannot dissipate energy. In fact, as Award
capacitor is derived from this element's capacity to store energy in an
electric field. Now it's a voltage
applied to a capacitor. What will happen when we apply
a voltage in the form of DC or AC voltage. So what will happen when
we connect a DC supply, a DC supply, or a DC voltage, which is a constant,
constant DC supply. Which means as the
voltage, for example, equal a constant value, let's say for
example, five volt. So this one is a constant, the value of the voltage. So if you look at this equation, you will find that the
current is equal to c multiplied by dv over d t, the rate of change of the
voltage with respect to time. So finds that the voltage does the voltage change with time, nor the voltage is constant
equal to five volt. So d v by d t equal to 0. That rate of change
of the voltage with respect to
time is equal to 0. Okay? It is a constant supply. So the current will
be equal to 0. So when we connect our voltage supply
that does not change. It is a constant. It can be a DC voltage, but it is an order
changing with time. It is, it is constant. One value of five volt, ten volt, whatever it is, it is a constant value, then the current
will be equal to 0. Now, in this case, when you think about this, if we connect our
supply, DC supply. A capacitor, it means that
the current is equal to 0, equal to 0. What does this mean? It means that the capacitor itself is acting as
an open circuit. No current but same
as an open circuit when we have a resistance
equal to infinity. So it means it
does not allow any current to Boston in
the DC condition. So it is blocking is I dc? Now what if we, if we connect a battery or a DC voltage across the
capacitor, the capacitor charges. So let's understands us. So when we connect a capacity of two our supply, it
started charging. It charges. Why? It does it a charge? Because it is at the beginning. We are supplying current
adds the beginning. At the very beginning. Sensors or source are moving from here and accumulating here, and moving from here
and accumulating it. So we have positive
and negative charges. So adds the beginning. We have accumulation
of Q here and ML accumulation from
overcharges here. So during this process
that charges are moving, moving through the wire or
the electrons are moving. Okay? So when the electrons
are moving, it means there is an
electric current. When, when we are charging the capacitor
at the very beginning, when the capacitor is
completely charged it we have here at Ball stiff
and the negative charges, no charges are passing
through the wires. It means that the
electric current is 0, no change in the voltage. So that change of the volume, DV over DT, is
representing water, representing the change in the, in the voltage here
across the capacitor. The capacitor when
it is a charging, when it started to charging, it's a voltage is not constant. It is a changing. Until fully charged. It will have a constant
value, a constant voltage. So at the beginning there is
a current, charging current. When it is completely
actual returns the current, the voltage will
be constant value, which means that the
current is equal to 0. I hope it's clear. Again at the beginning, it charging means that we are supplying current
to the capacitor. Why is it as a current? Because the voltage of the capacitor is
not fully charged, it is steadily changing. When it's completely charged, voltage will be constant
across the capacitor, so that current will be 0. Okay? Now, you have to know that the voltage on the
capacitor must be continuous because
the voltage on the capacitor cannot
change suddenly. Now, what does this mean? If you look at this
figure, for example, you will find that here
we have this case, this is a supply, okay? There's a supply which is
connected to the capacitor. And this is another supply
connectors that capacitor, you will find that this
supply is acceptable, it can be done. This exists, supply
is not possible. Now why is this? You'll see that here. Here is starting from
here at time equals 0, you will find that the
voltage of the supply increasing or the voltage
across the capacitor increasing until
reaching a maximum value is going to start
decaying, and so on. So here you will find
that d v over d t is the rate of change of the
voltage across the capacitor, is an acceptable value. Very small value as equal
to the slope of the line. So DV over DT. What does this represent?
The slope of this line? Now this line or this, the slope is accept
or can be any value, for example, 108,
whatever it is, the slope of this line. So when we have an
acceptable slope, we will have an acceptable
current in our circuit. Now, let's look to
the other case. At this point exactly. So if the voltage across
the capacitor changes from 0 to any value, let's say for example, two volt. Okay? And how much time at
t equals 00 time. Okay? What does this mean? It means that dv over d t, the rate of change of the
voltage with respect to time is equal to a V final
minus V initial, which is two minus 0 divided by time taken to move from
here to here, which is 0. So as you can see,
it will be two over 0, which means infinity. The rate of change of the velocity was a
voltage with respect to two pi in voltage with respect to time is equal to infinity. So this one is
equal to infinity, which means that the
current will be infinity. Very, very large
amount of current. And I would like to ask
you, is this possible, is this possible to
change suddenly, the voltage will exist
across the capacitor. Know, this case,
which is not real. It is not allowable, not allowable and not possible. That suddenly change in the
voltage is not possible. That voltage should increase gradually as you can
see, go down gradually. We cannot just change it
like this suddenly, okay? Because the capacitor
does not allow this. Okay? That's why the capacitor is
used as a voltage limiter. It limits as the voltage or
the limits to be more exact. Voltage limiter for
the rate of change. It limits the rate of
change of the voltage. Okay? Now, since we discussed
Zach capacitor, we need to give a small id about the ideal and
non ideal capacitors. So the ideal capacitor is a capacitors or does not
dissipate any energy, which is the ideal case. We said that the capacitor
is used to store electrical energy and does not dissipate any energy. However. Or to be most effective against
the ideal it takes apart from the circuit when storing energy in its electric field, then return as a privilege seen previously stored energy when delivering power to the circuit when we connect
two to eight reel, which is a practical case, are non-ideal capacitor has a parallel mode and
leakage resistance, every larger resistance
connected to parallel to it, which dissipates
electrical energy. This resistance can reach a 100 mega ohm and is neglected for most
practical applications. So let's see. So this
is the ideal capacitor. The capacitor like this, connected to our supply, connected to a
circuit like this. However, this is the ideal case. The non-ideal or
rail capacitor is having a resistance
in parallel to it. This resistance is almost
equal to 100 mega ohm, can be highly to a 100 megaohm, very, very large resistance. So if you think about this, if we have a supply connected to the capacitance like this. Okay? When the capacitance, when the capacitor
is fully charged it, you will find that the current, is there any current
passing through the capacitor after it
completely charged? No. 0 current busing here. But here there is a
very small current, very, very small
current passing here. Why? This grant is like this
moving from here and goes through the
resistance, then gets back. Now why is this current
is very, very small. So it dissipates very, very small power. Why? Because the current,
this current is equal to 0 supply or the voltage
across the capacitor, V capacitor divided
by our resistance. Okay, Week or buster or V supply divided by
the resistance. This little sun is very large. So the current passing through this resistance is very small. That's why is the effect of this resistance
can be neglected. We neglect it for most
practical applications. Okay? So I hope in this two lessons, you understand more
about capacitors. And now we are ready to have some examples on the capacitor.
66. Solved Examples on Capacitors: Hey everyone, In this
lesson we are going to have some examples
on capacitors. So the first example is
that we have this capacitor connected to supply
at DC supply. Now we have this capacitor is
three picofarad capacitor. This is its own capacitance and voltage applied across
it is that when ti volt. So we need to find two
requirements in this problem. The first requirement
is that we need to find the queue or the charge
stored on the capacitor. Second requirement
is that we need to find the energy stored. Okay? So let's just start. We have here as three
picofarad capacitor with a 20 volts across it. Now, so here is, this is our v and this
is our capacitance. Now remember that that cue, or the amount of the
charges is equal to C, the capacitance
multiplied by voltage. So it will be like this. Q equals CV. So C, which is the
capacitance, is three. Since Be cool, pico is ten to the power
negative two, whatever. This is, cooper and
multiplied by the voltage, which is 20 volt. So this multiplication will
give us 60 be equal column. Okay? So here as you can see, we can, we can say that this
13 multiplied by 20 is 60 multiplied by ten
to the power negative 12. Okay? So we can say is that the
amount of the charges is 60 multiplied by ten to the
power negative one column. Or we replace this ten to
the power negative 12 width. So we have 60 Pico column. Now the second
requirement is that we need to find the energy stored. So the energy stored inside
a capacitor is equal to half c v square,
if you remember. So we have the capacitance. We have the voltage. And you will need to
substitute to the capacitance, which its own value, three multiplied by ten to
the power negative 12th. So we have that the energy
stored is half CV squared. So capacitance is three
multiplied by ten to the power negative 12 multiplied by
v squared is a voltage. A square, D square is
20 multiplied by 20, which is four hundred,
four hundred volts. Here, what will be
400 volt squared? Because we squared the voltage. Anyway, you will find
the 400 volts squared. So we will substitute
it in this equation. So we will have that
energy stored is 600 pico. Again, we multiply it by four, multiplied by three, multiplied
by a half gives us 600. And the ten to the
power negative 12. We added it as a. Okay. Now let's have another example. So the voltage across R5 microfarad capacitor
is V as a function of t, is equal to ten cosine, 60 thousand volt, find
the current through it. So first, this five microfarad
is our capacitance. V as a function of t. This is voltage, okay? Now then cosine six t, What does this representing? This sort of presenting
a cosine wave or an AC supply, AAC supply. What does an AC supply mean? And what's the difference
between AC and the DC? Dc means DC. Dc is a unidirectional. It means that it has
only one direction is or the values are
positive or negative. So some singular
exists is a DC supply, something like this,
is also a DC supply. Why? Because all of the
values are positive. Also if our supply is like, like this, negative only, then it is a DC or it
is something like this, whatever it is variable. But it has one
direction negative, so it is a DC supply. Now, AC supply like
this for example, if we have something,
something positive, negative, positive, negative, positive,
negative, and so on. So this is called an AAC supply or alternating
current supply. Alternating current means it is changing from
positive to negative, negative to boast of and so on. So you see all step negative, the negative like this. It can be build boast
of only or negative. So in our example here, this one is an AC supply. Why? Because as you can see here, cosine wave, it
will be like this. Okay, cosine wave
and its peak is ten. So as you can see here, this is ten volt as
a function of time, then this one is negative ten. So as you can see, this is an easy supply
alternating, okay? Okay, so what does
the requirements, this problem we need
to find that current. So first, what does the value of current
inside the capacitor? What is the equation? If we remember that current
inside the capacitor is equal to C DV over DT, or the capacitance, which
is five microfarad, five micro as y multiplied by ten to
the power negative six. As you can see here. Ten to the power negative six is Mike multiplied by d v over d t. What does this mean? That differentiation, differentiation of the
voltage with respect to time. So the voltage is ten
cosine 6 thousand t. So we need to differentiate
then cosine 6 thousand. Okay? So how can we differentiate
this function? So we have d over d t, then cosine 6 thousand t. So then there's a
differentiation. We have a constant multiplied
by a cosine function. So we will leave this
constant as it is, then multiply it by that differentiation of
cosine, 6 thousand D. What is the
differentiation of cosine? The derivative of
the derivative of cosine is negative sine, negative sign six thousand. Six thousand multiplied by the derivative of
the angle itself. So the derivative
of 6 thousand t, The derivative of
velocity is 6 thousand. Okay? So ten cosines, because
I was anti derivative, is this function, negative
ten multiplied by 6 thousand, multiplied by sine six t.
So as you can see here, negative then 6 thousand signs, 6 thousand t. And here, what does this representing? This representing five Mottola
buttons while Nick six, this part is capacitance. So when we multiply all
of this by each other, we will have negative
0.3 sine t and bear. This is the equation
of the current. Now, let's have another example. So in this example we have, we have the voltage across, we need to find the voltage across R2 microfarad
capacitance. It was a current through, it is I as a function of t equal to six multiplied by e to the power negative 3 thousand,
mainly Ambien. So this is a equation of the current that flows
through this capacitor. And we have that the
initial capacitor voltage, V initial is equal to 0. You will understand how can
we are going to use, okay? So let's start. So what is the equation
of the voltage with respect to two as well? Remember that the voltage
is equal to one over C integration of
the current with respect to time plus
the voltage initially. So as you can see here, V, the voltage is one over C. Integration of the current as
a function of t from 0 to t from 0 plus the initial voltage. Okay? So we have this bot integration from 0 to any time
t of that current. Plus the initial words are starting charging or the
short and the forest, the voltage, or the
voltage at time equals 0. Okay? So here as you will see, the initial capacitor
voltage is 0. So v as a function of 0 or at time equals 0
and y equal to 0. So this part is equal to z. Okay? So now we will have in the
equation only this part, one over C integration from 0 to t r as a
function of d t. Now, one over C, the capacitance
is two microfarads, so one over two multiplied by ten to
the power negative six. As you can see here. Then
integration from 0 to t, integration from 0
to t for the current as a function of time six e to the power negative
three cells and T six e to the bounding
three cells and T, d t. And remember, here we have
what we have milli and bear. So you can make the output
equation millivolt or zoster, take them mentally and convert it to ten to the
power negative three. Okay, so we convert this into an bear by multiplying by ten to the power negative three. Okay? Okay, now what,
Now we have here, one divided by two, spawning 610 to the negative 36. So take this six voted outside here and ten to
the power six outside. So we will have only an
integration from 0 to t of e to the power
negative result thousand. It's the results and
t dot d t. Okay? So we only have the
integration of this function. So how can we integrate something like this,
the exponential? If you want to know
the integration of this equation, it
is very simple. First, e to the power
negative results and T, it assembly that
exponential as it is. As it is. Then divide that by it by the
derivative of this power. This power is
negative 30 thousand t. Its derivative is
negative 3 thousand. 3 thousand from 0 to t. So this is what we done. You can see your
negatives resultant e negative three t from 0 to t. And you will find that this part is six divided by
two gives us three. Then to the power negative three divided by ten to the
power negative six, it gives us ten to
the power three. Okay? Okay. So as you can see, this is our final equation. So as you can see, three selves and divided by negative results. And so we'll have here
what is this gives us, gives us negative one multiplied by this exponential from 0 to t. So this can be written
as like this exponential. The finite explanation
negative series cells and t minus e initial, which at time equals 0. E to the board next resolves. And T or multiplied by 0 gives us e to the power
negative 0, which is one. Okay? So we have exponential
minus one. But remember that we
have here negative one. So all of this is
multiplied by a negative. So it will be
negative goes here. So it will be negative e to
the power negative 3,018. And negative negative
becomes a plus one. So we will have one minus e to the power negative
results and ti volt. Okay? So this is a voltage
equation in this example. Now let's have another example. So example number four
on capacitors finds a current through a 200
microfarad capacitor. Voltage is shown in figure. So as you can see
here, we need to find the current through
a 200 microfarads. So this is our capacitance and this representing the
voltage equation. Our voltage is an AC supply. A part of it posted
another part and negative. So it's going up, down, up, down and so on. Okay. So what are we going to do
in order to get the current? Remember that the current
is equal to C DV over DT. So what are you going to do? The capacitance is 200
microfarads, 100 microfarad. And d v by d t assembly. That differentiation of
all of this function. Or if you would like
it much easier, it assembly equal to
the slope of this line, slope of this line,
slope of this line. So first, let's try that difficult method and then I'm going to give
you the easiest measure. Difficulty method is
that we need to get v as a function of time at
this different times. So as you can see from 0 to one, we have this straight line. From one to three. We have this straight
line from three to four, we have this straight line. Okay? So we have three straight lines. So we need three equations
representing this regions. So let's start. First. Here we have from 0
and this part is 50. So how can we write
this equation? Remember the equation
y equal to m x plus c. This is the equation
of a straight line. So here y is our
voltage equal to. M, what is our x? Our x is time plus C. Now for us to m here representing
the slope of this line. So the slope of this
line is equal to m, or the slope is equal
to y two minus y, y1 over x2 minus x1. This is from what? From the mathematics. Okay? So anyway, he would, Y2 is the final y here. So we can say m is
equal to y two, the final y is 15. The initial y, y, y1 is equal to 0 here minus 0. Then finally x, which is one, minus the initial x, which is 0. So it will give us 50. Okay? So we will have v equal to 15 multiplied by time
plus that constant t. So as you can see here
from this equation, at time equals 0,
when t equal to 0, the voltage is also equal to 0. Okay? At time equals 0, the voltage will be equal to 0. So c, what is the value of C? C will be equal to z. So the equation of
this straight line, the straight line here, would be equal to
V equals to 50 T. Okay? So as you can see, at time
equals 0 is a voltage will be 0 at time equal one, so volts will be 51,
voltage becomes a 50. Okay, so this representing
the first straight line. The second straight line here, we need to find its own slope. Also. We have here, this is the initial a point, this is a final point. Initial X, final x m, or the slope of the
line is equal to Y2 final y negative 50
minus y1, which is 50. Okay? So as a final point, negative 50 initial 0.5050
minus the initial point, then x2, x3 minus x1. So we'll have negative
100 over two, which will give us negative 50. Okay? So you can say is that here, voltage is equal to
negative 50 t plus sin. So we can choose any
random point here. As an example, at this point, we have our time equal to
when this one becomes two, the voltage will be 0. So negative 100 plus c equals 0, so c will be equal to 100. So the second equation, V equal to negative
50 de los hundreds. This is the equation
of the straight line. Okay? Okay. Now, the last one, which is this a straight
line from here to here, we will have the slope
of the line while to Y2 minus Y1 over X2 minus X1. Y2 is the final y, which is 0 minus
the initial watch is negative 50, negative 50. So it becomes all stuff
50 divided by final x, which is four minus
initial exogenous ring. So it gives us 50. This
is a slope of this line. Okay? Now, here, why? Or the velocity v
will be equal to 50 t plus constant again. So what is this constant value? For example, at time equal for the voltage
will be equal to 0. So 01, when this one becomes a 44 multiplied by
50 gives us 200. So c will be given
as negative 200. Okay? So the equation v will be
equal to D t minus two. So this is a question of what
is the equation of here, of this one of the
last straight line. So as you can see, we
obtain the three equations of this 33 straight lines. As you can see. First 1, second 100 minus
15 as the last one, is negative 200 plus 50. Okay? For now, what are
we going to do? We have the voltage equation. We need to get the current. So the current is equal
to C dv over dt eating. So we will
differentiate this one, this one, and this one antebellum told it
was a capacitance. So current is equal
to C d v by d t. So that capacitance
200 microfarad. So to see, which is 200
multiplied by ten spawn negative six and multiplied by
derivative of this 50 T. What's the derivative
of t is 15. What's the derivative of
this one is negative 15. Derivative of constant is 0, and derivative of negative
50 is negative 50. Negative 200 becomes
050, t becomes 50. Okay? So we multiply this
by this equation. We will have, our current
will be ten milli and bear negative 10 million
bear and ten milliampere. So it can be drawn like this. Okay? Okay. So this is a difficult semester. We'll have a difficulty
method is that we need to get the equation of every line. Okay? Now, if you understand well, that definition of d v by d t. What does d v by d t? If we have a voltage equation
and we need d y d t, the derivative of this equation. What does this mean? D v, d t in mathematics, means that slope of the line, if you will remember
that the slope of this line is what is 15? Y2 minus Y1 over X2 minus X1. The slope of this
line is negative 50. Slope of this line is 15 dB
Y2 minus Y1 over X2 minus X1. So that dV by d t, as you can see here, is 15 negative 50. And as you can see, 15 negative 5050,
which represents slope of the line slope
of the line slope XY. We obtained it easily without writing all of this and
negating the constants. We assembly get the
slope of the line. And we already know now
that d v by d t, Okay?
67. Series and Parallel Capacitors: Now let's discuss that series
and parallel capacitors. So how can we combine in series
and parallel capacitors? So as we know that from resistive circuits
as a series and parallel combination
is a powerful tool for reducing our circuits. This technique can
be extended to the series parallel
connection of capacitors, which are sometimes encountered
in our electric circuits. So we would like
to, we like or we desire to replace
this large circuit containing a large
amount of capacitors by a single equivalent
capacitor C equivalent. Let's say for example, we have this circuit. We have a current source. We have group of capacitors
in parallel, C1, C2, C3, C4, until CN,
number of capacitors. So we would like to convert to this larger circuit into
something glycolysis, one equivalent capacitor
with only the supply. Okay? So in order to do this, we need to know what is the formula for combining
is a parallel capacitors. And what is the formula for
combining series capacitors? Let's start in the
parallel capacitors, as you can see here, we
have a current source. This current source provides
current to a capacitor C1, C2, C3, and C4 until
number of carbons to see. And so as you can see here, what you can notice
from this formula is that the current
I is equal to I1, I2, I3, I4 IN like this. So from KCL, the current, supply current is equal to the summation of all
of the currents. Now, if you remember, what is the value of
I1 or I2 or I3 or i and u i1, or the current, current inside our
capacitor is equal to C dv over editing the capacitance of the capacitor multiplied
by the voltage across it. Dv is a derivative of the
voltage across the capacitance. For example, I one will
be equal to c one, its own capacitance multiplied
by voltage across it. So as you can see,
the voltage here plus minus is called V. This voltage across C1 is similar to the
voltage across C2, similar to the voltage
across history and so on. So I1 will be C1
D v over r d d v, d v is the voltage
across the capacitor. And i2 will be equal to c to d v over d t. Okay? So the capacitance multiplied by the derivative of
the voltage across it. So as you can see,
C1 dv over d t, c to d v over d t, C3, d v over d t y is the same voltage because
all of them are battery. So you will have this
equation like this. So sensors are
current is equal to the capacitance DV over DT. So you can see as result set. So from this equation you
can see C1 dv over d t, C2, dv over C3, DV over DT, and so on. So we'll find that
we can take d v over d t as a common factor. Dv over details are common
factors and then multiply it by C1 plus C2 plus C3 until Cn. So it will be the summation
of all of the capacitance. So as you can see
here, that that current flowing in this circuit, which is the same current here, is equal to what is equal
to the C equivalent, equivalent capacitance
multiplied by the derivative of this voltage. Or you can see that that equivalent capacitance is what is the summation
of all capacitors, since they have
the same voltage. So we take the DV over
DT as a common factor. So the current will be
summation of all of these capacitors multiplied
by the derivative. So what do we learn
from this is that the equivalent capacitance
of a parallel capacitors, equivalent capacitance
of a parallel capacitors is equal to the summation
of these capacitors. So the equivalent circuit
in order to transform this parallel into one
capacitor is equal to C. Equivalent is C1 plus
C2 plus C3 until cn. Okay? As you can see, it's
the equivalent of n parallel connected
capacitors as the sum of the
individual capacitance. Now what if we have
series capacitors? We have C1, C2, C3 until cn. And we have a supply voltage. V is the voltage
across C1, C2, C3, C4, and not equal to each other
since they are in series. So we have the voltage
across V1 as C1V1, C2V2, S3, v3, and so on. Now, as you can see here, what is that common in
the series circuit? The common is the current. The current flowing
through all of these capacitors are similar to each other because
they are in series. Okay? Remember this second thing
from KVL in this circuit, we know that the
supply voltage is equal to the summation of
all of these voltages. From KVL V, or the
supply voltage is equal to V1 plus V2
plus V3 plus v n. Now what is the value
of the voltage, each of these voltages, okay, So we have here is a current I flowing through the circuit. Now, what is the
value of voltage V1? Remember that V, V1 is
equal to one over C1. Integration from 0 to
t of that current. The team plus the initial
value of the voltage V0, V1 at time equals 0. And v2 will be one over C2. Degradation of the current, same current because
they are in series plus v2 at time
equals 0 and so on. So we'll see like this, we have voltage is
equal to one over C1 integration of current
plus the initial current, plus one over C2 integration of current plus the
initial voltage, plus, and so on. So we'll find the year, what
is common in all of this. You will find that this
part, It's common. Okay? So one over C1 integration
of the same current, one over C2 integration of
the current one over C3, degradation of the same current. And so we can take this
current integration of this kind of common factor
between two brackets, one over C1 plus one over C2
plus one over C3 and so on. Outside plus all of
the initial voltage. Will find here is that in
this circuit, for example, that the voltage V across
the capacitor is equal to v, is equal to c equivalent, or one over C equivalent
integration of the current plus the initial
voltage across this capacitor. Okay? So we'll see that this is a voltage
of this circuit, which is this one,
this equation. So if you compare
this equation of the forest circuit with the equation of the
second circuit, you will find that the
equivalent capacitance, one over C equivalent
is equal to one over C1 plus one over C2 plus
one over C3 and so on. So we'll find that
by deleting this, the equivalent is
a reciprocal of the equivalent capacitance is a summation of that reciprocal
of each individual. Capacitance. Finds that the equivalent
capacitance of a senior is connected
capacitor is reciprocal. What does this
phrase broken mean? It means a one over,
over something. Okay? So here one over C equivalent is equal to the summation
of the reciprocals of the individual capacitors, one over C1 plus one
over C2, C3, and so on. So if would like C equivalent, it will be equal to one
over this summation, which is the reciprocal
of the sum of the reciprocals of each
individual capacitors. So what we learn here, well, owners that insert
parallel capacitors, the capacitance will be
summation of all capacitors. In series capacitor, that the equivalent capacitance is
that risk broken of the sum of the reciprocals of the individual capacitors
will find that that parallel capacitors are treated similarly to the
series resistance. And the series
capacitors are treated similarly to the
parallel capacitors. So in that if we have two
capacitors in series, we will have one
over C equivalent equal one over C1
plus one over C2. So as you can see, it will
be c equivalent will be C1, C2 over C1 plus C2. If you remember this equation, you will find that similar
to our equivalent R1, R2, R1, R2, R1 plus R2. So when did we use
this equation? When we had two resistors, R1 and R2, going to toss a ploy when they
were parallel to each. Awesome. Okay? However, this equation is when c1 with c2 connected to supply. So you can see that the C1, C2, when they are in theaters, they are treated
in battery as if they are parallel
resistors. Okay. Let's have some
examples on this.
68. Solved Examples on Series and Parallel Capacitors: So the first example on series and parallel capacitors
that we need to find the equivalent
capacitance seen between the two terminals a
and B of z circuit. So we have C equivalent. We would like to find the
equivalent capacitance. When we look at here, we
have a 60 microfarad, 20 microfarad, six
micro four out 520. So how can we do this? Really, really easy. So first, you will find
that in this equation, we will have number one. You can see that six microfarad and 20 microfarad are what? Z have the same initial
and same final node. So in this case, six a
microphone route and 20 microfarad are in parallel. So what is the
equivalent of this two? What is the equivalent of
this to the equivalent? Is there some
mission 20 plus six. So we will have here 26. Why? Because they
are in parallel. So they're equivalent is
at summation 20 plus six, which is 26 microfarad. Now we will see here we have five microfarad
and 20 microfarad. They are in what? Zr series with each ours. Okay? So since they are in series, they will be treated like this. Z equivalent one over C, equivalent of this to, let's say C equivalent
one is equal to one over 20 plus one over five, okay? Or C equivalent is 20 multiplied by five
divided by the summation. So if I, if my
calculation is correct, I think it will be four
for microphone. Okay. Now we have this 60
micro far out like this. Okay? You will find that we have
four and the 26th or what? Our battery, their
combination is four plus Twenty-six series
as if the series resistor, so it will give us certainty. Series with 60. Like this. So sick and uncertainty
are in series, so they're equivalent
is 60 multiplied by 30 divided by summation
will give us, as you think it will
be 20 microphone. Okay, I see. So let's see what
does the steps again. So first, as you can see here, that when two microfarad
and five microfarad are in series, are in series. The equivalent capacitance is 0 multiplication
divided by 0 summation, or one over C equivalent is equal to one over
20 plus one over five. The artery as if they
are parallel resistors. So it will give us an
equivalent of four microfarads. We have here. Instead of this, we have four microphone. Now we have a form microfarad, six omicron furrowed,
and the 20 microfarad. So it will be this part. It's equivalent is what
is a capacitor like this? Equal to four
microfarad capacitor. This capacitor and this
capacitor are all in, sorry, this capacitor and this capacitor and this
capacitor are all in parallel. So it will be summation. So we have 20 plus
six plus four. Plus six plus four
gives us 30 microfarad. So we will have as
30 microfarads, which is equivalent
of all of this. See it as with a microphone. So it will be 30 microfarads
series with a 60 microfarad. So they're equivalent. Is there multiplication
multiplied by the some mission? Why? Because they are in series. The equivalent capacitance
of this circuit is 20 micro. Okay? Now let's have another example. So in this circuit we have 30 volt at 20 microfarad at 20, mainly for odd, for odd
40 millifarad, 20 milli. So what do we need? We need to find the voltage
across each capacitor. We need to find V1, V2, V3, v3 is a voltage across 40 milli for far-out
and 20 milliliters. So we need to find
this voltages. So how can we do this? What does that voltage? Voltage is simply one over C integration of current dot d t plus the initial voltage. You remember this? So the question is, though we know the
initial voltage, know we don t know
the initial voltage. So we cannot use this equation because we don't know
the initial voltage. So what does the
second solution? The second solution is
that we know that Q, or the quantity of charge equals capacitance multiplied
by the voltage. So here we can get the
voltage is equal to Q over C. The voltage V1 will be q across that 24 divided by
20 millifarads. So as an example, V1 will be
Q divided by 20 millivolt. So the question is, how can we get this Q? So first, you have to know that the Q similar to the current. So when we say that, when we think about with that, you can treat it
similar to the current. So the current or
the Q quantity of charge on this plate is the amount of
current flowing here. The current is equal
rate of change of q in Z and Z have
the same direction. They are moving
towards this plate. So the Q here, this is the amount
of Q going out here, going to the plate. So the key here is
similar to the current. If we think about it. So we'll find that the queue here should be equal
to the Q here. Why? Because they are in series. They have the same
current sensor senses at 24 out and certainly for all
having the same current. So they have the same queue. Again, same current,
flowing, same q. And this current,
or this Q will go and be divided here and here. So we will have here Q2, for example, and the Q3. They're summation is a Q here, which is the input queue. Okay? Now, in order to find V1, we need to find the skew. So how can we get this
order, finds a current. How can we do this assembly? We need to find the
equivalent capacitance first, and we'll tell you more now. So first, we need to find
equivalent capacitance. So we have 40 mainly
far out and 20 milli for odd as they are parallel. So they're equivalent. Is there some
mission 20 plus 16. So we have equivalent of
cyclicity mainly for OT. Then we have a 20-minute for, mainly for OT and six
similarly for all in series. So they're equivalent is, or one over C equivalent
is one over 20 plus one over 30 plus
one over 16, like this. So again, 40 plus 20 is this
summation is millifarad, then the equivalent is
equal to zeros broker reciprocal one over that risk
broker of each of these, a summation of the reciprocals of each individual capacitor. So we have one over
60 plus one over 70 plus one over 20 here, 203626. So the equivalent
capacitance of this system, It's then mainly for OT, the queue or the current going
through this capacitance, similar to the current or
Q going from this supply. So you think about this is
that if we get the Q here, we can use it to
get the voltages. So how can we get this Q, Q assembly equal to the capacitance
multiplied by voltage. So the total charge is equal to capacitance equivalent
multiplied bys or supply. So ten mainly far-out. The blood supply
voltage two volt gives us 0.3 column, okay? Now we have the total
charge Q equal 0.3. This q is equal to the queue across V1 and the
queue across V2, because they are in series, so they have the same current, so they have the same charges. So if we need the voltage here, it will be the
quantity of a charge divided by its
capacitance, like this. So you can see V1 is equal
to the Q going of this sub, out of this supply,
which is 0.3. V2 is the Q going
down will supply 0.3 divided by the
capacitance of each one. V1, divided by 29 different V2. V1 and V2, certainly metaphor. So we'll have the voltage here, 15 volt, the voltage
here ten volt. So we have here 1510 volt. Now, the last requirement
is that we need V3. So if you apply KVL here, you will find that 30
volt is equal to V1 plus V2 plus V3, or V3. V3 is equal to 30 volt minus
V1 minus V2, like this. So V3 is supply minus the voltage drop
across the capacitor, which is 15 volt, minus the voltage drop across the second capacitor, ten volt. So we'll have five volt. Is the voltage across
that 40, mainly for it. Okay. Now, is there any
other solution? Yes, there is another one. If you think about this, we have q here, same queue traveling here. And it does the
same queue across. The equivalent of this to the equivalent of this
soul is 60, mainly for. So Zach queue across their equivalent is a queue across their equivalent is
similar to the queue here, because the equivalent is
Sierras with this part. Okay? So the equivalent,
their equivalent of this has a quantity of charge q. So we can get it. So voltage. So the voltage will be q, which is 0.3 divided by z, equivalent to a 60
million different. So this will give us the voltage here
across the equivalent, which is the same voltage
across that 20 milli fraud, and across the 40
millifarads like this. Okay? So Q divided by z equivalent
gives us five volt, which is similar as the value which we just obtained here. Same as dismissal. So both of these methods are, can give us the
same requirement. Okay.
69. Introduction to Inductors: Hey everyone, In this lesson
we are going to discuss the third element also serve the passive element in our
course for electric circuits, we discussed first
the resistors, then we discussed
the Zan capacitors. Now we need to discuss
the inductors. So as you can see here, the inductors is this one, this one, and this one. All of this representing
an inductor. An inductor, what does it form? You will see that this is
an iron or a conductor. This one is a conductor, but with several
number of terms. So as you can see here, we have a conductor like this
one, this conductor. And as you can see, it is coiled or form it around
the air cooler. Okay, like this one. You can see here, if we have a code like this one like this, it can be an iron core or
it can be and can be air. So we bought our
wildlife exists, then keeps rotating
around it like this. Like this. So this
is called the query, which is of course an inductor. Okay? So it is very important to
have this number of turns, this number of
donors and the code, there can be a cola exists, an iron core for example, a code like this. Or we can have air inside it. This formation is
called the inductor. The inductor is a
passive element designed to store its energy
in its magnetic field. So the capacitor stores electrical energy in the
form of electric field. The inductor stores it in
the form of magnetic field. So there is a theory
which you will learn in the course of
electrical machines. If you go to our course for electrical machines and learn
about magnetic circuits, you will understand
that when a current, when a current,
alternating current, variable current going through
an inductor like this. What will happen is that a magnetic field inside its
so-called will be formed. So as an example here, you can see the current
entering glycolysis, entering like this
and going down. Our board like this. This according to
a certain city, will provide magnetic
field like this. Okay, sorry, it's
so cool, like this. So where did we learn
about this from the induction machines
or the induction, see here this is a
phenomenon which happens in nature when we have a variable current going
through an inductor like this, it will provide or produce magnetic field
inductors that can be found in various applications in electronics and
power systems. They are used in power
supplies, transformers, radios, TVs, radars,
and electric motors. Any conduct, any conductor of electric current has
inductive properties. So any conductor like this, if we have a wire and the
current passes through it, variable current, then we
have an inductive property. And inductive property. It can be like this
rotating like this. Number of donors,
as you can see, a number of donors,
1234 and so on. So as we increase the
number of donors, that induction or the inductive property
increases inside this inductor. So the inductive itself
or the inductor itself, as you can see here,
this property is found in this shape and can be
found in our wire like this. But the difference is
that this inductance of the wire is low compared
to something like this with larger
number of donors and an iron core or conduct
a conducting material. So it is different depending
on the construction. So in order to enhance
the inductive effect, a practical inductor
is usually form it into a cylindrical
coil like this. You can see here we have cylindrical coil
like this one is the form of cylinder with mini tunnels
of conducting wire. You can see we have
1234 and so on. So if you would like to draw it like this,
we have a wire, then 12345 and so on. Inside it we can have an iron core or we can
have like this one, or we can have air. Okay. Now an inductor consisting of
a coil of conducting wire, a coin of conducting wire. Now, before we go and understand
more about inductors, I would like to explain
something about induction. So you will ask me, what does inductive mean or
what does induction mean? You have to know is that scientists found
that an engineer or a scientist found is that when we have a coil
like this one, I call it like this one. Okay? And connected to a resistance, for example, any loop. So this conductor
representing what? Representing a coin. Okay? Now, is there any
current flowing? Knows there is no current flowing because we have
only a resistor and a wire. Wire in the form of number
of turns like this one. As if we are connecting
this one to our resistance. So there will be no current, no reason to have any current. Now that scientists found
is that when we have, for example, a magnet
like this, okay? A magnet like this. Okay? And do we start north and south and we start
moving this magnet. Those are left and to the right, moving it like this. Okay, close to this coin. What will happen
is that there is a variation in the
magnetic field. The magnetic field seen
by this coil is varying. So due to the presence of
variation in magnetic field, there is a voltage which induced between
these two terminals. Okay? So this area is called
the deal was the EMF E, or a voltage forming between
these two coils equal to negative and d phi over d t. Okay? So n here is the number
of turns of the coil. And d Phi of Z t, d t is a variation of the
magnetic flux always time. Since we are moving this magnet left underlies the
left and right, we are causing a variation of magnetic field
close to this coil. So what will happen is
that this coil will start producing induced EMF at its terminals to produce a current that produces
magnetic fields. Why, in order to oppose the effect of the
original magnetic field. So why do we have induced EMF? Because this induced EMF
produces a current that produces a magnetic field that tries to oppose the original
effect of this magnet, of the variation
of this magnetic. So this property
is called Desert. This voltage is
called the induced, induced EMF, or electromotive
force induced the voltage. That's why this property, this property which
happens here, is called the induction. The induction. That's why here, as you can see, when we increase
number of turns, we have more induced EMF or
the induction increases. Okay? So this is not related
to the course. You will learn about induced
EMF in the right course, which is the
electrical machines. Now, if a current is allowed
to pass through an inductor, it is found that
the voltage across the inductor is directly proportional to the rate
of change of the current. If we have a current flowing
here and this inductor. So the voltage induced
at this terminal, at the terminals of the coil is directly
proportional to what? Those are rate of
change of the current. So as you can see here, is that this equation
is that the voltage V, which is for MIT at the
terminals of this inductor, is equal to L d over
d t. L is called the constant of proportionality or the induction inductance
of the inductor. L is called the inductance. So as you can see, we
said that when we, when a current passes through current buses through this coil, there is a voltage formed
between its two terminals. This voltage is dependent
on what is dependent on the rate of change of this
current and inductance. So v is directly proportional to the rate
of change of the current. The current must be variable. It must change. If this current is
a constant value, then there is no
voltage induced. Okay? So v is directly
proportional to D over DT. Now we can replace
the constant of proportionality by constant of proportionality, which is l. So we say v equals
L di over d t. Now, L is inductance which
is measured in Henry, the owner of the American
inventor Joseph Henry. Now, inductance. Inductance is the property
whereby an inductor exhibits opposition
to the change of current flowing through
it, measured in henrys. So when we have a magnetic
fields that is varying, we will have here a
change in the and use the math that tries to keep this magnetic field constant. And another time when we have a varying current flowing
through this inductor, we will have what? We will have an induced EMF. What is the benefit
of this induced EMF? It tries to oboes, is this a change in the
current inductance of n inductor depends on
the physical dimension and construction
of this inductor. Now, how can we get
this inductance? We can get this inductance. We have different shapes which are derived from the
electromagnetic see, okay, So there are many, many inductance for different
types of inductors. For example, is a solenoid. So light is one of the famous
inductors which are used. Its equation is that
the inductance equal to n square mu multiplied
by area divided by L. Now, first n squared, what is n? N is the number of turns. So when the number of donors increases, the
inductance increases. The second property,
which is mu. What is mu? Something which is
called permittivity. What does it's a benefit, it allows the flow of magnetic
fields inside its score. Okay, it's something related
to magnetic circuits. To win this increase
inductance increases the area, which is the area
of the core itself, when it increases
and hence increase as inductance increases
and the lens, the lens increases
and decreases. So this is the length
of the core itself. So n is the number of donors. A is the cross-sectional area. Mu is permeability of the core. But in my devotee is epsilon. Permittivity is epsilon of the capacitance,
which is Epsilon. Here mu is the permeability
which allows the flow of magnetic field inside its
magnetic core itself. So as we said now, the inductance can be increased by increasing the
number of turns. Using the material was
high permeability, increasing the
cross-sectional area or reducing the
length of the coin. Now it's a practical inductors
have inductance values, which can range from
a few microhenries, such as in communication systems to tens of Henle
is empower system. So as you remember
that we said before that the capacitance is in the range of picofarads
or micro farads millifarad. One for odd or tool for rod
is very large value here in in inductance we
have also microhenry, millihenry, similar
as capacitance. Sometimes we can have
tens of analysis. It's okay. When do we use this? There is another type of Doctors are called
the super inductors, such as the supercapacitors. We have super inductors or superconducting
coil as I remember. So this is, has a
large number of Henry to store large amount
of magnetic field. And we can use this
when we need it. Okay? Now, types of inductor
similar as capacitors, it can be fixated value
or a variable value here, as you can see here, those are different shapes
for the inductor. As you can see here,
we have two wires. And do we have here a
core and around it, wire coils or the
number of turns around does this
similar here around it? So the inductors, you can have a fixed value or
can be variable, we can change it. The code can be made of iron, steel, plastic, or air. The terms coil and Joe are you also use the for inductance. So we can say inductor
or a coil or that joke, whatever it is, all of
them represent the same. Okay. Now similar as the
capacitance or inductance is
independent of current. Same error as the capacitance, independent ohms, the voltage, or the quantity of charge. So the inductor is known
as a linear inductor. This type, which is not
dependent on the current, it's a value does not
change when the current flowing through it and whatever the current
flowing through it. So it's called the linear, okay? It does not affected
by the current. However, there are other type
which is called nonlinear, which its inductance are
affected by the current. Now what does this
circuit symbol for? The inductor? So as you can see here,
that this is a coil. As you can see here,
are Khoi exists. So this is an inductor. When we don't have anything
or two lines beside it, it means that this, this
coil is made of air core. Okay? So it will be like this, something like this,
and around like this. Okay. So inside it, there is no cool. When we have these two lions, it means it is
made of iron core. So as you can see,
it will be inside. This would be a goal like this, made of iron, like this one. You can see inside it. Cool. When we have this score, we will have this two lines
representing the presence of iron inside it or
a core made of iron. If we remove this core, then we will have only air, which means that we have ARCore. Now when we have a line
like this exists as usual, what does this mean? It means variable. So it means it is a
variable or an accord. It is inductance or a
changing can be changed. Now, let's discuss the current
equation of an inductor. We know that the
voltage is equal to L d over d t. From
this equation, we can see d over d t is
equal to one over l v, d over d t equal one over
L V. Now we have this, we take this to the other side, v dot d t and L
becomes one over L. Now what is an ecosystem? So we have here d, We have d t, So we will integrate both sides. So it will give us
i integration of d I is equal to one over L. Integration of the
voltage with respect to time plus what is
the initial current? Similar to what? Similar to the capacitance. When we said before that the current is
equal to c d v over d t. So when we
get the voltage we integrated and plus
initial voltage. So as you can see
here, current one over L integration of the voltage. So it will be like this
plus the initial voltage. Okay? So as you can see here, one over L integration of T naught to T V as a function of t d t
plus the initial current. So we are starting our
time equal t naught. So we will get the initial
current plus the integration. Where I as a function
of t naught, is the total current negative
infinity to t naught. And of course is a current at
negative infinity equals 0. So because it is
practical and reasonable, because at negative
infinity it means at a very, all the time. So at the very old the
time means is that the inductor is not
a charge it, okay? So this current
will be equal to 0. We can assume it is 0 unless
it is given a value for it. Okay? Now we have the voltage,
we have the current. Now we need to find
the equations. We need to know the
power and energy. So power is equal to the
voltage multiplied by current. So we have current and
voltage L d over d t, l d over d t. Now we need to find the power. Power is energy, energy
stored in the inductor. Energy is equal to the integration of
the power, like this. Energy equal to integration of the power with
respect to time. So the power is
equal to L over D, D or L DI DT. So d t, We'll go with D T.
We will have L di L I D. So the integration with, will have the
integration now will it change from negative
infinity to t? It will be from, for example, 0 to any current. Will be I to any current. Or you can simply sensor
it is a function of time. We can keep it as it is. So here we have our AGI. So we will have half. L I squared as a function of t. One important note
here is that correct, we should make this 10. And make this one is fine. Okay? So it will be
l integration of AI, it will be half square. Okay? Then from what? From
current equal to 0 to any current or so we will have half
L square minus z. So it will have half
LI squared. Okay? So as you can see here, this current can be
current at anytime t. Here can be at
negative infinity. So as you can see here, this one and this one. So it is the same formula. So if we assume that the
current starting at 0, so we will have the
famous equation, which is energy store to buy
an inductor half LI squared. This is a very important rule which you will face a lot
in electrical power system. So inductor energy stored in
inductor half LI squared, energy stored in capacitor
half c v square. Now important nodes. The first one is that
the voltage across an inductor is 0 when
the current is constant. So if we have a DC current
source, DC current source, the current is equal
to a constant value, for example, five and
bear as an example. So in this case you will
find that voltage equal to L d over d t. D over d t.
What does this represents? The differentiation of
current with respect to time. So that current is
a constant value. So this differentiation
will give us 0 or no change in current. So the voltage will
be equal to 0. Voltage across an inductor will be equal to 0 plus
minus will be equal to 0. So if you think about this voltage equal to
0, what does this mean? It means it is a short circuit. That's why an inductor l saw
like a short circuit to DC. Now when this happens, when it is fully
charged it, okay? So when we put a DC supply,
it starts charging. So the current is not
constant at the beginning. Then when it reaches
a steady state, when it's fully charged, it, it will give us 0 current. Because the 0 current, so the voltage will be 0 and this will
become a short circuit. Now an important property
of the inductor is that each opposition to the change in current flowing through it, the current through an inductor cannot change instantaneously. So we cannot, the current
cannot change. Suddenly. Saw it as a circuit. So if you remember, we had this figures, but for the voltage, we said that in capacitors, the voltage cannot
change instantaneously. It can not change the
form here to hear from 0 to maximum or any value
in a very short time. Why? Because in capacitors that we have the current
equals infinity, which is not possible. In the inductor
waves are currently changes from 0 to maximum. It means that d over d t
is changing current from, for example, here, five here
and here z in as 0 time. So final minus initial
divided by the time taken 0. So d over d t and d is
a variation in current, variation of voltage
or delta I over delta t. So changing current five
minus 0 and the time is 0. So this will give us infinity. So is this possible? It means that the voltage
equal to infinity, which is not, of course
practical and does not happen. That's why that inductor limits
the variation in current. So the capacitor limits as a variation or the sudden
change in the voltage. The voltage cannot
change suddenly. In the inductor, the current
cannot change suddenly. Okay? Now finally, we need to
understand the differences between ideal and the
non either inductors. So we have an inductor
like this one. Okay? So this inductor, is
it like this or not? The ideal inductor does
not dissipate energy. It is stored in form of magnetic fields that
can be retrieved later. The inductor also takes
apart from the circuit when storing energy and
provides power to the circuit, we're returning
previously stored energy. So let's understand this. So if we have an
inductor like this, we connected it to our supply. It will start charging or storing energy until
maximum value. When we, we have
here our supply. Let's draw it like this, for example, like this. Then after fully charged, if we disconnect the circuit, it will store the
magnetic field. It has a magnetic field
has energy inside it. So when we start connecting
it to a load like this, it will start to providing current will provide
power to that. Okay? So that's why it's called
the energy storage, similar to the capacitor. Now however, the practical
case or the rail inductor, however, dissipate
electrical energy. So the idle does not exist. The practical nonideal has a significant resistance,
resistive component, so it has a small
resistance called the Tsar winding resistance in CSOs at this resistor dissipates some energy consume somehow
it is a very small resistor, but it should be added in our analysis in this
course analysis, but in general, in power
system for example. This is census. The
inductor is made of a conducting material
such as cupboard, which has some resistance. This resistance is known
as the winding resistance, and it appears in series with an inductance of the inductor. The presence of weight
mixes that makes it as an energy storage system and an energy dissipation device. Dissipation because
it's consumed energy, insulin resistance, and energy storage because
stores energy in solids, this inductance, it is usually very small
and can be ignored. However, in power system,
in electrical machines, when we both represent our
inductance and resistance, when we present our inductance, we have to add that
as a student in electrical machines
and we will have inductance series with
the winding resistance. Okay?
70. Solved Examples on Inductors: Now let's have some
examples on the inductors. So in the first example, find the voltage across the inductor and
the energy stored. If the current through
R 0.1 Henry inductor is equal to i as a function
of t equal to ten t, e to the power negative
five t and bigger. So the first step,
we need to find what we need to find
voltage and energy. We have the equation
of the voltage. We have the equation of energy. So the voltage is equal to L, d over d t. So the voltage
here is equal to L. What does the inductance? Inductance is given as 0.1
multiplied by d over d t, which is a derivative
of current. This current then g e to
the power negative five t. Okay? So take them outside here. So ten multiplied by
0.1 gives us one. So we have here one multiplied by the derivative
of this function. So we have here t, e to the power negative five t. So this is a multiplication, differentiation or
a derivative of a multiplication
of two functions. So how can we get this? If we have fuel onto know, I will just give you
the formula right now. Let's say we have two functions, X and Y function in,
function in time. Let's say for example, I would like to get the
derivative of this. So it will be derivative
of the first, the blood by the
second as it is, plus derivative of second multiplied by the
forest as it is. Okay? So this is why what
are we going to do? Derivative of t is equal to one multiplied by
the second as it is. Plus derivative of the second, derivative of e to
the power negative five is negative five. The blood by two
e negative five t multiplied by e is
a forest as it is. Okay? So as you can see here, we can take e to the negative
five t as a common factor. And we will have
one minus five t multiplied by e to the
power negative five t. Okay? So let's see it. Now. Like this. E to the power
negative five t and one minus five t As with that. Okay? So now we have
the voltage equation. Now we need to find the energy. So energy is equal
to half a square. So half L is the inductance, which is 0.1 Henry that current is a square
of this function. So we'll have like
this half LI squared, 0.15 I squared is the square of the current square of this
function is a square, means ten to the power two, which is a 100 t to the
power two is t squared. E to the negative five. T to the power two
is negative ten t. So we'll have the energy
stored is equal to five t squared e to the power
negative ten, the usual. Okay? Now let's have another example. Find the current through
A5 Henry inductor. If the voltage across it is V as a function of time
equal to 30 t squared. When t greater than 0
and when t less than 0, the voltage is equal to Z. And Y is the energy stored
at time equals five seconds, assuming that the current
is greater than 0. Okay? Now let's start. So we have this equation of the voltage and do we need
for us to that current? So the current, as
we learn is one over L integration of the voltage
of plus the initial current. So here we have, if we look back again, here, as you can see, is that the voltage, okay, at bind less than 0, the value of voltage equal to 0, so that current at
time equals 0 to z. So this part equal to 0, starting from time equal to 0. Okay? So what we do is that we
integrate from 0 to t, which means this equation. So we will have the current
equal to one over L. L is equal to five, Henry. We can say here 51 over
five multiplied by the integration of from
0 to t of this function. So routine t squared d t. So t divided by
five gives us six. D square integration
from 0 to t. For t squared is t
cubed over three. This will give us two t cubed. Let's see, As you can see here, that two t cube
is a final value. Now, the second requirement
is the energy stored. So the energy stored, as you can see, how alloy, which is five multiplied by
the square of this current. So we have two t q squared. We'll see that we have
half multiplied by five, multiplied by two. Square is equal to four
multiplied by d q. It will be t to the power six. Okay? So we will have five
multiplied by four is 20 divided by two is ten. So we will have done
T to the power six. So this is the energy
stored at any time. Now, what we need
is that we need to find the energy stored
at time equal five. Okay? So we will substitute in this
equation by T equal five. So the energy stored is equal to ten multiplied by five
to the power six. Okay? Let's see if it is
correct or not. Let's delete all of this. So we'll have ten t
to the power six. Okay? So as you can see,
60 divided by six, that gives us then
t to the power six. And we said substitute
with t equal 5 second. Okay? So we'll have ten d
five to the power six, which gives us
156.25 kilo joule. So as you can see
in this equation, this one is similar to this one. So as you can see, half
multiplied by the inductance, which is five, multiplied
by the current squared, which is four d
to the power six. So four multiplied
by 520 divided by two gives us ten
to the power six. Similar to this equation.
So what is the difference? No difference. It
is the same ID. Here we use this a power equal to voltage multiplied
by current. Voltage is 30 t square and
here to t to the power three. So they're multiplications
gives us 60 t to the power five and energies
integration of power. So we integrated this
function with respect to time and from 0 to five. So we will have
the same equation which is half L I squared. Now as you can see, this is another solution which
is how will I squared, which is also obtained
half LI squared. Now, let's have another example. Now, let's see this circuit. We have a 12-volt as a supply. We have one ohm five onto Henry, one for art and for one. Now what does that
requirement in this problem? It says under DC condition, under DC conditions,
find the current. I, find the voltage
across the capacitor. Find the current L, which is the current through
the inductance number to find the energy stored in
capacitor and inductor. So remember here is that the
condition is DC conditions. Dc conditions. What does this mean? It means what will
happen if we have a DC supply across
the capacitor. We said before that when we have a DC supply at steady state, after a long time, you'll find that
the voltage here, or this capacitor is
acting as an open circuit. And what will happen
to the inductance. So we said under DC conditions, it will become a short circuit. So we can draw our
circuit like this, an open circuit
and short circuit. So the first step, as you can see here in the
circuit sensor, we have an open circuit here. So the current passing through
here will be equal to 0. Since it is an open circuit. So the supply current
is equal to the current passing through this
inductance or I l. So I will be equal to IL. Now as you can see,
we have a supply 12 volt, one arm five. So what is the value of current? Current, as you can see here, very easy circuit.
As you can see. The world were volt
divided by one ohm plus five for y because the
current will go like this, goes through the five ohm here, then get back to the supply. From giving 12 volt divided by six ohms gives us
two and bear. Okay. So now we obtained
the current I and II. Now what we need now
is a voltage V C. Now, how can we get Vc? Okay? We can apply KVL in this loop, or by applying KVL in the slope, or apply KVL in any way. Or there is a very simple
and easy method is that the voltage between this point and this point is equal to what? The voltage between this
point and this point, right? So the voltage here
across these two points is what across the five
ohm is equal to what? Equal to phi multiplied
by the current IL, which is five multiplied
by two gives us ten volt. Ten volt is the voltage across here for all n Zach capacitor. Now, let me ask you, what is the voltage drop across the four ohm is a volt drop, is the current multiplied
by that resistance. So is there any
current passing here, knows that current
is equal to 0. What does this
mean? It means that the voltage drop across the
four ohm is equal to 0. So the voltage
difference between this point and this point
is the capacitor voltage, which is at ten volt. So as you can see,
that voltage of V C is the same as the
voltage across that five ohm, since they are parallel
to each other and no voltage drop
across the forearm. We have VC equals ten volt. Now lost the requirement
is the energy stored. Energy stored is
really, really easy. The energy stored
in the capacitor is half multiplied by
its capacitance, multiplied by the voltage
across eight squared, which is W1 square. That energy across the
inductance assembly half L, I square half L, which is the
inductance two Henry. And the current
squared is two square. So as you can see, half Cv
square and half LI squared. Okay? So this was another
example on inductors. I hope these examples
were helpful for you to understand more
about inductors.
71. Series and Parallel Inductors: Hi and welcome everyone to this lesson in our course
for electric circuits. In this lesson we
are going to discuss that series and
parallel in inductors. Okay? So if we would like
to combine inductors in series and in parallel, what are the equations
should we use? Okay? So for example, if
we have an electric circuit law exists
consisting of L1 and L2, L3 on that LAN. So these are in series. For example, I would like
to combine this into one inductor or one inductance. So how can we do this? In order to find the
equivalent inductance of a series connected
or a better connect to sit of inductors. In practical
circuits, we need to start analyzing these
circuits, same as capacitors. So let's just start in
the series inductor. So we have here voltage source or a current source
or whatever it is, there is a voltage here
between these two terminals. And do I have here
V1, V2, V3, V4. And we know that that current in a series circuit or in a, in an electric circuit
in series components, they have the same current. So by applying KVL in this loop, you will find that the
voltage source is equal to V1 plus V2 plus V3
until V and legs, this supply voltage is a voltage is equal
to v1 until v n. Now the question is, what is the value of V1? What is the value of
V2 and V3 and so on. So first, as you know
that in inductors, in inductor,
inductors, inductors, young ones that the
current as our voltage is equal to L D over DT. Okay? So the voltage across any inductance is equal to
the value of this inductance. As an example, V1, then it will be L1 multiplied by the derivative of the
current flowing through it. So as you can see here, this one is a current flowing through L1 is similar to l2, is similar to L3 until LN. Okay? So we can write this
equation like this. We find that the voltage source
V1 will be L1 d over d t, v2, y2 d over d t, and so on. So as you can see
in this equation, we have d over d t, d over d t, d over d t, and so on. So we can take d over d
t as a common factor. We have here d over d t, and it will be L1 plus
L2, L3 and so on. My exists. So we'll find that the
voltage source here or here is equal to the
equivalent inductance L, equivalent multiplied
by d over d t or the L1 d over d t plus
l2 DIY DT and so on. So finds that this is a second, third, and this representing
that furrow circuit here. From here you will find
that J L equivalent is equal to L1 plus
L2 plus L3 and so on. So the equivalent inductance in a cmos circuit is the
summation of all inductance. Okay? So like this, so
the L equivalent, their equivalent of a
series inductors is equal to L1 plus L2
plus L3 and so on. Okay? So similar to the
resistance in series. So if you have a
resistance R1, R2, R3, and so on, R1, R2, R3. Then the equivalent of this resistance will
be R1 plus R2 plus R3, similar to the
inductance L1 and L2, L3, then it will be
a loan plus l2, L3. Okay? So as you can see, it is
equivalent inductance of a series connected inductors is sum of
the individual inductance. So the inductors in
series are combined in the same way as a
resistors in series. Now what will happen if we
have parallel inductors? In parallel inductors
that we have a source I will be divided into i1, i2, i3 until I n. Okay? Now you will find that
again in parallel circuits, the supply voltage across
L1 is equal to l2, is equal to LS3 equal to n. Then the voltage here, similar to the voltage here, similar to the voltage here, and so on. Okay? So from this LFO, apply KCL at this node. Here, you will find that
the current entering, which is I equal to the summation
of all output currents. So I will be equal to I1
plus I2 plus I3 until I n. Now, Zach current, what
is the value of current? Remember that we said
that the voltage is equal to L d over d t. Okay? So the current is one over N, currents equal to one over L. Integration of the voltage
with respect to time. This plus the initial voltage
plus the initial current, of course, at time equals
0 or at any instant. So as you can see here, the current, for example, L1 will be one over L1
integration from t naught at any, at any starting the time
until finished time towards a voltage d t plus the initial current of
our y1 at time t naught. Okay? So since we started
at t naught is the initial current
should be at t node two, because it is a point at
which we start charging our inductors plus one over
ln2, the same voltage. So as you can see,
the same voltage because all of them are in parallel plus the initial
current and so on. So as you can see here, we have this term. Here. Here is a common factor. So we can take
integration T naught to T v d t as a common factor. So we'll have one over L1 plus L2 plus one over LC
until one over LN, like this, plus the
summation of all currents, all initial currents will
find that this part, this part is our L equivalent, the equivalent inductance
in this circuit. The equivalent L equivalent, one over L equivalent, one over L equivalent
is equal to one over L1 plus one over l2 plus
one over LC and so on, since they are in. Okay? So what you learn
from here is that the inductance and the
resistors are treated the same. So if we have
inductors in parallel, then we will use
the same formula of Z resistors in better. If we have inductors in series, then we are going to
use that formula of inductors in series or
resistors in series. So the equivalent inductance of a parallel inductors is the reciprocal of the sum of the reciprocals of the
individual inductances. Note that the inductors
in parallel are combined in the same way
as resistors in parallel. So we have resistors, we have inductors,
we have capacitors. These three elements. You will find that resistors are treated the same as inductors. However, the capacitors
are different from inductors and resistors. If you remember that in
series, for example, the resistors in series
summation is the summation of the resistors and inductors in series is summation
of inductors. However, capacitors in
series is the reciprocal of the sum of the reciprocals of
the individual capacitors. So that capacitor
is the one which is different for two
inductors in parallel, similar to resistors in
parallel, like this. Multiplication over
the summation. Okay? Similar as if we have two
resistors in parallel, then we say R1, R2
over R1 plus R2. Here, similar to if we
have two inductors, L1, L2 over L1 plus L2. Now you have to know that
delta star transformation. So we discussed
here and this part, series and parallel connections of capacitors and inductors. And we found is that
each of these has its own formulas or own
equations or own role. Now, what about delta star? We learned about Delta star transformations
in the resistors. And we found that we have a conversion from delta to star or star to delta to
simplify our electric circuit. What about the inductors
and capacitors? You will find that
that delta star transformation of
the capacitors, inductors and resistors, all
of them have the same rule. So if all the elements are of
the same type, for example, if we have a Delta
format of resistors, Delta format of inductors, Delta format of only capacitors. Then we can use
the same formula, delta star transformation
of resistors to get the equivalent
capacitors and inductors. So as an example
here is the rules. If you remember, we
had a delta to star. So we have here RCRA are BRC. Then we would like to convert it into a store, for example, like this, three
resistors like this. So we said before
that this resistor is our CRB over the summation. As you can see here, RC
or B over the summation. This resistor, for example, RCRA over the summation, RCRA over the
summation, and so on. Okay? Now what if we replace
this resistor with LLC? I'll be LA and we
have L1, L2, L3. Then you are going to have LA
LB equal to this equation, this equation, this equation
replace each are by L. You will find that here. In order to convert
from this to this, you are going to use
the same equation. L1, for example, L1 will be LAB LLC over L a
plus lp plus LLC. Okay? So it is the same equation,
nothing is changing. All what you have to do
is that, for example, L1 equal to RB RC, it will be, l will be LLC RA, RB RC. So it will be L loss
L being plus LC. Very easy. Okay? So it is a same equation, but we replace the variable L or R with its own requirement. For example, if we
need a capacitance, then it will be C B C, C, C a plus C, p plus C, C. Okay, If we need the capacitors. Now, a summary folds as
three basic elements. So here is a summary for all of the elements
which we have discussed, that resistance or resistors, capacitor, inductor and so on. So as you can see here,
the voltage equation, the current equation, the power, the energy
stored power or energy. Here we have the
series combination. You can see that the
series for the resistor as similar as a series
of the inductance. The power of the resistor, similar to the power
of the inductance. However, that in
series it will be as if we have a
parallel resistors. Here, as if we have
series resistors. So as you can see here,
at DC, if we apply a DC, nothing happens that
resistance will behave the same in the series
and z DC or AC. If we apply a DC voltage source, then it will become
an open circuit. If we apply a DC
source to an inductor, it will become a short circuit. Now, can we change the variable? Suddenly? Can we change the variable
of any circuits such as the voltage or current
is a resistor. It allows any change
in the circuit. However, sensors that capacitor is the current is equal to d v over d t. So it limits
the variation in voltage. So as you can see, the
voltage is limited. You can't change it
suddenly in the inductor, as you can see here, d over d t. So the inductor limits is
a variation in current. So the current cannot change, suddenly, engage
from the inductor. Okay? So in this lesson, we discussed the different series and parallel combination
of the inductance. We also learned about
the basic elements. And we learned about the delta transformation
is that it is the same in the capacitors
or inductors.
72. Solved Examples on Series and Parallel Inductors: Now let's have an example on the series and
parallel inductors. So in the first example, find the equivalent
inductance of this circuit. We are having this circuit. And for Henry, Henry 8201210. Okay, so let's just start. So it is very easy. Treat the inductance as
if it is a resistor. So as you can see,
that 20 Henry, that well we Henry
that then Henri, All of these are in series. The equivalent of 20 plus 120 plus ten gives
us a branch here. So we have here a
Ford henry like this. And do we have here seven Henry? And we have here eight hundred
eight hundred, like this. We'll find that 201210, all are in zeros, so 20 plus ten is
30, plus 12 is 42. So we have this 42, Henry. Okay? Now what does an ecosystems and accept is that we
have here seven, Henry and 42 Henry. So these two are
parallel to each other, so they're equivalent is seven multiplied by 42 divided
their summation. And the output of
this part is series with the four Henry and a ten. So we now have the
equivalent inductance. So as you can see
here, the ten Henry, that, well we Henry, Henry are all in series. So combining them gives us
our four t2 henry inductance, as you can see here. Now, this 42 Henry is parallel
with the seven Henry. This branch is parallel
to this branch. So they can combined as 0 multiplication
divided by 0 division. So we will have six
centers which are representing the
equivalent of this part. Now we will find
that our circuit will be like this. Afford Henry. See here is with Henry series with a tannery like this. Okay. So 468. So their submission
will be 1800. As you can see. This is the equivalent
of what of Z is sucked. Let's have another example. In this example, we have a circuit like this
to Henry Ford. Henry, Henry. We have a supply current. We have a current I1 going into this inductance
and the current I2 going into this inductance. And then we have voltage V1, which is the voltage
across a two Henry. And voltage V2, which is the
voltage across 12 vanity, or the voltage across the 49. Now we have here is the
equation of the current. Four multiplied
by two minus e to the power negative
20 milliampere. This is the equation
of the supply current. Now, as you can see here, are two at time equals 0, equal to negative 1 million, then we don't have the
equation of this curve. We have, what do we have the current value of
current at time equals 0. This value is equal to negative one millihenry
milliampere. Okay? Now, what do we need
in this circuit? We need here to do some ports or solver
some requirements. The first one is that we
need to find the value of current i1 at time equals 0. So we need to find the
value of this one, I1, when a time equal to 0. The second requirement
is that we need to find the value of the supply
as a function of time. We need to find the
value of V1 and this voltage as a function of time and V2 as a
function of time. Hi, E1 and E2 as a
function of time. Okay, Don't worry, it
is a very easy problem. Just do step-by-step. Okay. Let's start. We have here in this first requirement
is that we need i1 at time equals 0. So we need to find
the value of this current at time equal to z. Okay? So if you look at this
circuit, we have the current. We have I1 and I2. The current I is
equal to I1 plus I2. So this has a larger equation, which is this equation. And i2, We don't know i2, i2 at time equals 0. So what we need now is that we need the current
at time equals 0. So what it will be, it will be this current. Time equals 0 will
be equal to I1 prime equals 0 plus i2,
our time equals 0. So the current supply, current at time
equals 0 is equal to i1 at time equals 0 plus
i2 or time equal to 0. So i2 as prime equals 0, i2 prime equals 0
here, this equation, this equal to negative
1 million and Bayer and I1 is this
one is what we need. And the current
at time equals 0, we have an equation as
a function of time. So we say that at time
equals 0 is equal to four multiplied by two minus
e to the power negative ten. Now, negative ten, negative
ten multiplied bys of pi, we need a time equals 0. So we say z here. This, so it will give us two minus e to the power 0 is one. This will give us one. So we will have four
multiplied by two minus one, which is four and b. So we will have four equal
to one at time equals 0. So from this equation takes
us to the other side. So y1 prime equals 0
would be equal to five. As you can see here, equal
to five milli and bear. Now, here we have this
current time equals 0 and we have discount at time equals 0 and this
current at time equals 0. Now what does the
second requirement? Second one is that we need the voltage as a
function of time. What voltage? This one. Now, as you can see here, we have what is the voltage
inside the circuit? The voltage in inductance
equal to L d over d t. Now, you can see this is the
voltage, supply voltage. And we have this total current. We have this equation which
is representing our current. So we can use this in here. Now, this current d over d t should be
multiplied by what? By L equivalent in order
to get those supplies. So this circuit can be
like this plus minus v. And then we have here
like this L equivalent. So the current flowing here
is similar as this boy. So the voltage is equal to
L equivalent d over d t. So all what we need is the
equivalent inductance. So how can we get the
equivalent inductance? We have four Henry,
better to 1200. So it will be four multiplied
by 12 divided by 0. Summation four plus 12 is 16. So it will give us
three, I think. So too Henry series with a 300, it gives us five henrys. So this is equivalent, let's see five. Okay? So as you can see here,
L equivalent is to Henry plus the parallel
combination of 412. Okay? So now the voltage will be
L equivalent d over d t. This equivalent, which is
five Henry d over d t. Now, d over d t is the
derivative of the current with respect to time. So if you look at the current, we can write it like this. Four multiplied by this bracket. So we can say
format the blood by 24 multiplied by is exponential. So we'll have eight minus
four e negative then t. So d over d t, the derivative of this part, the derivative of a constant
is equal to 0 minus four. Multiplied by derivative of the exponential is what is the
derivative of this part? The derivative of
negative ten is negative ten multiplied boys exponential
itself negative ten K. So we'll find that
we have five Henry. Then the four,
then negative one, negative one, and
the negative ten, negative ten e to the
power negative ten. So multiple location
of this gives us 200 e to the power negative ten. So we have this
equation which will be presenting our supply. Okay. Now what does
an extra step? Now is an extra requirement
is that we need v0, v1 as a function of t. So V1 is really, really easy how V1 is
equal to L1 d over d t. Okay? So why do I? Because the current I is
the one which is flowing through this
inductance will be L1, which is two Henry multiplied by d over d t is the derivative of current
with respect to time, which we just obtained
here, this part. So it will be like this. V1 l, which is two
n-ary d over d t, which is this spot. Okay? Okay. So it will have the voltage V1 as equal to e to the
power negative ten. Now, the next requirement
is that we need V2, which is the voltage across
the inductance here. This inductance, or
this inductance. Okay? So this voltage can
be obtained easily. How, as you can see from KVL, you will find that the
supply voltage equal to V1 plus V2. So we need V2. V2 is equal to the voltage
minus V1 my exists. So we subtract this
from this gives us V to easily, as you can see here. Okay? So now we have the
three equations. We have V1, V2, and the supply voltage
and the current equation. Now what is the remaining part? The remaining part is
that we need to find I1 and I2 as a function of time. The current RE1. How can we get it under i2? It's really, really easy. So first, remember that
the current inside an inductance is
equal to one over L, integration of the voltage
with respect to time plus the initial
current like this. So I1 is equal to I1 is
equal to one over L, which is one divided by four. Integration from 0 to anytime t of the
voltage across eight, which is V2, plus the initial
current at time equals 0. If you'll remember, we have v2, which we just obtained
in the previous slide. And we have all u1
at time equals 0, which was obtained also in
the forest requirement. Why integrating and adding, we will have this equation. Now, i2, how can we get I2? You have two options. I2 is equal to I minus I1 or
I2 is equal to one over L. Integration of the
voltage plus the initial current, like this, i2 as a function
of t one over L, which is 12 Henry integration of the voltage across
it, which is V2. V2 is the voltage across
that wall where Henry, Ford henry plus the
initial current. So this will give us
the final equation, negative e to the power
negative ten t. Now, how can we check if
you obtained I1 and I2 using the formula of one over L integration
of the voltage, then you can do a
very small check to make sure that you are
solving it correctly. You have to find that
if you add i1 and i2, it will give us the
initial current i. So as you can see here, eight minus three e to the
power negative ten. Multiply it by plus negative
e to the power negative ten. We have eight minus negative three and negative
one gives us negative four. So 0 summation is eight minus four e to the power negative t.
So as you can see, eight minus four
e to the bonding. So this is correct,
as you can see. Okay? So this was also the example on zeros and the
parallel inductors.
73. Application Integrator: Hi everyone. In this lesson
we are going to discuss the first application on the
inductors and capacitors, or to be more specific
Zach capacitors. Okay, so the first
application is integrated. Now we have to know that the important op amp circuit that use energy storage element, including integrators,
differentiators. This op-amp circuits often involve resistors
and capacitors, sometimes inductors, which makes it bulkier
and expensive. Now, the first type which
we are going to discuss, or the first application
is called the indegree, is an op M whose output is proportional to the integral
of the input signal. So simply the output is considered as the
integration of the input. That's why it is
called integrate. Remember that
non-inverting amplifier. We had in the non-inverting
air resistance, resistance, resistance
as a supply. And the resistance
adds a feedback which connects between the
output and the input. Okay, So we have the
inverting amplifier. Now, in order to convert this circuit into an
integrator, very easy. How can we do this? Simply replaces feedback
with a capacitor. That's all what you need to do. So we have the same circuit
of the inverting amplifier, but we added a capacitor and instead of the
feedback resistor. Now let's understand
the relation between the output
and the input. We have here two currents, i, r, which is a current flowing
through the resistor R. And I see which is the current flowing
through the capacitor. Now, let's understand first
each voltage of point. So here we have this point
which is the supply voltage. This point is called the V, which is equal to 0 volt. Okay, if you'll remember
from the analysis of the ideal amplifier
or the ideal opamp. Ok. Now this point
is equal to V out. So the current ir is
equal to IC. Why? Because we said before that the current through an op
amp is equal to 0, if you remember from
the previous lessons. So now i are equal to IC, so i r equal to IC. Okay? What is the value of IR? Ir is the current flowing
through the resistor, which you can be the
difference between this voltage and this voltage
divided by that resistor. So it will be V minus
0 divided by R. Now, what is the
current of a capacitor? Now remember that
the capacitor is a current is equal to C.
Current is equal to C, which is a capacitance, d v over d t. Okay? And what is dv or what is the voltage differentiated
as a voltage, if you remember, is the
voltage across the capacitor. And since there's a current
is entering our capacitor, so the voltage is plus
minus, like this. So the current
entering plus, minus. Okay? So here, what is the voltage
across the capacitor is a voltage between this
point minus this point. So this point is 0 minus
this point which is V out. So finds from this equation
is that V input are equal to c negative c DVR. So as you can see, you
can take this one to the other side also
it will be V m, both of our RC and take the
negative to the other side. So we'll have negative
V input over RC. And integrated gives us the wow. So let's see again
this equations. So at node a is a
current i article. I see the current
IR equals V over R. And I sequence
negative c d v over d t y negative scene because you remember
that the current, again, if you didn't take notes, dV over V T. And that potential difference of the voltage difference across the capacitor is this
voltage minus this voltage. So it will be 0 minus V output. So this difference
is negative V out. So taking this outside
gives us negative c, d v out over d t. Okay? Now equate this to Vm over r equal negative c d v over
d t. So from here you, by integrating both sides, you will find that the voltage V out is equal to
negative one over RC integration of the v m d
t plus the initial voltage. Plus the initial voltage, assuming that the initial
voltage is equal to 0. So we will have this
final relationship. Okay? Now, as you can see
that the output, what's the relation
between output and input? As you can see, the
relation between them is an integration. It is called an integrator
because integrates the input. Now, let's have an example. So we have this op amp. We have V1 equals ten
cosine two t millivolt, and V2 equals 0.5 millivolt. Finds the V output in
the op-amp circuit, assuming that the voltage across the capacitor is initially 0. So what does this circuit, as you can see, this
is summing amplifier. Summing amplifier. We have this input and this
input going to the negative. So it is summing inverting
amplifier, okay? But you will find that
there is a small change. And instead of having here a resistance, a
feedback resistance. So we have our capacity. So we can say is
that this circuit is a summing integrator because
we have more than one input. So we will have the
circuit like this. The V output in an integrated in one integrator
is negative one over RC integration
of the volt V input. So we have two inputs. So we will have
what we will have. The output will be, the output
will be the first input, negative one over R1 C, because we have one
capacitance integration of the first input, V1 plus negative one over R2, which is the effect of the second voltage
integration of v2 d t. So it is a summing integrator. Okay? So as you can see here, summing integrator and v would equal to
negative one over R1C. Okay? So this V input V1, then we are talking about R1, V2, V2, talking about R2. As you can see here. So when we substitute, we have negative one over R1C. R1 is three mega multiplied by capacitance
two microfarads. Integration of the V1. V1 is equal to ten
cosine two t dot d t minus integration of
the second 11 over r2. R2 is 100 kilo
ohms multiplied by the capacitance
integration from 0 to t of the second voltage. Now, here, as you can see, the integration of 0.5 t
is the integration of t. Integration of t is
t squared over two, t squared over two. Now what is the integration
of cosine two t. Integration of cosine is sine. Okay? So integration
of cosine is sine two t divided by the derivative of the derivative of two
t is equal to two. So we'll have sine
two t divided by two. And that then becomes here. And one over six or
presenting this part. So by doing this, you will find that the
equation will be like this. After simplification. This was assemble example
on the integrator. So you can understand
how it works.
74. Application Differentiator: Now let's discuss another application
which is differentiate. The differentiator is
opposite to the integrator, as it seems from its name, it is an open circuit. The output is proportional to the rate of change
of the input signal. Or to be more specific
or much easier, the output is a
derivative of the input. So if you look at the
inverting amplifier, as we said before, now in order to get
the integrator, we replaced the feedback
resistor with a capacitor. Now, in order to convert this, In this, to convert this inverting amplifier
into a differentiator. We replace R1 with a
capacitor like this. And every sink as it is. As you can see, it
is really easy. So as you remember from
the previous lesson, the current I, R, and IC. Now this point is
equal to 0 volt, and this point is V out. This point is V. Okay? So the current IR is equal to the difference between these two
voltage divided by R. So the current is in this direction going
out from this node. So it will be 0 minus
V out divided by R. And I capacitor
is equal to what, c d v over d t. So the voltage
is a difference between, since it is input here, so it will be positive
and negative. So the difference between
the voltage across it, which is V m minus
0, minus 0 is Vm. So ir is equal to IC as before. So negative V out
over r equal to C dv in both over d t.
So from this equation, you will find that
V out is equal to RC negative RC dv over d t. Okay? So let's see. So again, by applying
KCL IC equal to IR values of IR and IC
as it was also obtained. Equate these two equations. And then we will have
V Albert equal to negative RC divi
in both over d t. Very easy equation. Now, let's have an
example on this. So we have, in this example, we have this input signal,
triangular signal. A triangular signal,
as you can see in the form of a drink. So sketches the output
voltage for this circuit. This one. Given that the input
voltage is like this. And take v out equals
0 at time equal 0. Now, if we look at this circuit, we have V M, what we have. If we have a resistance here
and the resistance here, then we will have an
inverting amplifier. But we replace the input
resistor with 0.2 microphone. So now it becomes
at differentiate. So we will start, what
is the first step? We need to get V out. We need to get V input
as a function of time. So we need to represent the signal in the
form of times sigma, or the inner form
of an equation. So the first step is that here you can find we
have a straight line here. From here to here. This is our first straight line. Then we have a declining
straight line from here to here. So this one is repeated here. So if we obtain this equation, it will be enough, and this one will
be similar to it. So let's start the
first one here. Here we have 0, and
here we have four. Time equals 0, time equals two. Now again, y equal to mx
plus c. Y axis is our v, m equal to m is the
slope of this line. The slope of any
line is equal to Y2 minus Y1 over X2 minus X1. Our final y is for
the initial y is 0, so it will be four
minus 0 divided by x, x final minus x initial to minus 0 multiplied by the x axis, which is our time plus c. So
we will have four divided by two gives us V
equal to two t plus C. Okay? Now we need to find
the value of a scene. So when the time equal to 0 and this one is equal
to 0 at time equals 0. Every input is equal to 0, so c will be equal to 0. So the first equation
is V input equal to T, as you can see here,
equal to two t. Now, what does that point here? I will tell you right now. Okay? So let's return of x like
this here using this happen. Okay? Okay, So now the
second equation, this line, this line. Now as you can see, we have here as this one, which is y equal
to m x plus c y. Here, y is equal to V m. And
m is the slope of the line, final y minus final y, Y2 minus Y1, which is, what is the value of y here? 0. What is the value for year four? So it will be 0 minus four. What does the finite x four? What is the initial x? Two. And x is equal to the time
plus c, which is constant. So V input is equal to negative four
divided by two gives us negative two t plus c. Now at time equal to 0 to find singing
at time equal to one, this one becomes a to the
value of the V input is four. The input is four. Negative two multiplied by
two gives us negative four, takes us to the other side, will be four plus four. So C is equal to eight. So we can say V input is
equal to eight minus two t. Okay? Now, here, this is the equation. Okay, let's see here, eight minus two t. Now, someone will ask me, here, you said that eight
minus two t, But here, eight minus two t. The first equation is 2 thousand
t. And instead of two t, What is the difference? The difference is that
here, when I substitute, I substitute with the
time as it is to mainly, for example, when t
equals two milli. So I substitute with t equal
two to get the voltage. But here if you would
like to substitute with 2 second in
the second unit. So here, when I would
like to say at time equal to then I2 multiplied by ten
to the power negative three. Okay? So this multiplied by
three zeros gives us the value for the difference
between these two. They are same, similar
to each other. The difference is
that you substitute with two or substitute with ten to the power negative
City as you would like. Okay? So now we have the equation, we have our scene. So what we are going to do
then we say negative RC, derivative of the voltage. So the derivative
of the voltage here multiplied by negative
RC, like this. Okay? So the voltage V output RC, RC is equal to ten to
the power negative three multiplied by the
derivative of the voltage. So the derivative is derivative
of 2 thousand t gives us 2 thousand derivative of
this 2000s negative two. So we'll have negative
two volt and two volt. Why? Because this derivative
will give us 2 thousand. And this one gives us
negative two sounds. Okay? Then multiply this
by negative RC, RC. So multiply this by negative ten to the
power negative three. So negative here gives
us negative two. Negative three, texas
is three zeros and this one becomes
two, as you can see. So when we say this is a
V output in this range, which is similar to this range. So we will draw it like this. From 0 to two millisecond
from here to here. The value is negative two. From two to four. From two to four is two
volt, as you can see here. And the signal will be repeated. Negative two, then two volt. Okay? Now, something which
is really important, which one is correct? Should we use to t as our exhaust obtained
or 2 thousand D? The correct one is 2 thousand t. Not to t. Why? Because as you can see, the time is substituted
with its own SI unit, which is two multiplied by ten to the power negative three. Okay? So I showed when I obtain
the Zan in the beginning, when I use the
slope of the line. For example, this one, it will be Y2 minus Y1. So it will be four minus
0 divided by x2 minus x1. X2 should be like this. Two multiplied by
ten to the power negative three minus x one. So we will have two sounds, and t and z. Okay? So this one is more
correct than what I did. Okay? So difference is that
when I substituted as if each of these multiplied by ten to
the power negative c, ten to the power negative
three, and so on. Okay? So this was an example
on that differentiate.
75. Introduction to First Order Circuits: Hi everyone. In this section or this lessons, we are going to start discussing that first
order circuits. First, we need to
understand what is the meaning of first
order circuits. So as you know, is that in the previous sections
of this course, of that electric
circuits course, we discussed three
passive elements. We discuss resistors,
capacitors, and inductors. Okay? So each of these elements
where discussed alone, we have resistance only, we have capacitance
and inductance. Now in order to form is
a first-order circuits. We are going to combine two
or three passive elements. Okay? So usually we have
circuits which have, does not have only one element, but have two or more elements. Circuits which have
two or more elements. The circuit swatch you have
to passive elements such as a resistance and capacitance
or resistance and inductor. They are called the
first order circuits. However, the circuits
which have resistors, capacitors, and inductors,
all with each other. It's called
second-order circuits. So here in this section
we will discuss the tool symbol circuits of
the first-order circuits, which is, which is a
circuit comprising a resistor and capacitor and a circuit comprising a
resistor and inductor. So we have our C and R L. So
we have RC and RL circuits. These circuits which we are going to discuss
in this section. So what's the
difference between zap purely resistive
circuits and RC and RL. So remember that when we apply Kirchhoff's law to the
purely resistive circuits, we have algebraic equations. Purely algebraic
equations, which does not have any differential equations. However, when we start
discussing the RC and RL, we will have
differential equations, which means we have derivative, which is much more difficult to solve algebraic equations. This one which is differences RC and RL from the pure
resistive circuits. So the differential equations
resulting from analyzing the RC and RL are
of the first order. Okay, so we have RC and RL, which are first-order circuits. Why is it called
the first-order? Because they form equations, differential equations
of the forest order. Okay? Hence, the circuits are known as the first-order circuits. So you have to know
that there are two ways to excite the circuits. Number one is by initial conditions of the storage elements
in the circuit. What does this even mean? It means is that
our capacitor or our inductor is
already a charge it, or it has stored
energy inside it. So the first one,
which is a source, a free or the natural
response of the circuit. It means is that our capacitor or our inductor was
initially charged. Then we are going to
connect it to a resistor. We will see its response. This response is the natural
response of the circuit, or it's called the
source of free circuit. Okay? So the energy causes the
current to flow inside the circuit and gradually
dissipated inside xy resistant. So what does this
mean as an example? As an example, let's
say for example, we have a capacitor. This capacitor was initially
children by using a supply. So, So it's all
did, for example, with a voltage V naught,
completely charged. And we have another
resistor like this. Now, when this capacitor
is completely charged, it suddenly we
removed our supply. We don't have any supply now. Our combustor is a charge it. Then we start connecting it
to a resistor like this. And the current will flow
through the circuit to be dissipated or the cause
power dissipation in the resistor here. The current flow here
inside the resistor. This response is
known as a source of free circuit because
we don't have any supply connected now. And at same time calls and
natural response of a second. Okay? So this type is called
the source free circuit. Okay? Although the source
of free circuits are free of independent sources, we don't have any supply, but they can have
dependent sources. The second way of writing is a first-order circuit is by
using independent sources, connecting it to us. Apply the two types of the
first-order circuits as a tool to ways of
exploiting them adds up for four possible situations
in our course, what are we going to do? We are going to discuss first, we have two circuits. We have RL, we have RC. Okay? This are the two types. These two types have two
types of excitation. First, they all have, we have salsa free circuits
which are initially stored, have initially stored energy. And the other way is by connecting it to
independent source, such as a voltage source
or a current source, similar to the RC, which have two cases. So first, our course will
be divided like this. First we will discuss as a
source of free circuits. We will discuss RL and
RC source free circuits. Then in the second part, we are going to
discuss the excitation using the independent sources. And this response is known as the forced response
of the circuit. When we connect it to supply. The forces response of RL and RC will be discussed
the two, okay, so we will discuss as a
source of free or LRC than the force at the response of RL and RC in another section. Okay?
76. Source Free RC Circuit: So now let's discuss
forest as a source, a free RC circuit. First, you will
have to note that the source of free RC circuit occurs when a DC source
is suddenly disconnected. So you have here a capacitor and you have here a resistor. Ok? So let's say we have
here our supply. Okay? We have a voltage source, DC voltage source V, for example, plus minus. And this will be connected to a capacitor parallel
to a resistor. So what will happen here is that this voltage source will start providing electrical
power to the resistor. And at the same time, we'll start a charging the capacitor. After a very, very long time. The voltage across the
capacitor will be equal to the supply after
a very long time. Okay? Now, what will happen or what we would like to discuss here. We would like to discuss this capacitor is now having
an initial voltage V naught. It is completely charged
with a value called AV node initial value. Now, let's say for example, we disconnected the supply legs as deleted this
from the circuit. What will happen to
the circuit when we suddenly disconnected
this supply? Okay? So our capacitor is now, is completely a charge it with a value
called the V naught, and now connected with
a resistor in series, which is R, and supply
is completely removed. Now, we would like to see
the response of the circuit. This response is called
the natural response of this circuit is
known as a source free RC suck. Source
of free means. It does not have any source. Now finds that the energy, since this one is initially, have initially stored
power, electrical power, it will start to providing electrical power
to the resistor. Is the energy stored
inside a capacitor started being dissipated
in the resistor. So we would like to
analyze the circuit. Okay? So after time equal to 0, time equals to 0 is the time at which we disconnected
the supply. So the initial voltage
of the capacitor V 0, which is the initial
voltage equal to V naught a certain value. It can be given or according
to the analysis of our electric circuit, V naught. So we have here our capacitor with a voltage V naught, okay? Now, after removing the supply, we have capacitors in
series with a resistor. Now that when was our
supply was connected, we have a current IC was
going to Zach capacitor and the current IR from KCL at
this node, for example, KCL here, you will find
that IC plus IR equal to 0, I c plus I r equal to 0. Okay? So this summation is
dependent on what? Dependent on the
direction we proposed, we say is that for example, we assume that the current IC is the current entering
is a capacitor, and the current IR is the
current entering the resistor. So Forest, since our voltage is initially chose was v naught. It means that the energy stored, corresponding energy
stored is half CV square, half c v naught square. This is the initial amount of energy stored
in our capacitor. Remembers this value. Because we are going to prove something from our circuit
here at this node, you will find that all
C plus IR equal to 0. Now what is the value of IC
and what's the value of IR? So you can see that
the voltage here, we have a voltage V
between this torn, between this point and
the ground and this point and the ground voltage. So the current IC or only
knows that the current of the capacitor is equal
to c d v over d t. The current across a
resistor is equal to the voltage divided by
resistance like this. So c d v over d t plus
v over r equal to 0. Okay? Now from this equation, we can rewrite like this. We can say d v over d t
equals v over c all over RC. Okay, let's delete this. So now we have dv over v t
plus v over RC equal to 0. Now I would like to, so you can get this
to the other side so we can write it d v over d t equal to
negative v over RC. Okay? Then we can take d t here, and we take the
voltage back here. So we can have dv divided
by the voltage equal to negative one over RC
d t, like this. Okay? So let's. The neatest things
sport, sports. So we have this
equation like this. Okay? So we have dv over v
equals negative one over RC d t. So that's why our circuit is called the
first order circuits. Why? Because as you can see, we have a voltage equation or an equation with
the first order, first order
differential equation. Okay? So what are the next steps? And step is integration
of these two sides. So the integration of this and the integration of
this will give us what? We have dv over v, The integration of one over x. Integration of one over x from differential equations or
derivative is equal to, or integration of one
over x is equal to what? Equal to ln x. So the one over V
gives us Len V. And the integration of
negative one over RC d t gives us
negative t over RC. Now since we have integration, it means a plus a
certain constant. So we say integration
plus C. Okay? Our constant here we said
it is Lenny assumption. Okay, Lenny. Now, what does the next step? The next step is that we need
to find the value of Len. How can we get Lenny
or the value of a, which is our constant? So assembly, this can be
done by initial conditions. So we know that at
time equal to 0, the voltage will be
equal to V naught. So we can say at
time equal to 0, the voltage will be V naught. Okay? So we will have Len V naught will be equal to
negative 0 over RC gives us 0 plus a. So what
does this mean? It means that the V
naught equal to a. So we can rewrite our equation as this
part becomes v naught. So as you can see,
like this, okay, from the initial
conditions, as you can see, a will be equal to V naught. So we can say Len v equal to negative t over RC
plus ln v naught. Okay? So now, as you can see here, we can take, we can
rewrite this equation. We can take the land to
the other side here. It can be written
like this, Lynn v. Then v minus a equal
to negative t over RC. Now Len something minus
something gives us len. The first one, v minus
ln means divided by E equal to
negative t over RC. In order to eliminate the, remove this land, we have to take the
exponential of two sides. So we can say e to the
power ln V over a, and e to the power
negative t over RC. So we will have V over F equal to e to the power
negative t over r c. Okay? So the voltage will
be equal to a, e to the power negative t over
r c, As you can see here. Okay? Okay. So now we have a equal
to V naught, as we said. So our voltage will
be equal to V naught e power negative t
over RC, like this. So this is our response. Okay? So as you can see, it is
a decaying exponential, e to the power negative. It means our voltage is decaying
from V naught like this. So this shows that
the voltage or response of the RC circuit is an exponential decay
of the initial voltage. Since there's a response is due to the initial energy stored, ends up physical
characteristics of the circuit. And the not due to some
external voltage or cancers. Hence, it's called the natural
response of the circuit. Now, there is another element which is called
the time constant. You will always
hear this element. You have to know about it. It is denoted by this
symbol, like this. What does this symbol mean? Or how it is pronounced or And written it's called Tau. Okay, So this Greek letter
is pronounced it as tau. Tau. So tau here is what does tao would represent, representing
time constant. What does also time constant? You can see here
we have v naught e to the power
negative t over RC. So this R c can be replaced
by something called tau. So tau in our circuit here in the R C circuit is equal to R
C called the time constant. Okay? Now we have V equal to V naught e to the power
negative t over Tau. Now, okay? So tau is what does
this even represent? So when time is equal to Tau, what will happen when
t is equal to Tau? You can see Tau
will go with style. So we will have v naught e
to the power negative one. Okay? So the voltage will be
equal to 0.368 V naught. You can see e to spawn
negative one gives us points as 368 and V naught. What does this mean? It means that wins time. Once a time breaches
that time constant tau, which is our c, what will happen in this case, the voltage is dropped from
V naught, 2.368 V naught. Okay, So as you can see here, the time constant is
the time required for the response to the k to
a factor of one over e, which is e to the
power negative one, or 6.8 per cent of
its initial value. So as you can see here, this is our response. This equation can be
represented like this, starting from V naught, then exponential decaying, decaying, exponential like this. Theoretically at infinity, at time equals infinity,
we will reach 0. However, we don't
reach 0 at infinity, which is 0 at
approximately five Tau. Okay? So if you chose, if you look at the circuit
at time equals tau, your funds as the voltage
value will be 0.368 V naught. Okay? Now, the voltage response, as you can see here. So V naught equal to e
negative tau over RC, and we know that RC is tau here, so it will be V naught
e to the one gives us 0.368 V naught z equation, which I just wrote or I just obtained in
the previous slide, that we call RC
negative t over Tau. Now, the circuit
was a small time constant or small tau gives a faster response to
reach the steady-state. In our case here z, which means we have quick
dissipation of energy stores. Energy stored. However, was a circuit
with a large time constant means that we
have a slow response. It takes longer time
to reach steady-state. Okay? So whatever,
whatever is the time constant is small or large, circuit will reach a steady
state in five time constant. Okay? So if you look
at the circuit here, V naught, V as a
function of time over v naught equals e to the
power negative t over tau. That issue between the
voltage with respect to the initial voltage
at time equals time, we reach that 68%, which is six 0.8% of the initial voltage we
drop at time equal tau. We dropped 2.876 per cent. What about to tau? If we substitute with two tau, you will reach a certain
0.5 per cent, three tau, four per cent for tau, one per cent, five Tau, approximately almost 0 or 0. Here, one to 0.6 per
cent of the voltage. This value is
approximately equal to z. We assume it is 0. Okay? So usually, when does
the circuit reaches a steady state after five
time constant, okay? Now, this curve shows you the different values of del or the effect of
different values of tau, a tau, or time constant. When it is large,
what does this mean? Larger time means slow response. It takes a longer time
to reach steady state. As you can see, Tau equal to, you can see long time
to reach low value. However, tau equal one faster
response that we call 0.5. Very fast response. The larger RC. So the larger or see, the longer the response. Okay? So by controlling the
resistance and capacitance, we can control the
time constant, which means that we can control the response of our circuit. Finally, the power
dissipated in the resistor. What is the power in a resistor? A resistor, what is the power? Power is equal to the voltage multiplied
by its occurrence. So we have the voltage, which is a voltage
across that resistor, and the current flowing
through the resistor. So assembly, what is the
value of the voltage? The voltage is the
voltage obtained, which is V naught e to
the negative t over tau, multiplied by what does
the value of current? Current is equal to the
voltage divided by what? Divided by the resistance. So we have V naught e
to the power negative t over tau divided
by the resistance. So we will have or here, V naught multiplied by V
naught gives us V naught squared e to the power
negative t over tau. E to the power negative t
over tau is 0 summation. So it will be e to the power
negative two t over Tau, as you can see. Okay? Okay, so what does that energy, so the energy stored or
not the energy stored, but the energy absorbed
by the resistor at any time t is equal
to what assembly. We know that energy
is equal to what? Energy is equal to the
power multiplied by time. Is that what we learn it? However, since our power
is a function in time, power is a function in time
e negative two t over Tau. So it means that we
cannot use this relation. We have to use integration. So we integrate the power
from 0 to any time t. We have V naught squared over r e to the power
negative t over tau d t. The integration
of this function, this part is a constant,
so it will be like this. V naught squared over r, e to the power
negative two over tau. It will be, as it is. Integration of exponential is integration of e to the power, let's say for example, e to the power negative a. As an example, the
integration of this part will be equal to e to the power negative as this divided by the
derivative of this part. So let's say for
example, this is an x with respect to
integration with respect to dx. Okay? So e to the power negative
x integration of this is, this part is
constant negative a. So as you can see, we are integrating with
respect to two times. So we have negative
two t over Tau, so it will be negative
two over tau. You can see here. So when we divide like this
e to the negative two t over tau divided by
negative two over tau. You will find that divide, divide means that this
tau will be above here. So it will be exponential multiplied by tau
divided by negative two. So you can see negative two
and tau was becoming here. Then we substitute with
our limits from 0 to t. So we wouldn't have finally
half CV square one minus e to the power negative two
over tau and tau equal RC. Okay? So what we can learn here is this representing energy
absorbed with boy resistor. Okay? So if you look at this
equation at time equal to z, at time equal to 0, you can see it will be
equal to half CV square. One minus at time equals 0. Exponential of 0 gives us one. It means this part will be 0, so the energy absorbed by the
resistor at time equals 0, equal to 0. No energy absorbent. Okay? What about after a very, very long time, Let's
say infinity, okay? After a very long time, e to the power
negative infinity, it will be equal to 0. So this exponential after a very long time, it
will be equal to 0. So we will have half c v naught squared
multiplied by one. So that at time equals infinity, or to be more
specific, five Tau, our energy of insides
that resistor will be equal to 1.5 c v naught square. So if you remember
that this power is what is the
initial stored power inside the initial stored energy inside the capacitor.
So what does this mean? It means after a very long time, all the energy stored inside the capacitor
will go to the resistor. So as you can see, as
time goes to infinity, you will find that half
CV square is the energy stored in energy of
solo advisor resistor, which is similar to the
energy inside the capacitor, or initially energy stored in the capacitor
at time equals 0. So it means that the energy that was initially stored inside a capacitor is eventually
dissipated in the resistor. Okay? So this, in our lesson, in this lesson we discussed is a source of free RC circuit. Now we would like to discuss
some examples on this.
77. Example 1 on Source Free RC Circuit: Now let's have an example on the source of free RC circuit. So as you can see
in this circuit, we have five ohm, better two-point one far-out. We have eight ohms
series was at 12. So this capacitor was initially
charged it with 15 volt. You can see that the
voltage of the capacitor, Vc at time equal to 0
is equal to 15 volts. So this is our initial voltage. Okay? Now what we would like to
get is that we would like to get the voltage
of the capacitor, voltage across this
resistor and the current i x for t
greater than 0. Okay? So anyway is a forest
important thing to get is Vc. From VC you can get vx and
from vx you can get IX. Okay, let's start first. As we remember from
the previous lesson, we said that the
voltage is equal to V naught e to the power
negative t over Tau. So we have here, our voltage is this one. V naught is equal
to what, 15 volts. So this value is 15 volt multiplied by e to the
power negative t over tau. So what is the
remaining thing here? The remaining part is
that we need to find tau. So tau is equal to
R multiplied by C. The capacitance is
equal to 0.14 out. Okay? What about the resistor? What is the value
of the resistance? Resistance of that all
the time constant. So what is the resistance
that resistance is that seven and resistance where exactly the seven resistance adds the terminals
of the capacitor. What is the resistance? This capacitor is C. Okay? So if you look here carefully, is that this capacitor, this is a two terminals
of the capacitor. It sees a five ohm battery
to eight plus 12 ohm. So the equivalent capacitance, or the equivalent
resistance of this circuit is a capacitor with are
equivalent, they are equivalent. Resistance of this circuit
is five own butter to 21. Okay, is this is all equivalent? So as you can see here, we have eight plus 12, gives us 20 butter
tools or five ohm. So the R equivalent
is a product. The word ball is
awesome mission. So 20 multiplied by five
divided by the summation, which is 20 plus y
gives us four or so. This is our equivalent. Now, we have the
capacitor's capacitance. We have the initial voltage
and we have our equivalent. So from here we can get tau, which is a time constant. Our equivalent, which is a for all multiplied by
the capacitance, which is 0.14 out
equal to 0.4 seconds. Now we can write our equations. The voltage is equal to 15, which is the initial
voltage, or time equals 0. E to the power
negative t over tau. Tau is 0.4 seconds. Okay? So this equation, one over open for can be equal
to negative 2.5 t. This equation is
similar to this one. So what does n steps? So we have here the voltage
is the first requirement. Now we need the current
and the voltage. Now, as you can see,
VC voltage of V C, which is this value,
is equal to what? Is equal to the voltage
across eight and the 12 volt. They are parallel to each other. Okay? So the voltage here
as the voltage across 812 ohm is Vc, okay? So what does the voltage
across V x, okay? There's a voltage
across V x is equal to the total voltage,
which is V c, multiplied by its resistance, divided by the total resistance. Okay? Why is this? Because we have two
parallel elements. So the voltage here
and here is Vc. And by using voltage division, we can get the voltage here as dwelled with the
water boys or summation. As you can see here, 12 divided by 120 plus
eight multiplied by the voltage 15 multiply by e
to the power negative 2.5. So this will give us nine multiplied by e to the
power negative 2.5. So this is a voltage Vx, which is a voltage
across 12 volt. Now as a final
requirement is that we need to find the current x. So how can we get IX? Ix is simply equal to
any current. Okay? Any current is equal to what? Are any current is equal to the voltage divided
by resistance. Okay? So I need IX, okay? We can say V x, which is this voltage across that voltage divided
by resistance. So the voltage vx
divided by 12 volt gives us the current IX, okay? Or you can do it
as another method. We can say V c divided by eight plus 12 gives us
also the same current. So as you can see, I x
equals vx over 12 gives us 0.75820 negative 2.5 T and bear. So as you can see that
the current is also decaying across the resistor. Ok. So in this lesson, we had an example on the
source of free RC circuit.
78. Example 2 on Source Free RC Circuit: Now let's have another example. So the switch in this
circuit has been closed for a very long time. And it is opened at
time equal to 0. Find the voltage V as
a function of time, which is the voltage
across the capacitor and the initial energy
stored in the capacitor. Okay? So how can we do this? Pretty, pretty easy. Okay? So as you can see,
at time equals 0 is a switch before time equal to 0. The switch was closed
for a very long time. And our time equals 0. Boom, we start
opening our circuit. Okay? So here as you can see, Before switching our circuit, our circuit was like this. Close the circuit. So we have 20 volt, three arm, nine ohm, one ohm ends
at 20, mainly far-out. Now when the switch is
closed for a very long time, it means is that the capacitor
will reach a steady-state. So what is the steady-state of the capacitor? If you remember? The steady-state of the
capacitor is an open circuit and the steady-state of the
inductor is a short circuit. So as you can see before,
time less than 0. Before switching, you
will find that here, supply three arm mine own, as you can see here, one arm and the
open circuit here. Okay? Why open circuit? Because at steady-state, when Zack capacitor
is fully charged, it becomes an open circuit. So now what I need, I need to v is equal to V naught e to the power
negative t over tau. So this is our equation for
the voltage of the capacitor. So the first, we need
the initial voltage, which is voltage right
before switching. Which is the voltage
at steady state when the switch was closed
for a very long time. So the voltage here, when this capacitor
becomes open circuit, how can we get this? It is a voltage across as
our mind on, why is this, because this resistor does
not have any current, no current deposits here. So the voltage between
here and here, similar to the voltage
across the capacitor. Because this open circuit means is that there
is no current here, which means there
is no voltage drop. The voltage across
Vc at time equals 0, or the initial voltage is the voltage across the line arm. Now we have here three ohm. Do we have here nine on? So how can we get this
using voltage division? So the voltage across the
nine Ohm is equal to 20 volt, multiply it by its
resistance over the total resistance from
the voltage division. So as you can see, the voltage Vc as a
function of time, we can make it more
accurately at time equals 0, equal to nine over nine
plus three multiplied bys at 20 volt gives us 15 volt, which is t less than
0 before switching. So this voltage is the
initial voltage when the switch was closed
for a very long time. And the capacitor
reaches a steady state. It's a voltage becomes 15 volt. So we have now with
that initial voltage, now what do we need? We need Tao, which is our scene. So the capacitance
is a 20 millifarads. What about the resistor? So we need to draw
our circuit after switching or after
openings as well. So here you will find that the voltage 15 volt is the
voltage across the capacitor. As we know that the
capacitor is used as limits, is that variation in voltage. Remember that the voltage across the capacitor is equal to c, or the current across that
capacitor is equal to c, d v over d t. So the current across
the capacitor cannot, as a voltage across
the capacitor cannot change instantaneously. Or d v over d t
will be very high, means that the current
will be very high. Okay? So the capacitor, it limits the dv over d t.
What does this mean? It means that the voltage Before switching is after switching, the same after switching,
exhaust after switching. So when this was switched it for a long time There's
volts or 15 volts. Okay? So when we open this switch, it is still 15 volt, just right after switching, still 15 volt, because it does not change
instantaneously. So you can see that the initial
voltage becomes 15 volt. Okay? Now, when you draw the
circuit after switching, Making the Switch or wind, okay? So when does part becomes
obeying the law exists, it means that all of this is cancelled out of our
circuit, open circuit. So we will have only is online on one arm and the capacitance. So you can see 20
millivolt millifarad. One arm ends are
minor and the rest is open circuit because
we open the switch. Now the question is, what is
that equivalent resistance? We said that we need
Tao, which are C. So R is R equivalent or R7. And so between these
two terminals. So if we look at our circuit, we have one arm
series was a line on our R equivalent
is that ten ohms. From here we can get tau. Tau is our equivalent, which is a ten ohm multiplied
by Zach capacitance, which is a 20 milli for OT, gives us a Dao of 0.2 seconds. Now we can write our
equation like this. The voltage across the capacitor
for time greater than or equal to 0 is equal to VC at 0, which is the initial voltage, 15 volt, multiplied by
negative t over tau. Tau is 0.2 seconds. Now, the initial stored, okay, forget back here, we need the voltage and the
initial energy stored. So for you get back here. So what does the
initial energy stored? We know that the initial energy stored at time equals to 0, equal to half c v naught square. Like this, have CVC square
or the V naught squared, as you can see,
half C capacitance, which that 20 milli fraud. And V square, which
is 15 squared, gives us 2.25 jewelry. So that's representing
the initial energy stored inside our capacity. So this was another example on the solar cell
free or see, suck.
79. Source Free RL Circuit: Hey everyone, In
this lesson we're going to start
discussing is a source, a free RL circuit. So in our course for
electric circuits, so we discussed as a source of free RC circuit and
we had two examples. Now in this lesson we will start discussing the RL circuit. So as we remember the source, a free RL circuit. What does a source
of free means? It means that we don't
have any supply. It means that the source
of free RL circuit occurs when a DC source is
suddenly disconnected. So the energy stored inside the inductor is released
to the resistors. So in the RC circuit, we said that we have
connect to supply, a DC supply like this. Okay? Vdc like this, and we start providing current
through the inductor is a resistance, okay? So this inductor
will be shorted. It will have energy
stored and sold it. Then when we suddenly
disconnect is a supply, we will see the
behavior of our suck. So the behavior of our circuit is known as a source of free. We don't have any
supply in this case. And we know that we have now is that natural response
of the circuit. Okay? Now another thing which is important when you
have like this one, you ought to charging
an inductor. You can see the current is
flowing through an inductor. So the voltage forming
between plus minus v. Since the current is
entering the inductor. Now we'll remember that here, this is the direction
of the current. And the inductor. If you remember that
the inductor voltage of the inductor is equal to L D over DT. So what
does this mean? It means that the current
cannot change instantaneously. So what does this mean when
we disconnect as a supply? That current does not
reverses direction. If the current is moving
like this, from here, from this point to this point, it will remain moving
in the same direction. That's why you will find that
when we remove the supply, you'll find that the polarity of the voltage becomes the
legs as plus minus. What does this mean? It
means that the current will be going out of the inductor. Okay, that's all you will
have a positive here and negative here when the
source is disconnected, okay, to represent that Zack current is going
from the positive, going in the same
direction as before. Okay? Now as you can see,
we have this voltage across the inductor and
we have the resistance, the voltage across the resistor, and the current flowing
in the circuit. So forest at time equals 0, the instant of disconnecting. So supply, the inductor
is initially charged. So it has an initial current. I naught. Okay? So if you remember that when
we connected the supplies, there is a current going through the inductor at a certain moment that we have an initial current. This current cannot
change instantaneously. Why do to the patients of what? Due to the presence
of an inductor which prevents that
representatives large change of ELF
Zach current or d over d t. So it limits
the I over d t. So it means what
it means that the current before
switching exhausted before switching is equal to the current after switching
zoster after switching, which is similar
to the capacitors. A capacitor. If you remember as a capacitor, we said that it is the voltage
Austin Before switching, which is V naught, is equal to the voltage after switches,
just after switching. Because capacitor does
not allow that dV by d t or z does not allow a large change in the voltage
with respect to time. Okay? So we have here
current I naught, which can be obtained
from the current before switching, exhausted
before switching. Okay? And the energy stored,
similar as before. We said that the energy
stored of an inductor is half L i square,
half LI squared. So the initial energy stored in the inductor is half
L I naught squared. Okay? This is the energy stored inside the inductor when we
disconnect switch. Okay? Now are blind KVL
in this circuit, you will find that we have
V L plus V are equal to 0. The voltage across the inductor
here from KVL, like this, for example, plus VR
and VL equal to 0. Okay? As you can see. So what is the voltage
across that inductor? If you remember, we
said before that the voltage across an
inductor is equal to L d over d t. And the voltage across
the resistor is R multiplied by the current. So as you can see,
the voltage across the inductor is L d over d t. And the voltage across the
resistor is equal to what? Equal to the current multiplied
by the resistance R. Okay? So if we assume, for example, the current
flowing like this. It will be the same as
the current flowing. It will be Ld over d t plus r i. Or if you assume like this, it is a same id as it will
be a negative sign common. So two will give us
the same equation. Okay? So now what, so we
will rearrange this. So you have D over DT, lots of divide by L.
So we will have d over d t plus r over l equal to 0. So now what? So now
we would like to separate each of these
equations as we did before. We said here, for example, over DT, same as what we
did in the RC circuit, equal to negative or
L. So we would have dy over Zach current or equal
to negative R over L d t. Okay? So by integration of both sides, integration of one over I
is less than or equal to. Integration of negative out over L with respect to time gives us a negative r t
over L plus Lenny, similar as what we did
in the capacitance. So as you can see, we
will have likes us. So as you can see, len, the current equal to
negative RT over l, negative RT over L. As you can see,
Lynn I and negative l. Lin I-naught is the one
which is like this plus linen. Okay? So the a is a constant as
you remember that when we, in order to get the
value of a similar as what we did in the capacitance
at time equal to 0, the current will be I naught. So we'll find that the
equation will be land or inode equal to a. So a is equal to I naught. Okay? So then I equal to negative
RT over L plus ln naught, which is similar as here. You can see they exist
to the other side, gives us negative LN I-naught. Okay? Taking exists here. Then I minus linear or learn
all in minus ln or not. As you can see here. Similar as we did, exactly as we did for that
solves a free RC circuit, the total visa same
equation, okay? So difference is that we remove this aka voltage and we
replaced it with the current. So we have here as
this subtraction. And then I minus I naught
is Len divided by I-naught, then all divided
by I-naught from the Lyn equations or formulas equal to
negative RT over L. So we will take the
exponential of both sides. So we will have in the end, which is a current equal to I naught e to the power
negative RT over L. From this equation, you
will find that tau, which is a time constant in
RL circuit, is equal to what? Equal to R L over R, L over R. Now why is this? Because if you remember here, or equal to I naught e to the power negative
t over tau. Okay? Similar to what we did
in the RC circuit. So tau here, we have negative t. You can see negative t. So we need to take
this r down here. So it will be L over R likes us. So it will be negative t
divided by L over R. So L over R here is our
tau, as you can see. Okay? As you can see, I
equal to I naught e to the power
negative t over Tau. Now remember that,
that there was a time constant in the equation of the capacitance walls. Our c tau equal to RC
in the RL circuit, it is equal to L over R. Okay? It's similar as before. Tau is a time constant
that reaches as, that wins our current time taken so that the current
reaches such 6.8. But I wasn't present
of I-naught. Similar to 6.8 per cent of V naught inside the
RC circuit. Okay? So as you can see here is
the response of the circuit. So as a time of switching, you can see that at
time equals 0 when we switch or weeds
connect the supply, you can see that the
current before switching is equal to the current
after switching to I-naught. Starting from here,
you can see that the current is starting to decay, exponentially decaying e to the power negative t over tau. Okay? And I'll tie
me call tau at here. You can see up an equal tau. Find the value of the current
is equal to 0.368 or. So. As you can see,
the voltage across a resistor in this case, here, volts across the resistor will be the current multiplied by resistance or a multiplied
by r. And the resistance R. And the current is equal to I naught e to the power
negative t over Tau, naught e to the power
negative t over tau. So this is a voltage
across the resistor. Now, what is the
power dissipated? Power dissipated in the
resistor is equal to the voltage multiplied
by the resistance. Or we can say another equation, which is the power
dissipated inside any resistance is equal to
I squared multiplied by r. So r squared is the
current squared square of this is square root
of this equation, r squared e negative t over tau, square root of e to the power negative t over tau is negative two t over Tau, okay? Multiplied by the resistance. Okay? So we have i-node square r, square r e to the power negative two t over
Tau, as you can see. Okay? So this is the power dissipated. What about the energy stored? Energy stored is also we'll
get through integration, integration of the power
with respect to time as we did inside that circuit
of the RC circuit. Okay? Now why is this? Because simply that power is a
function in time. So we cannot just say energy
is power multiplied by time. We need to integrate
it because we have the power as a
function in time. So the integration of this equation with respect
to time, similar as before, integration of e to the power
negative t over tau is the exponential divided by
negative two over tau. Similar as the previous, what we did exactly in the previous equation
of the RC circuit, similar as before, okay? So funds that in the end, we will have that energy stored, our own energy absorbed
but not stored, absorbed by that resistor
is half L I naught squared y minus e to the
negative two t over Tau. Now if you look carefully about this equation and time equals 0, this exponential will be 11. Answer will be one, which means one minus 10. So the energy stored, energy absorbed by the resistor at time equals 0 is equal to 0. This is a time of switching. 0, energy absorbed
by the resistor. Now at time equal to infinity, when t equal to infinity, this exponential
will be equal to 0, which means that WOR will be half l naught squared.
What does this mean? It means that all of the energy
at time equals infinity, theoretically, at
time equal infinity, all of the energy stored inside the inductor is transferred to the resistor or dissipated
in the resistor. Ok. Now of course,
similar as before, the current reaches a steady
state after five Tau, five times the time constant, similar as RC circuit. Okay? So now let's have
some example on this to understand this circuit.
80. Example 1 on Source Free RL Circuit: So in the first example on the
source of free RL circuit, we have here is a circuit, we have an inductor, we have a two ohms for all, and we have a dependent source. Remember the bend
and the source. Now here we have i
at time equals 0, equals ten, and Beta, which
is the initial current. Initial current stored
inside this inductor. The initial current
of the inductor. Now what we need is
that we would like to find the current as
a function of time. And we need to find the current i x as a function of time. Okay? So let's start. So we
have the initial account. We have, we need the equation
of the current I is equal to I naught e to the power
negative t over tau. Okay? Now, what does the
initial current I naught, I naught is given as ten m. What is the remaining parts
are many important is tau. Tau is equal to L over R. The inductance itself
is equal to 0.5 Henry. Okay? Now what is the remaining part? The remaining part
is the resistance. What is the resistance
of this circuit? Okay? So remember, remember what is
this resistance in the tau? It is a seven and resistance, or the equivalent resistance
seen by the inductor. So how can we do this? You can see that here, this circuit, as if you
have two terminals here. At this point. You have two ohms here for
all that we have here, our dependent source,
remember it is a dependent source and
connected like this. Okay? So that's seven
and equivalent is seven equivalent seen by
the circuit like here. Or R equivalent is a
seven and resistance. Now, what does the
problem here is that the problem is that we
have a dependent source. We have this dependent source. Okay? So how can we solve with this? Remember that in our previous
lessons when we discuss the ZAB equivalent resistance
or seven and serum, we said that in order to get R7 and inside a
circuit like this, we need to do what? To add an independent source. Okay? We, as an example, we are going to add an independent source like
this plus minus I exist. It's a value, for
example, one volt. So our supply of one volt, and this is a current
flowing out of it. Okay? Or you can do the
reverse makes up or step down and the
negative upwards. Now, then by doing that
analysis in this circuit. By doing the
analysis in circuit, you can get the value
of the current. Then all seven will be equal to one volt
divided bys account. Okay? We have here a dependent source. So let's see. So as you can see, the equivalent resistance
is the same as seven and resistance adds
the inductor tetanus. Because of the dependent source, we will insert a
voltage source V naught one volt adds inductor's
terminals a and B, these two terminals
here and here. So as you can see here, plus, minus one volt. And do we have here
a current going from it called the I-naught, okay? And we have here is that to own four ohm and that
dependent source. Now remember, look at
the dependent source. You can see three
times the current. What guarantee current or current I flowing in that
direction, in this direction. But this I naught is
flowing like this. Remember, these two
currents are not similar. They are opposite to each other. I naught equal to negative. So how can we solve with this? We need the current
in this circuit. So we are going to do
KVL or mesh analysis. We have all U1 and current I2, I1 representing this loop, and i2 representing this loop. Okay? Now as you can see, the current I1,
here are E1, sorry. What is the direction
of the current? Like this? What is the direction
of I1 like this? So the i1 is equal to
the original current i. So what we can say is that it
will be three or E1 because it is a same direction
of the current in either direction of the
current flowing like this, as you can see here. And here, it will be three i1. Okay, is that as the first part? We're going to do KVL
here and KVL here. So the first one in this loop, how can we do the assembly line? We have a loop like
this, this direction. So we will meet first plus
one volt plus one volt, then go down here like this, resistance at two ohm. So we have a
resistance to all what does the current flowing
through the 2-ohm. We have all U1 minus I2. So it will be I1 minus
I2 from this caveat. So from this equation
we will have I1 minus I2 equals negative off. In the second loop here, we have a current like this. Negative three i1. Okay? And then go up here, the forearm plus four. What current flowing
is i2 only A2. Then we go like this. We have the two ohms, so plus two i2 minus y1, y2 minus y1, as we did in the lessons of
the mesh analysis. So you can see the equations. You can see six I2. Here we have for i2 and the
two i2 gives us six items. And negative 31,
negative three or E1. And then we have negative
21, negative two ion. Okay? This will give us
negative five I1, I1. So we will have I2
equal to 05 or six I1. So we have now two equations, i1 and i2 from
this two equation. So we can get i1 and i2. As you can see. So what do we need here is
only one from this equations, i1 is equal to negative
three and bear. Now is I1 that require
the voltage current. Know why? Because what we need is the current going out
from this supply, which is all in OT. So i-node going
like this or not. But I E1 going like this or E1. So as you can see, they are
opposite to each other. So I naught equal
to negative I1, I naught equal to negative y, one equal to three, and bear. Okay. So now we have the current,
we have the voltage, we can get resistance and seven and which is the
voltage divided by current, V naught over I naught, which is one divided by three. So this is the equivalent
resistance of the circuit. So we have the inductance
and we have that resistance. So we can get tau, which is the time constant. As you can see, L over R, l is 0.5 Henry half or 153
divided by R equivalent, which is one of our three, gives us three over two seconds. Okay? So now we have time constant, so we can write our equation. So the current is equal to I
naught the initial current, e to the power
negative t over tau. So you can see all
inode is at ten AM. Bear is the initial current
given in the problem. And e to the power
negative t over tau. Tau is three over 23 over two. So we can say takes us to sense
it is division, division. So we can take to here it will be negative two t over three, negative two T over C. Okay? Okay, so now we have the
current, first current required. Now what do we need also, we need IX, we need i x. So as you can see
that this inductor is parallel to the two ohm. So it means that x is
equal to the voltage of the inductor divided
by Z2 ohms, right? Voltage divided by resistance. The voltage of the
inductor here, similar to the voltage V. Okay? So what does the value
of the voltage here is the value of the voltage
is simply equal to L, which is the inductance
0.5 Henry d over d t. So it will be inductance
multiplied by d over d t, or the derivative of the
current with respect to time, derivative of this function
with respect to time. So the voltage L di over DTLS, 0.5, d over d t is
a current is ten. Derivative of the exponential
is negative two over three. Negative two over three multiplied by e to
the power negative two over three t. Okay? So this multiplication will give us negative ten over three, negative two over three t. Okay? So this is our voltage. Now what about the current IX? It will be voltage
divided by that too. Okay, So it would be
voltage divided by two, which is the voltage divided
by the resistance R, V over R. So we will have to
exist one divided by two. It will give us this equation. Okay? So now we had an example on
the source of the RL circuit. We understand that now
how can we get that current through
an inductor of n, initially charge it inductor. And we can now understand how
can we analyze our circuit.
81. Example 2 on Source Free RL Circuit: Now let's have another example on the source and free or Elsa. So in this circuit,
we have a switch has been closed for
a very long time. This switch was closed
for a very long time. And so when the time equal to 0, find the current of the
inductor as a function of time. So first, what we need, again, we know that the current of
the inductor is equal to I naught e to the power
negative t over tau. So we need the initial
current all naught, and we need as a
time constant Tau. Okay? So how can we do this first, the initial current can be
obtained before switching, when the switch was closed. And tau is R is L over R L to Henry and R is the
equivalent resistance. So let's take it step-by-step. So first I naught when the switch was closed for
a very, very long time. So what does this
mean? When the switch was closed for a very long time, it means that the inductor
reaches its steady state. And if you remember, if I ask you, what is the
steady state of an inductor? The city-state is at this
inductor will act as a short circuit after
a very long time. So we have here as a student
on barrel to short-circuit. So this 16 arm will be gone. So we will have a twelv for all, for all with a short
circuit and the tool. So as you can see, like this. So we have a short-circuit. And do we need this
initial current, fore arm, 12 or two ohm and 40 volt. Now as you can see, what
we need is this current, which is I naught. How can we get this
by current division? Okay, So we have 40
volt and we have here to all zeros with the
parallel combination. So first, what we
need is that we need the equivalent resistance
of this circuit. So as you can see,
when t is less than 0 or searches closer
for a very long time, the inductor will act as
a short circuit to DC. Norm is short-circuited and the resulting circuit to get I1, we combine four
ohm and 12 ohm in parallel to get
this value stream. First we have for all better to 12 all this parallel
combination, which is this equation, gives us three oh, okay? So we have a three ohm series. Was that to all three
ohms series was to own, which is our equivalent
of our circuit. Before switching. Supply current will be equal to 40 volt divided by the
equivalent resistance. 40 volt divided by the equivalent resistance.
So it gives us eight. And this current is current at steady state
before switching. So this current is
eight and bear, this is our one current
we need nodes or current we need is the current
as a function of time. So we need this current. How can we get it using
is a current division. So the current here is equal to eight and bear, multiply it by. We have four on
butter to 12 ohm. So there'll be the
other resistor divided by the total resistance. Like this from what we learned in our course for
electric circuits or in the voltage division or
that current division lessons. So we have the current I as
a function of 26 and bear is the current node,
initial current. Okay? Okay, so the second part is
that we need after switching. So when the switch is opened, as you can see, this part
of circuit is canceled. So we have four ohms, 12 ohms, Christine
arm like this. Okay? So we need the R equivalent, equivalent in between the two
terminals of the inductor. When we look at a
circuit like this, we have a 16 arm to arm. Okay? So our equivalent is, I think it all, okay. Which is 120 plus
four parallel to 160, give us eight ohms. So this is the odd equivalent
that required the four tau. Tau equal to L over R. L is two Henry and R is the equivalent resistance
after switching, which is eight Ohms like this. So I have the time or the time
constant to one over four. So we can get right
now our current I naught e to the power
negative t over tau. One over tau, which is one divided by one over four
gives us four negative 40. Okay? So now we discussed
another example on the source if Ri RL circuit.
82. Step Response of an RC Circuit: Hey everyone, In this
lesson we are going to discuss the step response
of an RC circuit. So we discussed before as a
source of free RC circuits, source of free RL circuit. Now I would like to discuss is the step response
of an RC circuit. So when DC source of an RC
circuit is suddenly applied, okay, so ends our
source of free. We had a DC supply which
is suddenly disconnected. In the step response, we have a DC supply that
is suddenly applied. The voltage or current source can be modelled as
a step function. And the response is known
as a step response. Okay? The step response
of a circuit is, it's a behavior when
excitation is a step function, which may be a voltage
or a current source. As you can see here. So what does this mean? Let's say we have
a switch which was opened for a very long time. Okay, And do we have
here a resistance? We have a capacitor
and we have a supply. Then suddenly we start
applying our DC source. So we close this switch. Okay? The First Estate was
it, it was opened. So the capacitor may have an
initial voltage or may not. Okay. Then we suddenly applied it. So this soul, so let's start
a charging our capacity. Okay? So this behavior is step
response for an RC circuit. Why it's called step response? Because the voltage itself
is a step function. So it will be like this. Okay? So at time equal to
0 at any instant of switching when
closing, the voltage, which is applied
to the circuit was 0 and then suddenly
becomes V naught. Okay? So it was 0. Before the time of switching, the voltage was 0, then it was suddenly applied. So it will be a constant value. Okay? So this is a step
function can be represented by U as
a function of t. You can see we have removed the switch and we added
u as a function of t multiplied bys V s.
So what does this mean? Let's see here now,
you can see this is called the step function
as a unit step function. Okay? So it means that at time 0 or time before
0, the value is 0. Okay? Then after time greater than 0, at t less than 0, the value
of the step function is 0. And our time greater than 0. In this range, you will
find that the value is one, unity, as you can see here. At time equals 0 itself, at t equal to 0,
it is undefined. We don't know why, because as you can
see it as a stamp, it changing from 0 to one. So what is the value of the voltage in this
part, I don't know. It is unknown, undefined. We can not know it because it is changing from 0 to one
deodorant in 0 seconds. Okay, so we don't
know the value here. Okay? Bub, before switching a time less than 0
as the value is 0, and after switching
is the time is equal to after t greater than 0, the value is one. So as you can see,
this step function is a representation of this one. So let's say if we say VS
multiplied by a step function, which is V S, for example, any value such as, for example, 15 volt. What does this mean?
It means we multiply this curve by 15. So it will be instead of one, we will have 15 volt here
and we will have here our V S. So our supply, will it change it from 0 to 15 at a state of switching
before switching 0. And after switching it
reaches 15 volt. Okay? Is this is called
the step function. Okay? So this step function
produces a step response in our RC circuit. Okay? So Zionist step function
is 0 for negative value of time and the one for
both the value of time. Okay? Now before we go
to the next slide, I would like to mention
something important. Let's say, for example, I don't want to make
this one starting at 0. I would like to make this the function remain
as the 0 and the, for example, of
prime equals three, I would like it to
step like this. So before 30 and
after three is one. How can we do this assembly? You can type u as a
function of t minus three equal to 0
and the one when t. Less than three and
greater than c. Okay, How did we do this simply by making
the step function t minus the phase shift or another phase shift
that has a time shift. Okay? So if I would like
it to start stepping from time equals three
ion, make this three. If I would like it at
five, for example, then I will make this one
t minus five, and so on. Okay? Okay. Um, he lets us to delete all of this and get the pen like this. So let's just start again. We select the capacitor voltage
as the circuit response, similar to the RC circuit, that source of free RC circuit. We selected the voltage
response is the one which is
important for us to. We will assume that our capacitor
has an initial voltage, V naught, which is
determined from what? From Zach conditions
of before switching. Although it is not necessary
for the step response, sensors or a voltage of the capacitor cannot
change instantaneously. So as you can see here, v before switching equal V after switching equals v naught, which is the initial voltage. So when we, when the
switch was all wins, then we close it like this. The voltage of the
capacitor will not change. It will be the same
after switching. Why? Because as you remember, that's a capacitor
limits and DV over DT, as we discussed before. So 0 plus means just
after switching. 0 minus means exhaustive
before switching. This, what does this,
what does this mean? Then? What is the next step? You can see here, v 0 is the voltage across the capacitor
just before switching. 0 plus means it is a voltage or immediately
after switching. So if we have this
circuit like here, we can remove the switch, which means a step function. With this function v, u as a function of t, It means at time
equals 0 is a voltage will be applied. Okay? So what does this mean? It means that this voltage are timeless and 0,
voltage will be 0. Our time greater than zeros
of voltage will be V S. Okay? This, what does this mean? Now if you look at this, this circuit is the same. Before switching, this
circuit was opened. So it's a voltage
applied here is 0. Since it's silicate is all bend. So the voltage applied
across the capacitor is 0. Okay? Then after switching is a voltage applied
will be V S. Okay? So as you can see, same, same thing here, okay? So this representing
this one plus this one. Okay? So let's just start using
this one below here. Okay? So by applying KCL
in this circuit, we can get the
following equation. So as an example, I will assume that
currently exists current, the current and the
current flowing here. Same current flowing through that resistor is
the same current flowing through the inductance
suzanne capacitance. So that current source
as the capacitance is c d v over d t. The voltage across the capacitor equal to the current
through that resistor. What is the current
through the resistor is the voltage divided by R, voltage across that
resistor divided by R. So the current is going from here to here and this direction. So it means it will be plus, minus the potential difference
between these two points divided by R plus is with v, u as a function of time. So we'll say VS you as a
function of time minus v. Okay? So as you can see, we
could write it like this. You can see C dv over d t plus v minus v u as a
function of t over r. Now, this equation is
similar to this one. How assembly? You can take this one
to the other side. So we will have C dV
by d t minus v s, u minus v over r. Okay? Then we take the
negative sign to here. So we'll have
negative plus Here, we have so positive sign. So we'll have v minus v, v minus v s you, okay? So this equation, similar
to this equation. Okay, let's delete this. Then. What Zemo are
going to divide this policy and rearrange it? I feel divide by C will have
dv over v minus v SU over, over RC, as you can see, and takes us again
to the other side, we will have this
equation. Okay? So it's as simple
as just what I, what I use Austin written. If it is a same equation,
nothing has changed. Then what's the next
step we have here? Voltage. And do we have here d t? So we will take d t to the other side and Texas
voltage back here. So you will see is
that here we have dv over v minus v S equal to
negative d t over r c. Okay? So let's continue. So we have here this equation, then we are going to integrate
both sides like this. So the integration
of the site and integration this side
with respect to voltage, with respect to t. This will
give us a Len V minus V s, Len V minus V s, Since it is one
over v minus v S, or one over x gives us ln x. So one over V minus V S
gives us learn v minus VS, similar to what we did in
the source IP three circuit. And we have here, the limits are from, we start from the
initial voltage to the voltage at any instant. Okay? So we are integrating
what voltage, voltage over the capacitor. This capacitor, if we
do the integration, it will start from the
initial voltage V naught to any voltage at any instant, not just at infinity,
but at anytime, so that I can get the voltage of the capacitor at
any given, okay? And here's our equation was
sort of from 0 to anytime t. As you can see. So you can see len V minus V s, we will substitute with V
as a function of time minus the substitution of V naught
minus substitution of vino. Here, we replaced HEV with V as a function of
time and v naught. Okay? So any exhaust, it is just
a normal integration. And the integration of
negative t over RC. That substitution,
we will have t minus 0 or minus minus
0 means a plus z. So we'll have
negative t over r c. Now since we have two
Len minus each other, we can make IT
division like this. Len v minus v over
v naught minus v s. Okay? Remember all of this is V, this V naught, this V, sorry, not this V. This V S is similar as V S u
as a function of t. Same as we've just written. We will divide it
later. So it is this. They are the same. Okay? So we have here negative t over RC. Then we are going to take the
exponential of this part. Exponential of this part, we will have a v minus
v over v naught minus v s gives us e to the
power negative t over tau is our
RC, as we learned. So the voltage V minus V S equal to V naught minus v s
multiplied bys exponential. Then we can rewrite our
equation to be finally, law exists to be V as
a function of time equals v plus v naught minus v s e to the power
negative t over tau, a time greater than 0. Okay? So finally, we will have
this voltage equation. So before switching,
before switching, okay? What is the value of the voltage across the capacitor,
which is V naught. Before switching.
After switching, when we close the switch, we will have this equation here. This equation are presenting before switching and
after switching. So as you can see it is, the voltage is starting from V naught and the bank
charges boys supply voltage. Okay? So as you can see as this response before
time equal to 0, before time equal to 0, this equation, we
will have V naught. As you can see, V naught. Then after switching,
starting after switching, we will have this equation. We are suppose V naught minus v s e to the power
negative t over tau. So as you can see, is
this representing what's representing that our
capacitor will be charged. It will be charged exponentially on the racing is a steady-state, which is V supply. Now, as you can see, how can we prove this assembly? Here likes us. Okay? You can see is that at time
equal to 0, for example, at instant of switching
for the capacitor voltage, we said that the
capacitor voltage does not change instantaneously. So the voltage before switching is equal to the
voltage after switching. So as you can see,
V naught and after switching at innocent of
switching V naught as it is. Okay? How can we prove this
at time equal to 0? This exponential
will give us one. So it means that we have V
S plus V naught minus v s. So VS will go with
V S. So we will have only v naught
at time equal to 0. Similar as you can see. Now, what is the
steady-state value? If you take time
equal to infinity, it means that this
part will be 0. So v naught minus v s
multiplied by 0 gives us 0. So NZ end up by m
equal to infinity, our voltage will become V
supply, as you can see. Okay? So what does this mean? Similar as before, when the
capacitor is the charge it, for a long time, it's a value or reach
of the supply voltage, or the voltage at time
equal to infinity. This of course not always the case depending
on our circuit, as we'll see in the
soul with examples. Okay. Now what about Zack current? The current of the capacitor
is equal to c d v over d t. So we will differentiate
this function. Okay? So if we differentiate V naught, it will give us
zeros are currently before switching is equal to 0. Because d v over d t is the derivative of the voltage
with respect to time. This voltage is a constant
value, so it will give us 0. After switching, we will
differentiate this equation. So differentiation of this
equation will give us finally, V S over r e to the power negative t over tau u
as a function of time. You as a function of
time here representing Zach t less than 0
and t greater than t. For t greater than 0 only, we will have this equation,
which is this part. So what does, what is
the benefit of you? You make us divide
it into two parts. T less than 0 and
t greater than 0. It means that when we are
less than 0, it will be 0. When y are greater than
one or greater than 0, it means that this
part will be one. So we will have V S over r e to the power
negative t over tau, which is this equation. Okay? I hope that is clear. So if we plot this one, you can see before switching, current is equal to 0. And after switching
suddenly, boom, increases to V over
R. As you can see, at time equal to 0. It will be V S over r
reaches V, V S over r. Okay? Boom to V over R. Then due
to the exponential decaying, it will start decaying
until reaching 0. Now if you look at this
two curve as you will find that this voltage is
continuous, okay? V naught. Starting vino, you
can see the function is continuous after switching, is still continuous, v naught and increasing over this
function is discontinuous. Why? Because at before
switching zeros then add the innocent of
switching can see this part, which is undefined. Jump is from 0 to V s. So it means this
function is not continuous. Okay? Now, here is an important part. Now, the RC or RL circuit, instead of going zap
previous method, we have made that KCL or a KVL in order to find
the step response. Is there any other method or any other a way to get
this equation? Yes, there is another way. We found that as our voltage. For example, in the RC circuit, we found that the voltage
has two components, can be divided into
two components. The first one is
dividing it into a natural response
and forced response. The second approach
is divided into our transient response and
steady-state response. So for example, we will start
with photos to composition, which is the natural response
plus the forced response. So the natural response, as we learned before, is the response of the circuit. So the source of free is
called the natural response. Remember that when we had stored the energy
in our capacitor, and then it will start dissipating
power in the resistor. So we had that
decaying exponential, V naught e to the power
negative t over Tau. This decaying
exponential response is called is a free source of free response or
natural response of circuit misrepresenting
the first component. Second component
is the response of the supply or the effect of the supply or the
independent source effect. Which forces our circuit to
reach another steady-state, which is the supply voltage
as an example here. So as you can see, V equals V n plus vf v. And natural vaulted,
natural response to voltage and also the
response voltage. So the natural response V naught e to the power
negative t over tau, which is one we
discussed before. Plus v f, which is the
effect of the supply, only the force response
voltage on it. So how can we do this? It is written as v S1 minus e to the power
negative t over tau. So what does this mean? Okay, So how can we get this? Now remember that this supply, when the supply
is applied, okay? Let's say for example, this voltage was only 0. Okay? So if we look at our
circuit like this, as a voltage was 0, V as a function of time. So it was 0. Okay? So this is a starting point. Now when we apply is
the independent source. What will happen is
that this voltage will start increasing exponentially until reaching
steady-state value, which is V supply. Okay? So as you can see, at time equal to 0, the voltage will be equal to 0. And at time equal to infinity
by m equal to infinity, v will be equal to V supply. So if you're looking
at this equation, this equation satisfies
these two conditions, or this waveform, VS one minus e to the power negative t over
tau at time equal to 0. This part will be
given, will give us 0. At time equal to infinity. It will give us V supply. So this equation representing is the effect of the
independent source only. This part representing
the effect of the stored energy
of the capacitor. Okay? So this summation
will give us as the previous equation or the
response of the RC circuit. Okay? So the natural
response is the one which we discussed before. False. So the response, it is a one which
is produced when an external force is applied. At representing words, the
circuit is false it to do by the input excitation. So you'll find that
the natural response eventually dies along those
are transient component of the force response is
exponential part is leaving only the steady-state component
of the force response. Ok? So as you can see here, at time equal to infinity, this part will be equal to 0. This part will be equal to 0. So you will find that
the only component remaining is V supply, which is steady-state. Okay? The second one is that we can
assume that our response is divided into a
transient response and steady-state response, permanent and temporary port. So you can see that the transient Robertson things are transient two components. Okay, So if we look back
here at this equation, you can see is this part and v as negative e to the
negative t over tau. Here is this parts are thrones and the
embroidery component, or they're not
permanent component. And the steady-state represented because a permanent component, which is V supply. As you can see
here, the V supply is a permanent component. Okay? So this is another
representation of this. Okay. It's a transient,
is temporary. It means that the portion
of Zach complete responses that decays to 0 as
time reaches infinity. And the steady-state
response is a portion which remains after the transient
response has died out. Okay? Okay. So anyway, anyway, this previous to a composition which is
steady-state and transient, There's a force it and
the natural response. I would just, I would like, wanted to mention
it because it is important to understand
this concept. So what are we going to do
in our solvent examples? So we are not going to think about what is a transient
or steady state. We can just get it
using an equation. So this equation is
representing like this. You can see V as a function
of time as equal to V infinity plus v at time equal to 0 minus v at time
equal to 0 to infinity, e to the power
negative t over Tau. This is our general equation, which will help us to
get this equation. So V infinity, it
means the value of the voltage at steady-state
or at time equal to infinity. Okay? For example, in our
previous circuit, it reaches V supply. And V 0 is the initial voltage and tau is a time constant. So as you can see, via 0 is the initial voltage
at time equal 0 plus, which is a problem after
switching, or the voltage after, just after switching, V infinity is the final or
the steady-state value. So now let's have some
examples in order to understand the step
response of an RC circuit.
83. Example 1 on Step Response of an RC Circuit: So now let's have an example on the step response of RC circuit. So the switch has been closed. So the switch at position a for a very long time and at
time equal to zeros and switch goes to be fine desire voltage of the
capacitor as a function of time and find
its value at time equals 1 second and time
equals four seconds. So let's start first. In order to, as you can
see here, at switch one, this capacitor was
connected to that 24 volts, 24 volts supply for
a very long time. And then when it switches to be, you will find that it
has now a step response. It is suddenly connected
to another supply, which is a 30 volt. That's why this is a step
response of an RC circuit. So what we need is some
elements for us too. We need the voltage 0, which means the voltage
exhausted before switching. We need the voltage at time equals infinity after switching, steady-state voltage
after a very long time. And when it tau,
which is our scene. So let's just start
step-by-step. So first one, we
need the voltage V at time equal to 0
or before switching. So this is connected to a
switch a for a very long time. What does this
mean? It means that our capacitor reaches
a steady state. And if you remember that this is a steady-state of a capacitor connected to a DC supply
is an open circuit. So we will have here
an open circuit or v between this point
and says void. And we have 24 volt, three kilo-ohm and
five kilo-ohm. Okay? So we will have like this. Okay, Let's, let's draw it. Okay? So we'll have a 24
volt line exists. Okay? We have a three
kilo-ohm, three kilo ohms. We have a five kilo
ohm, five kilo ohms. And we have here
this two terminals, which are representing
the capacitor voltage V 0 or before zoster,
before switching. So the voltage across
the capacitor is the voltage across R5 kilo-ohm. And the, by using the
voltage division, you can see that we
have a 24 volt supply. We have a three kilo-ohm
and five kiloohm. So by using voltage division, the voltage across the
five kilo ohm is equal to 24 volt multiplied by five
divided bys or some measure. Like this. 24 volts multiplied
by its resistance, which is the five kiloohm, divided by the summation
of the two resistors. So the initial voltage
of the capacitor after, before switching for a very
long time is 15 volts. This voltage is of course, is a voltage before switching. Just after switching. Because the voltage cannot
change instantaneously. Okay? Now, the next step is that
when switch goes to B, we will have a
circuit like this. We have a voltage of 0.5 millifarad capacitor connected
to our four kilo ohms. And the salty volts plus
minus searching vote. Okay? So what will happen is that this voltage at V
equals infinity, at time equals infinity. So the steady-state voltage, what do you think
is the steady-state voltage of this circuit? As a steady-state voltage
is assertive volt. After a very long time, that capacitor voltage will be charged with the
same as the supply. Now if you don't know
why 30 volt assembly, you can do like this. You can think about it like after reaching a steady state, we have a four kilo-ohm here. We have our 30 volts. And at time equal infinity, what will happen
to the capacitor since it's connected
to a DC supply, it will be an open circuit. Plus minus V infinity. The voltage here, what
is the value of voltage? The voltage V infinity
is the supply minus the voltage drop
across a four kilo-ohm. However, since
there's a capacitor will become an open circuit. So the current will
be what will be equal to 0 at steady-state. So the voltage drop across
a four kilo-ohm is 0. The V infinity will be equal
to the 30 volt supply. Okay? It is the same
idea as we did here. So as you can see here, the switch is in position
beam before the time course. And you can see V infinity is certainty voltage because
it acts as an open circuit. Now the last element we need, we have v 0 and v infinity. We need that tau, which
is a time constant, is all 70 multiplied by Zach capacitor when this
one is connected here, as you can see, we have
an independent source, capacitance and resistance. So as you can see
from this circuit, we have only one resistance, which is a four kilo ohm. So it will be our 17 is
the art of the resistor, four kilo multiplied
by the capacitance, which is 0.5 millifarads, gives us a time constant
of two seconds. So we have our three elements, so we can type our equation. V as a function of time is
equal to V infinity plus V 0 minus V infinity e to the
power negative t over tau. So it will be 30
plus 15 minus e to the negative t over
tau is a 2 second, as you can see here, V infinity
is adds a steady-state, which is assertive volt. And V, which is the
initial voltage, which is also obtained 15 volt. Okay, So this is
our final equation. Now what's the next step
we need at time equal 1 second and time
equals four seconds. So we'll substitute with t
equal one and t equal four, as you can see. Okay? So as you can see, as time
increases, as time increases, the voltage increases
and approaches what approaches the 30
volt or V infinity, which is a supply voltage or
the voltage at steady-state. So this was a very simple
example on the RC circuit.
84. Example 2 on Step Response of an RC Circuit: Now let's have
another example on the step response
of an RC circuit. Switch has been closed for a very long time and
solvent at time equal to 0. Okay? So the switch was closed
for a very long time. Then at time equals
0, it is open. Okay? Now we need to find
the voltage and we need to find the current
inside our circuit. Okay? So the first step is, as you can see here, we have a salty u
as a function of t. What does this mean? Remember that this
is a step function, which means at time less
than 0, it will be 0. And our time greater than 0, it will be one. Which means that
the voltage before switching a time less
than 0, it is equal to 0. And after switching
it will be two volt. Okay? So here we have a condition
before switching. When this is closed, the voltage will be 0. And the when it is opened, the voltage will become 30 volt. Okay? Okay. So as you can see here, as our resistor
current cannot be, can be discontinuous
at time equals 0. Once the capacitor
voltage can not be. Hence, it's usually
better to find the voltage then
obtain the current. Okay? So what does this even mean? You know that here in the
requirements of the circuit, we need voltage and
the current voltage across the capacitor and
the current flowing here. So we will get this two by using the equations one
before switching. We will have current
when time is less than 0 and the time when
the time is greater than 0, similar to the
voltage t less than 0 before switching
and after switching. Now using is a voltage, we will get our count. Okay, So let's do this. So before switching,
before switching, it will be like this. For us, is our voltage here, as we just said, 0 volt and 30 volt. Now, before switching,
before switching, how can we represent
our circuit? You can see here
are timeless and zeros before switching means
that the voltage is 0. It means it is a short
circuit like this. We have a ten ohms, we have a 20 on lenses. We have the capacitor like this. We have that before switching. This is closed, so we will have the supply voltage
plus minus ten volt. Okay? So you can see the
voltage across the capacitor is equal to what? Equal tools or supply voltages. They are parallel to each other. So the voltage before switching, V 0, before switching
at time less than 0, it will be equal to the
supply voltage ten volts, because they are parallel
to each other. Okay? What about the current? You can see here we
have our current here. We need this current. Now if you look carefully, you will find that
the ten ohm here, this turn on is parallel
to that wins you all parallel to the capacitor. So what does this mean? It means the voltage
across 20 ohms, ten volt. Voltage across the
genome is also ten volt. So the current before
switching will be equal to voltage divided
by resistance. Current here is ten, all equal to ten ohms here
and divided by the voltage. Okay? But remember, remember something which is pretty,
pretty important. The current is
from here to here. From here to here. Flowing like this. However, is a
voltage ten volt is between here plus,
minus ten volt. Potential difference
between this point and this point is ten volt. And we need the current which
is opposite to this sign. So it means that our
voltage is negative ten. Our current should be equal
to one and negative one. So as you can see here, a voltage of the initial voltage of the capacitor ten volt. The current is equal to
negative v over ten, negative one and V, which is a ten volt. Now if you don't
understand this, it is really, really easy. As you can see here,
our circuit plus minus, this is ten volt plus minus also ten volt plus, minus ten volts. Okay? So here, the current flowing
like this, current flowing, Let's say for example, I x is equal to ten volt. Divided by our resistance, which is ten ohms
gives us one and bail. However, I is opposite to i x, so it will be negative one. Okay? Okay. It is very important to realize that the
direction of the current, so the capacitor voltage
cannot change instantaneously. So the voltage before switching equal to the voltage be after
switching what ten volt. Okay? Now what will happen
after switching? After switching? This one will be an open
circuit, as you can see here. So we have open circuit and supply this part as
if it does not exist. Okay? So we'll have a capacitor. We have at 20010 ohms, and we will have our voltage. Now, remember when prime
greater than 0 after switching, this supply will be
what will be 30 volt. So we will have a capacitor arm, arm answer two volts. So we will have
another step response. Okay? So what we need is two parts. First one, we need the
resistance and we need the voltage across
that capacitor. So as you can see
here, now, operative, you can see by using voltage division as
you can see here. Okay? So let's delete this. So this is open, so it will be removed. This part will be 30 volt,
as you can see here. Now we need the voltage
across the capacitor. The voltage across the capacitor
is a voltage across the 200 at time equal to infinity v infinity at
time equals infinity. This will become
an open circuit. So we'll have a 30 volt in ohms. And the voltage across the capacitor is equal to the
voltage across that 20 on. The voltage across
the two ohm is 30 multiplied by 20
divided by the summation. Sorting multiplied by 20 ohms
divided by the summation, as we'll learn it in
the voltage division. So the voltage at infinity
equal to 20 volt. And then we have the
initial voltage. Now, the last element
is what is R7? So if you look at
this circuit between these two terminals,
look at here, are seven is 20 ohm battery
to attain on. Okay. So as you can see,
ten on parallel to the 20th home as a
capacity of terminals. So we'll have 20
over three ohms, same as what we learned
in the serum lessons. Now we will write our equation after getting is
a time constant. Time constant is all seven multiplied by
Zach capacitance. All 720 over three. Capacitance is one over four. So we'll have five
over three seconds. So as you can see that
V as a function of time V infinity V 0 minus V infinity e to the power
negative t over tau. Tau is three over five
over three as I remember. Okay, five over 35 over three. Here, five over three can become three over five, as you can see. And V 0 is the initial voltage
Austin before switching. And V infinity is a
steady-state voltage. Okay? So the loss the elements, so we have the voltage
now we need that current. So we need the current
flowing through our circuit. So how can we do this? That current here
is equal to what? If you look carefully
at this circuit, you will find that,
Let's delete this. One is that the current is equal to the summation of
the two currents. Current flowing here, the
current flowing here. The current flowing
here through that when two ohm is the voltage of the capacitor
divided by the 20 ohms. And the current flowing
through the capacitor plus c d v over d t. So the summation of
these two equations will give us the current. So as you can see here, I equals V over 20
plus c d v over d t. So the voltage here divided
by 20 gives us this part. And see, which is
capacitance 0.25 for odd multiplied by d v over d t is a differentiation
of this equation. So it will give us finally one plus e to the
power negative 0.6. So we have our voltage
equation and current equation. What does this two
equations that represent, representing a time
greater than 0? So as you can see,
this one voltage here, current here after switching. And they will find
something important here. T less than 0, t less than 0. This has a value which we
obtained before switching V 0. And this has a current
before switching, which is just before switching. Now, you will notice here
something which is important. You will see here t less than 0 and t greater than
or equal to z. This one is t less than
0 and t greater than z. What is the difference here? You can see there is an equal here. So what does this mean? It means that the
voltage equation, enzyme capacitance, RC
circuits that step response. This Walter equation
is continuous. Now since we don't
have here an equal, it means that the current
is discontinuous. So you can see here
at time equal to 0, Let's substitute with
time equal to 0. You will find that
when two minus ten e to the negative 0, it means 20 minus ten gives
us a voltage of ten volts. So you can see is that t less than 0 or equal to t equals 0, equal to ten volt. This value equal to the
value at time equals 0, which means it is continuous. Now, if you look at this
equation of time equal to 0, here, you will find that that current is equal
to two and bear, however, the current
exhausted before switching negative one and after
switching to ambient, just after switching,
which means that the current is discontinuous. The value or not equal to each. Awesome. Okay? So this was another example on the step response
of an RC circuit.
85. Step Response of an RL Circuit: Hey everyone. In the previous lessons, we discussed the step
response of an RC circuit. Now we would like to discuss is a step response
of an RL circuit. So as you can see, we
have this circuit, we have a resistance, we have our supply, and we have our inductance. And we would like to get, since we are talking about RL, then we're going to get
I as a function of time. When we had all see, we get the voltage
as a function of pi. Okay? So before switching,
this inductor can have an initial stored energy or
cannot have any store Danish, depending on the case of
our electrical circuit. So then when we
close this switch, what will happen is
that this supply will start to providing
current through this inductor and the
start charging gate. Okay? So what do we need here
is that we need to get current as a
function of time. So again, our goal is to find the inductor current I
as a circuit response. So let's say that our
response is considered as a summation of the
transient response and steady-state response. So instead of doing KCL and KVL as we did in the RC circuit, we will do the easiest
method, method, which is classification or dividing our current
into two components, the transient and
steady-state component. Okay? Similar to that capacitance
or the capacitor, when we divide it into transient and
steady-state response. Or we can divide it also into natural and forced response. So first, our current equal to i transient blas,
or a steady-state. So what is a transient state? Remember that when we
discussed before that dr, source of free RL circuit, we said that the transient
response I is equal to I naught e to the power
negative t over Tau. This is a decaying
exponential decay. However, we don't know
the initial current, so we will say it is a. As an example, we will say it
is a constant, call the a. Now why is this? Because if you're
returning back here, first, this response is
that natural response. To be more specific
natural response. Here we are talking
about the transient, which includes what
includes the Zan, natural response and the
charging of the inductor. So as an example, let's take it back so
you don't get confused. Okay? So here this is a false
and then natural, okay? So transient here you can
see is that the transient for the voltage consisting
of two components. One which is v-naught
negative t over tau, which is a natural response. And the other component is VS e to the negative t over tau, which are representing
the transient response of the application of
the voltage source. Okay, So we have
natural response, the natural response
and forced response, which has a transient, the component of
the force response. Similar to this we will
have in the current, we will have I naught e to the
power negative t over tau, which is a natural response. And another part which is
the force of the response. Okay? So in the end you can see that this equation can be
written as VT, for example, equal to cornerstone
to call the a, for example, e to the
power negative t over tau. Okay? This constant in the RC
circuit is V naught minus v s. Now it's the same
idea for the RL circuit. I can say is that the
transient component or transient is equal to
a certain constant a, which I don t know, e to the power
negative t over Tau. Okay? So that constant a in this circuit for the RC
v naught minus v s for RL circuit. I don't know it yet. Okay, So we will get it
later in this lesson. So if we get back here to the RL circuit, like sense here. So here you can see that we have that transient response a e to the power
negative t over tau. And tau is L over R,
similar as before. Now is the steady-state
component. What is the
steady-state component? It is a component when the
circuit reaches steady state. So if we look at this circuit, when we close the
switch and the inductor reaches a steady state due
to the application of what, remember, DC source, not AC. But DC, this one will
become a short circuit. The inductor will become
and short circuit. The current flowing
inside our circuit. What it will become, it will be V supply over r,
the steady-state current. So as you can see here, is
a steady-state response is the value of current
after a very long time, after the switch is closed, the inductor becomes
a short circuit ends and tire source. Voltage VS appears across
are all of the voltage here. We'll go through our, because we will have here
a short circuit. So in this case, the steady-state
response or the current at steady-state conditions
will be VS over R. Okay? So as you can see, if you apply the same principle here
to the RC circuit, you will get the
equation required. Okay? So here's a transient eight e to the power negative
t over tau as a steady-state VS over R. So we can say is that
the total current, a negative t over
tau plus V over r. Now what we would like to get is that a or the constant a. How can we do this? We know that at current, at time equal to 0, which is innocent of switching, we know that the current
will be equal to what will be equal to I naught, which is the initial
currently before switching. Okay? So the current before switching equals the current
just after switching. Why is this? Because if you remember, if you remember that
our inductor does not allow the I over d t does not allow that sudden
change in the current. So the current cannot
change instantaneously. So it means that the currently
before switching I equal to I 0 bolster equal to
initial current I naught. So before switching equal to, after switching equal
to the initial current. Okay? So from this condition, at time equal to 0, we will have current
equal to I naught. Okay? So let's delete all
of this like this. Okay? So before
switching quality, after switching, or the
reverse equal to I naught. So from this we can
say I naught equals a e to the power negative
0, which is one. So we'll have a plus VS
over R. So from here we can get that our constant a is equal to I naught minus v s over r. Okay? Okay. So we will take this one and substitute it here like this. So our current as a function of time is equal to V S over R, which is this component. Plus is a transient component, e to the power negative t
over tau multiplied by a, which is I naught
minus v S over r, naught minus V over R. Okay? So if you look
carefully at this, this equation, you
will find that this one can be
represented like this. All as a function of time
equal to i infinity plus I 0 minus i infinity e to the
power negative t over tau. Similar to the RC
circuit, the RC, we said V as a function of time equal V at infinity equals v 0 minus v infinity e to the power negative towards that was
similar to each source. And here you can see
this i infinity here, which is a steady-state
value VS over R. And I naught, which is
the initial current INR, which will be obtained
from the circuit itself. Okay? So of course, is
your online infant or the initial and final values of the current respectively. Now something which
is important here, someone will tell me, okay, is this circuit is
an open circuit. So how do we have any current or why is the initial
current exist? Okay, is this equations
represent the general case, not just as this circuit
only, but in general. Okay? So what does this mean? It means that for example, if we have a resistance
like this, okay? So it means that this,
if it has stored energy, it will provide that current
through this resistance. Okay, so we will have an initial current
depending on our circuit as we will see inside the examples of this
part of the course. So what about the voltage? Voltage, as you remember of the inductance RL circuit or the inductance to
be more specific, L over d t inductance
multiplied by the derivative of this equation will give us l over tau r e to the power negative
t over tau with this, when time is greater than 0, when this switch is closed. So this one can be
written like this. V as a function of time. We as L over R, which is equal to tau. This part is equal to tau. So tau, tau will cancel. The jaws are so we'll have e to the power
negative t over tau. As you can see here. And you as a function of time,
what does this mean? It means before it's switching, it was 0 and after
switching will be one. So it means that V
as a function of time before switching, it is 0. And after switching it will be V S e to the power
negative t over tau. Okay? So as you can see here, the current before
switching, it was 0. Here is this graph representing
what I naught equal to 0. You can see step response of an RL circuit with no
initial inductor current. So we assume that our initial, this part is equal
to 0, equal to z. So as you can see before
switching, it was 0. Then after switching it will start increasing
exponentially. The reaching i infinity, which is V over R. Okay? Now what if we have I-naught? What if we have initial currents and we will start like this. We will have i-node like this. Then switching it will be
increasing exponentially. Okay? So if we have INO that will
be just shifted, okay? You can see that
the function here, or this graph, or both
of them are continuous. However, what about the voltage? You can see that
the voltage before switching at t less than 0, v as a function of
time is 0, okay? And after switching
V as a function of time would be V S
negative t over tau when prime greater
than 0 from this equation. So before switching as this
voltage was equal to 0, why? Because the inductor
reaches a steady state. So it was a short-circuit, which means that the voltage
is 0, as you can see, and then suddenly boom
after switching it, it changes from 0 to V S. That's why this function
is discontinuous. You can see t greater or less than 0 and t
greater than 0. There is no equal here. Because there is a
step part here in the voltage response. Okay? So we discuss disaster
response of an RL circuit, which is similar to the same
idea of the RC circuit. Now we would like to have some examples to
understand this.
86. Example 1 on Step Response of an RL Circuit: So the first example on the step response
of an RL circuit, we have a switch which was closed for a very
long time, like this. Okay? And that time greater than 0, at time equals 0, we
open this switch. Now what I would like
to get is the current flowing through the inductor
as a function in time. Okay? So first, how can we
do this assembly? We need 0 or infinity. We need what are seven to get? Tau. Okay? So we can get all as
a function of time for us to before switching
or 0 or initial current. So when the switch is closed
like this, what will happen? You can see a short circuit
parallel to is a three ohm. So what does mean? It means these three
ohm will be removed. So we will have here in the
end short circuit like this. Okay? So we have a ten volt at
two ohms and the inductor. Now we said that the
initial current here, after a very long time, we open the switch
after a very long time. It means that this inductor
reaches steady state. So the steady-state is that this inductor will become what will become also a short
circuit like this. So again, as this
switch was closed, so it is a short-circuit
better two or 30, it will cancel the three. And then we have our inductor, which reaches a steady state
before opening the switch. So it means it will
become a short circuit. So we have a ten volt ohms and all of this
ashore the circuit. So what does the initial
count or a naught, which is the current day before switching equal current
after switching equal, what was a ten volt divided by the total
resistance, which is a tool. So it means it will
be five and Ben. So as you can see here is
a current before switching and after associates
equal to each other called ten over two ohms, because both of them
are short circuit. So we have the initial current. As you can see, r is 0
equal Powerball stiff, your eyes, your narrative
equal five and bears. Okay. Okay, now what we need, now, we need to open the switch. The switch opened, and
they get all infinity, which is steady-state current after a very long time
of opening this switch. So we have a ten volt, we have that two or three
ohms and inductance, the switch as if it does not
exist since it is opened. Now what i infinity
it means that this inductor reach it against
the steady-state. So it means it will become what becomes a short circuit like us. Okay? Okay. So we have a ten volt, we have a two ohms, we have a 33 ohms and current. So what is the current
infinity equal to the ten volt divided by
the total resistance, which is two plus three
equals five ohms. So it will be equal
to two and bear. Okay. Now the last element or
the last parameter we need is resistance or seven at the terminals of
the inductor here, look at this circuit like this. So this does not exist. So if we look at the circuit, this will become
a short circuit, or seven become a short-circuit. So when we look at our circuit, we will have a 30 plus to omit means that we
have five ohm or 70. So as you can see, the two
ohm and three Omar series so that i infinity equal to m
bear as we just explained. Then R7 and will be as a
five ohms because they are the two elements as R7 at the two terminals
of the inductor. So we can get there,
which is L over R. Okay, so it will be one over 15. Now we have all of our
parameters such as always 0 or infinity and Zotero. So we can write our
equation like this. I infinity plus 0
minus infinity e to the power
negative t over Tau. So we will have our final
equation representing current.
87. Example 2 on Step Response of an RL Circuit: Now let's have another example. In this example we
have an extra point. So we discussed before one
switch or wind and the closed. Now we have two switches. Okay, so let's see how
can we deal with this. So at time equals 0, the switch is closed. So this is the initial
position of S1. This is the initial
position of S2. So at time equals 0, this one will be closed. This is switch two is
closed at time equal four. Okay? So let's close that
time equals 0. This closed at time equals four. Now what we need is that
we need the current of the inductor as a
function of time. Then we will get the
value at t equals two seconds and t
equals five seconds. Okay? So how can we deal with, as you can see, we
have three regions. So our current will be divided
into three sectors, okay? The first one is before
switching t less than 0. Okay? So before switching,
what we have here, we will have, we will see
our, our current, okay? Before switching. Then from t equal to 0
to t equal four, okay? We will obtain our current. And then from T greater than four seconds
equal to four seconds, we will get our equation. So we have 123 regions, okay? Okay. Second thing
you can see here, less than or equal.
Less than or equal. Why? Because our current
is continuous in RL, the current is continuous. In RC, the voltage
is continuous. Okay? So let's just start. First step is a time
less than z, okay? Okay, so as you can
see, from 0 to four, t less than 0 and t
greater than the, from 14 less than 0. What will happen? Okay, let's see. So this one is opened. You can see there is
an open circuit here. Okay? So what about this
part is open circuit. So this part is eliminated. It is removed from our circuit. Now, this switch is also opened. As you can see, this
part is also deleted. Okay? So as you can see, we
will have only as six. So it will be like the six ohm series with our
five henry inductance exist. This point open circuit. And this part will be also
an open circuit. Okay? So what does the
current flowing here? What do you think is the current is flowing through the inductor? All of the sources
does not exist. All of that system
is open circuit. So the current flowing
through that inductor before switching any of these
two switches is equal to 0. So the current before switching equal current after
switching equal to 0. So at time less
than or equal to 0, the current will be equal to 0. Okay? So the first box, okay? Now, what about the second part? Second part here, we
close this switch, okay? Time greater than 0, from 0 to four seconds. We close this switch. Okay? So we have a 40 volt, like this, 40 volt. Okay? We have czar, four ohms. Then we have here the switch. This one is still open, open, so it will be
an open circuit. So we will have the six ohm. And then we'll have
five henry. Five Enter. Now we need to do the equation. Write the equation of this part. Okay? So what does the equation
of this part you can see is that remember the equation which contains 0 or infinity
and tau. Okay? So first, what is tau? Tau is L over R. So the inductance is five
Henry and a resistance. As you can see, if you
look at this circuit, you will see that
there resistance is four ohm and six ohm,
which is a tenon. So this part is turned on. So here we can get tau. What about 0's that
initial current, of course, from the
initial condition here it is equal to 0. Now what about infinity, which is a steady-state current? If the switch is closed
for a very long time, what will be the current here? It will be like this, this
five Henry will become a short circuit after
a very long time. So the current flowing
here will be infinity, will be 40 volts
divided by ten ohms. So it will be four and bear. Okay, so we have
infinity or 0 and tau. So we will write our equation. Okay? Just ignore this switch. Don't, don't think about it. Just treat it as a
separate circuit. Okay? Okay, so as you can see here, is a current infinity equal 40 volt divided by four plus six, which is ten ohms. Four plus six, or 74 plus six, which is ten ohms. So we will have tau L over R, as we just said. So we will write our
equation as we normally do. Okay? This equation is valid
forward from where from 0 to four seconds, okay? Because after four or
starting from for, this switch will be closed. Okay? Okay, So we have
here this equation. Now what do we have to do next? So we have the second equation. First equation is that
alpha t less than 0, the current is equal to 0. And from 0 to four
this equation. Now the next step
is that we need to, we have this switch
closed, okay? Now we have this
switch also closed at time equal four, same as before. I need 0 before zoster, before switching and infinity just after switching or
at steady-state and tau, which is the time
constant. Okay? So the currently before, just before switching is equal to the current
adjust after switching. How can I get this
by substituting t equal to four
in this equation? Because the current at
time equal four, remember, current is a
continuous waveform. So at time equal to four From this equation and
substituting here was always t equal four. We will get our current, which is the current
just before switching, which is equal to the current
just after switching. You can see here. This does not
affect the inductor because the current cannot
change instantaneously. Therefore, the initial
current are equal four. We will substitute in this
equation with t equal four. So four multiplied by negative two gives
us negative eight. So in the end, our
current will be approximately equal
to four and bears. Okay? Okay. Now what does an extra step? The next step we have
sort of presenting I 0 of t greater than or equal for the equation which
represent d greater than four, I is 0, is four and bear. Now we need our infinitive
after a very, very long time. So after a very long time, you will see that here we
have a source, 40 volt. We have a four ohms. We have this switch
closed to arm. We have a ten volt plus, minus ten volts. I exist. We have the six ohm. Now, I infinity. It will be, this Henry will be, or this inductor will
be short circuit. Okay? So what I need is the value
of the current flowing here. The current flowing through
the resistance six. Oh, okay. So how can I do this? You can see that here
we have voltages. We have voltages. So we can apply KCL, apply KCL at this point. Okay? So let's do this first. You can see here KCL,
we have this current. We have three current
to current entering and one current leaving the
current year entering. As an example, we will assume this to entering and
the current leaving. The current entering
this node here. The current flowing
here is equal to what? Equal to the 40 volt minus
the voltage will be, let's say is s1 is denoted by V. So it would be 40 minus
V divided by four ohms. Similar as what we did in
the KCL part of the course. Here we have another
current which is then volt minus V divided bys a tool. Then volt minus v divided by the two ohms equal to
the current going here. The current going
here will be equal to V minus this point, which is the ground,
divided by six ohm. So it will be V, V minus 0 divided by the SEC. So as you can see here. It would be V over six. You can see this equation is
an equation in one unknown, which is V, which is
voltage of this node. So here we will have
v equal 180 over 11. Now I need the current. So the current will
be this voltage V, which is V minus 0, divided bys at 60. Like this, like this, v over six from
the KCL equations. Now, before I close this, the next step is
that we need Tao, which is L over R. L is five Henry and the R is R7. We have here our
inductor in this point, our inductor here between
this point and this point. Okay? So let's just read this. Okay, so we have this
a short circuit. This one is also
a short-circuit. And I need all seven in between this point and this point, between the terminals
of the inductor or 70. So if you look at the circuit, we have how many? We have two independent source, two independent sources to get R7 and we will
deactivate them. This will become
a short-circuit. This will also become a
short-circuit similar as what we did in the seven lessons. We will have the six ohm series with for Own butter to two. So this combination
for parallel to two is series with a six or so, as you can see, for
butter to the two, series was a six. So we'll get our 722 over three. Now, tau is L over R seven, and so we will have
15 over 22 seconds. Okay? Now, let's write our
equation for the final time, I infinity plus i4
minus infinity. Now, something important
here you can see negative t minus four over Tau. Now why is this? Because this equation starts
from four, so it's shifted. You can see t minus four
is exponential because of the time delay it is shifted as this occurs
at time equals 0. So t will be e power
negative t over Tau. Now, if it occurs at
anytime it will be E negative T minus, let's say T naught over del t node representing
the time of switching. If it is at four seconds, it will be minus four. Like this. If it is a 6 second, it will be t minus six. So I as a function of time, will give us this equation. A minus two minus four over tau. We have this infinity
and the initial current. So this is the equation which is representing the T
greater than four. Now finally, we will
put all this together. We will have all as
a function of time. When t less than
0, it will be 0. When t between 04, it will be this equation. And T greater than four, it will be this equation. And finally, what
we need is at t equal to and t
equals five seconds. So at t equal to, which one of these we
will use at t equal to t equal two is in this range. So we will use this equation. As you can see, the
time equal five, which is greater than four, we will use this equation. As you can see. Why, because five is greater
than 42 is between 04. Okay? So this was another example on the step response
of an RL circuit.
88. Introduction to AC Electric Circuits: Hi, and welcome everyone
to this port or this section in our course
for electric circuits. In this section, we
are going to discuss the AUC or the alternating
current electric circuits. Now you have to know that AC electric circuits are
really, really important. Why is this? Because you
will find the AC circuits in power electronics,
in electrical machines. And the AC system is
the one which we are using in transferring
electrical power. So it is really,
really important to understand what does easy mean. Ands are different concepts
related to AC circuits. So first, we have to know
that there are two main types of electric current or
electric voltage in general. Current or voltage. First one, which we are, which we have discussed in our course for
electric circuits, is called the direct current, or DC, which is a current that remains constant with time. If you look at the current
with respect to dy, as time pass, you will
find that the value of the current is constant and
it is a positive value. So when this current has
a constant value with time or constant direction, and it's really,
really important to understand that
direction concept. So e.g. if we have
something like this current with
respect to time, and this current is negative, negative all the time. In this case, this was also called DC current or
a direct current. Why? Because it has one
direction or unidirectional. Unlike the EC, which is a changing its direction
all the time. Okay? So let's see AC. If you look at the AC
or alternating current, it is a current that varies
sinusoidally with time. Or it is in the form of a
sine wave or a cosine wave. So you can see that that current is changing the whole time. And most importantly, that, that current, it
changes its direction. If you look at this waveform, you'll find that we have
a part of the time, this part, the current
is our bolster value. And other times you'll find
that the current is negative, then positive, then
negative value. So you can see that
this current value of the current is
changing with time. Virus was time, sometimes
positive, sometimes negative. That's why it's called
the alternating current. It alternates or keeps
changing its direction. Unlike the DC or direct current, you will find it is always
positive or always negative. So if you look at these
two currents in real life, you will find it like this. You will find that for
DC we have a bowl safe terminal of the battery and negative terminal
of the battery. The battery is a source
of a DC current. You'll find that
the current itself has uni, direction,
one direction. That's what's called direct
current goes from Paul, stuff like this to negative. Okay? Unlike the AC current, which we usually
denote by this sample, you can see the sample. The sample which is
almost like a sine wave. Okay? So when you see a supply
with this sample, it means that this supply is an AC current or
alternating current. You can see that the current
is switching into direction, one going positive than
other times goals like this, you can see sometimes like
this part of the time, and then it switches its
direction in the other time. Unlike the DC, which is always
in one direction. Okay? Okay. So you have to understand
that AC voltage it can be, or AC voltage or current. You can see that if we
have an AC voltage, you have to know AC voltage
will produce AC current. So when I say AC
voltage or AC current, they are the same. It means that the supply
itself is alternating. So you have to know that AAC itself or the
alternating current, it can have many, many different waves or different wave forms
or different shapes. If you look at a, C e.g. you will find that
AAC can be like this, can be sine wave, or it can be a cosine
wave like this. Cosine wave like this. Okay? So both of these wave
forms are called as a sinusoidal voltage or
an AC sinusoidal wave for both of them are
called xyz sine wave and cosine waves used to
represent the AAC system. Okay? Now, if we, if we have, is there any other waveform
is for voltage or current? Yes, there are other forms. You can see is that we can
have something like this. Like this triangular
waveform like this. This is called also AC because it has a power to pull
step and Paul to negative. Now also we can have a square waveform like
this, square like this. Okay? So all of this can be produced using different power
electronic circuits. Okay? So in the end we have AC, which means it is alternating. Now we usually, when we
are talking about with AAC system or AC generators, we usually have this
waveform, that sine wave. Usually, if you look
at this waveform, which is a sine waveform, it is the one which is generated from our electric generators. So the sinusoid is a signal that has a form of sine
or cosine function. So when we say sinusoidal,
the generated voltage, it means that it
is in the form of a sine wave or a cosine wave. The sinusoidal current is
usually referred to the AC, referred to as AC or alternating current is
switching its direction. So such a current is called AAC because it reverses
its polarity. Sometimes all steps,
sometimes the negative at a regular time intervals. Now the circuits
which are driven by sinusoidal current or
a voltage sources, are called the AC circuits. Now of course, as a power generated and transmitted
to our homes. Or is a form of an
AC, sinusoidal wave. Because it is an easy to
generate and transmit. If you'd like to learn how do we generate this
waveform is you can refer to our course for
electrical machines. Now, what I would like to understand here is
that you will have to, Since the same
waveform is but in, as a function of two
different parameters, you will find that
this waveform, which is a voltage as
a function of time, voltage as a function of time. You will find that here we're representing
here as a function of omega t. Here we are representing as
a function of time. Let's first understand
what does that tie in? With respect to time? You will find that
we have like this. You can see from zero keeps
increasing until peak value. Then it starts
decaying to 01 small. Then it starts going
into the negative part, then starts increasing
back to zero. So you can see
from here to here, we have a part which is
a negative pollster, and another part
which is negative. Now, this is called one cycle. We have one cycle. Okay? Now, this cycle, this one cycle occurs
in a time called t. Okay? So T representing what representing the time
taken to make one psych, which is formed with the
football stuff and connect. Now, you can see after
time t we have again one positive and
Amazon negative, which is takes time T. So we will reach to t. We have one t plus t gives us two t. So t is aperiodic time, periodic time is
the time taken for the waveform to make
one Psych, one cycle. Now, you will see that it is corresponding to when we
are talking about omega t, which is the angular frequency, Omega t. Omega is the angular frequency
multiplied by time, which is omega T
representing angle. If we would like it as
a function of angle, you'll find that from zero. Two Pi means half cycle. And from pi to two pi, the other half or one complete cycle is
corresponding to two pi, or in radian or in degrees, it will be zero hundred
and 60 degrees. Okay? So we say that two pi
is one complete cycle. Two pi is one complete cycle, which is corresponding
to our time equal t. Okay? So if we would like to represent as this
wave form assembly, we say v as a function
of time or the voltage, or the sinusoidal voltage is equal to V maximum sine Omega t. What does V maximum represent? Each row representing the
magnitude or the amplitude of the wave or the maximum value reach it in the post of
cycle or negative cycle. So if you look at here, we
will see that this bar, this point is a maximum value. That's why it's called the Vane V maximum, maximum voltage. And then the negative cycle, we have negative V maximum. So the maximum point is
V maximum sine Omega t. Omega is called the angular frequency
and t is our time. So you have here VM
is the amplitude or the magnitude of the sinusoid
or the sinusoidal wave. Omega is called the
angular frequency. How many radians per second? And Omega T is called
the argument of the sinusoid or the
angle of the sine wave. The current angle as
a function of time. So finds that the sinusoid
repeats itself every t, and that's why it's
called the periodic time or the period
of that sinusoid. From the two plots,
as you can see here, at two pi, after
one complete cycle, we have the periodic
time t or the two pi, which is omega t
equal to two pi. Okay? Omega t. Now what is the time which
is corresponding to two pi? If you look here at two pi, we have a time called t, which is a periodic time. So he says that omega
t equal to two pi. So after time equal t, our angle will be two pi. So omega t will be
equal to two Pi. When we are comparing
these two figures, you will find that that
periodic time is equal to two pi over omega, okay? And you know that Omega equal to two pi multiplied
by the frequency. Frequency. What does the frequency means? So we have frequency, which means how many cycles
are obtained in 1 s? So after a time called 1 s, how many cycles are formed? This is called the frequency. How many cycles per second? So the relation between omega or the angular frequency is
that we take frequency, which is how many cycles per
second multiplied by two pi. So if you take this
to this equation, you will find that the
periodic time will be equal to one
over the frequency. So here you will
find that t is equal to two pi divided by two Pi f, which is one over f. Or the frequency itself
is equal to one over T. So finds that the frequency, or how many cycles
is one over T. Okay? Okay. So you will find is that you
will always hear this point, is that the electrical system is operating at a frequency
of safety hurt us. Or other country is
operating at 60 hz. So what does this mean? It means that after
atomic while 1 s, e.g. for the 50 hz, after
time equals 1 s, we will have 50 cycles. Okay? So in just 1 s, the supply is a switching
positive, negative, positive, negative
50 times in 1 s. So you can see it as
really, really fast. Okay? So what does this, what does a frequency mean
in our electrical system? So here is an example. If you look at the
alternating current in the AC system or in
our electrical system, you will find that when
we connect AC supply, such as the one which
is in our socket, when we connect it to the
pulp or the lambda e.g. what will happen is that you
will find that the current, since it is an AC
current switching, sometimes positive,
sometimes negative. So you can see it's switching
all the time like this. So what will happen is
that supply starts from 00 means that this bulb has zero voltage or no illumination. Now, as a voltage increase, the illumination of the NAM starts to increase.
As you can see here. Stuart is illumination starts to increase until we
reach maximum value, then the illumination
source to decrease or the light of support
starts to decrease due to, due to decrease in voltage. Then starts to increase again
in the opposite direction. So the light will increase. Then light will start decreasing again
until reaching zero. So you can see that
the pulp itself, its light, is it changing, increasing until reaching
the full illumination, then starts decreasing, reaching zero, and
then increasing, maximum value,
decreasing and so on. However, when we
look at any pole, but in our home, we don't disease as
we see it always on or always giving
the full illumination. Now why is this? Because this cycle happens
sick is t times in just 1 s. So our eyes cannot see
this fast exchange. So that's why you
will always see some bulb with
full illumination. Okay? Now, general expression
for an AC sinusoidal wave. So V as a function
of time is equal to V maximum sine Omega t plus Phi. So we have omega t, which are representing
that changing with time, change with time. However, we have an
additional element, which is called Phi. Now what does phi mean for
representing the phase angle? We say that phi is
called the phase. Now we have to understand
that we can represent this in the form of
radians or degrees. Now what does this mean? It means that we can say
sine pi or we can say e.g. sine hundred 80 degrees,
as we would like. According to the representation
of this equation, you can say phi
in-degrees on, in radian. Now, what does five mean phi
representing a phase shift? We say it phase shift. Now, why does this even happen? This phi or this
phase shift occurs due to the presence
of different loads. As an example, you
will find that when we have an inductive load, we can have this phi
as plus 90 degrees. If we have a resistive load, it can be zero. If we have a capacitive load, it will be negative light. We will see this in
the next two lessons. But for now, So Phi in general is dependent
on the load itself. Okay? So how can we represent
this phase angle? So let's say we
have two voltages. V1 was out with any
phi or phi equals zero and v2 which has plus phi
or an phase shift Phi. So if I would like to
represent them on a graph, how they will look like, they will look like this. You will have V1, which is V maximum sine Omega t. So it means that at
time equal to zero, value of the voltage
will be zero. So it will start from
this point, from zero. And at Omega t equal Pi
over two or 90 degrees, you will find that we will
reach the maximum value. And we will zero to pi. Zero will be zero. And here at three pi over two, it will be negative V-max. Okay? So this is the original waveform which are exhausted
discussed before. Okay? Now, if we look at V two, now we have V maximum
sine Omega t plus Phi. Now omega t, This one at time
equals zero, it is zero. Like this. Now, what about
this one at omega t equal to zero
in this equation, v2 will be equal to
V maximum sine phi. So it is not equal to zero, it has a certain value. So at omega t, at omega t equal to zero, the second waveform will have a value here and
certain value here. And you will find
that at omega t equal to negative
fi, at this point, negative four, you'll find that the
value of the voltage will be equal to zero sine negative
y plus y equal to zero. So you will find that the
voltage is V2 A-star. Before V1. There is a phase shift, a lag between them. This lag or lead is known
as phi or the phase shift. So all finds that this
phi lead to what lead to V2 starting earlier than V1. Okay? So, and so this phase shift occurs
due to different loads. So we'll find that
starting point or veto occurs first and time. So we say is that
the voltage V2, this way waveform is
leading, leading V1. Now, why it is leading? Leading by an angle Phi? Now why does leading it
means that it is faster than IT or it started
earlier than it. Okay, so we are saying
it is a leading V1, or we can say is the reverse. We can say that V,
V1 is lagging or behind V2 by an angle Phi. So lagging means behind or late. However, leading means elderly, or it means it is
leading or before. Okay. So when you hear
lead and the lag, you now understand it means
there is a phase shift. One of them is behind the other one or started
earlier than the other one. Now in this case, when
zeta is a phase shift, when there is a difference
in angle between them, we say that they
are out of phase. They are not having
the same phase. However, if both of them, if both of them
have the same phi, Let's say this one is plus
phi and this one is plus phi. So we say that z
have the same angle. So we say that the in-phase, or if we have phi equal to zero, and this one is
the sine omega t. So it means that they
are also in phase z are flowing goes each of us. In phase means that they
reach their maximum and the minimum values at
the exact the same time. Now, this comparison is valid when when both of them
have the same frequency. So both of them should be having the same frequency so we
can compare between them. And not necessarily have the same maximum
and the minimum. But the most important
thing is that they have the same frequency. So as you can see here, we say that this is v2. So we say V2 is leading
V1 by an angle Phi. Or we can say V1 is lagging
or behind V2 by phi. In this case, both of
them are out of phase. And if the angle is zero, it means that both of them are in-phase or moving with each us. Now, here are some rules of sine and cosine because
it is important if you would like to convert two waves or finds a
phase angle between them, both of them should be sine
waves or cosine waves. That's why you need
to understand how to convert from sine to
cosine or cosine to sine. So here are some rules
which she can help you to convert from sine to
cosine or cosine to sine. Now in the next lesson, we are going to have some
examples on sine and cosine functions or to be
more specific, phase shift. And how can we obtain the
phase shift between two waves?
89. Solved Examples 1: Hey everyone. In this lesson, we
are going to have some solvent examples
on czar sinusoids. So in the first example here, we have this one which we
need to find the amplitude, the phase, period, and
frequency of the sinusoid. Which sinusoids is one. We have V as a function
of time equal to 12, cosine 15 plus ten. So the first step is that we
need forest the amplitude. So if you look here and
compare it with V as a function of time equals two me maximum sine
or cosine here, cosine omega t plus phi. So if you convert this
equation with this one, you will find that
the amplitude, which is the maximum
value V max, is equal to 12. So the amplitude is
equal to 12 volt. Second one is the face. Now, if you remember,
phase is our phi. If you look at this equation, we have 50 t plus
the phase shift, which is our phi. The phase shift will
be ten degrees. So the one which is a
period, what is the period? The period is how much time it's taken to form
one complete cycle. So if you look at here we have
50 T and we have omega t. So from this equation, we can find that omega is
equal to 15 radian per second. From here, we can find that
omega equal to two pi, multiply it by the frequency. From this equation. If we combine these
two equations, we will find that f, or the frequency is equal to the 52 is omega divided by two pi, which is the frequency required. Now what does that periods
or period assembly, which is t, is equal to one
divided by the frequency. So it will be one over f, which is two pi
multiplied by 50, right? Or two pi over omega. So find that the
angular frequency, which is 50 radian per second. So the period T is equal to pi over omega
or two-pi over 50, as you can see,
which is 0.125 7 s. What does this mean? It means that 0.125 7 s is the time required to
form one complete cycle. So if you look at
here at our graph for the voltage as a function
of time with respect to, let's say T with
respect to time. So we have one complete for, one complete cosine
for this cycle is completed at the
time equal 0.125 7 s. So in order to move from here, this whole time is one
period which is 0.125, 7 s. Now what is the frequency? Frequency is one over T. So it will be equal to y exists equal to one
over t, 7.958 hz. Now what does this mean? It means that in 1 s, we will have almost
eight cycles. Okay? So the frequency here is 7.9. 8958 is equal to how many cycles in 1 s. So
it is almost eight cycles. We have eight cycles
in a time of 1 s. Okay? Now, let's take
another example here. We have these two voltages. We have V1 and V2. And we would like to
get the phase angle or the phase shift between
these two voltages. We would like to know
which sinusoid is leading. So first, in order to compare
between two voltages, they must have the
same frequency. So if you look at here, we have omega t and we have Omega t, which means that they
have the same frequency. And we said before, the magnitude is not important. It's the same magnitude
is not a condition. The most important
thing is that they should have the same frequency. So if we, if we would like
to convert between them, we need to do
another thing which is both of them should be sine waves or both of them
should be cosine waves. Okay? And both of them should
be positive or negative, have the same sign. So the first step, in order to convert
between them, we must express them
in the same form. If we express them in a cosine form with
positive amplitude, then we will have like this. Okay, let's delete all of
this so we can express them in a cosine form
or in a sign for. So here e.g. I. Will express them in sign for. So if you look at here, we have sine omega T plus -90 degrees is equal to
plus minus cosine omega t. So here you will
find that for V1, you see v2, both
the value and sine. Now I would like to convert
this to a positive ten, both the value with a sign. So what we need is we
would like to convert as a cosine N2 sine. So you'll see that
sine omega T plus -90 degree is equal to
plus minus cosine omega t. Let's use this rule. So you can see that we have here negative cosine omega
t plus 50 degrees. So the first step
is that you will find that the sign
here is negative. So plus, minus, plus, minus. So we select a negative. So we'll have here is the angle negative negative 90 degrees. The first step, second
step is that we will replace each Omega T
by Omega T plus 50. So we'll have cosine
Omega t plus 50. So it will be, this one will
be omega t plus 50. Okay? So we'll find is that n, z and cosine negative
cosine omega t plus 50 is transfer the two sine omega
T plus 50 minus nine. As you can see,
sine omega t plus 50 -90 degrees using this rule. So you'll find that
when we remove this, 50 -90 is negative 40 degrees. So we'll have ten sine
Omega t -40 degrees. Okay? Now, you can see that here we have sine omega t minus ten. So what I'm going
to do is that I will divide this into two parts, Omega t minus ten
and -30 degrees. So this summation is
omega t minus four. Now why did I divide
it like this? So I will consider this one. Similar to this part. We'll find that the phase
shift is the negative third. So the phase shift, if you look at these
two waves in the end, we have a phase shift
of negative 30 degrees. Okay? So we'll find that
V0, V1 is lagging. V2 by 30 degrees, or V2 leading V1 by 30 degrees. So I'll find that v2
leads V1 by 30 degrees. Why? Because if you take this one, omega t minus omega t
minus ten, same angle. But you will find that here, the difference between
them is a negative. So V0, V1 is lagging
by 30 degrees, or V2 leading by
this salty decrease. So V2 leads V1 by certainty. Now we have to
understand that here. This one is sine four weeks. A breast exam in assign form, not a cosine form because
you can see sine and sine. Okay? So this was another
example on the sinusoids.
90. Phasor Representation of AC: Hey everyone. In this lesson, we will
discuss another concept in AC electric circuits
called the phasors. Phasor is a complex
number that's representing the amplitude
and phase of a sinusoid. So the goal here, and instead of using the
voltage or the current as a function of time in
the form of sine or cosine. I would like to accept, press it in the
complex number form or in the phasor form. It is much easier to express our voltages and currents
in the form of phase. Okay? So let's understand
more about this. So here, if you remember, if you remember from
complex numbers, okay, if you don't know
about complex numbers, please send me a
message and I will send you a free course
on complex numbers. If you don't know
about complex numbers, then you will not
understand this. You need to know about
complex numbers. So if you remember
in complex numbers, so we have three main forms. We have a complex
number form, which is, which is our complex numbers
equal to x plus j y. Or to be more specific, it is consisting of two parts, the real part and
the imaginary part, the real part of the complex number and imaginary part of
the complex number. This form is known as the rectangular
form, a complex ion. We have another form which
is called the polar form. And this form we'll
use the magnitude of the complex and non-power
and the phase angle. And the last one we
have exponential form, which is a magnitude e
to the power j and phi, which is a phase angle. So we will find that r is
the magnitude of that. The phi is the phase angle
of the complex number. Okay? Now, how can we obtain
r and phi simply, if you have x and y, then the magnitude
assembly equal to root x square plus y square is a square of the real part plus the square of
the imaginary part. And the phase angle
assembly equal to ten minus one y over x, or the imaginary part
over the real part. And also we have another form. If I would like to
get x component only, then it will be r cosine phi. If I would like zap
imaginary part y, it will be r sine Phi. So we will have this final
form of our complex number. Now, if I would like
to represent on axes, and this is really, really important
because you will find this in electric circuits alone. You all find that usually
when we are talking about real part and
imaginary part, we usually talk about
also the rail power, real power, which
are represented by the power absorbed
by that resistor. And the imaginary part, usually we talk
about with Zach Q, or the active power, or the power that we store in our inductor
or our capacity. We will learn about this
concept later in this course. So we have here z
equals x plus jy. So real part, the real part is x and the imaginary part y. The summation of this two
vectors that gives us z, which is a complex number. The magnitude of z, you can see from this
triangle 90 degree triangle R is equal y
squared plus x squared. And the angle Phi, which is measured
from the real axis, remember it is
measured from here. So if we have vector like this, that is like this, it means that phi
is equal to zero. So as a phi positive, then it is measured the
legs, this poster phi. If phi is negative, it will be measured the
Lexus from the other side. Okay? From here you can see is that phi or tan phi
is equal to y over x. That's y phi is equal to
ten minus one y over x. Okay? Okay. And from here you can see that if I would like extra component, it will be R cosine Phi. Cosine Phi. From mathematics, cosine
Phi is equal to x over r and sine phi is opposite
over the hypotenuse. So sine Phi will be
equal to y over r. Okay? From these two equations,
we obtained this. Okay? So this is from basics or refresh it as a
complex numbers. Okay? So what do we
mean is we would like to convert that
V as a function of time equal to V max
cosine omega t plus phi into complex phone
V-max and the angle Phi. That is what we need. So we will learn how
can we do this, okay? So first, if you remember
the Euler identity, which we learned in the
complex numbers e to the power plus minus j
phi z exponential form. It can be divided
into two parties. In real and imaginary part, we have cosine phi
plus minus j sine Phi. Cosine Phi is simply the real part of
this complex number. N sine Phi is the imaginary
part of this complex number. So if you look at here, it was bar plus minus j phi is, this is the real part. This is the imaginary part. That's why we say if we
would like cosine Phi, we take the real part of e. If we would like sine Phi, we take the imaginary part of e. Okay? Now we have V as a function
of time equals to V max cosine omega t plus phi is the one which
we discussed before. Now what if I would like
it in the complex form? Okay, if you look at here, at this one, here, we said cosine Phi is equal
to rail e to the power j phi. Okay? So first, you can see
here v as a function of time equal to leave
V-max as it is. Instead of having phi, we will make it
omega t plus phi. And this one will be
omega t plus phi. So we'll find that cosine
Omega t plus Phi is the real part of e to the power j omega t plus phi,
which is this one. You can see rail of
e to the power j omega t plus phi e to the power
j omega t plus phi V max. You can add it here or
you can keep it outside. Most of them are correct. Okay. So you have this form, so you can divide
it into two parts. E to the power j omega t multiplied by e to the
power j phi, like this. Why? Because if you remember
e to the power a plus b from
exponential equations, it is equal to a, e to the power a multiplied by e
to the power p. So we can divide this
into two multiplications, okay, to multiply it values. Okay? So from here we can say is
that V as a function of time, is equal to Re L of v, e to the power j omega t. You can see e to
the power j omega t. We'll leave it as it is. And we will consider this part, which is V max e to
the power j phi. We will consider
that as V capital. So we'll find that
V capital will be Vm e to the power j phi, which you can be written in the complex form as
V-max and the angle phi. Okay? So you will find that
in the end we will able to convert the
V as a function of time and to v-max and
angle phi like this. So find that in order to convert from time domain to
the phasor domain, we have two points
which we may consider. First one and we should
have cosine not sign. Because if you remember here, we take the real part
which is cosine. So this must be cosine. Then V max, which
is a maximum value, and the Phi is our phase angle. So you can represent
it like this. And this is what we use
in electric circuits. We use to represent
our AC parameters, such as voltage and current in the form of vectors or phasors. So you can see that we have the real axis and
the imaginary axis. And we have V max angle Phi. So we will have V-max. The length of this
vector is V-max, which is the magnitude
of the vector. And the angle Phi, which is measured from the real axis, as
you can see here. If it is positive, if it is negative phi, then it will be measured
from the opposite side. Okay? Okay. Now you can see here
we have two vectors, V equals V-max angle Phi and the current equal 0 or a maximum and the
angle negative seat. So you can see here this vector is the first one
which we discussed. And I maximum negative theta. So it is measured in
the negative direction, since it is negative seat and the magnitude of the
vector is I max. So here is a small
representation. If we have V maximum
cosine omega t plus phi, then we'll say, say, you can see cosine in the time domain. You can see time domain, Faisal, you can see
magnitude and the ANC. Okay? If I would like to convert
from this to this assembly, you can see V-max as
it is maximum value. And remember cosine. So we will take this
phi will be phi. However, if we have some
singular exists, we have v-max, which is V max sine
omega t plus phi. Now you can see it
as falling -90. Now why is this? Because we said before, to convert from time to face or you need this one in design for, if you make this
one in cosine form, it will be V maximum cosine Omega t plus Phi -90 degrees because you will convert
from sine to cosine. So subtract 90 degrees. So we'll have the angle now is falling -90, which is this one. Same idea for current,
current maximum cosine. So it will be the same
angle Theta here. And the sine will be converted into cosine with
the same concept. It will be a Sita -90 degrees. Okay? So finally, before
we go to solve it, the examples to understand
how can we deal with this? Phasers. Here are some rules
of complex numbers. So let's say we have z
equal to x plus j y, which is a general formula, r and the angle phi. And this is from rectangular
to the phasor form. If we have x1, x1 plus y1, R1 and angle Phi,
that to equal x2, y2 plus j y to equal R2
and the angle Phi term. Now the first one, if I would like to add
two vectors, Z1 and Z2. So in order to add two vectors, you need them in the
rectangular form. Now why is this? Because
it is very easy. Assembly you a new order
to add these two vectors. You take the rail was
real and imaginary. Imaginary. So we'll have X1
plus X2, Y1 plus Y2. If you would like to
subtract the door minus two, it will be X1 minus
X2, Y1 minus Y2. Assembly. You subtract the real part
and subtract imaginary part. Okay? Now, let's say I would
like to multiply two vectors or two
complex numbers. So we can multiply them
in the rectangular form, x1 plus x2 y1
multiplied by X2, Y2. You can do this. Or the easiest way
is that you have R1 angle phi, R2 angle phi. So if I need that, the
one multiplied by z d2. Then assembly you multiply
the magnitude R1, R2 and add the two angles, phi one plus phi two. If you are going to
divide this two vectors, then you will divide
R1 divided by R2 and subtract the two angles. Okay? The rest broken, it
means one over z. That is broker of anything. Let's say if we have z,
then its reciprocal, it will be like this one. And what is that? So if we have z, which is x plus j y. So it will be, if we would
like that is broken, it will be one over r and
it will be negative phi. Okay? So that is broken, which is one over R
and the angle phi, it will be one over R. And the angle will be
negative since it is here. Okay? The square root, if you'd
like the square root of z, then you take the square of the magnitude and
the angle by half. Now why is this? Because root two, it
means to the power half. So you take half, multiply it by the
angle, so we'll get 5/2. Finally, we have the
complex conjugate assembly. We have z star, which is a conjugate.
What does this mean? It means you just
reverse the sign of j. So assembly, if you have, if you would like
the star of Z12, it means that we will
make this one negative. And instead of plus j, it will be negative j. And if it is already negative, you will make it positive, okay? Your reverse as a sign of j. So I have x minus j, y equal to r. And since we reverse this one, we will reverse also
the angle here. And finally, one over j
is equal to negative j. Okay? So now we discussed Faisal's and we'll discuss
the roles of complex numbers. Now in the next lesson, we will have some examples
on the phasors is to understand how can
we deal with them.
91. Solved Examples 2: Hey everyone. In this lesson we
are going to have some solver the examples
on the phasors. So we have this complex numbers. We have 14 and the
angle 50 degree plus Duany and angle negative
cell two degrees and all to the power half. Second one we have done and angle negative salty
plus three minus J4 divided by two plus
j four minus three multiplied by three minus
j five and conjugate. Okay? So we would like to evaluate, we would like to find the
final values of this. So first step, since we are talking here about some mission, we have a Faisal
plus Mozart Faisal, or polar form plus
another polar form. So since we have some Michigan, we need to convert to this
one and this one into what? Into this form. We need it x plus j y. So the first one
is equal to what? Equal to 40, which is
a magnitude or design. Cosine plus j for cosine 50 or sine, sine 50. Remember that x
is equal to what? Equal to the magnitude
r, which is 40, multiplied by cosine, the angle, which is 50 degrees, plus j sine phi. So that is the first one. So the first one here, 40 angle 50, 40 cosine 50. So the first board
plus four t j sine 54, t j sine 50. So you will get finally, this forest rectangular form. Second one which is 20 and the angle negative
salty assembly as 20. Cosines are first plus 20 sign negative k
and j. This form. So we'll have finally
70.32 minus a j. Then the second step is that we are going to add
these two vectors. Okay? So adding these two
vectors will be rail, rail and imaginary
plus imaginary. So we'll have this rail plus
rail for 43 and imaginary, imaginary is negative
is plus 20 j. Okay? Now since we are
talking about which is the square root, square root. So we need to convert this
form into polar form again. So converting this to the polar form or before
T7 and angled 25. Where did you get this? The magnitudes for
seven is came from x squared plus y squared
under the square root. So it will be 43 squared
plus 20 squared, all under the square root. Ends angle is ten minus one
y over x, which is 20/43. So we will get 25 degrees. Okay? Now final part, which is
getting the square root. So the square root
of this part is a square of this square root of 47.72 and the half of the 25. So it will have like this. If you take the square root, it will be square root
of 47, which is 6.2. 91.5 of 25.6 times
three is 12.81. Okay? So this is the first one. Second one here
which we have ten and negative 30 plus
three minus jail. Now divided by this one. So first, the easiest
to sing is a conjugate. So we have here conjugate, which means that this
part will be post. Okay? So we will remove
this conjugate like this and add plus here. Okay? Now second step is we
will need to convert this one into rectangular
form to add these two, which is ten cosine
negative 13 plus j, then multiplied by
sine negative 30. So you can see like
this first one, which is 8.66 minus J5.
Where did we get this? It is this one is ten
cosine negative 30, and this one is tan
sine negative search. Okay, so we have this part. The conjugate here became plus. Second step is that we will
need to add these two. So it will be posted
was posted or railroads ran on post office
boast of railroads variable, which will be 11.66
and imaginary, imaginary which I negative J. Now divide it by this one. This one, how can we get at
it is really, really easy. Simply, you can
multiply this two. So it will be like this. Forest multiplied by first. So two multiplied
by three is six, then second
multiplied by second. So we have for J and
J gives us negative 24 multiplied by five is two n. And j multiplied
by z is j squared, which is simply j
is root minus one. All squared gives
us negative four, which is negative two. And then you multiply. That means and
extremes means here is three multiplied by
four j gives us 12 jn. And five j multiplied
by two gives us ten j. So you will find that six
or -20 is negative 14.12. 0 plus ten is 22 j. Okay? Then what is the
next step converting rectangular form
into polar form. So first as a magnitude
which is r will be this one is squared
plus 22 squared, all under the square root. So root of 14 squared
plus 22 square. The angle phi is tan
minus one y over x, which is 22 over negative 14. Remember negative 14, okay? So we'll have an angle 122. Similar idea for this one. So this division will be
14/26 gives us 0.565. And the division of
this is negative 77.6 divided by this means -122. It will be this angle
minus this angle. So there's some measure or negative summation will
give us negative 160. Okay? Okay. Now let's have another
example on this. So we need to transform this
sinusoids into phasors. We need to convert
them from this for the time domain form or the presentation into the
phasor representation. So how can we do this? So first, we have, our current first step
is that we need cosine. You can see we have cosine. Then second step, we
will look at y here. So you can see we have
six cosine 50, t -40. So what does this mean? In order to convert this, it will be I will be maximum
value since the V-max, V-max are two will be our Emax, which is six, the
angle negative four. So we converted the first
one into polar form. Pretty, pretty straightforward. Second one is V equals
negative four sine t, t plus 50 degrees. So how can we convert
from this, this form? We need first to cosine and
the sine to be pollster. So how can we do this? Remember that
negative sine angle equal to cosine angle
plus 90 degrees. So this means that this one
will be same magnitude, but negative sign,
negative sign. It will be cosine. This angle plus 90 degrees. So it will be sorted t
plus 50 plus 90 degrees. So as you can see for
cosine t plus 50, so two plus 50 and
adding 90 degrees. So we'll have this final
form for cosine of t plus 140 degrees. Now, if we look at here and
convert it to this form, so v will be maximum
value, which is four. And the angle, what is the
angle Phi is 140 degrees. So we'll have like this. Okay? Now let's have another example. You can see here we need to find the sinusoids representing
buys these phasors. We have this complex forms or this phasers and would like
to put it in the form or the time domain form of V-max cosine omega t plus phi or
Imax cosine omega t plus C. Okay? So first step is that we have
current and what do I need? I need two parts. I need the maximum value, so I need this sport
to become equal to Imax sine omega t
plus phi, right? I would like it in this form. How can I do this?
I need first is the maximum value and I
need the phase shift. So assembly, you can see we have a vector negative serine, which is real plus J4, which is the imaginary. Which is similar to this form. If I would like to get the
magnitude and the phase. The magnitude here is the
R. Phase, is this phi. So first to get R or
the maximum current, it will be root three squared or negative three squared
plus four squared. So finds that the magnitude
will be equal to five. And the angle Phi will
be ten minus one. Y over x, which is 4/3. However, don't forget,
we have a negative sign. So it will be negative c. So it will be like this. So all you will be
negative c plus J4. So it will be in the phasor
form or in polar form. We have maximum
value or a maximum which is three squared plus four squared
or negative three, all squared plus four
squared, which is fine. And the angle
hundred 26 came from ten minus 14 over
negative three, or the imaginary part
over the real part. Now to convert this into the
time domain for assembly, I will be imax, which is five, as you can see, and the fall you will
be hundred and 26. So it will be like this. Okay? Now second one,
which is a voltage, you can see we have
j e negative j 20. So first I need the magnitude, second, I need the phase. First. As you can see from
this equation, it's clearly that eight
representing what? Representing v-max or the
magnitude of the voltage, maximum value of the voltage. Now you can see we have j and we have e to the power
negative j2 end. Okay? So I would like here
to get the angle. How can I do this? Simply, you have to know that e to the power negative
j 20 is this one can be represented like this
can be represented as one. E to the power negative Z. 20 can be angle negative two. And j can be represented
as magnitude to one. And G itself representing
90 degrees, checked. Okay? So if you multiply
these two together, you will get 21
multiplied by one, which is 1.90 degree
plus negative 20, which will give us 70 degrees because it is multiplication. So we'll find that 70 degrees is our Phi and V-max is eight, its value is eight. So let's see. You
can see that here j equal one and
angle 90 degrees. So J eight and negative 21 and negative 20 and multiply by eight gives
us eight and negative 22. And j is 1.90 degrees,
as we said here. So they're multiplication
will give us eight and the angle 70
degree as our obtained. Okay? So from here you can say is
that voltage is equal to eight cosine omega
t plus 70 likes us. Let's have another one. If we have this two phasors, I1 and I2, I1 is four
cosine omega t plus 30. And I2 is equal to five
sine Omega t minus twice. Now I would like to
add these two vectors. So first we have to
convert them into what? Into rectangular form. Into the rectangular form. So in order to do this, they must have cosine. So the first one here is
four cosine omega t plus 30. Second one is assigned. So second one, I would like
to convert it into cosine. So how can I do this? It will be cosine Omega
t -20 -90 degrees. Okay? So the first one, I1 will be four and the
angle 30 clear forward, which is the magnitude and
the angle sorted degrees. Second, one will be five, cosine omega t -20 -90 degree, which is five cosine
omega t -110. So it will be five and
the angle hundred 1,010. Okay. Now why did I convert
it to cosine? Because if you remember, this phase or form is
using cosine, not sign. So we need to convert
as a sine to cosine, similar to this one. Okay? So now we have i1 and i2. So in order to add
them together, we need to convert this into
rectangular form x plus j y. And this one to rectangular
form x plus j y. So x plus j y for
the first one will be x will be four cosine t. And y will be for signs. For this one it will be x will be five cosine
negative 110. And the wall you will be five. Sine negative hundred
ten will have like this. First one is the sport, and second one is this part. This one is four cosine t, and this part is four sine t. This one is five
cosine negative 110. And this part is five sine
negative hundred entity. So we will add imaginary, imaginary and real with Israel. So we will have this final form. So we will convert this
into Faisal like this. How assembly, this
value is root, this one squared
plus one squared. And the angle is
tan minus one y, which is negative 2.678, and x which is 1.754. So in this lesson, we had some solvent
examples on the phases. Hope is clear. Now for you, how can you deal with voltages and currents
in the phase or four?
92. Phasor Relationships for Circuit Elements: Now let's discuss the Faisal relationships
for circuit elements. So we now know how to
represent as a voltage and the current in the phasor
or the frequency domain. Now, you may ask how can we
apply this to Zach circuits which involved with R
and L and C or RLC. How can we deal with the circuits that
contains these elements? So we need to transform is our voltage current
relation from the time domain to the frequency
domain for each element. So first, let's say we have
a resistive load like this. So let's say we
have a supply and AC supply that provides
as certain current. Let's say this current is
equal to I m cosine omega t. And this current is flowing
through a resistor R. So what we need to know
is that we would like to find the voltage V across it. So as you know that
the voltage across any resistance is equal to R, the resistor, multiplied by the current flowing through it. So we'll have voltage
equal to IR or equal to r i m cosine omega t plus phi. So you can represent this
enzyme Faisal formula exists RIM as it
is and angle phi. So we can say is that this
value is V maximum of what? Of the voltage
across the resistor. So you can see is that AI itself is equal to
i m and the angle Phi k. So we can say is that the voltage is equal
to r multiplied by I. So what does this mean? It means that if we
draw the vector here, we have the real part
and the imaginary part. And we drove e.g. the vector
of the current, like this. The voltage itself will be
the vector itself multiplied by the resistance R. So
it will be like this. And the angle of them is
suppose equal to five. So in this case, both of the voltage and the current have the same angle phi. So we say is that the current
and voltage are in phase. So here, as you can see, here is a voltage and in time domain and in the
frequency domain. And when we draw
the phasor diagram, which are representing the
vector and its phase angle. You will find that the
voltage equal to the current multiplied
by resistance R. Okay? And the angle
which is phi for current and voltage z or z have
the same angle phi. Now, let's say we are
dealing with an inductor. So we have again our supply
and AC supply, a current, AC supply that
provides a current called I Am cosine
omega t plus phi. And this goes to an
inductor by exist. This inductor has
an inductance L. So what we need is we
would like to find the voltage across the inductor. So the voltage across the
inductor will be equal to, if you remember that V equal to L d over d t from the
circuit analysis, which we have
discussed it before in the inductor section of our
electric circuits course. So we will take L
as it is and get the derivative of i
will look like this. So you have v equals
L-O-G I over d t. So we have L and the derivative of cosine Omega t plus Phi. So the derivative of
cosine is negative sine. So we have negative
sine Omega t plus Phi multiplied by the
derivative of the angle. So we are differentiating
with respect to time. So the derivative of omega
t is omega will have voltage equal to
negative omega LI m sine omega t plus phi. So you know that negative sine omega t
plus phi negative sign can be converted
into cosine like this by adding 90 degrees. Now, you will understand
why are we doing this? We're doing this because
if you remember that as the magnitude or
the magnitude and the angle Phi is
corresponding to cosine. So we need to transform
this sine to cosine to be able to convert it to
into our phasor form. So negative sign will be cosine angle boss
90-degree like this. So we will have omega L I M cosine omega t plus
phi plus 90 degrees. Okay? So what you can see
here is that we can transform this into phasor as a voltage will be
omega L I M and cosine omega t is an angle
phi plus 90 degrees. So what we can learn from this, we can learn if we look at this current I is equal to maximum. And the angle phi. If we look at the voltage
across the inductor, you will find that the
voltage has a magnitude v m, which is omega LI M, and then angle phi
plus 90 degrees. So it means that our voltage
in case of the inductor is leading the current
by 90 degrees. So we say that the inductor makes the current lagging from the voltage or mixes or voltage leading the current
by 90 degrees. So as you can see here, we can say I am and
angle Phi, the sport. So you can see here omega L I
M and angle phi plus 90 is, this can be corresponding to i m angle phi omega
L i m Omega L I M the spot and the angle
phi multiply it by J because J itself is equal to one and
the angle 90 degrees. So if you combine this together, you will get this form. Now why did we do this? Because I would like to type
it in the form of j Omega L. So you can see we
have IM and angle Phi is our current and omega L, omega L as it is. And the leading value is
90 degrees, which is j. Okay? That's why you will learn. You will learn is that when we are talking about a resistor, when we represent it
in the phasor form, you will find that
we say are as it is. However, when we
are dealing with the inductance, inductance L, you will find that we say j omega j omega L. And
for Zack capacitor C, you will find that
one over j Omega C. You will learn all of
this in the next lesson. Now y j, because this j causes and leading in the voltage and the lagging in
that capacitor. So we have voltage here
and we have our current. So let's say represent
this on the phase diagram. So we'll find that here. We have our current
with an angle Phi, and the voltage is
leading by 90 degrees. So it will be phi plus 90 degrees will give
us the voltage. So you can see is that the
voltage leading the current by 90 degrees or the
current lagging, the voltage lags,
voltage of 0.90 degrees. Now what about that capacitor? So let's say we have a supply
which is V voltage source. And this voltage
source is connected to a capacitor like this. So the voltage across the
capacitor will be V supply, which is the AC voltage, which is Vm cosine
omega t plus phi. Now what do we need to find the current flowing
through the capacitor? So if you remember from our
lessons in electric circuits, so we said that the current
of the capacitor is equal to c d v over d t. Like this. So we'll have current
equals c d v over d t. So we will get the derivative
of the current like this. Let's delete this. So if you get the derivative
of the current, so I will b, c, d v by d t is the
derivative of the voltage. So it will be Vm. Cosine will be sine
Omega t plus Phi. And we have here
also negative sign. So if you do some
analysis similar to what we did in the
previous slide, you will find in the end
that the current is equal to j omega C V j Omega CV. So if you get back
here like this, here you will find that
V is equal to j omega L. Okay? In the, in this case, in the other one here, in that capacity,
you'll find that the current is equal to j omega C V. The same steps that
we did before. So what we will learn
here is that the current is leading the
voltage by 90 degrees. Or we can say is that
the voltage is equal to I over j omega c. Ok? Now, as small hint for you, as small hint just as
little width hint. If you look at here, you will find that the
voltage across the capacitor is equal to I over j Omega C. And if you know that the
voltage drop is equal to current multiplied
by the resistance in that pure resistive circuit. So what about width? The capacitance. So the voltage across
the capacitance, we can say current multiplying
something called ecstasy, which we will learn later. Okay? Now what is the value of x is c, It is one over j omega c. Ok? Similar to here. If you get back to
the previous one, we can say is that v is equal to x L multiplied by the current, which is equivalent
resistance of the inductor. Now we don't say resistance. We say for x l and x c, We call them reactants. Okay, don't worry, we will learn about this in the next lesson. So you can see that x
L will be j Omega L, which are representing they're
not the resistive effect, we can say the storage
effect or the resistance. I don't want to say resistance, storage element effect
inside our circuit. So we'll find here
is that voltage equal to I over j Omega C. So we can say it is equal to one over j is equal to negative j. One over G is negative
j over Omega C. So we'll find that our voltage
is lagging by 90 degree. Negative j means negative
minus two degrees. So you will see
like this that when we draw the current and voltage, you will find that
the voltage itself lagging from current
by 90 degrees. So you can see voltage equal to a cosine omega t plus phi, so it is V and the angle Phi, as you can see, V
and the angle phi. And at the same time, it is lagging from the
current 0.90 degrees. So adding 90 degrees we
will get the current. So what we will learn from this, we learn that in the
resistive circuits, that current and
voltage are in phase, they are following each other. In the inductance
or the inductor, you will find that the
current is lagging. The voltage. If you look at the capacitor, you'll find that the voltage
is lagging from the current. The inductor has an
effect of making, because our current late, the capacitor has the effect
of making the voltage late. Okay? So a summary of all of our loss. This is important
because when we analyze circuits which are having
capacitors and inductors, we use the frequency domain, or j Omega instead
of the derivative d over d t or d v over d
t or the integrations. We instead use this
method because it is much easier
to convert a from current to voltage or voltage to current by just the
multiplication and adding ankles. So let's have an example on
this to understand the idea. Don't worry, we will have some circuit analysis examples which will help us
such as KVL, KCL, which will help us
learn how to deal with this elements in the
real life, in real life. So finally, here we have a
voltage supply, voltage, we have a voltage source, which is an AC voltage source applied to an
inductor like this. Okay, now what I
would like to get, I would like to find the current flowing or the AC current
flowing through this inductor. So what we know is that, remember that for the inductor, the voltage is
equal to j omega L. Okay, if you get to the
previous slide here, you will see that for
L or the inductor, V is equal to j omega LI. From here, if I need a current, it will be V over j Omega L. So that's what I'm going to do. It will be V over j Omega. Now what is the value of V? If you look at here, it
is cosine and post them. So we can say it is the
magnitude and the angle 45 degrees, like this. Okay? And omega, which is a frequency, is 60 radian per second
angular frequency. You remember that
this one is omega t. So omega t radians per second. Now let's substitute. So we have like this, the current equal to voltage
divided by j Omega L, V is equal to 12 and
the angle 45 and j, as it is, read, the omega is 60 rad and
L is Henry given 0.1. Now, we will take all of
these magnitudes together, 12 divided by seconds
t multiplied by 0.1. It will give us two. What about the angle? We have angle 45 and j is equal to one and
the angle 90 degrees. So this means that
45 -90 degrees gives us negative 45 degrees. So this is a phasor
form of the current. Now if I would like
to convert it into the actual value or the
sine, sinusoidal value. It will be two cosine seconds, d t -45, like this. So I as a function of time
to cosine 60 t -45 degrees. Now I would like you, if you'd like to get the
current in another way, how can you do this? You know that voltage
equal to L d over d t. Okay? So you can get the current by integrating the voltage
soon you will get this value and the Baltic here and integrates a
voltage and blah, blah, blah up to
get the current. Okay? So you will see that
using just j Omega L, very small abbreviation help us, us or the frequency domain Halloween us to get
the current very fast. That's why when we do a circuit
analysis in AAC systems, we use the frequency domain.
93. Impedance and Admittance: Hi everyone. In this video we will talk about the impedance
and admittance. So in the previous lessons, we obtained the voltage
and current relations in the frequency domain for the three passive elements
for that resistor, resistors, inductors,
and capacitors. So if you remember the
relations that we said that for the pure resistive
circuit or for that resistor, the voltage across it
is simply equal to the resistance
multiplied by current. And for the inductor, we set for the inductor, the voltage is equal to j omega L multiplied
by the current. For that capacitor. We said that the
voltage is equal to the current divided by j omega C. So this equation can be
written in the form of a ratio between the
phasor voltage, phasor current like this. So we can say V over I
is equal to R and V over I is equal to j omega L and V over R is equal to
one over j Omega C. Now why is this? Because if you remember
that this relationship, which is v over i equal to
r is our Ohm's law. Right? So in a circuit which has
a resistance only V over I representing the
resistance which prevent us the flow of current. If you look at this circuit, which includes the inductor, you will find that in
instead of having our, we have j Omega L. So we can say this one
is the one which is by preventing the flow of current in the form of the inductor. This one representing
the effect of the capacitor or the resistive
effect of the capacitor, or the one which you present
as a flow of current. So you can see that H1 in the frequency domain has
its corresponding value. Now from these
three expressions, when you obtain the Ohm's
law in the phasor form for any types of
elements as following. Or that impedance. So we have impedance, which is called Z, is the ratio between the
voltage over current, or the voltage equal
to the impedance, multiply it by current. And here is a frequency
dependent quantity known as the impedance
and the measured in ohms. So this value or this
resistance is in ohms. J omega L is in ohms, one over j Omega C is in ohms. Okay? So the impedances of
any electric circuit, it is a ratio between
the voltage or is that fails or voltage to be more specific phase
or voltage to the phasor current and it
is measured in ohms. So the impedance here, what does the impedance
to represent it representing the opposition that the circuit due to the flow
of the sinusoidal current. Although the impedance is a
ratio between two phases, it is not a face and does not correspond to a sinusoidally
varying quantity. What does this mean?
So as you can see, is that is the ratio between
voltage and the current. But you have to
remember that voltage and the current in
phasor form like this, V-max and angle Phi. And the current is all
Emacs and the angle Phi. Okay? Soul finds that this one is corresponding to cosine
omega t plus phi. And this one is cosine omega t plus phi or theta,
whatever the angle. However, the ratio between them, which is V maximum
over i maximum. And the angle for which is
angular voltage minus c tau, which is the angular of current. So this one is Sita. You will find that this one
is not corresponding to cosine omega t naught
corresponding to this, it is a constant dividing. That's why we say is
that the impedance, despite being a ratio of
two phasors, V over I. It does not phasor itself because it is
not varying quantity, it is a constant quantity. So what does this mean? For us? Okay? So we know that the voltage
is something like this. It is a sinusoidal wave. And I for current, also add a sine wave or a
cosine wave, whatever it is. However, if you look at the, which is the ratio
between voltage of our current, let's say e.g. we will say this one. V over I is equal to j omega L. You can see that omega is a
constant value and the ALU, which is inductance,
is a constant value. And j exhaust mine to degrees. So it means that our z here is a constant value similar to the resistance here,
which is equal to Z. This one is a
constant, the value, it is not a sine wave, it is a constant value. Okay? So here if you look
at each element, each element in the
frequency domain. So if we have a resistance,
inductance and capacitance, if we will have a resistance
in the impedance or in the form of impedance or
in the frequency domain. Or the impedance
will be equal to R. And the L, which
is inductance, will be j Omega L. And the
capacitor will be with, which is z equal to
one over j Omega C. So you can see that here. This is a resistance. This one and this
one are called, what are called in electric
circuits, the reactants. Okay? So when you hear
the word reactants, we are talking about
the inductance is the equivalent impedance
of the inductance and the equivalent impedance
of the capacitor. And sometimes we say
is that J Omega L, we denote it like this, x l. And the one over j Omega C, We say it is x c. Okay? So here we have the
three elements. So if we consider two extreme
conditions, let's say e.g. we have omega. You can see that omega itself, which is a frequency
omega itself, effect as the value of L and C. However, the resistance
is constant, it is not affected by omega. Now, let's consider two
cases when omega equal to zero and omega
equal to infinity. Okay? And let's see what will
happen to l and C. Let's refer us to
consider omega equal to zero for a DC sources. Now why omega equal to zero is corresponding to DC sources, it's really, really easy. So let's say we have V equal
to V maximum cosine omega t. Let's say we don't
have phi here. We don't have an angle here. So we have V maximum
cosine omega t, which is our sine wave, sine wave, or AC wave. Now, let's say we are
talking about omega equal to zero or zero frequency. When omega is equal to zero, we have cosine zero, which is corresponding
to a value of one. Cosine zero is equal to one. So our voltage will be V max. It will be a constant
value like this. Okay? So what does a
constant value mean? It means that we
have a DC source. So again, if you have
a frequency equal to zero or angular
frequency equal to zero, it means is that our
supply is DC supply. So let's see what will
happen if we apply omega equal to zero to the
inductor and the capacitor. So you can see that when
omega equal to zero, z will be equal to what? Equal to zero? Omega equal to zero. So j Omega L, it will be zero. What about the capacitor? Will be one over j
Omega C. If it is zero, then it will be equal to 1/0, which means it will
be equal to infinity. So that corresponding impedance, corresponding impedance of
an inductor is what is zero? What does this mean? It means it is a short
circuit like this. So add D, C. That's why, if you remember in our course
for electric circuits, we said that when applying
dc source to an inductor, we say that at steady
state conditions, we will have that inductor
as a short circuit. So now we understand
why this happens. Because omega equal to zero, it means that the impedance
will be equal to zero. So it will act as
a short circuit. It does not have any impedance or any opposition to the cat, so it becomes a short circuit. Now for the capacitor, we say is that when we are applying a DC source
to a capacitor, it will become an open circuit. So you can see that it
becomes open circuit at DC. And we proved this how
when z equal to infinity, very large impedance is corresponding to very
large resistance. So it means that we
have open circuit. So it means that by applying
DC to AC capacitor, we will have an open circuit. Now, let's see, use a
different condition. Let's say we have
very high frequency, omega tends to become infinity
very high frequencies. So if omega equals
to infinity here, we will have z
equal to infinity. If omega here equal to infinity, then z of the capacitor
is equivalent to z will be one over
infinity, which is zero. Okay? So it means that our inductor, when we have very
high-frequency, our inductor will behave
as an open circuit. Here. Open circuit at a very
high frequencies. The capacitor will act as a short circuit at
high frequencies. Now you have to understand
that this method, this method of becoming open circuit and short circuit
at different frequencies, are used in filters. Okay? If I would like to eliminate or remove certain frequencies
from our waves, such as radio signals
or radio frequencies. We use filters. Filters are used to removing or eliminating different
frequencies or unwanted frequencies. Okay? So we use the idea of capacitors and inductors to
do this function. Okay? So let's say we
have this circuit, we have elements here, each element and
its own impedance. And we would like to analyze
this electric socket. So forest as a first
step to analyze any electric circuit which
contains an AC supply. Okay, let's say this one is
an AC supply, like this. Ac, AC supply. So when we have an AC supply, what are we going to do? We are going to put each one of these elements
in each impedance for. So you will see that
for the resistance, the equivalent impedance is R. This one will be R as it is. For the inductance
or the inductance L, you'll find that it's a
corresponding impedance is j Omega L. So we say
that this element is j omega L. Then Zach capacitor here will be one
over j Omega C. So this capacitor will
be one over j Omega C. So we added all of our
elements in the impedance form. Now, if I'd like to get the total impedance
of this circuit, it will be R plus j
Omega L plus j omega L plus one over j omega C. The impedance of the effect
of each of these elements. Now, you will notice
something here that we have all as it is, plus j Omega L. Okay? And do we have here one over J? Now, if you remember that
we said in complex numbers, one over j is equal
to negative j. Okay? So one over j will be
equal to negative j. So I can say is that it is equal to negative j one over omega C, or negative j over Omega C. So I can say negative
one over omega C. Okay? You can see here j Omega L and minus j over Omega
C, one over omega C. Okay? So you can see is that we have our impedance consisting
of two components. Rail part, which is R, and the imaginary part, which is j omega L
minus one over omega C, is the imaginary part, omega L minus one over Omega C. Now, this part of the circuit, omega L minus one
over omega C is, can be, can be written as x. Or the reactants of us. Ok. Ok. So we'll find that
we can express this z in the complex form. The impedance equal
to R plus j X, where X is subtraction
of these two elements. Or if we have zoster
inductance, e.g. then it will be omega L. If
we have capacitance alone, it will be minus one
over omega C, and so on. So you will find that R or the resistance is the real
part of the complex number z, which is a resistance, and x is the reactants or
the imaginary part of z. Okay, so we call
this part resistance and this part is
called reactants. Reactants can be
positive or negative. So if you remember x here, which I just said, it is equal to omega L
minus one over omega C. Assuming that we have
a circuit like this, if we have inductance or
loans and we will type omega, if we have capacitance alone, will type one over omega C. So if this x is positive,
what does this mean? It means the effect
of omega L is much higher than the
effect of the capacitance. So what does this mean? It means that if
you remember that the capacitance or
the inductance here, inductance matrix as a
current lagging the voltage. So we say is that when x, when impedance is inductive, when x is positive. So the impedance is inductive
when x is positive. And in this case, when we have inductive circuit, we say it is
inductive or lagging. Current, lagging the voltage because the effect of the
inductance is much higher, the effect of capacitance. Now when this one is
negative or x is negative, it means that we having at capacitive system
or a capacitive, or it means that the effect
of the capacitance is much higher than the effect
of the inductance. And in this case we say that
the capacitive or leading. Why leading? Because the
current leads the voltage. Because if you remember
in the previous lessons, or in electric
circuits in general, we said that resistance current in phase
with the voltage, the inductance
current, leg voltage, that capacitance current,
lead voltage. Okay? So when the effect of
the inductance higher, it means the current will lag. Effect of capacitance is higher than the
current will lead. Okay? If they are equal to each other, then we will have a
pure resistive circuit. They cancel each other. And in this case, we will have a condition
which we call resonance. Resonance in electric circuits, which we will discuss in our
course of electric circuits. So the impedance can
be represented in the polar form as
magnitude and phase, since we have real and
imaginary component. So you can private lungs
us that z is equal to R plus j X equal to a
magnitude and angle. The magnitude is
the square root of r-squared plus x squared. And Sita, which is the
angle phase shift, is tan minus one x over r. And you have to
understand that Sita here, representing the phase shift
or the phase angle between voltage and current are, in this case, r is our z
multiplied by cosine Theta. And x is assigned seat, as we discussed before
in the complex numbers. So we learned about impedance. Now, let's see what
does admittance mean. It is a reciprocal
of the impedance. So if you remember
before we discuss this at resistance had an
inverse, one over r. We had the inverse one over r. This inverse, or the
reciprocal of the resistance, was known as the conductance. Similar to the impedance Z, we have an inverse
called one over z or y, which is called the
admittance. Okay? Now why do we study? Admittance, or why do we study the reciprocal
of impedance? Because it is, the
admittance itself is very helpful in the analysis
of parallel circuits. Okay? So that's why we need to
understand the admittance Y. And it is measured in Siemens, okay, it's a company,
Siemens came from. The admittance. Admittance Y is equal to
one over z or I over V. Okay? So we can write it in
that complex form, since we said that is
equal to x plus j y, okay? Over z is equal to
resistance plus j X, which is our reactants. We can say is that y is equal
to component g plus j b. And you have to
know that G is not, that is protocol of R and the P is not that
is broken off x, not the reciprocal of x. You will learn how can
we do this right now? So you can see is that
y equal to g plus j b. And g is a real part
of the admittance, and b is the imaginary
part of z admittance. G is written as or called the Zak conductance and b
is called the symptoms. Okay? So admittance, conductance
and symptoms are all expressed in the unit of
Siemens is called Siemens. Okay? So how can we find the
relation between that? We know that y is
equal to one over z. So we have y, which
is g plus j b, and that is one, z is r plus jx, as
you can see here. So how can we find the
relation between these two? Simply will do like this. First, we have this
complex number, one over r plus jx. So we will multiply
by the conjugate. So you can see the conjugate
of R plus j X is r minus Jx, someone to blame
here, auto minus Jx and the r minus j x like this. So r minus j x
will be like this. And R plus j X multiplied
by r minus g x is r squared x
squared, like this. Okay? So if we divide this into
two components like this. So we can say this part is
equal to r squared plus x squared plus r squared
plus x squared. That first part. And do we have
here negative j x. So if we convert this
part with this part, you will find that g is equal
to r over r squared plus x squared and b is equal to negative x over r
squared plus x squared. You can see that from here that g is not the reciprocal
of resistance, such as in resistive systems. And if x equal to zero, then g will be one
over R because we will have only the resistance. Okay? So finally, all of this
representing our load impedance and admittance of each element, of each element, not the whole circuit,
each element, the admittance of r is one over our admittance of j Omega
L is one over j Omega L. And that maintenance of C is one over j Omega C is j Omega C. Now, let's have a quick example on this in the impedance
and admittance. And we will learn
how to use KVL, KCL, nodal analysis and more and when we apply it
to the AAC systems. So in this example, we need to find the
voltage as a function of time and current as a function
of time in this circuit. You can see we have the
voltage equal ten cosine 40. So if I would like to convert this part into
the complex form, you can see this one is
V-max cosine omega t. And the angle is zero. We can say is that the
voltage itself solve V supply as a
magnitude and phase, magnitude and phase is
magnitude than the angle zero. From here, we can find that
omega is equal to 4 rad/s. Now, the resistance
itself is 5 ω, as it is all in the
frequency domain, or equal five, it will be
as it is for this element, which will be, this is
the capacity, right? So we said that the x or
the frequency domain. The capacitance is one
over j Omega C, right? So it will be one over j Omega, Omega is four and the
capacitance is 0.1. Forearm. So you have here r and the x. So from here you can get the total impedance and
you can get the current. So let's see step-by-step. So first, as we learned voltage
in the frequency domain, the supply voltage is ten
and the angle theorem for the impedance, you can see impedance
is equal to a real part which is five plus
one over j omega C, which is one over J
four multiplied by 0.1. So one over j is negative j 1/4 multiplied by
0.1 gives us 2.5 ω. So we have our impede. Now, the first step
is that we have supply and we have the
equivalent of the circuit, the total impedance
in the circuit. So you, if you remember that
z is equal to V over I, or the current required
in the circuit will be this voltage divided by the
impedance five minus j 2.5. So we will convert this one
into magnitude and phase. So we can see like
this ten and the angle 0/5 minus j 2.5. You can see you have
two options here. Okay? You can multiply by its
conjugate as we did here. The lawyer conjugate
five plus j two point 5.5 plus j 2.5. Then we convert this
into magnitude. And then the other way
is that you can take this one and make it
maximum or not Z, maximum magnitude of z. And the angle sit like this. That is a square of five
squared plus 2.5 squared. Theta is tan minus one
negative two point 5/5. Okay? The same solution. Then you subtract and divide. You will get finally is
the same answer. Okay? So now we have the current, current flowing
through our circuit. Now what I need is the voltage. What is the value
of the voltage? Voltage here inside our circuit
is simply equal to what? Equal to here, voltage will
be equal to the current. Multiply it by accessing. Okay? Or we can say that
current I multiplied by one over j Omega C. So we have the
current which is 1.789, and we have one over j Omega C. So we can write it like this. We have current which is 1.789. And then between six and j, Omega, Omega is
four and C is 0.1. Now, j is corresponding to what? Corresponding to 90 degrees. We'll find that
the subtraction of these two and division will
give us negative six to 3.43. So this has a voltage. So finally, you can type them in the time domain because we
need it in the time domain. So the current will be our Emax, which is 1.789 cosine omega t, which is 14, plus
that phase shift, which is 26 degrees. For the voltage,
it will be V max. Sine omega t minus six is three point
for this angle here. Okay? Now a small check in order
to make sure that you are getting the
correct resolution. Simply if you remember that the current flowing
through a capacitor, what is the relation
between I and the voltage? Voltage across the capacitor? What is the relation? That current is
leading by 90 degrees. What current and what voltage the current is flowing
in the circuit, leading the voltage across
the capacitor by 90 degrees. If you look here, adds
a current and voltage, you will see that 26.57. And this one is negative three. So the difference in angle
between them is 90 degrees. So your solution
is correct. Okay? So in this lesson, we
discussed the impedance and admittance and then we have a quick solvent example on them.
94. Kirchhoff’s Laws and Impedance Combinations in the Frequency Domain: Hi, and welcome everyone to this lesson in our course
for electric circuits. In this lesson, we will
talk about with the KVL and KCL or Zachary Sharp slow
in the frequency domain. We know about KVL and KCL. Kvl, which is, hey, is that the voltage inside
a loop is equal to zero. The summation of all voltages inside a loop is equal to zero. Or the KCL, which says
that the summation of the current entering a node is equal to the summation
of the current leaving. So here is the same idea, the same idea in the
frequency domain, similar to the time domain. So we cannot do a
circuit analysis in the frequency domain
was our KVL and KCL. So we need to express them
in the frequency domain. We'll find that our KVL, which is a Kirchhoff's
Voltage Law, the summation of all voltages in the frequency domain
is equal to zero. And summation of
all of the currents in the frequency domain
is equal to zero, similar to the time domain. So if you have a circuit
like this and AC voltage VS, and do we have here a
resistance and e.g. an inductance like this. So we'll say that the
summation of all voltages, we have v plus the voltage
across the resistor plus the voltage across the
inductor is equal to z. Okay? Summation of all
voltages equal to zero. For the current,
summation of all currents inside the nodes that
will be equal to zero. We will see all of this when we have some soil with examples. So again, for the
impedance combination, if I would like to combine several impedance in
series and in parallel, how it will look like, it will be similar
to that resistance. So you can think
about the impedance similar to any resistance. So if you have a group
of resistance in series, the impedance will be summation
of all of the impedance. So if you look at this
circuit in using KVL, if you apply KVL, you will find that
the supply voltage is equal to summation of all
voltages in solid or SEC. So the voltage v equal
to V1 plus V2 until v n. And we know that the
voltage drop V1, e.g. it will be multiplied by V1 and V2 is I multiplied
by z2, and so on. So I'll find that
the total voltage inside our circuit is the current multiplied by the
summation of all impedance, which is equivalent to z. So you will find
that z equivalent of this circuit is
equal to V over I, which is the summation of all of these impedance Z equivalent, equivalent impedance of group of impedance in series
is summation. Okay? Now, what about the
voltage division similar to that resistance
is the same idea. If I would like to get
the voltage V2, e.g. or voltage V1. Let's say I would like v1. V1, what will be
V1 will be equal to the supply voltage, okay? Multiplied by the
impedance that the one, since we are talking about V1. So it will be one divided by the summation of the two
impedances, Z1 plus Z2. Like this. For V2, it will be V supply multiplied by z2
divided by summation. This is what, what,
what we have done in our course for electric
circuits, okay, as a voltage division, as a summation of
resistance, the same idea. Nothing change it at all. Okay, except instead of using
the time domain values, we use the frequency
domain values. Now if we have a
current in parallel, you will find that
from nodal analysis. Again, you have a current
source, provides current i1, i2, i3 to the elements
that do 12 until n. So what do we need to find the equivalent of
all of these system? So again, it's a nodal
analysis is applicable here. So we can say is that here, this is our node here. And this one is the
current entering, which is current I is equal
to the total current leaving, which is I1 plus I2
plus I3 until I am. So we'll find that
the total current. Entering is equal to the
total current leaving, which is i1, i2. Until now, you can see
that in this circuit, that's the voltage across R1, across the supply IUIE supply. Current source is equal to V, which is similar to the
voltage across that one, equal to the voltage
across R2 and so on. We can say is that
the current I1 will be V divided bys and one. And current I2 is
v divided by two, because all of them
are in balance. So in the end you will
have v multiplied by 1/1 plus one over
z two until one over n. So from here you can
find that the equivalent, that the equivalent one over
z equivalent is equal to 1/1 plus 1/21 over n, which is all your overview
you can see here, I divided by V gives us
one over the equivalent. Because you know that
here voltage equal to I, multiply it by the
equivalent in general. So z equivalent is
equal to V over I. So I over V will be one over z. So i over v, which is, this part, is one
over z equivalent. So as you can see, similar
to a resistance in parallel, we said that one over R
equivalent is equal to one over R1 plus one
over R2 and so on. Okay? So the impedance will
be the same idea. Okay? So you can think again of the impedance rules or laws similar to the resistance
and the force at mittens. You can see one over the
equivalent is why equivalent? And one over that one is y
11/2 is y two and so on. Now for Karen to division
similar to the same idea of the DC circuits or the
resistance and so on. The same idea, if I would
like the current I1, I1 will be equal to what
will be the total current. Multiply it by the
other impedance divided by the total impedance. Impedance is at two divided
by the total impedance. And I2 is equal to
the total current I multiplied by 1/1 plus two. Okay? Now, if you don't know where
did we get all of this, you need to get back to our
course for electric sockets. You will find there
is a DC circuit, so with the current division, voltage division and KVL, KCL and so on. Okay? Now, finally we have the
y and Delta networks. We discuss them before, or Y delta transformation or
star delta transformation. So this transformation
is helpful in simplifying our
electric circuits. So let's say we have a, B, C, which are representing adult as this one representing delta. And I would like to convert to this formation into a star four. So we have a P that scene. And the star formation is Z1, Z2, and Z3 with Zan
neutral point n. Okay? Now why, why delta and start
connections are important? Because you will find them
in electric circuits and specifically the
three-phase systems. Okay, don't worry,
we will discuss this later in our course
for electric sockets. So here let's say I would
like to convert from y to that. What does this mean? It means that I have that one. I have the two ends at three. And I would like to convert
this into a scene, e.g. ends at B and a. So e.g. that n will be equal to what will be equal to
the a will be equal to x1 multiplied by Z2
plus Z2 multiplied by z, three plus three multiplied by one divided by the
impedance Z2. Okay? So simply what we are going
to do in the three cases, that is it B, is it C? You will find that
the first part is the same in all of them. Simply we say Z1, Z2, Z2, Z3. And since three is at one, multiplication, then
divided by one, if you are talking about the farthest away
impedance, which is one. Okay? If you are talking about zy, zx and zy away
one, which is two. If you are talking
about z, these three, then you see you are talking
about the three lines. This similar to the rules when we had your
resistive circuits. If I would like to
do the reverse, Let's say I have a star delta connection and I would like to
convert this into a store. Let's say I need that one. So it will be p multiplied by z seen over the summation
of our summation. If I would like to e.g. then it will be seen that
a divided by summation C, a multiplied by summation. If I need e.g. x3,
then the closest one, which is a P divided
by summation said, is it paid want to
poison missions, which is similar
to what we did in the first two sections
in our course. Now, you have to
know that Delta or why Circuit said to
people and said, What does that mean? It means that if z have equal impedance in
all three branches. So what does this mean? It means that here we have e.g. if we're talking about with
the Y connection here, balance, it means that one
equal to Z2 equal to 3.4. The delta connection, this one, that a equals b equals c. Now, if we take all of this and substitute in this
equations here, you will find that in order to convert from delta to
star or star to delta, you will find that Delta
is equal to three times y, or y is equal to 1/3. Okay? Okay. So in the next lesson, we will have some examples
on this methods in order to understand how to apply these
rules for the AC circuits.
95. Solved Example 1 on Impedance Combination: So in the first example on
the impedance combination, we have here this
circuit consisting of a capacitor resistance
series with a capacitor inductor
series with a resistor. And we would like to find the
equivalent of this circuit. Windsor frequency omega
is equal to 50 rad/s. Okay? So in order to find the
equivalent input impedance, we need to convert each of these elements into
the frequency domain. Okay? So first, what is our Omega? Omega equal to 50 rad/s? Now, if I would like to
find that equivalent, it will be this element. Let's say it is x1. And you will find that this part is parallel
to this part. So we have Z2 and Z3. So the equivalent
impedance is equal to x1 plus x2 parallel to x3. Okay? Now, what do we need to do? Is that why you need that
one in the frequency domain. I need to in the
frequency domain and z3 in the frequency domain. So first you can see that one impedance of
the two millifarads, z2 is as serious combination
of 3 ω and ten mainly for out that three is a series combination of 0
point to Henry and 8 ω. So let's say that one x0, x1, or the impedance
of a capacitor. We said it is one over j omega c omega 50
radian per second. And capacitor is tune. So it will be like this. 11 over j Omega C, one over j is negative j. That is a forest plot. Second part, we have the two, which is a series of 3 ω status was that
then millifarads. So if you look at the sport, you will find that
that is equal to r pluss plus j Omega L. Right? We have real part and
imaginary part resistance, which is 3 ω plus j Omega, which is 50 radian per
second, and inductance, which is ten milli milli ten
to the power negative z. So it will be like this. So we have two, which is okay, here are not omega L, Okay? I'm here. R plus j Omega L is for
this one, for this branch. And this one is a capacitor. So it will be like this, okay? So it will be, that
will be equal to r plus one over j Omega C, because we are talking about
capacitors and inductors. So resistance will be 3
ω plus one over J omega, which is 50, and capacitor, which is ten millifarads. So we'll see that three, which is a real part
of the resistor, plus one over j Omega C, one over j 50 multiplied
by ten, mainly for odd. So this will give us
three minus J2 ohms. Okay? This one is, this branch was three is one which
I am talking about, R plus j Omega L because L, because we have Henry
here or an inductor. So resistance is 8 ω plus j omega 50 rad/s and
inductance is 0.2. So we will have the
impedance like this, eight plus j Omega L. So we will have a block j. Okay, So we have the
three elements here. Now what does the next system
you can see here we have Z1 plus Z2 parallel
tools at three. So this one is
parallel to this one. So x1, which is negative j ten and Z2 parallel to z three, okay, is this two are
parallel to each other. So similar to, similar to what did we get
this you will understand now. Similar to a true
resistance in parallel. So if you have two resistance
or resistors in parallel, R1 and R2, their
equivalent is what? R1 multiplied by R2
divided by R1 plus R2. Same idea for the impedance. It will be the one multiplied by z2 divided by x0, x1 plus two. So we have here as this
impedance, this impedance. So they're multiplication
is this one. They're summation is three
plus eight, which is 11. Negative j2 and plus j
ten gives us plus j. So you will have
this equivalent. Now you need to multiply
these two together. And remember that j is equivalent to root negative
one imaginary number. So j square, which is
j multiplied by j, is the square of negative
one root negative one. So it will be negative one. Okay? Okay. So when you simplify
this, like this, you will have this
final ambivalence. Okay? Now, something here
which is important too, is that you will
find that 11 plus J converts it to 11
squared plus x squared. What happened here simply we multiply it by Zach conjugate. So we multiplied here by
11 minus j, 8.11 minus j. Now why did we multiply
by the conjugate to eliminate this j two. So I can remove
this way and they have 11 squared
plus eight squared. So when you multiply
these two together and divide by 11 squared
plus Eta squared, you will have this final form. So we will add negative j
ten plus negative j 1.07. We will have this
final impedance. Impedance is 3.22 minus j 11.07. You can see this ****, but also zap reasons or the conversion of
the elements such as the inductor and
the capacitor in the frequency domain help us to analyze our
circuit more easily. You can see we can now
find the equivalent of a circuit that contains
several elements. Unlike the DC case or
the previous cases, when we had only resistors or only inductors or capacitors. Now we can deal with different
elements in one sec, okay?
96. Solved Example 2 on Voltage Division: Hey everyone. Now let's have another example on the impedance combination. Or in this example
we'll talk about with the voltage division. So we have this circuit, which we have an AC supply, 20 cosine 14 -15 degrees. We have a 60 arm, then mainly furrowed
and five Henry. What we would like is to get the output voltage
on the circuit, output voltage, which is a
voltage across the inductance. So how can we get, as you can see, is that if
we look at this circuit, we can get the voltage across R5 Henry using
voltage division. Okay? So first step is that we need
to convert our elements, such as the supply, is our resistance inductance, and the capacitance in
the frequency domain. So first we will start
with our supply. We have a 20 cosine, 40 -15. So from this week alone is
at omega As equal to what or the angular frequency
equal to 4 rad/s. And we have here
is a maximum value 20 and the angle negative 50. Okay? So this representing the
representation of our supply. Okay, so we have
20 and the angle negative 15 and
omega equal to four. For the sake of stay home, it will be as it is because
it's a pure resistance. For the capacitance,
we know that the frequency domain
representation representation is one over j Omega C, Omega is four and C, which is our capacitance
than merely for odd. So from here we can
get negative j5 ohms. And the four-five Henry. We can get it using what? Using j Omega L, which is four multiplied
by the inductance five. So we have now with
that representation of each of our component. So how can we get the voltage across this part or this part? So what we need is
that we need to get the equivalent circuit
of this two parts. So if you look at the frequency domain equivalent
as this is our circuit. Now, let's go here. So what we are going to do, we have x1, which is the impedance of
the second still on? And we have z2, which is the equivalent of
this two parts. Okay? So from here is
the equivalent of this two parts as they are
parallel to each other. So we will get the equivalent. So two, which is equivalent is negative j 25 ω
parallel to j 20. So we have two elements
parallel to each other. So they're equivalent will be their product over
their submission. So you can see product
negative j 25 multiplied by j 20 divided by summation. So this equivalent
will give us j hundred or so now we have that one. So we have like this, the equivalent of this
circuit like this, okay? 20 and the angle
negative 15 by exist, we have 60 ω, which is x1. And we have the equivalent
of this part, which is what? Jay hundred. So we will have here like this. We can make it up look like
this and say J hundreds. Now what we need is that
you have to understand that the voltage across j hundred is the same voltage
across negative j 25 and the same across J2 because all of them are
parallel to each other. So the voltage here is V output. So from voltage division, we can get V out. V out is equal to V supply, multiplying by j hundred
divided by summation. Using voltage division, we have a V supply 20 and
Dangun negative 15. And we have our zipped, which is j hundred
divided by the summation. So what we can do is that
we will convert this into by multiplying
by the conjugate, this part, by multiplying
it by the conjugate. Or we can convert
this into phasor form and get the combination
of these two terms. Okay, in the end, these are
all different messages. To simplify this. Then after all of this, we will get the final voltage, which is 17.15 and
angle 15.296 degrees. So if I would like to represent
this in the time domain, it will be 70.15 cosine omega t, which is 40 plus
15.96, like this. So you can see maximum
value cosine omega t, Omega here is the same
omega of the supply. T plus the 15 degrees, which is the phase angle. So in this example, we learned how to get the voltage division
in AC circuits.
97. Solved Example 3 on Impedance Combinations: Now let's have another example. So in this example we need to find the current in these suck. So if you look at this circuit, we have our supply 50
and the angle zero, and we have the equivalent
frequency domain of all of our components. You can see here J6, 8 ω negative j three, and so on. Okay? So how can I get the
current in a circuit, in any circuit like this one, which is a supply
current is equal to V supply, which is 50. And the angle zero divided
by the equivalent, which is the voltage
divided by z. So what I need here is
that I would like to get the equivalent impedance
of all of these. Suck. Okay? So you will
find that here. If we look at here, we find that these two parties are
series with each other. And these two parts are
series with each Awesome. However you will find here, I'll star connection, the spot. Okay? So in this part, you
don't know if four ohm parallel to negative
J3 or series was at. This formation is known
as the star formation, which we have discussed before. So we need to convert an order
to simplify this circuit. We need to convert
this star into delta. So the delta will be like this. We will have an impedance
like this here, another impedance
here like this, and another impedance like this. Okay? Which is a Delta equivalent. Now, you can see when
we convert to delta, we will be able to
simplify our circuit. So you can see that we
have this spot, okay? You will find that,
let's say e.g. this is a star connection. You can see it's
pretty, pretty clear. Okay, so let's say
we have that one. And do we have to and the three. So when you convert
this star into a, into delta, you will have this part parallel
to that, the one, you will have three parallel to this part and to well-being parallel to the equivalent of the sack will find that here. E.g. if you take the parallel
of this two parties, we will have like this circuit, like this, plus -50
and angle zero, 12 ω. Okay? Like this. So we have 2 ω
negative j for which is this branch
parallel to that one. So we will have, let's say, let's call it x1 equivalent, like this one, equivalent, which is a parallel of
z one and this branch. Okay? Then we have that three
parallel to this part. So let's call it that
three equivalent lines as we have another branch which is between this
point and this point. We will call it like this. Okay? So we have Z12 here and the equivalent between
these two is connected to it. One is connecting between
this point and this point, will find that in
the end we have that the one series was
z3 equivalent. And they're equivalent
is parallel to z d2. And the equivalent of all of this is serious residual volume. So from here you can
get that equivalent. And then you can divide the
voltage, apply this value. That is the first message. You have to understand that. And circuit analysis, there are different methods to
obtain the same thing. Okay? Let's see another method here. The second method is that
you will find that we have delta network,
another delta network. Now, whereas this delta, you will find that
the delta network is consisting of a three points, which is a, B, and C, which is
this branch. Okay? You can see it's forming a belt. Now. The Delta network can be
converted into a war in network. And it will relax us. Okay, So we'll find
that we have a, B, and C between
one neutral point. So you can see we have this belt similar to a triangle like this. Not pretty clear, but the
hint is our triangle. So we can take, we
have a and B and C. You can see a, b, b, c. And then between ca like this. Okay? So we will convert this delta formation
into a store like this. Like this was one neutral point. So we'll take from a
and B and C like this. And, and see. Okay, so we will have this neutral point and we
have one coming from a, one coming from B, and the one coming from. See, if you look at the
circuit here we have a, B, and C, One coming from C, one coming from beam, and one coming from me. You will see we have
now a star connection. What do we need to get
is the value of z a n, value of bn, and value of Z C. Okay? So simply you will
find that a n, which is the first one here. This one, you will find that
it is close to 4 ω J4 arm. Multiply it by this branch. If you look at this one as our closest one is
this one and this one. So you can see multiplication divided by the summation
of all three branches, it ohms plus J4
plus 2 ω minus J4, which is this part. This gives us, then
this part will give us j for Motorola advisors if
you distributed that J. So you can see that
here j law exists. We will have to j and j multiplied by negative j is opposed to one, as you can see. Okay? So in the end you will
have this first Z, which are representing
that AN second, which is that b n, which is this one, this branch. You can see that
if we look at it, what is the closest
to resistors that are closest to is the 8 ω and for j. So you can see it. And for g divided by summation, which is done for the last one, which is this branch. You can see it's a
closest one is this one. And this one which is
8 ω and two minus J4. So 8 ω and two minus j four. Okay? So now we have the value of Z n, c n, and b n. Okay? Now what does an
ecosystem you can see we need the equivalent z. So you can see is
that we have 12 ω. Okay? 12. See here is with AN
series with the dan, like this plus c. This was what was that combination
of this parallel form. You can see two-port
is parallel to each ares p n minus j, three parallel to the
other branch here, which is C n plus
js six plus 8 ω. Okay? So we'll find the line exists. We have a 12 plus a
n plus b n minus J3. That the C N to C N plus
a j six plus eight. If you combine all of this, do some analysis, you will
get this final impedance. So the current will be this
voltage divided by this. So we'll find that
the final current is equal to 3.6 is six. And angle negative 4.204. Okay? This is in the phasor
form because it is given in the phasor form or in the
problem we have a phasor form. If I would like to convert
this into time domain, the current will be maximum
value cosine omega t, Omega T plus or -4.24, 0.204. Okay? However, in the problem
we don't have Omega. So this was another example
in the impedance combination. Hope it's helpful for you to understand more
about AC circuits.
98. Solved Example 1 on Nodal Analysis: Hi and welcome everyone
to this lesson in our course for AC circuits. And this lesson we are
going to have a solver, the example on nodal analysis. But in this case, and instead
of having DC circuit, we are going to
have an AC circuit. So we would like to know
how can we apply and nodal analysis to AC circuits. So forest in this circuit, we would like to
get the current i x using nodal analysis. So you can see we have a
supply here and supply 20 cosine for t. We have 10 ω, we have one Henry, we have 0.1 for odd
to IX and 0.5 Henry. Okay? So the first step you have to do is that you need
to convert all of the elements into
the complex domain or in the polar form. So forest, we will convert
to the frequency domain. Have first, we have 20, cosine 14, which is 20. Or we can say V max cosine omega t plus phi. So here we have one supply, one AC supply, which will determine the
other values of Z. So maximum value is 20
and Omega itself is four. And the phase shift
is equal to zero. So here if we convert this to the frequency domain
or the polar form, it will be maximum value
and the angle which is z. And we can see here that omega, or the angular frequency
is four radian per second. Now we have our first element. Second element is that we
need to convert the one Henry and the 0.5 Henry
and the 0.11 Henry. How can we convert to
the frequency domain? We know it is j omega L
and this one is j omega L. And this one is one over j
Omega C. So we have Omega, which is four radian per second, and we have capacitance
0.1 for odd inductance, which is one Henry inductance, which is 0.5 Henry. And by substituting, we can get the three frequency domain form. Now what are we going to do? We will make our
circuit likes us. So we would have changed it
all of the elements into the frequency domain or
into the polar form. Now, what does the next step, since we are talking about
with nodal analysis? If you remember that,
know that analysis, what are we going to
do in this analysis? We simply assign for each node. For each node, we will assign
for a certain voltage, such as V1, V2, and so on. So first, if you look
at this circuit, we have this large node
which is grounded. So this one has a
voltage equal to zero. The second dynode here, which is this one. You can see is that here
we have zero voltage. So what does the value
of this voltage? You can see we have a supply
which is representing the difference between
this point and this point. So if this one is zero, then it must be this
one is 20 and angle z. Now we have this node
which we don't know. We will say it v1 and we
have this other node, we would call it V2. Now, all of this
of course is v2, and of course all
of this is V one. Okay? So we are going to apply
nodal analysis for V1 and V2 in order to
get the current IX. Okay, let's start first for the forest node,
node number one. Here. You can see we have
this one here. So any element, we have
a current entering here, a current going here, and a current entering
this note, okay, coming from the
supply and one going through this loop
and ongoing here. So we will have, if you remember from KCL, nodal analysis is based on KCL. And then KCL we say
that the summation of all the currents
entering the anode, Let's say entering equal to summation of all
currents, leaving. Okay, so what is the current entering and
what is the current leaving? So if you look at this circuit, you will find that the
entering current is the current coming from the supply. What is its value? It will be the difference in
voltage divided bys at 10 ω. So let's delete this. So
we say current entering. So the current will be
difference in voltage between here divided
by the 10 ω. So they France involve should be 20 minus V1 divided by the 10 ω. So 20 minus V1 divided by ten. Then we have a current leaving, which is our x. We will not, we will neglect
these are IXL for now. We will say is that it is V1 minus zero divided
by negative 2.5. So it will be V1 divided
by negative 2.5. Then we have a
current going here. So we have current
going like this. So it will be difference between these two voltages
divided by four. So it will be V1
minus V2 divided by J for V1 minus V2
divided by J four. Okay? Okay. So that is a forest
nodal analysis. So if we simplify this equation, we will have this
final equation. Okay? Now, let's apply again KCL
at node number two here, you can see that we have a
current entering like this and current going into this one and the current coming
from the other supply. So by applying KCL here, you will find that the entering, which is what are
the two entering? It will be the current
coming from here. And two i X, we have X
which is entering current, then the secondary current
coming from the supply. It will be V1 minus V2 over for j equal to the current leaving, which is V2 minus
zero divided by J2, which is V2 over J2. Ok? So from here, you will
find that we have to I x. So what is the value of I x? If you look at this equation, I x is equal to V1 divided by negative j
2.5 from this part. So we can take this and
substitute it here to have one equation with v1 and v2. Okay? So if we simplify this, we get this final equation. So finds that we have
this first equation. We have this second
equation, okay? Okay, now, what does the next step we are
going to solve this. We have two methods, is to, let's say make V1. We will say from this equation
that V1 will be equal to negative 15 V2 divided by 11. Okay? We want e.g. so
we will take this and substitute it in
the first equation. So we will have one
big equation with V2. So we can get the
value of V2, okay? And from it, we can
get the value of v1, such as a forest. My second method
is that we can use the grammar method which we
have discussed in our course. So the Cramer method inside DC circuits or in
the DC analysis, we said that we can put these two equations in the
form of matrix like this. You can see that
equation as a matrix, we have V1 and V2. So we have V1 and V2. Okay? Now, v1 as the first
element is one plus j 1.5. Second, 111 V1. So it will be like this. Second part, j 2.5 and 15, or the second coefficient
equal to 20 and Z. 20 and Z. So how can you make sure that these two equations is similar
to this, simply like this. You can say that the
first equation will be one plus j 1.5 multiplied by V1 plus j 2.5
multiplied by V2 gives us 20, which is the first equation. Second equation, 11 V1
plus V2 gives us z, which is the second equation. Now by using Chrome
or muscled, we can, we can solve as S1 and obtains the value of
v1 and v2 directly. So if you remember from the Cramer method in
the linear system, we have two equations that we put in the form of a matrix. You can see here a, b, c, d, which is four coefficients, 1234, okay, equal to e and f, which is 20 and z. Now, how can we get? The value of x, which is V1, and the value of y, which is V2. You can see that if
the determinant of a does not equal to zero,
what determinant? This determinant, if this one, if you get the value
of this determinant and those not equal to zero, it means that the system
will have one solution, which is x r x will be determined divided
by determinant of a. Determinant of a and y equal to determinant divided
by the determinant of a. So what does the forest one
and what's the second one? Simply. You can see here we have the coefficient
matrix, which is this one. Now, let's say I'm
talking about x. So x is a and C. A and C is a spot. So what I'm going to do
is that I will take E and F and put it in a
set of a and C. So we will have E,
F, BD, like this. Okay? Now, if I'm talking about y, which is b and d, Then I'm going to
take E and F and replace B and D with it. So we will have
ACEF, ACEF, okay? So the first element or
the forest variable, we will replace its column
with the output values. If we're talking about
the second variable, we will replace that second row or the second column
with E and F. As you can see here. If you have three
elements as in the soul, the one we will
replace the soul of the column with E
and F, and so on. Okay? So let's, uh, get forces
that determinant, the determinant of this one. How can you do this? It will be simply multiplication
of these two variables. So from here, that
determined to be a multiplied by d minus
b multiplied by c. Okay? So we multiply these two minus Zomato
application of these two, it will give us 15 minus j phi. Now, then we are going to get that force, the determinant. We said that we need delta one, which is for x or for V1. So we will replace the
first column with 20.0. So it will be 200 j 2.5 and 50. As you can see, if you get, its value will get
several hundred. For the second delta two, we will replace the second
column with 20.0 likes us. So by obtaining that
determinant of value, we will have negative 220. If you don't know this or
if you don't remember this. You can get back to our
lesson for Cromer method in the DC circuit analysis in the beginning of the
electric circuits course. So we will get V1 as delta
as delta one divided by delta and V2 as delta t divided by delta,
as you can see here. Okay? So from here we
obtained V1 and V2. Now what do we need
in this example? We need the value of I x, which is V1 divided
by negative j 2.5. So we will have like this, we will take v1 and divide
it by negative j 2.5, which is negative 90. We will have this final value. So our current would be 7.59 and the angle 108 point forward. So if you convert this x, which is 7.592 time domain, it will kill our exists
7.59 cosine Omega t. Remember omega t is similar to the supply plus angle 108.4. So in this example, we learned how to apply nodal analysis to
the AC circuits.
99. Solved Example 2 on Nodal Analysis: Now let's have
another example on the nodal analysis
in AC circuits. This circuit, in
this AC circuit, we have a supply here, three and angle zero. And we have another supply
tenant, angle 45 degree. This is a voltage source, and this is a current source. Now we need to find V1 and V2. In this example. The first step is that
we need to have all of our elements in the
frequency domain or in the polar form. So as you can see is that
all of the elements here, negative j, tan, and angular
45s three and angle zero. All of them are in the complex
form or in polar form. Okay? Okay, so what are
we going to do? We need now to do
the nodal analysis. So we have V1 which are
representing all of this voltage, the voltage of
this node, and V2, which is a node and this
node is equal to zero. Okay? Now, what does an extra step? Next step is that we will start applying KCL to each node. And you have to remember
that if you look carefully here we have
a voltage source. Okay? So what does this mean when
we have a voltage source, it means that we
have a super node. So as you can see here, if you look at this circuit, you will find here we
have our super node here. Okay? So sober node here, you will, you will see here
is that V1 and V2, we have an equation
between them. You can see this one is a positive and this
one is a negative. So you can see is
that then the angle 45 degrees is equal
to V1 minus V2. V1 minus V2. This is the first equation
from where you can see here, supply difference
between this point, which is V1 minus V2, is ten, and the angle 45-degree. Now, since we have this voltage, we can not do nodal
analysis here. We will have a super node
between them. Now why is this? Because if I would like to find the current year, it will be V0, V1 minus V2 divided by the resistance or the
supply, which I don't know. So here we are doing a supernode as we did
in the DC circuits. As if all of this one big note. Then in this case we
are going to apply. Zack ECL normally finds that three ampere is
entering negative J3, leaving six, JSX leaving and V and 12 or leaving from
v2 going into that ball. So let's apply that model. And so we have a
three ampere equal to all of the other counts. So the three and Beta
will be equal to V1 divided by negative j3, V2 divided by j six, V2 divided by 12 volt. So we have this first equation. So by simplifying it, we will have this
second equation. The first equation is obtained
from the nodal analysis. Second equation is obtained
from the super-node here, or from the difference in
voltage between V1 and V2, which is done on the
angle 45 degrees. So we'll find that
V1 minus V2 equals ten and the angle 45-degree or V1 is equal to ten and the
angle 45 degrees plus V2. So we have this two equations. Now what are we going to do? We can do the Cramer
method or you can take one and substitute it
in the second equation. So what we are going to
do is that we will take this equation and substitute
it in instead of V1. So we have assault
six as it is. Okay? Now, let's see what
will happen here. So we have J four
multiplied by V1, which is a v2 plus ten, and the angle 45 degrees, V2 plus ten and the
angle 45 degrees. And here we have J four. First part is we will
have J4 multiplied by V2. So we have J four multiplied by V2 plus then multiplied
by four is 40. And the J is converted to plus
90 degree plus 90 degree. 90 plus 145 is 175 degrees. Okay? So this is a forest plot. This part representing the spot. So plus one minus J2. So let's say it will be
one minus j, two V2. All of this will
be equal to six, which is the spot. So we have served six
years is this one. Okay. Now we have plus 40
and angles certified. So we will take it
to the other side. It will be negative 42
and angle 135 spots. So we take this to
the other side. Now we have here one
minus two j or J2 v0, v2, and we have J4, V2. So you can see we have plus four j and we have minus two j. So there are some mission is
to J, as you can see here. And the one will be, as it is, all of this
multiplied by V2. Okay? So now we have one
equation in V2. So we can obtain the value
of v2 and form this. We can get V1 by substituting
in the reverse equation. We said that V1 is equal to
V2 plus ten and angle 45. So it will be, V1 will
be the summation of V2 plus this voltage. Okay, So we will have
this final form. So now we discussed another example on the
anode and analysis, but this time we wouldn't
have a supernode. And we understand now how can we use nodal analysis
in AC circuits? You can see that this is pretty, pretty similar to that. Dc circuits.
100. Solved Example 1 on Mesh Analysis: Hey everyone. In this lesson, we are
going to have a solver, the example on the
AC circuit analysis using mesh analysis. So in this example we have this circuit and we need to
find the current I naught, which is the current
flowing like this, and set our circuit
using the mesh analysis. The first step in mesh
analysis is that we will assume a current in each loop. So you will find
that here we have for this loop, we have I1, for this loop we have i2, and for the slope
we have our string. So let's start with our u1. If we look at I1 here, you can see that I1
is consisting of this follow-on components
in this slope. And this slope, we have I1. And I1 is flowing
through 8 ω j ten ohm, and negative j to j
ten minus J2 plus A2. You can see eight J
ten negative j. Okay? Is there are three
elements in which is a current I1 flows. Now, what are the second
elements for this loop? For the, for the
rest of the loop, you will find that we have ice cream flowing
through as j at n, and we have i2 flowing
through negative J2. So in this case it will be negative ice-free
multiplied by j ten and negative I2
multiplied by negative J2. Ok, so you can see negative
I2, negative ice cream. And for i2 flowing negative J2, as you can see here. And for all three j, then you can do this message or you can do as we did in
the DC circuit analysis. We said that we will
do a loop like this. So we will have eight lines us. And if we go like this, then we will have J ten. So we'll say plus j. Then we have I1, which is flowing to the right, and I3 in the
opposite direction. So it will be minus I3. Okay? Then we go to this
element negative j two multiplied by I1 minus I2. All of this equal to zero. This will be equivalent to this. Also of the two
methods are correct. This is a method
which we discussed in the DC analysis of
electric circuits course. If you would like to use this, it is okay if you would like
to use the other method, is that we will see, Let's say we are
talking about I1. So we will see I1
flowing through eight J ten negative J2. So we say I1 plus
the summation of all of these components. Then we have I3 and I4. So we'll say negative
ICT, negative I2. And the two elements in which
each of them are flowing. As you can see here. For the second loop, the same idea we
have plus 20 and angle 90 degree by exist plus two inch
entanglement in green. Shows this loop like this. And we have negative J2, negative J2 and the force, so we will have I2 multiplied by four plus negative
j2 plus negative J2. So we'll have i2 four
minus J2, minus J2. Ok? Now we have, in this lobe
we have I3 and I1, I3. So we will have negative i3
multiplied by this element, negative i3 multiplied
by this element. And we have negative I1, negative I1 multiplied by this element in which
is flowing negative J2. Okay? Then the last loop, it's pretty, pretty clear. We have I3 flowing in the same
direction of this source. So i3 will be five
and bears like this. So what is the next step simply, you can see we have
three equations, 123 with three unknowns. But we already know i3, i3 equal to five. So we will substitute
with I3 equal to five in this equation
and in this equation. So we will have two equations with two variables, I1 and I2. So what is the next step? We are going to add
this in the form of a matrix as we did before. In order to use Cromer muscles. You can see we have i1, i2, i1, i2 equal to j FFT
and negative j 20, negative j ten,
which is negative j. So J FFT and the negative j. We have eight plus J1 and J2. As you can see here,
the first two elements and J2 four minus J4. As you can see. What's the next
step, we will find the determinant for this matrix, which is determinant
of this part. Delta. As we did before. Then we're going to get
delta one and delta two. Delta one is taking because S1 and replacing the first column, delta two is taking this column and replacing the
second one like this. So we have delta two. Now, why did we use delta two? Delta one? Because
we need I-naught. Okay? I know the assembly
equal to negative i2, i2 flowing like this, and I naught flowing likes us. So I naught is equal
to negative I2, so I will obtain i2. I2 is delta two and delta
two is eight plus eight. And J2, which is a
sport and taking the Scollon and replacing
the second one. So we have J FFT
negative j, sorry. Okay. So this will give
us 340 minus J 240, which will give us
this final value. Okay? Now this value, what
is the value of i2? I2 is simply equal to Delta
to the vowel is by default. So we will take this
one and while it by 68. So we'll have this final value. Now, I know there
will be negative I2, which means we will add the 182, this nk by exists. So where did we get this? Again, I naught equal
to negative I2. So it will be negative 6.12
and the angle negative five. So in order to eliminate, this, eliminates this negative,
making it posted, we will add 180 degrees. 180 plus negative
35.22 gives us 144.78. Okay? So in this example
we talked how to, or we analyzed how can we use mesh analysis in
the AC circuits.
101. Solved Example 2 on Mesh Analysis: Now let's have another
example on the mesh analysis, but in this case we
will have a super mesh. So we will see how we are
going to deal with it. You will find that till now, every sink in DC circuits
is similar to AC circuits. That same analysis, nothing
it change it at all. Okay, so let's start. First. We need to find the
voltage V naught across this capacitor using
the mesh analysis. So what are we going to do? Our first step is
that we will assume current in each loop, current in each row, like this. Okay, so we have i1, i2, i3, and I. For now in this loop, we don't have any
current source. And this loop we have one current source which
is three and bear, which means that we have i2 like this and three umbrella axis. So it means that i2 is equal to negative three and they're pretty, pretty
straightforward. Now second equation, if
you look at this loop, you will find that we
have here so per mesh. Now why is this? Because if we try
to do KVL here, we have a four and pair. However, we have also I3. So it means that all
four minus r is three, is equal to four amperes. I4 in the same direction of the four and pears and I3
in the opposite direction. So you can see that here. We can not do a KVL
here or a KVL here. We can do a KVL owns a big loop. So the supermesh,
it means that we have two loops between them. Account sold similar
to what we discussed in our course of DC circuits. So let's delete
all of this. Okay? Like this. And what are we going to do? We are going to do the mesh analysis or a
KVL in each of the loops. So first we are going
to do a KVL here. So if you looked like this, gave you a like this, you will see negative 10 v
in the clockwise direction, negative ten volt. So in the first loop, you
can see negative ten. Now, if we look at
I1, I1, like this, what elements it
flows through it, it flows through 8
ω and negative J2. So it will be eight minus
J2 multiplied by I1. So eight minus J2
multiplied by RE1. Now what are the
two other currents flowing through the
elements inside this loop? We have I3 flowing through 8 ω, and we have i2 flowing
through negative j 2 ω. So in this case we will have
negative i3, negative I2. You can see negative three, negative I2 and I2 flowing
through negative J2 here. And isolate flowing through
the 8 ω like this. Okay? So we have this equation
function in i1, i2, i3. What about the second row? Second row, you can see i2 is equal to negative
three and beta one, which we obtained here. This is a simplification
of this one. So you can see for mesh two, I took one negative three
and bear as we obtained. And for the supermesh, we are going to do a big
loop here like this. And we have the equation
which we obtained iPhone minus IC equal to four and
pair, which is this one. Okay? Okay, Now that supermesh KVL, if you look at here
for this big loop, you can see we have i4 flowing
through six ohm and j phi. So it will be six plus
j phi multiplied by I4, six plus j phi
multiplied by IL-4. And we have also I3 flowing through eight and
negative J four. So it will be n minus j for i3. Now what are the two currents
which are not in this loop? Out of this loop, we have J5 and I2. We have i1 and i2. So it will be negative I2, negative I1 multiplied by
eight multiplied by J5. You can see negative one, negative I2 and I2 J5. So we have J5 and I want eight, so we have it here. Okay? So now finally we
have four equations. So what are we going to do? We will take first, the first equation, we will take i2 and substitute it here. Okay? So substituting this one here, we will have like this. Combining these two together, we will have eight minus J2, I1, and I3, and we have negative
three substituting it here, we will have this
final equation. So we have equation
in I1 and I3. Okay? Now, what are
the next system? You can see we have
the second equation here are three i1, i4, i2 for us to take i2
and substitute it here. We will take I4 and
substitute here. So we will have one
equation in I1 and I3, as you can see here. So finally we have our
two equations in I1, I3, I1, and I3. So we are going to
form a matrix again using Chrome or method in
order to solve with us. So eight minus J2, negative eight, then we
have negative 814 plus j. Two parts. Here is a two coefficient,
as you can see here. Then we are going to watch
one we would like to obtain, we would like to obtain E1. Now why is this? Because if you look
at the requirements, we need V naught, which is the voltage
between here and here. So this voltage is zero, current flowing like
this and this direction entering this capacitor
multiplied by negative two. So negative J2 multiplied by the current entering
the capacitor. So what does that
current entering? We have I1 like this, and we have i2 like this. So it will be I1 minus I2. So y1 minus two. Okay? So I chew, we already know
i2 equal to negative three. So what we need is our E1. So we are going to get
from here delta one, delta and delta one. So here delta will be equal to this determinant of this matrix. Then delta one, we will take this part and replace it
with the first column. So you can see that
the first column is the sport became ten plus j six and the negative 24
minus j 5 s colon as it is. So we obtain this final value. Now, be careful
that all of this, this all is a one
coefficient, let's say a. And all of this is another
coefficient called B e.g. so this is our two
values, a and B. They are not a, B, C, D. Okay, don't get confused
with the signs here. All of this is one big block representing
one part of the matrix. Okay? There are not four coefficients. Z are only two coefficients. So I1 will be delta
one divided by delta. So now we have I1, we have i2, so we can get
the voltage ally exists. The voltage will be
negative J2 I1 minus I2, as we said before. So we will have this final
value of the voltage. So in this lesson, we had another example
on the mesh analysis. Hope it's helpful for you
to understand how can we apply the mesh analysis
to AC circuits.
102. Solved Example 1 on Superposition Theorem: Hi and welcome everyone
to this lesson in our course for AC circuits. This lesson we are
going to discuss the super position
theorem or application of the superposition
theorem on AC circuits. So superposition
theorem, we discussed it before in DC circuits. So in this example, we need to find the
current I naught flowing through the circuit using
the superposition theorem. So how can we do this? The first step is
that we will find that we have two supplies. We have an AC supply at current source and
a voltage source. In order to find
that total current, we need to find the contribution
of these two sources. So we have our inode,
which is required, is consisting of our inode dash plus I naught double dash, which are due to the voltage
source and current source. So the first step is
that we will find the contribution of
the voltage source. How can we do this
by elimination of the current source or
deactivating the current source. So that the activation
of current source means it will be an
open circuit like this. Okay? So that first the current
is I naught dash, which is the contribution
of the source, which is 20, and the
angle 90 degrees, which is j 20. So how can we get discount? We first need to simplify this
circuit to find a source. So you can see is that this
branch and this branch, you can see negative
J2 is parallel to j ten plus 8 ω. As
you can see here. You can see here negative J2, this part or this component, and the eighth plus j ten
are parallel to each other. So we need to combine these two together to have our exam, which is 0.25 minus j 2.25. Okay? So this branch will
be equal to this value. Now what is the
value of current? Current will be
equal to the supply divided by the total
impedance like this. So R dash will be
supply which is j 20/4 or plus negative j2 plus the resultant of
all of these components. So we have four minus J2 losses or resultant, which is that. So this will give us this ratio. So our current will be equal
to this value or inode dash. Now for I-naught double dash, what are we going to do? We are going to deactivate
the voltage source. So we will make this one as
short circuit like this. And it will give, of course
is our five and bear. Now in order to find the
current I naught w dash, we will do the mesh analysis. One lobe here, another
here, another here. You can see that I know double dash is equal
to negative I2. So our concern or our
important request is that we need to
find I2 forest. We will do the first mesh. You can see using
the mesh analysis, we have the first lobe here. So we have i1 and what
are the components? 8 ω plus j, ten plus negative J2. So it will be eight plus
eight j, like this. J m plus negative J2
gives us eight plus j. Then we have current
I flowing through J, J ten, and the current I2
flowing through negative j2. Okay? So this current will
be equal to what? These two components will be. Negative y is three
and the negative I2, you can see negative three. And then we have here
negative I2, okay? I is three multiplied by j ten. So can see I3 negative
three multiplied by uj ten. Now second part, negative
I2 will be multiplied by negative three to negative two. So negative and the negative
gives us positive J2, E2. Okay? So that is the first
loop, second loop i2. In this loop, you
will find that i2 multiplied by four negative J2, negative j two,
which is negative j. So it will be four, negative four, as you
can see here, i2. And we have this two elements. In the first element here
we have negative three, negative J2, so it
will be plus J2, I3. This part we have negative one
multiplied by negative J2, so it will be plus J2. As you can see, the
last loop here, you can see in this
loop i3, like this, is in the same direction
of the five pairs. So in this case, i3 will be
equal to five and bears. So what will happen now? We will take I3
and substitute it here is three and
substitute it here. So we will have from this equations by substituting
ice cream here and here. And the first one here, by substituting it here, we will have five plus J2 gives us plus ten J,
as you can see here. And expressing I1 in the
as a function of i2. Taking this equation
and voting here, we will have I1
equal to this value. Now why do we do this? Because I would
like one equation which contains all
of the values of i2. Since we need only to remember we need I2
to get the current. So I don t need I1, I need only i2. So what did I do simply, I took I3 substituted
here and here. Okay? So we have this
equation became this one. Then I will take I1
and they make it as a function of i2, as
you can see here. Then we will take this
equation and get back to the first one
here, like this. Okay, so we replaced
each one bys this value can see only
y1 bar in this big value. So we have a large
equation in i2. So from here we can get I2
as equal to this value. So know, double dash will
be negative I2 like this, i2 like this, and I know
double dash going upwards. So it will be negative
I2 as you can see here. Okay? So the total current
will be I naught dash plus I naught
double dash, like this. So this is a total count. So in this lesson when we had the first example on the
super position theorem, applying it to the AC circuits. Remember, if you are, don't know about with the
superposition theorem or you have forgotten about
the superposition theorem. You need to get back to our DC circuit scores or the electric circuit scores
to understand this one. Okay.
103. Solved Example 2 on Superposition Theorem: In this lesson, we
are going to have another example on
the superposition. Now, this example is
really, really important. Now why is this
example important? You will understand now. So in this example,
we need to find the voltage V naught the
voltage across the 1 ω. And as you can see, we have two Henry. We have 0.14 and
how many supply? We have IDC supply. We have our current source
with an omega equal to five. And then we have another source
with omega equal to two. Okay? So as you can see, we have different supplies
with different frequency. If you remember that
the frequency of the DC supply is equal to zero. So we have here
omega equals two. We have omega equal five
and omega equal to zero. So the question is, how can I express to Henry in the frequency
domain soda use, if you remember,
it is j omega L. So what omega should I use? Should I use that too? Or five or zero? Which one should I use? You don t know actually. So we can not use mesh analysis or nodal analysis or
any of these theorems. So what are we going
to do in this case? We have to use a superposition. We can find the effect of
each of these supplies. Okay? Okay. So sensors are, circuit operates at as three different
frequencies. We need to find
in order to break the problem into single
frequency problems. So we have the V
output will be equal to V1 plus V2 plus V3, which is a contribution
of H of z supplies. So V1 is a contribution
of DC source. V2 is the contribution of the voltage source,
AC voltage source. And V3 is our contribution
of that two sine phi of t. So we will start
by the DC source. Dc source means
omega equal to zero. Okay? Like this. So omega equal to zero
first before anything, we will deactivate
those supplies. So this one will become
a short circuit. This one will be
an open circuit, so we will cancel it as if
it does not exist at all. So we have Joe Henry, see here is with one arm
and that we have here 4 ω and we have 0.1 far out. What our power to Henry, if you remember that
in when we apply a DC source into inductor, it will become eventually
a short-circuit. So if we would like
to make sure that you have to know that to Henry J, Omega L. And the omega of the
DC source is equal to zero. So it will be J zero, which means that the impedance of this element will
be equal to zero, which means it will become
a short circuit like this. So you can see this
is an open circuit and short circuit here. You can see we have
only one ohms. As you can see, it ends
a short-circuit similar to hear what our
PowerPoint one for art. So if you remember that
the capacitor will become an open circuit in DC system, if you would like to make sure
we have one over j omega, one over j Omega
and Omega is equal to 01/0 means infinity. So that Z equivalent of that
capacitor will be infinity, which means we have
an open circuit here, as you can see here. So we'll find that
our circuit is now simplified into this
pretty simple circuit. And we would like to
find the voltage V1, which is the voltage
across the 1 ω. So how can we do this using
the voltage division? You know that V1
is equal to supply five volt multiplied by its resistance over
the total resistance. You can see it's
resistance which is one divided by the
total resistance. So this would give us 1 v. Now we have to understand that this is negative
v1 naught V1. Now, why is this? Okay? You have a current coming
out of this supply, 5 v coming out of the supply. So it will suffer. It will cause a voltage
drop across the 4 ω. Like this plus, minus. And then when this current goes through this second resistor, it will suffer another
voltage drop plus minus. So this voltage drop is
called, let's say e.g. let's call it x. The voltage drop x is equal to 1/1 plus four
multiplied by five. Okay? But if you look at
the polarity of V0, V1, it is opposite to the
original polarity of x. So it means that V1
will be negative x. That's why v1 will be
negative of this value. So V1 will be
negative 1 v, Okay? Why? Because we are measuring it in the opposite direction
of the voltage drop. Then we're going to see the contribution of
the second supply, which is a voltage source here. So first we will make
this one a short circuit, and this one will
become an open circuit. Then what I want
to express all of the elements in the
frequency domain like this. So you can see is
that ten cosine two t is ten and the angle zero. And omega, which is omega T, is two radian per second. Now, the two-handed
will become J Omega L, which end our Omega which
you are using is two here, which is a contribution
of the supply. So it will be j for two
multiplied by two gives us four. We have 0.1 for audit
will be one over j omega c omega equal to and
this equal to, point to one. Okay? Now, let's take suppress our
circuit with these values. You can see short-circuit for all the fraud is
negative j5 ohms. The voltage here is
V2 for us, okay? Which is the contribution of the second supply
to Henry is J four. Now what do we need? We need to find
the voltage V two, which is the voltage
across the 1 ω. So how can we get this
voltage assembly? We first get the
equivalent of this branch, the spot four and
the negative J5, It's equivalent
to be this value. So as if we have a
block like this, it's a value is 2.439
minus J 1.2 951. And series was it is a
one-arm and J for home. Now what I would like to
get is a voltage drop across 1 ω, and you can see it. So polarity, similar
to the polarity of the original voltage
drop coming due to this current flowing
through 1 ω. So the voltage V2 will be
ten volt multiplied by its resistor divided by the total resistance in
the system like this. So we have ten and dangles here, which is the total voltage
multiplied by its resistance, which we need the voltage
drop on, which is one, divided by the total
resistance of the system, which is one plus J4. And the plus the branch
which we obtained, which is 2.439 and so on. Okay? So v2 will become this value. We obtain the V0, V1, and we obtained a V2. Now what about v3?
Order to get v3? First, we need to find the
contribution of the supply. So we'll make this
one a short circuit. We will make this one
also I short circuit, and we will convert all of our elements into the
frequency domain. So you can see here our
two sine phi of t here, omega is five, okay? And that current
source will be two. And the angle, what
negative mind? Now why is this? So this one is angled zero, but this one is angle might. Now we will have
to remember that this complex form,
V maximum e.g. and the angle is equivalent to V maximum cosine omega t plus c2. So you have to get cosine. But you can see we
have here sign. So two sine phi of t
is equivalent to two cosine five t -90 degrees. So this cosine phi of t -90
is similar to sine phi of t. So when I saw this one can be converted into the
frequency domain. It will be two and the
angle negative light, as you can see here. Okay, The Ford henry here, it will be j Omega L, which is omega 5.2. It gives Austin J 10.1 over j
Omega C, similar as before. Now, when we simplify
our circuit, we have a short circuit
here and Joanna, it will become J ten. So we will have this branch. This branch will be J,
as you can see here. And two sine phi of t will be two and the angle negative
line voltage drop as it is, it will be V3, which is the
contribution of the supply. And then we have 0.1 fallout
will become negative J2. And this one is a short circuit, so it will become 4 ω. Now, what I would like to get, I would like to get v3, which is a voltage
drop across what, across the 1 ω. So how can I do this
using current division? Using current division,
we have a current here, I one, this is our supply
and this is our current. So how can I get y1? First, you will need to
simplify this branch. These two are parallel
to each other. As you can see,
they are simplified as one block like this, one big block like this. 0.8 minus J 1.6 ω. Okay? So you will see that the
current source here, giving current, part of the current will go
through this resistor. And the other part
will flow through the resistor in series
with this branch. So in order to find the current
I1 which is flowing here, it will be the other resistor divided by the total
resistors like this. And see, I1 is equal
to the supply current, multiply it by the
other resistor, this resistance and j ten
divided by that total resistor, J ten plus one plus this branch, j m plus one plus this branch. Here we can get the
value of R1 and from your nose and V3 is equal to I1 multiplied by one arm. So we will get V three. Now, if you don't know
about current individually, of course you can get back to the DC Circuit scores or the electric circuit
scores to understand this. Okay? So as you can see here,
multiplication by one, which is the same gives
us finally this value. Now we have V1, V2, and V3. What is the next step? We will add all of
these three together to get our final
voltage like this. So the voltage as a function of time consisting of
three components, 123 for us to component or 0.498 and the angle negative
30, which is cosine. This part is due to what
new to this source. Remember, you do the
source and omega equal to. So it will be 2.498
cosine omega, which is two T plus the
angle here, negative. Second power, which is this one, will be due to the
contribution of this supply, which have omega of five. Okay? So you can see 2.33, which is the maximum value here. And sine phi of t, which is the Omega
T plus ten degrees. Now we had, did we get this? Okay, This one is equal
to cosine Omega t, which is five t minus eight. Now, I would like to convert this cosine to become
sign like this one. So what I can do assembly, you can say it is sine phi of t and add to this
line to degrees. So you will have ten degrees. So it will be -90 plus
90 degrees, okay? To convert the from
cosine to sine or similar to sine angle plus 90 degrees. Okay? So my tip -80 gives
us plus debt. Okay? So now we obtained
the value of voltage, which is the contribution of
a three different sources. Okay?
104. Solved Example on Source Transformation: Hey everyone, In this lesson, we are going to start
learning how can we apply the source
transformation on AC circuits. So forest, if you remember, the source transformation
is based on transforming a voltage source with a resistor in series into a current source with
a resistance in parallel. And if you remember in the previous lessons
of DC circuits, we said that the
voltage source is equal to our series resistor multiplied by the current IS K and the current
bias is V over R S. Okay? So here we have
this one like this. So let's say we have
a current source and resistance in series. And I would like to form a voltage source with a
resistance in series, which is equivalent
to the system. So assembly is our
voltage source will be the current multiplied
bys, that is a store. Okay? As you can see here, ends in series with
the resistance which is in parallel with
that current source. Okay? So we'll find that here. In DC circuits we
had only resistance, however, in the AC circuits. So we have a zed or impedance
and instead of resistor. So it is a same idea, but instead of
having a resistor, we have an impedance z. Okay? So let's have an example on
this to understand this idea. So we would like to
find the voltage V x in this circuit using the
source transformation. So we are going to do source transformation
several times. So we can simplify
our circuit into a very simple form to
obtain the voltage. So first, as you can see here, we have our supply 20 and
the angle negative 92, which is negative j 20. Okay? We have a five ohm 4 ω 3
ω J4 negative 13, 10 ω. Now, the first step
is that we have a voltage series
with a resistor. So what we can do is that
we can convert this into a current source parallel
to the 5 ω y-axis. You can see we have
voltage source, see it as was 5 ω. So we can have a current
source parallel to the 5 v. Now what is the value
of the current source? It will be Zara
supply divided by the resistance to winter
Dangun negative 90/5, which gives us negative j four, which is negative j 20/5, which is negative J four. Okay? So as you can see here, we have a current source
parallel to a resistor. So as you can see, we
have this parallel form. We can take five parallel
to three plus J4. So we can simplify this
into parallel form, which will give us
2.5 plus j 1.25 ω. Okay? So we'll have one block like
this of 2.5 plus j 1.25. So what we can do
next is that we have a current source to zero. Okay? So we can convert this
into a voltage source, again, series with the same. Okay? So how can we do this assembly? We have this
multiplied by that IS, it will give us the voltage
source, as you can see here. So if we take this z, which is equivalent
of this part, multiplied it by
Zachary on the source. As you can see here, we will get the voltage source, which is five minus j
ten, as you can see here. And zeros. Was it the equivalent, which is 2.5 plus j 0.125, which is this part. Okay? And the rest of the system for Ohms and negative
three is 13.10 ω. Okay? Now what is the next step here? You can find that we
can say is that all of this is one big. And we need the voltage
across the 10 ω. So what we can do is that we can use the voltage division. So the voltage across R 10 ω or V x will be v x will
be the supply voltage, which is five minus j ten, okay? Multiplied by its resistor,
which is at 10 ω, divided by the total resistance of the system or the total
impedance of the system. So it will be 2.51, 0.25. This one for all negative
j 13 and the 10 ω, which will give us
this final value. Now we will have to
remember where did we get the voltage division if you
don't remember, simply, if you know that, what is the voltage drop
across the 10 ω, it will be the current
multiplied by 10 ω. So 10 ω multiplied
by the current. So the current is a supply divided by the total impedance. So that supply divided by the total impedance,
it gives us, this part, gives us the current flowing
through the circuit. So when we take this current
and multiply it by 10 ω, we will get the voltage vx. So in this lesson, we
took a solvent example or a very simple example
on how can we apply the source of transformation
in AC circuits.
105. Solved Example 1 on Thevenin Theorem: Hey everyone, In this
lesson we are going to discuss the seven and the
Norton equivalent circuits. So if you remember in the previous lessons
of the DC Circuit, so we said if we have a linear circuit with a
two terminals, a and B, I can change this
linear circuit into R1 voltage source
and one impedance, which is known as the Seven
and equivalent circuit. And we can take the same circuit and we can transform it into a current source or in Alton
parallel to it, Norton. And if you remember
that n is similar to that Norton and V7, n is equal to Z naught and
multiplied by I naught. But if you remember
in DC circuits, and instead of having that, we had our service, then we have our
Norton and so on. So in AAC and
instead of using R7, R7 AND, and, OR noting, we say that seven
and North and South. So we have now several
components such as capacitors, inductors, and so on. Okay? So we are not having
just one element, we have several elements. Okay? So let's just start by
learning how can we apply 7.0, similar as what we did before in the previous
lessons of seven. We are going to apply the
same steps in the AC system. So first, we have this large circuit and we
have two terminals, a and B. We would like to find seven and equivalent as the two
terminals of this circuit. And if you remember what does a seven and equivalent to mean, It means that seven and V seven. And so the first step
is that we need to find that seven N. And
what does that mean? It means we are going
to look at our circuit. So you can see it is for myth of independent sources only. What are we going
to do in this case? We are going to make a show or to deactivate all
of our supplies. And then what I want
to see the equivalent. Okay? Now, if you look
at this circuit, you can see this component or
this element is a capacitor and a resistor having
two common nodes. You can see here, one node here, and another node here. This one is a big node here. Okay? And the second node, which is common between
them is this one. So they have two
nodes in common. So it means that they are
parallel to each other. Similar to the 4 ω and j at this point and this point
are common between them, which means they are
parallel to each other. So we can write our
circuit like this. You can see negative j 6.8
ohm parallel to each other, 4 ω and J2 parallel
to each other. And we have the two
terminals a and B like this. So we will be forest, we will get the equivalent
of this branch, which is parallel to each
other, as you can see, the equivalent of that
second branch here, which is also parallel
to each other. And if you look at this circuit, we will have some
single leaves us. One component then a here, B, and almost our impedance like this and the all
connected together. Okay? So if we have a current
flowing like this, it will be the same
current going through the second resistor
or second impedance. So it means that Z1 and Z2 are in series
with each awesome. The equivalent result
is that both of them are in series
with each other. Now we obtained the server name. Now what I would like
to get is V. So V is the voltage between
this point and this point. Okay? So you will find that our circuit can be
written like this. We have our supply like this. We have this branch,
first branch, and a secondary branch, and p and d between
them is V seven. Now the question is, what does the value of V seven, V7 and can be
obtained like this. We can apply a KVL
here, this loop. So what we need is
we need to find the voltage drop across
all nodes of volts drop. First we need the
current I1 and I2. So how can we get
these two currents? If you look at the
voltage source, we have this voltage source
a parallel to this branch. To this branch. So the voltage here is
120 and dangled 75, and voltage here,
120 and Engels 75. Okay? Okay, So how can I get I1? I1 will be the supply voltage divided by the total
impedance in this branch. Why? Because they are in series. Current flowing here or
current flowing here is equal to zero because
it is an open circuit. So this two elements
are in theaters. This two elements are in series. So I1 will be this voltage divided by
the total impedance here. And I too will be a
voltage source divided by the total impedance
here, like this. Okay? So what is the value of V7? And we will apply KVL. Let's delete this. You can apply KVL. A lie exists in the slope e.g. so if you look at this lobe, you will find plus
v 7M, so plus V7. Then if you go like this, you can see that the
direction of the loop is inverse with or a
different from i2. So it will be negative I2
multiplied by the 4 ω, negative I2 multiplied
bys of 4 ω. Then if we go like this, our loop like this, it will be in the
same direction of I1. So it will be plus I1
multiplied by negative j six. As you can see, all of
this equal to zero. And we have R1, we have i2. So from here we can get that
we serve an N required. Okay? So this was the first
example on applying seven in serum on
the AC circuits.
106. Solved Example 2 on Thevenin Theorem: Hey everyone, In
this example we are going to discuss
how can we apply 7.2 AC circuits with dependent
and independent sources. So you can see in
this circuit we have this independent sources, independent source,
independent current source. And this one is a
dependent source. So it means that we
can not adjusted, deactivate and
look at the seven. And we need to add a source
in order to get the seven. And similar to what we did
in the R seven and serum or in czar seven lessons
with dependent sources. So first step is
that we are going to get the V7 and much easier V7, which is the voltage
across here, like this, V seven. Okay? So how can I get V7 and V7? And is the voltage here or
the voltage across here. Okay? So we can get it using a KVL by getting the
current flowing here. The current flowing here. Okay? So you can see we have current flowing through this element. Okay? We have a 15 and pair
coming from the supply. Okay? Then we have 0.5 I
naught flowing here, which is the same current
flowing here, like this. As you can see. Why? Because this two points or two terminals
are open circuit, the current flowing here
or here is equal to zero. So the current of the
dependent source, similar to this one. Okay? So here from this point you can find the z value of I naught
by applying KCL here. You can see we have our 15 and
pair entering and equal to 0.5 I naught plus
I naught y axis. Okay? Current entering equals a
total currently leaving. So from here we can get value of I naught equal ten amperes. So now if I would like
V7 and our blind big, in this big KVL or in the
small KVL, it is the same. Okay, So let's start. So V7, and if we go like this, we meet, we meet suppose step V7 and so
it will be plus V seven. And then we go down
like this and go here. You will find that our current
is opposite to our inode. So it will be negative I naught multiplied by two minus j four. So it would be negative I naught multiplied by two minus j four. Then if I go like this, you will find that
we have our car. Our current is similar
to 0.5 I naught. So it will be plus 0.5 R naught multiplied
by this impedance. As you can see, the
Einaudi equal ten amperes. So we can get V7 as
negative j 55 or 55 and the angle
negative 90 degrees. Okay? So now we have V7. Now why you need to
get the seven n? Now remember, if we
deactivate the source, as we will do exist and we try to get the sevens like
this, it will be wrong. Why? Because we have a
dependent source. We said before, if we
have a dependent source, we need to add our source, such as a current source or
voltage source to get the 17. So as you can see here, we
deactivated this branch. The activation is
an open circuit. Open circuits and cities
are current source. So this one does not exist. So we have 2 ω and negative J4. As you can see here,
I naught and the 2 ω and the negative and
the negative j four. Okay? So do we have this branch
4 ω plus J3, okay? We have this branch,
0.5 I naught. And here we added
a current source. So we select that current
source is three and angle zero. Now, you can choose any value. Any value would like a current source or voltage
source of any value. In the end the organ
to get that seven and similar to R7 and as before
will be V S divided by S. So if I selected
a voltage source, I will get the current. If I have a current
source like this, or you will get the
voltage across it. Okay? So what we need now
is a voltage source. Now why did I choose
a current source? Because it will help our
analysis become much easier. Okay, because we have a lot of, we have a current source here and that current
here and so on. Okay? So how can I get the current? Very simple. You can see here we have I S, which is three amperes. And do we have
here 0.5 or inode? And do we have here or not? So that current entering
equal to I naught plus 0.5. So I naught will be
equal to two m pairs. So how can I get
the voltage source? Again, we will apply KVL. So you can see if
we apply KVL in this big loop like this, okay? I exist. So direction is
clockwise, like this. So first you can see we have plus VS in this
direction, plus VS. Then if we go down
one like this, our current is always
a two I naught, so it will be negative. Note and multiply it by
the total impedance. You can see four plus
J3 and Joe minus J4. All of this will
be equal to zero. So from this equation, you will find that V S equal
to I naught multiplied by all of this as I node
multiplied by all of this. And the i-node equal two amperes so we can get the supply value. So what is the value
of the voltage value? So what is the value of z? Seven and it will be
voltage divided by three. As you can see, we
append that 70. So as you can see
in this example, we obtain also the V7 and
the seven of the circuit. In this case, in the case of
having a dependent source, we said we need to add
a dependent source, as we did in the DC circuits.
107. Solved Example on Norton Theorem: Hey everyone. In this lesson, we
are going to have us all with the example
on the Norton Theorem. So you can see in this circuit, we need to find the
current I naught. So in order to find i-node
using Norton's theorem, we need I naught and we
need the Norton of what? Of this big circuit. So we need to replace
this big circuit with a current source i n to the n. Okay? And we have this branch, I exist j 15.20 and
winning the current. So we replaced this big
circuit with IN parallel to n. Okay? Okay. So first we have here
independent sources. Okay? So in order to get that N
or the Norton impedance, so we will deactivate
the supply. So this will become
a short circuit. This one would be an
open circuit like this, okay, between these two
terminals here and here. Likes us. So you can see here we have an open circuit because we need the equivalent of all
of this open circuit. As you can see, this one became a short circuit and this one becoming
an open circuit. Now, you can see is that the Knowlton is equivalent
of this impedance. You can see short-circuit
battery to an impedance. So it will tell you this
impedance and go away. So we will have only 5 ω. As you can see, that
node n is equal to five. Now second step we
need are in Alton. What does All-in-all thing is? It is a current of the show, the circuit applied here, north. Okay? So Y short-circuit
because we have, we need the equivalent between this terminal and this terminal. So if I need I-naught, it will be short circuit. If I need V7 and it will be a voltage, open circuit voltage. Okay? Okay, so our in Norton like this short-circuit and
we have our circuit. So I need all in all ten, which is equal to the
current of this loop. So I'm going to
do mesh analysis. 123 currents, i1, i2, i3, ends our first loop here you can see if
we go like this, you can see I1,
forest I1 like this. We have negative J4. So we have negative J4. I1 is flowing through what? Through all of z's components. Okay? So it will be eight plus ten, which is 18 negative j2
and J4 it will be plus J2. All of this multiplied
by plus I1, I1. Why? Because its own loop. Okay? Now, what about I2 and I3? We have i2 and i3. Both of them will have a negative sign,
negative and negative. Why? Because it is not their own Do. It is all U1 loop. Okay? Now i2 flowing through
eight minus J2 and I3 flowing through ten plus j for all of this would
be equal to zero. Second lobe is super mesh. Why? Because you can see we
have a current source. This current source,
we have I3 like this. We have i2 like this. So i3 minus i2 will
give us three and bear. So we need another KVL, so we will use the
supermesh here. You can see mesh 2.3
form is as supermesh. So let's apply KVL
in this big loop. So this supermesh
contains I2 and I3. So it will be plus I2 plus I3, plus I2 plus I3. Okay? What does, what does what does the resistors or the impedance
that i2 flow with through? You can see i2 flow with
shrew, 58 negative J2. So it will be 13 minus J2. What about I3? I3 flowing through ten plus J4. What about I1? I1 is not its own. It is a loop of I2 and I3. Okay, So it will be
negative one, negative one. Okay, so here we have
eight minus J2, 10.14. All of this RE1 flow with
Sway eight plus ten, which is 18 negative
J2 and J4 is plus J2. So this is all y1 negative
I1 because it is, it is not its own Lu. Final equation which is coming
from the current source I is three minus two gives
us three and Beyers. Okay? So what do I need? I need to find what I
need to find only I3. Why? Because I Norton
is equal to I3. Okay? So how can I do this? I3 equal to I2 plus three. So I need i2 to
get the ice-free. So if you look at
these two equations, so we have these two equations, we will add them together. So if you add equation
one and equation two, you will get this. You can see negative
j 40 and you can see 18 plus J2 i1 plus, and the negative 18 plus J2 I1. So they're summation
would be zero. So this will go with this. If you look at here, we
have negative ten plus J4, i3 plus ten J4 ICT. So this will go with this. So we will have this
only these two parts, negative negative J2, which is plus J2, and then negative J2. So they will go with each other and we will have negative 8.13, which is five items. Okay? Okay. So from here, I two
will be equal to j. So we can get I3 equal
to three plus j, which is similar to I naught. So we will take this as a
current source like this, parallel, parallel
to it, z naught. This is the equivalent
of this big circuit. So we will replace this
with r naught and z naught. Then we will connect the 20
and J5 to get our inode. So how can I get on the
bike currently division? So I know the quantity, supply, supply value, multiply it by the other resistor divided
by total resistors. So you can see our
inode equal to I n, which is a supply multiplied
by the other resistor, divided by the total impedance. As you can see from here, we can obtain the
current required. Now, one important thing
you can notice here is that is that this method using
a Knowlton serum is much, much difficult than
other methods. Now e.g. if you look at the way this is our original
circuit, okay? So I need I naught. So what can I do? That simplest thing you
can do is that you can do a search image
slopes directly i1, i2, i3, the ICU
equal to I naught. So by doing this three
equations of mesh analysis, you can get all inode directly without the need
of if Norton's theorem. However, you can see
North and serum made us do several extra steps, made the equation
much more difficult. So the selection of the
method used in the analysis, well-defined how short or how long you will
solve the equation. So you can see mesh
analysis was much simpler than using the Norton's
theorem itself. Okay?
108. Introduction to AC Power Analysis: Hi, and welcome everyone to this part of our calls
for electric circuits. In this part, we are
going to talk about with the AC power analysis. So first, what are we going to achieve in this
part of the course? So what I would like to get
in the end of this section, I would like to tell you
the difference between what doesn't appear on the
power mean, apparent power. Apparent power denoted by S. What does this mean? What does an
apparent power mean? The act of power P is the active power P in the electric circuits
and the reactive power. Reactive power. And denoted by Q. I would like to know what
does an apparent power mean? What does an act upon it mean? What does it mean? What does a power
factor even mean? These concepts are very
important in the AC circuits. We would like to know
what does these terms mean and what are these values? So let's start by learning some basic concepts
about AC power analysis. So as you can see in the
previous lessons of the course, we focused on finding the voltage and current by
using that voltage KVL, KCL, by using all Sousa mesh
analysis and nodal analysis, that seven and
Knowlton and so on. Now, our main concern, or the major concern here in this class
or in this section, is a power analysis. You will find that power is the most important quantity
in the electric utilities. That electronic and
communication systems. Because such systems involved
with that transmission of the electrical power from
one point to another. And of course, any
electrical device, any industrial or household
electrical device, has a power rating. How many kilowatt
and so on, e.g. a fan or a motor, or a lamp, or a computer, any of these devices
have a power rating. The most common form
of electrical power, at which we transmit
electrical power, we transmit them at a frequency
equal to 50 hz or 60 hz. This dependence or a change
from one country to another. Now you will find that
some times when we are transmitting electrical
power from generation side. So let's say we have a generator that produces
the electrical power. And as we will know
in their courses, in our courses for electricity
or electrical engineering, you will find that the generator produces a three-phase power. Anyway, we don't, we are not concerned of what does
this even mean now. But what is important
for us is that when we transmit electrical power
from one location to another, we can use their AAC system
or we can use the system. So usually you will find
that most of our systems, our AAC, we transmit electrical
power using AAC systems. However, you will
find that sometimes, sometimes we use a
DC in transmission. Now when do we use
DC and transmission? When we have a very large
transmission lines, very, very large
transmission lines, in this case we use DC. Another case in which we use DC is that if we have a
country that uses a 50 hz and another country that uses a second
starters in order to transmit electrical power
from one country to another with different
frequencies, we use a DC transmission system. Okay? So in the end we need to
understand the meaning of AC power and the
types of AC power. So first we will identify what does an instantaneous
power mean? The average power,
the root mean square, root mean square, or the effective value of
the voltage or a current. And what does an
apparent power mean? And the reactive power. Okay, so we are going to
do throw all of this. So in the end we will understand
the root mean square, apparent power, reactive power, power factor, and active power. Okay? So in the next lesson, we will start by going through these definitions and understand what does that even mean.
109. Instantaneous Power and Average Power: So let's store to buy, meaning of the
instantaneous power. So let's say I have an
AC supply, AC supply. And this AC supply produces a voltage V as a
function of time. This is a voltage
produced by the supply. And we have a current coming
out from this source. So as if we have a
circuit like this, we have an AC supply, okay, With a certain voltage V and supplying current as
a function of time, e.g. and passive linear network, which is consisting of resistors, inductors,
and capacitors. So let's say we have here that. Okay? So the first thing
that I would like to learn is the
instantaneous power. Now what does this mean? So the instantaneous
power is defined as power as a function of time, is equal to the
voltage multiplied by current in DC circuits. In DC circuits, we
said that the power, power at any instant is equal to the voltage multiplied
by current, right? Same idea in the AC systems, we say that the voltage
as a function of time multiplied by the current
as a function of time. So this will give us n equation of the instantaneous power. So what does this mean?
It means that ZAP power at any given time. So as you can see that
the instantaneous power, which is measured in what is the power at any
instant of time. So let's say we have the voltage as a function of time and the current as a function of time
is equal to this equation. Each one is a cosine wave with a certain phase shift
C to C, V and C. Now you will find
that V, m and IMR, the amplitude or the maximum value of the voltage
and the current. And the angle Theta
v and c two are the phase angles of the
voltage and current. So if we substitute these
values in this equation, we will have
something like this. There's a power V
multiplied by current, which is this equation
multiplied bys this one, as you can see here. So it will be V MIM cosine Omega t plus Theta v
cosine omega t plus Sita. Okay, so what does
an extra step? The next step, we are going to use that from trigonometry. We can say is that if we
have two cosine waves, we can say it is equal
to half cosines. A difference plus cosines
are some mission. So if we apply this trigonometric
form to this equation, we will have
something like this. We will have zero
power is equal to half VMI m cosine Theta v minus I plus half V m by m cosine two omega t
plus Theta v plus c. Or you will find that
the instantaneous power, which is a voltage
multiplied by current, will be this equation. Now, if you look
at this equation, you will find is that it's
consisting of two parts. You will see that we
have half V-max imax. All of this is a constant
and the cosine Theta v minus c die, this is also constant. So you can see we have a port, this part which is constant. It doesn't change with time. And we have another part which
is a function of omega t. However, you will find
that it is j2 Omega t, which is double the
frequency of the supply. Okay? So we have this part, which is a cosine wave. So we have a DC part, which is a constant part, plus a variable part
or a cosine wave. So you can see that
it has two parts. The first part, which is a constant or the
time independent, you can see constant and
does not depend on time. And it's a value depends
on the phase difference between voltage and the
current Theta v minus Theta I. The second part is a sine wave. This part which is
a cosine wave or a sinusoidal function
whose frequency is twice the angular frequency
of the voltage or current. So if you remember that
if we get back here, you'll see that the
voltage or current, it's a frequency is Omega. However, if we look at that current here or the power here, you will find that
it's a frequency is two omega t, two Omega t. So it's, the
frequency is double, the frequency of the Voltage or current. So you can see that when we add a DC component to W cosine
wave frequency cosine wave, we will have this
final wave form. So what does this
waveform represent? This waveform representing what? Representing any
spontaneous power. So as you can see, one cycle
exists from here to here. This considered as one
cycle from here to here, and second cycle
from here to here. Now what you will find
is that what you will find that the time taken for two cycles is
equal to t y t, t is one over the frequency of the original supply of the
voltage or the current, frequency of the
voltage or current. Now that time, this
is T of this supply. However, sensors or frequency
of the power is double. It means it will make two waves in the
same period of time. So you will see that here. What we notice here
is that we have power sometimes equal to positive and other times
equal to negative. So what does this mean? It means that when the
power is positive, the power is absorbed
by the circuit. So our supply, when
the power is positive, it means it's supplying
electrical power to the load. Okay? When's the power
is positive above zero. During this period at which
we will have negative power. It means is that our supply is absorbing electrical
power from the circuit, the power coming from
the load to the supply. Now how is this possible? You will find that we
have a storage elements, we have capacitors
and inductors. And if you remember that
capacitors in one cycle, they absorb electrical
power and in other cycle, they provide electrical
power to the circuit. So the average power taken by this inductor or capacitor
is equal to zero. So they absorb electrical
power and store it, then supply it back
in another psych. The instantaneous
power it changes with time and it is therefore
difficult to measure. So what we're going to do, we are going to use another term which is the average power. Average power will help
us understand how can we deal with the power
inside the circuit. So instead of dealing with
the instantaneous power, we will see the
average power, e.g. the average power is the
power measured the visor. What meter? Meter is used to
measure the power consumed by any
electrical element. So when we connect
to this instrument, to the lute or any part, we measure the power. So the power measurement device, or what meter is called, is the average power. So this is what we are going to do in the electric circuits. We need the average of power in Eastern instead of
the instantaneous power. So this is the instantaneous
power equations that we obtained in
the previous slides. This power is the instantaneous
power at any given time. Now, what I need is
the average power. What is the average
power of this waveform? So as you know, is that
the average of any signal? So let's say I have y, which is average of
a waveform called x. Okay? So if I would like to get that average
of any waveform, it will be one over
is aperiodic time. One over the periodic
time integration from zero to this period of
the function itself, let's say x as a
function of t, d t. So if I integrate a function
and divide it by the period, we get the average value. Similar to a sine wave
like this so far, if I have a sine wave like this, and I would like to get
the average of this wave. So it will be, let
say is that period. Let's say that this
period is two pi, which is a period
of a sine wave. So it will be y, which is the average of this
wave will be 1/2 pi. Integration from
zero to that period, which is two pi multiplied
by the function. Let's say it is V-max. This is a maximum value of V max sine omega t d t, like this. So this will give us the
average of the wave. Okay, So what we are going
to do is that we will take this equation
and substituted here. So by substituting like this, we have two components, the first part and second part. Now, by doing this integration, you will find that
the average power is equal to the DC component. The average power is equal to half V-max Imax cosine
Theta v minus c dy, which is this part. Now the question
is, why is this? Because you will find here, we have this part and this part. We integrated the furthest apart and integrates
the second part, the integration of a
DC value is set or the average of a DC
value is that DC value. Let's say if I have
a voltage source, a DC voltage source of a
value equal to two volt. Since it is a DC supply
similar to this component, its average will
be also two volt. So the average is equal to the value of the supply itself. So the average of a DC supply
is similar to its way. Now for the second part, which is a cosine wave. Cosine wave is
something like this. Likes us. This cosine wave or a sine wave. Sine wave like this. A cosine wave or a sine wave. It's the average
is equal to zero. So the average of a sine wave or an average of a cosine
wave is equal to zero. That's why you will find that
this part is equal to zero. And we have only the DC part, which is this component, will find that if
we have, of course, we have the voltage equal V
max and the angle Theta v, or equal imax and angles
Sita in the phasor form. So have half of V. I conjugate is equal to half V-max imax
Theta v minus Theta. So if I take half
of the voltage in the phasor form multiplied by the current conjugate will
give us this function. Half V-max or Emacs. Voltage angle, which is theta v. And I conjugate means is that
zeta will be negative C dy. Now, this is a phasor form
of half VI conjugate. Okay? Now, if I converted
this phase or form into the royal plus
imaginary, like this, this one is equivalent to half
V-max Imax cosine Theta v minus c2 plus j half V-max
imax sine Theta v minus Theta. This phasor form can be
written as this form. So we can conclude, conclude that this
part, this part, similar to this part, This part is the real part of V i conjugate, right? So how VI conjugate is equal to this real part plus
the imaginary part. The real part of how VI conjugate is
similar to the power. So we can say is that
the real part is equal to the power, likes us. So power equal to half real part of the voltage multiplied
by i conjugate, which is how V-max Imax
cosine Theta v minus Theta. So you will find that if
we look at this equation. So we have our supply, supplying electrical
power to any load. Okay? So let's say we have a pure resistive load
up your resistive load. So in this case, the voltage
and current are in phase. They are in phase because
we have a resistive load. So there is no phase shift. So finds that C
times V is equal to c2c to be equal to C2. So in this case voltage and
the current are in phase, or we have a pure
resistive circuit. So when zeta V equal to C2, or is this part will
be equal to zero, this angle is equal to zero. So cosine zero is equal to one. So our power will be
equal to half V-max imax. Or since the voltage
multiplied by current or half I square multiplied by R, or the magnitude of the current square
multiplied by the resistor. All of them are the same. So what we can see that
all the power coming from the supply is equal to the power consumed
inside the resistor. Okay? However, if we have
another circuit, which is a pure
reactive circuit, what does this mean? It means that we have a supply connected to a capacitor
or an inductor. So in this case,
you will find that the difference in angle C, V minus C to 0 is the difference between voltage
and the current is 90 degrees positive
or negative depending on that capacitive or
inductive circuit. In this case, this angle will be 90 degree positive or negative. And the cosine 90
is equal to zero. So the average power
will be equal to zero. So what does this mean? It means that the average
of our consumed by a capacitor or an inductor
is equal to zero. And the average power
consumed by the resistor is equal to half
V-max all Emacs or have I squared R. So we'll find that the resistor
absorb power also time. However, a reactive load
absorbance zero average power. So what does this mean? It means sometimes it
absorbed or store it, a store or a store,
electrical power. And other times it will
supply electrical power. So sometimes positive consume, sometimes absorb, which means
sample, sometimes a supply. So it means it returns back the stored energy to the supply. So in this case,
the average power consumed is equal to zero because the store to power is returning back to the supply. So in this lesson, we talked about with Zara
in instantaneous power, average power in the electric
or the enzyme AC circuits.
110. Solved Examples 1: Hey everyone. In this lesson, we are going to have some solvent examples. On the instantaneous
power, average power. We have sinusoidal supply
voltage V equal to this value. And then we have a current coming out from it going to add passive circuit with a value of I as a function of
time equal to this value. Now what we need
in this example, we need the instantaneous power. We need the average
power that is absorbed. But boy is a passive
linear network. First, what is the
instantaneous power? Instantaneous power is
the multiplication of the voltage and the
current, right? So if you multiply
voltage and current, we have this equation. So we can say is this is
the instantaneous power. Now, let's do the same
trigonometry phone in order to learn
about something here. So you can see that
if we use that trigonometric identities that we have done in the
previous lesson, we can get finally is that power is equal to a DC component
and another component, which is cosine two omega, you can see omega 377. You can see we have here two omega plus the angle which is C, dv plus c to all, which is 45 minus ten to five. So what we can see
from here is that the average power is equal to the DC component, as
you can see here. So the equation of the average power that
we obtained is half V maximum Imax cosine Theta
v minus v max Imax, we have hundred 20 and
then as you can see, 120.10 and the cosine
Theta v minus C2. So C2 is plus 45 minus as a current which
is negative ten degrees. You can see negative
ten degrees. So this will give
us finally 144.2, which is similar to what? Similar to the DC component
of the instantaneous power z, DC component or the
constant value, to be more specific. Now let's have another one. So we would like to know the average power
absorbed by an impedance. So we learn is that the
average power is equal to half V-max Imax cosine, c to v minus c toy, right? So what we need is
that we need V-max, we need imax, we
need the two angles. So first you can see
that the voltage in phasor form is equal
to V max and the angle C v. So the maximum value
is 120 degree 120 v, okay? V-max hundred and 20 volt. The angle, this
angle is equal to z. Now what I would like to get, I would like to get the current. So as you know that the current in any electric circuit
is equal to the voltage. The voltage is boys. Okay? So we have a supply V, providing electrical
power to this impedances. So in order to find the current, it will be V over z. So that's what we
are going to do. Take this voltage
divided bys it. So we will convert this into phasor form as we
learned in the course. So in the end, we will have the value of
the current is equal to 1.576 and the angle
is 66.8 degrees. Okay? So our, our Emax is
the maximum value is 1.576. The angle is 66.8 degrees. Okay? So we will have
our average power equal to like this
half hundred 20 multiplied by 1.56 5276
cosine zero -66.8. So our average power
absorbed by the impedance, so that is equal to 7.24. What? Let's have another one. If we have this circuit, so we have a supply providing an electrical current to a
four ohm and negative J2. So we need to find the
average power supplied by the source and the average power absorbed by the resistor. So far as to what is
the average power, we know that average
power is equal to half, half V-max, or Emacs cosine Theta
v minus c. Okay? So what is the value of
the voltage equal to 5 v? Okay, what is the
value of angle? This one is so t degrees. Okay? So now we need the current. So the current coming
from the supply. So since we are talking
about average power coming from the supply
supplied by the source, then it will be the
voltage of the supply multiplied by the current
coming out of it. This is the instantaneous power. The average power will be u
is a voltage maximum value of Z Alpha supply multiplied by the current maximum
coming out of it. Like what is the value
of that current from KVL supply divided by the
total impedance. Like this. Supply divided by total
impedance will give us the value of the current
flowing inside our circuit. So from here you
can see 1.11 eight, which is Imax, 0.11, it exists. And the angle is 56, right? So you will have like this, average power will be
half five multiplied by 1.11 eight cosine -56.57. So this will give us that the average power provided by the voltage source
is equal to 2.5. What? Now, what is anycast
requirements? Requirement is that
we need to find the average of power
absorbed by the resistor. So if we look at this circuit, we have our supply, right? Providing an electrical power to the four ohm and a capacitor. So what do we learn it is that the average power of the
capacitor is equal to zero. Right? Average vowels through a capacitor or an inductor
is equal to zero. So pi logic, logic, the average power coming
from the supply is equal to the average power consumed
by the resistor. Right? So let's prove this. So in order to get the average power
absorbed by the resistor, it will be half V-max Imax cosine
Theta v minus C. Now, what is, what current is the current which is
flowing through it, which is a supply current, which is this value. Current from supply
is similar to the current flowing
through this resistor. Okay? What about the voltage? What voltage is the voltage of the supply or
any other voltage? So since we are talking about average power absorbed
by the resistor, it will be the voltage
across that resistor. Here. The voltage across the resistor. So how can I get this voltage? Simply, it will be the current multiplied
bys of 4 ω, right? So the accounts Rosa resistor
is the same supply current. And the voltage will be
the resistor which is 4 ω multiplied by the current. So it will give us this value. Then what we are going
to do is that we will substitute this value
in the equation. So as you can see that the
average power absorbed by the resistor half
V-max maximum value. What is the maximum value here? This value, or Emacs, which is 1.11 eight multiplied by cosine
Theta v minus Theta I. Now, biologic c to v
will be equal to C2 ice. And so we are talking about the voltage and the current
through the resistor. So in this case, z are in phase and the difference
in angle is equal to zero. So cosine z equal to one. Another thing you can see, see TV which is which is 56.57, this angle equal to C
to E, which is 56.57. So their difference
is equal to z. So in the end is about
would be half V max or max, which is 2.5 watt. Now, as I said at the beginning, by logic, the average power absorbed by
that resistor, 2.5. What is similar to the average power
supplied by the source, which is 2.5 volt. And the average pulse through the capacitor is equal to z. Okay? So you can see it's equal to the same average power supplied and zero average power
absorbed with boys capacitor. So in this lesson, we had
some Soviet examples on the average power and
the instantaneous power.
111. Maximum Average Power Transfer: Hey everyone. In this lesson we're
going to talk about the maximum average
power transfer. If you remember in
the DC circuits, we said that if we have a linear circuit
consisting of resistors, groove of resistors, and we connected to a resistive
load to it, Let's say r. And we said that from that maximum power
transfer theorem that we discussed
in the DC circuits, we said that the maximum
power transfer occurs when, when R L, the load resistance
is equal to r seven. So if we take the equivalent
of this circuit and we have plus minus v sub y exists and are seven
connected to our load RL. Rl. In order to have the maximum power transfer
that to this resistor, R L must be equal to R7. Now we would like to see the same process
in the AC circuit. If we have a circuit
like this one, a linear circuit
consisting of a resistors, inductors, and
capacitors connected to the load impedance. I would like to find the
value of load impedance that URL that will produce
the maximum power. So if I would like to transfer
the maximum power from the supply to this impedance, I will need to find
the value of z. Okay? So we need load impedance
that will produce maximum power transfer.
In the DC circuit. We solve with this
by the problem of maximizing the power
delivered the pie, supplying a resistive network to allude representing the circuit by its seven and equivalent. And the maximum power will be transferred when that
load resistance, or RL is equal to R7
and R7 and resistance. Now, the same
process we are going to do in the AC circuits. So first, let's start by
representing our circuit. So first we have zed 77, which is R7 plus j X7, which is the equivalent impedance
of that linear circuit. And we have our load, that URL, which is consisting
of RL plus j XL. Now our goal is to find
the value of RL and XL that will produce the
maximum average power transfer. If we look at the circuit
here, this circuit, the current will be equal
to V sub n divided by seven plus seven is equal to this value and L
equal to this part. It will be this equation. So we have this
equation of that. Now, as we know,
are, as we'll learn, is that the power or the average power max as
average power in general, is equal to half
the current square multiplied by the resistance. Okay, so here I'm
talking about what is the maximum power
transferred to the resistor, the resistor inside the cell. Because the average power across an inductor
is equal to zero. So as you know that
here we have the L, which is RL plus J exon. As our power transfer. The average power
transfer through an Excel is equal to zero. The maximum power transfer
or not the maximum, the average power
transfer is equal to the power consumed
inside the RL. So we say half I square
multiplied by the resistance. So I squared is the
magnitude of the current. So as you can see here, we have our current V7 and over r seven and plus j X7 plus R L and J XR. So I as a magnitude will be
equal to magnitude of V7. And so V7 and not a phasor form about the magnitude, okay, V7. And that this part
we have, it will be, its magnitude will be the root of the real part squared plus
the imaginary part squared. Okay? So the real part
is R7 and plus RL, or seven n plus RL. And the imaginary
part is X7 N and x, x one plus x. Okay, So this power to representing the
magnitude of the current. So if I take the square
of this current, it will be V7 and a square. And the square root will
be, will be removed. Since it will be square of the square root will be removed. So we will have
all seven m plus r squared plus X7 plus x squared. As you see here. The half came here
and R L came here. Okay? So what is our objective
is we would like to get the maximum
average power transfer. So you can see that
our power here, our power here, is equal
to this big equation. Now what does our unknown or what we would like to
get is the value of RL and XL that will produce
maximum average power. So we have two
parameters, RL and XL. So what are we going to do? We are going to get Zao, partial derivative of
z with respect to r m and the partial derivative of power with respect to x l
and equate them with zero. In order to get the value
of RL and XL that will produce maximum average
power transfer. Okay? Now why is this? Because if you remember
in DC circuits, in the DC circuits, in order to get
the maximum power, we get the derivative of
the power with respect to z resistance, right? And equate it to zero. However, here in the
power in AC circuits. So we have two parameters. We have our L and we have Excel. So we will get the derivative
with respect to partial P, partial r l and
equate it with zero. And you get partial P, partial x l and
equate it with z. Why is this? Because we
have two unknowns here, two parameters that
will affect the power. So we need the
value of RL and XL that will produce maximum power. So this is what we are going to do if you get the derivative of the power with respect to x, well, derivative of the
power with respect to RL. You will have this
two equations. And if you equate
them with zero, first one equate this with zero. You will find is
that the value of x is equal to x negative Excel. You can see XL will
be negative x7. And for the second
equation equated to zero, we will have our L equal to root R7 squared plus X7
plus x l all squared. So let's see this
equation really quickly. So you can see XL
equal to negative except negative except
r l is equal to root. This equation. You can see all seven
in X7, N, and X. Now you can see
that Excel itself, we said in order to produce
maximum power transfer, we have here x L equal
to negative x seven. So if I substitute
with Excel becoming a negative x seven wives
this negative x seven, unlike this, you will
see X7 N minus x over n. So this part will
be equal to zero. So the equation will be rude. Or seven and square, which will be equal to R7. So what we can learn from
here is that in order to produce maximum average
power transfer, Excel should be negative x7 and RL should be
equal to r seven. So you can see here
that of solute, which is RL plus j X, should be equal to r l, which is R7 and R7. As you can see, the
Excel negative X7, Excel negative x seven. So what we can see here is
that R7 in minus j X seven is seven and conjugate of Z
load required to produce maximum average power transfer is the conjugate of that 70. Okay? So for maximum power transfer, the load impedance
that must be equal to the complex conjugate of
the seven and impedance 70. So if you take this equation, if you take the lute or
if we take this values, excel equal negative x is
seven and r equal to R7. Okay? And substitute it in the
main equation of power. This equation, this representing the average power transfer. So the maximum power occurs
in what, in what case? When we take excel and to make it negative x7 and RL LR 7M. So if you take all
seven and plus RL, which is our seven, and x equal negative x is seven. This part will be equal to zero, and this part will be
r2, R7, and square. And we have V
squared L over two. You will get finally V7 and
square over eight or seven. And this is the equation of the maximum average
power transfer. Now, let's go and have
some examples on this to understand the idea of the maximum average
power transfer.
112. Solved Examples 2: Let's have some
solved examples on the maximum average
power transfer. We have this circuit here
and we would like to find the value of l that will lead to maximum
average power transfer to this immittance. First, we will start by getting
a V7 and end the seven. Okay? Because we need load impedance and we need maximum
average power, load impedance is 7M. Okay? Or is it 7M conjugate, which is all seven, n minus j, X is seven. Okay? The maximum power transfer
is V7 n squared as the magnitude divided
by eight or 17. Okay? So let's start First. We will get seven and adds
a true terminals here. We will make here
an open circuit and see what is the
equivalent circuit. So this one will be a
short circuit like this. Like this. And we will see
what is the equivalent. So as you can see, we will
have the circuit like this. So the seven and
will be equal to this two branches parallel
to each other plus j five. So you can see J5 plus four ohm battery to
eight minus six. Okay? So the equivalent
of this will give us 2.2 933 plus j 4.467. And then we know that
the load impedance, which is loot, will be
seven and conjugate. So we'll find that here, forest of it, V sovereign before we get
the load impedance, V seven, V7 and will be this two parts open circuit and we will find this voltage. You can see that this one will produce a current alliances. And all of the current
will go through this. And zero current will
go through here. Because here we have
an open circuit. So V7 and will be the voltage across this branch, this part. By using voltage division, it will be ten
volt multiplied by n minus j six divided
by the total impedance. Okay? So V7 will be equal to ten
volt multiplied by eight minus J6 divided by the
total impedance four plus eight minus six. So it will give us
this V Thevenin. So as we said before, that the load impedance will
be that 7M conjugate. So it is similar to that
seven and bought conjugate, which means negative here. So if you get back here, you can see instead of
velocity will be minus. Okay? Now the maximum average power is V7 and a squared
divided by eight or 17, or 7.2, 0.2, 933, and V7 and as a
magnitude, 7.454. So this will give
us this value of the maximum average power, transfer it to this impedance. So let's have another one. Find the value of
R L Z will absorb the maximum average power
and then finds is power. Okay? So first, in order to
get the value of R L is it will produce the maximum average power
in the AC circuits. Our L value should be equal to seven and conjugate or
to pee more specifics. And so we have our
resistor here. And this is a two components. So it means that our L as a magnitude should be equal
to z seven magnitude. Okay? So let's start First. We need the equivalent. So if we make this one a
short circuit like this, and look at our
circuit like this, you will find that
j 20 parallel to four t minus j three y-axis. So it will give us z equal to
this value for v sovereign. Let's delete this for V7 and
it will be the voltage here. V7, like this. So it will be what? It will be j 20 divided by the total impedance
J2 and J2 into plus 40 minus j sorting multiplied by the supply
voltage, voltage division. So it'll give us this value. Now, what is the value of RL will absorb maximum
average power. So we said before that RL. Should be equal to
seven n conjugate. Or the magnitude of R, L should be equal to the
magnitude of z conjugate. Why? Because seven is already are posted or a real
and imaginary part. So combining them together, we will have a positive
value, which is our L. So the magnitude of R
L should be equal to the magnitude of z conjugate, which is 9.412 square
plus 22.35 squared, all under the square root. So it will give
us 24.25 ω, okay? Okay. Now what is that? Maximum power? As we know that first, this is our equivalent circuit. And we know that the
maximum power is v squared divided
by eight or 70. That what do we learn it, right? However, you will find that this is not valid in this case. Now why is this? B calls? Because our Z glued was
consisting of two components, consisting of R7
minus j X7 N. Okay? So when we substitute with
this in this equation, in the main equation
of the power, we will get V7 and a square
divided by eight or 17. However, however,
here you will find that we don't have z. We have our l only. So our z is our L. Okay? So we have our L which
is equal to xy value of z as a magnitude, right? We don't have RL plus j XL. So what are we going to do simply we album to use
the main equation, which is half I square
multiplied by RL. So as we know that that average power across any
component or a resistor, it will be half the current
square multiplied by RL. So we have this equivalent
of the circuit, which is V7 and ends at 70, Ends a resistor, RL. So we have here a resistor, or the maximum power transfer
to RL is half I square RL. So that current, what is
the value of current? It will be V7 and divide
by seven and plus RL. So we'll have this
value of the current. Now is that magnitude have the magnitude square multiplied
by the resistor will give us the maximum
average power transferred or absorbed
by that resistor. So we'll see how I square RL, which is 1.8 square
multiplied by RL, which is a resistive
load. Okay? 24.25. Okay? So to give us 39.29, now remember, this is
a general equation. V7 and a square divided by eight or
seven is a special case. When that z, we have z del, which is RL plus j XR. When we have, however here you can see we
have only a resistor. So we cannot use this equation. Okay? We can only use this one. Okay? So in this lesson we had some solved examples on the
maximum power transfer.
113. Effective or RMS Value in AC Circuits: Hey everyone, In
this video we would like to talk about the effective or RMS or the root
mean square value. So we would like
to know what does an effective value
mean or a root mean square value mean for the
current or the voltage. So the idea of the
effective value arises from the need to measure
the effectiveness, the effectiveness of voltage or a current source in delivering
power to a resistive load. So the effective value
of our periodic current is the DC current that deliver the same
average power to register as the
periodic current. So what does this mean? So let's say we have
our supply here. We have an AC supply that
provides an AC current. This provide this electrical
power to the resistor. So we have a P average, average power delivered
to the store. Why is this AC supply? Now, if we have, if we have a DC source, dc source with a DC current provides also another
power, let's say P2. So what does an
effective value mean? It means that if I
have an AC supply, I would like to know
what is the value of D C supply that will
give us the same power. So B2 will be equal
to B average. So the power delivered in a DC circuit is equal to the power delivered
by the AC circuit. And I would like to know
what is the equivalent, equivalent souls of the periodic current
or the AC current, the equivalent value of the DC supply to
provide the same power. So that is what does a root
mean square value mean. Okay? So you have to understand
that the concept of the root mean square value or the
effective value really, really important in
electric sockets. Now, why is this concept
really important? Because we use it in
a power analysis. We use it in understanding
zen meaning of active power, reactive power, and
apparent power. Okay, you will find this
root mean square value in every electric socket in the power system
analysis and so on. So let's just start
by learning how can we get the root mean square. So first, we know that
in the AC circuit we have the power average
power is equal to one over the period integration
from zero to t, I square R, d t and d. We know that the
resistance is a constant, so we will take it outside. As you can see, we have the average power
absorbed by the resistor, all delivered by
the supply equal to the current square integration of the current squared d t. Now what about DC circuits? The DC circuit, the
power is equal to the current square multiplied
by the resistance, right? The power consumed or
absorbed by the resistor is the current squared
multiplied by the resistance. In the case of the DC supply, we have an AC current. Okay? We have D, C current. Now, I would like to get
the value of i effective representing the
equivalent value of the AC supply as a DC source. Okay? So if we take this
AC supply, Andrew, Bless the advisor
voltage source here, we will get the same power. Okay? So how can I do this simply, we are going to equate the
AC power with the DC power. So by equating this equation, we will find that the
current effective, which is a DC current, that will give us the same
power of the AC circuit. It will be root of one over
t integration of zero to t squared d t. So the same
idea for the voltage. It will be rho one over t integration from zero
to t v square d t. So what we can see here is that the effective value of current, or effective value
of the voltage, it is a square root of the mean of the square
of the periodic signal. Okay? So that's why we call it
the root mean square. So you can see here, if
you look at this equation, you can see we have forest
that we have our root, this root, so we say root. Okay? Then you can see we have here n integrations that we get from it, the average, okay? So we say mean or average. And we can see what average, average of the square of the current or the
square of the ball, so we can say is square. Okay? That's why we say that
this value of current is the root mean square value. Or the value of the voltage is root mean square of the voltage, or abbreviated as R, MS, MS. Okay? So when we say that we have
a root mean square value, it means that we have the root average of the
square of the signal. Okay? So how does this will help us? You will see that if we have
the power in the circuit, the power will be 0 RMS square multiplied
by the resistor. This is the power in
the AC circuits, okay? Or in the DC circuit. This power, the same
power can be obtained by root mean square multiplied
by r in the AC circuits. So anyway, what we
can see is that here, if we have a sine wave or a
cosine with a sinusoid wave. So if I substitute with
IM cosine omega t and this equation and square it and get the integration
under the square root. We will have this final value. So finds that the
root mean square of the current office I enjoyed wavy sinusoid or a sinusoid or a sine wave or a cosine wave. It will give us, in the end, it will give us our
maximum over root two. So if I have a maximum
cosine omega t, This is d, this
is an AC current. The equivalent root
mean square of it. The effective value of it is I maximum divided by root two. Similar to the voltage if we
have v-max cosine omega t, The voltage as an RMS value
will be equal to V max. Why did, why root two? Now, remember these
two equations. These two values are only valid. Only valid for what? For a sinusoid wave or a
cosine or a sine wave. If the waveform is a square, if the current is a square
waveform or any other wave, we can not use these formulas. Okay? So similar for the voltage, it will be like this. And we said that this
equations are only valid for the
sinusoidal signals. You will find that the
average power can now be written in the terms of the
root mean square values. So the power or the average power in
an AC circuit is equal to half V-max Imax cosine
Theta v minus Theta I. Remember this equation is four. What is four? The voltage and the current in a cosine wave, cosine omega t. And we know that V max our Emax. So we can say this part can be written as V
max or max over root two and root 21 over root two multiplied by one
over root two is half. So we divide this half into one over root two multiplied
by one over root two. And we know that V-max
over root two is V RMS and the imax
over root two is RMS. So it means that the
voltage as an RMSE value multiplied by the current as an RMS value
multiplied by cosine. The difference in
angles will give us the average power delivered. Or we call the average
power as the active mode, which is the power consumed
inside the resistance. Okay? So we'll find that the average
power can be equal to I squared multiplied by R
or V RMS squared over r. Okay? So how does this
equation help us? This equation, this
equation help us, us by dealing with the AC circuits as if
they are DC circuits. So you can see that in a
DC circuit like this one, Let's get back here. In this circuit. In the DC circuit, we know that the current, the power consumed
in the resistor is the current square multiplied
by the resistance. Right? Now, we converted this AC
circuit into DC circuit, the equivalent DC circuit, by getting the root
mean square values. Okay? So anyway, in any
electric sockets, in any electric socket in the AAC for if I would
like the power consumed, it will be I RMS square
multiplied by the resistor, or a V RMS square, which is the voltage
across there. Resistance, or a
resistor squared over r. Okay? So as if we are dealing
with DC circuits. Okay? So let's have some examples on the effective value to
understand more about it.
114. Solved Examples 3: So let's start by taking
this first example. In this example,
we have RMS value. We would like to
get the RMS value of the current waveform. So we have a current
as a function of time, has this waveform. If the current is passed
through a two ohm resistor. So r is equal to two ohm. Find is the average power
absorbed by the resistor. So Forest, I need
all you are a mess. The root mean square value
of the current waveform, the effective value
of this waveform that will give us the effective
value of the current, and that will give us the
same power as a DC circuit. Okay? So here, and the average power, the average power will be
equal to the current square multiplied by that resistor squared multiplied
by the resistance. So we have the resistor equal to 2 ω and we need the root
mean square current. So the first step that we
have to do is that we need to write our waveform in
the form of equations, equations that
representing the equation of the current for each time. So you can see that this
wave form of current, you can see starting 0-10, then from ten drops
down to negative ten, negative ten until four. Then it drops down to
zero and repeats itself. You can see that the cycle of the current cycle repeats itself every 4 s.
You can see 4-8, another cycle, 8-12,
another cycle. So the periodic time is
four, which is important. The period of the waveform is for it repeats itself every 4 s. Okay? Now we would like to
write the equation. You can see that the
equation of the current, you can see we have 0-2, we have a straight line, and 2-4, we have
a constant value. So we have 0-2 and 2-4. Okay? 2-4, it is really easy from here to here you
can see it's a value, is a constant value
of negative ten. From here to here.
How can I write this equation of
this straight line? So we know that y is
equal to m x plus c, which is the equation
of a straight line. So y here, like this, m is the slope of the line. So slope of xi1 is
equal to Y2 minus Y1 divided by x2 minus x1 multiplied
by x plus a constant. Okay? So our y is current. So we say I as a function
of t equal to Y2 minus Y1. Let's say choose any two points. Let's say this is
our final point, and this is our initial point. The second point, which is
why n2 is equal to ten and y, y1 is equal to zero. So it will be done minus
zero divided by x2 minus x1, x2, which is this final value, which is two and the
initial is zero. So it will be two minus
zero multiplied by our X, which is time plus constant. Okay? So 10/2 gives us five t
plus certain constant. Okay? So what's the next step? We need Zack constant, which is the intersection
with the y-axis. Okay, So how can
I do this simply, we will substitute here. So let's say at
time equals zero, when t is equal to zero, the value of the current
will be equal also to zero. It means that our constant
will be equal to zero. So the equation of the
current will be phi of t, which is this one. So I have this current. Now, what I need is the
root mean square value. So first, what is
the root mean square or RMS is equal to, you can simply remember
it as following route. We have a big root. Here. We have mean which is average. One over t, integration
of the team and square. So a square of this can I
squared from zero to t. So we will have this
integration lags as root one over t integration from
zero to t squared d t. Now, you can see that one over
t d is four, so 1/4. And this integration will
be divided into two parts. First part, from here to here, 0-25 t squared d t. Then integration 2-44
negative ten squared. By integrating the forest
and voting all of this under the square root and substituting with the
limits we will get finally, is at the root, mean square
value is 8.165 and bear. So what does this value mean? This AC current waveform
provides, let's say, a power equal to, Let's say as an example,
equal to, what? If I have a DC current, dc current like this of
8.165 and bear DC current, which is the root
mean square value. It will give us the same
power which is to work. This is to help you understand the meaning of the
root mean square. Okay? So what's the next step? We need average power. So the power will
be the square root of the current multiplied
by the resistor. Really easy. I square
multiplied by the resistance, which will give us 13031. So the effective current, this effective current
gives us this power, which is similar to the
average power delivered. The boy is a AC source. Okay? So the RMS helps us to simplify lots of equations
in our circuit. So let's have another example. We have this wave
form, this waveform. You can see it starts from
voltage as a function of t. Starts from zero, go with
to the peak which is ten, then drops down to
zero at angle boy. Then from pi to two pi is zero. You can see here we have a
zero, then repeats itself. So here we have also
zero like this. Here's, this part
is zero, and so on. So the cycle of this
waveform is 0-2 Pi. So the period is from, is equal to two pi. It repeats itself every two pi. Okay? Now, what does this even mean? What does this
waveform represents? This waveform representing a half-wave rectified sine wave. Okay, so what does this mean? So a sine wave, normally like this, like this. When this wave is passed through or provided to Eric to fire
a half-wave rectifier. We will have this negative
part will be eliminated, it will be removed completely. So we will have this
post apart than zero, boast of part, then zero,
as you can see here. Now this half-wave rectifier is used in many applications. You will understand
about rectifiers in our course for
power electronics. Okay, So when you
finish this course, go to our course for
power electronics to understand rectifiers
and much more. So what do we need
here is that I need the root mean square
value of the voltage. And the needs the average power dissipated in a
ten ohm resistor. So if I connected this waveform
to attain ohm resistor, what will be the
average power consumed? So first is that
root mean square. Okay? So in order to get
the root mean square, we need first to
write our waveform. So as you can see, we have a sine wave. A sine wave from zero to Pi, sine wave with a peak value ten. So it will be ten sine t. You can see it's a time
here is T naught Omega t. You can see it will be
ten sine t from zero to pi. From pi to two pi, we have zero and the
period is two pi. The root mean square
value is simply the root. So V RMS square, whatever it is. Or let's type it here. You can understand it. We are a mess is equal to root one over the period integration of v square d t from zero to t. Here we, instead of
putting this square root, we just added a square here
to remove the square root. Anyway, we will add the
square root in the end. So you will see that here
the voltage has a two parts, from zero to pi and
from pi to two pi. From zero to Pi we
have ten sine t. And from pi to two
pi we have a Z, Y. Integrating this
and y by 1/2 pi, we will get this
value. Like this. We will have a 25. So V RMS squared is equal to 25. So the value of the
root mean square, it will be the root of 25, which is 5 v. So V RMS square to 25. So V RMS is equal to root
of 25, which is five. Okay? Okay. So this sort of presenting
the voltage RMS value, effective value of
this wave form. Now what is the average
power consumed? The power through a resistor
is equal to V squared or V RMS squared
divided by resistance. Okay? This VRML square
divided by the resistance. So we obtained the average
power absorbed by the loop. So in this lesson, we had several examples on the root mean square value
or the effective value. And I hope you
understand the meaning of the root mean square value.
115. Apparent Power and Power Factor: Hey everyone. In this lesson, we are
going to talk about with the apparent power
and the power factor. If you remember from
the previous lessons, we had a voltage in a sinusoidal form or
an F cosine form. And the current in
Zack cosine form. And we had the average
power equal to half V-max Imax cosine
Theta v minus c toy. And then we said
before that half V-max Imax is similar to v root mean square I root mean square multiplied by cosine
Theta v minus Theta I. Now what we would like to do in this lesson is that we
are going to say is that v root mean square multiplied by all your root mean
squared is equal to S, which is the apparent power. S, or the apparent power
is equal to V root mean square multiplied by the
current Root Mean Square. The apparent power, which is
measured in volts and Bair, who's a unit of the
S is volt and bear, not what, but volt and bear. You can see volt and embed. It is identified as the
product of the root mean square value of the
voltage multiplied by the root mean square
value of the current. And the factor
cosine Theta v minus CTI is known as
the power factor. Okay? So the apparent power, why it's called the law exist, it is called that
because it seems apparent that the
power should be the voltage or
current product with the analogy with DC
resistive circuits. Because if you remember
in the DC circuits, the power is simply equal to the voltage multiplied
by current. So similar to here, similar to the DC circuits. For AC circuits, the
product of V RMS and I RMS is called
the apparent power, which is okay, because
it is apparent that the power should
be the product of the voltage and current. And it is measured involved
and bear to distinguish it from the average or the real power which
is measured in what? You have to understand
that we have three types of
power, three pipes. We have first is
the apparent power. Apparent power, which
is denoted by S, and it is measured in volts. And here we have two
other types of power, as we will learn in the lessons. First, we have the active
power BI, or active power. This type of power
is measured in what and what we will
learn in the next lessons. When we talk about
the complex power, we will find that we have
another type of power, which is called the Q, which is called the
reactive power. Reactive power. And the unit of
measurement is a var. Var. So we have three types of power. We have the apparent power, which is volt and pair. We have the active power
which is measured in what? We have the reactive power, which is measured in volts. So the apparent power,
apparent power, or the total power applied or
supplied by supply itself. So if I have a voltage
source, then I say, is that this voltage
source provide this S or an apparent power. Now, this apparent power
is divided into two parts. One of eight, which
is an act of power, and the other part is
the reactive power. The active power,
rubbers and things are power consumed
inside the circuit, such as in the resistance. The reactive power consumed, the power is stored, which is not consumed about stored and returning
to the supply. So active power is the type of power which is found
in the resistance. The reactive power
is founded due to the presence of a capacitance
or an inductance. Don't worry, we
will discuss all of this about with
reactive power and the relation between
them and zap parent power in
that complex power. So the power factor
here is dimensionless because it is a ratio of the average power to
the apparent power. So as you can see, it's a
power factor which is b over s equals cosine
Theta v minus C2. So if you remember in
the previous slide, we said that the power
is equal to V root mean square, root mean square, which is apparent power
multiplied by cosine Theta v minus cosine Theta v minus minus C2 is the ratio
between the power. And the apparent power, the active power to
the apparent power, active or the average
consumed about what we say is that this angle, Theta v minus Theta, we call it the
power factor angle. And you will understand
what we will use it or what is the importance of
the power factor angle? You can see that if we have an impedance which is
voltage over current, which is V max angle Theta v. Or Emacs see angles theta I. So V-max over iMac
Sita V minus C toy. Now what if I would
like to put this in the root mean square form? So if I take v root mean square, which is v over root two. So if I divide here by root two, and I felt my score, if I divide this by root two, then I will get root-mean-squared
error or root mean squared divided by root
mean square, like this. So similar to the electric, similar to the normal case. If we divide voltage
max the worldwide imax, it is similar to V
RMS divided by RMS. And you will notice that
the power factor angle, which is c minus c toy, is similar to the impedance, and the impedance angle is similar to the
power factor angle. So the power factor is
defined as the cosine, cosine of the difference between the voltage and the
current angles. So that is also the cosine angle of the
load impedance ANC. The word vector may be seen as the factor at which
is apparent power must be multiplied to obtains
a real or average power. So as I said in the
previous slide, that the apparent
power is divided into active power
and reactive power. So in order to find the portion or part
of the active power, we take S and multiply it by some artifact to get the
rail or the active power. Okay? You will find
that the power factor, it changes from zero to
humanity. It is 0-1. There is no negative
power factor. E.g. for a pure resistive load, what does a pure resistive load? It means that Theta
v is equal to c2. So their difference
will be equal to zero. So cosine zero will be one. So the power factor is unity. So when we say is that we
have a unity power factor, it means that we have
a pure resistive load. And in this case, you will find that
the apparent power will be equal to
the average power. All of the power generated
will go to the resistance. Because we have B over
S is equal to one, which means b equal to
S. So in this case, we don't have any reactive power for a pure reactive load. Or when we are connecting to pure capacitive
or a pure inductor, we haven't difference in
angle plus nine -90 degrees, which means that cosine
90 is equal to zero. So the power factor will
be equal to zero in the pure reactive load or
inductive or capacitive load. What does this also mean? It means that there
are no or there is no consume the average power. So you will find that winds
up our factor is zero. We have a pure inductive circuit or a pure capacitive circuit. So all of the
electrical power is stored and return that
back to the supply. Okay? We don't have any
consume the power. That's why we say that's
a power factor is zero and the average
power is zero. So in this lesson, we discussed the concept
of the apparent power, which is V RMS
multiplied by RMS. And we discussed the concept
of the power factor. Remember, these
concepts are really, really important in
electrical power system. They are really,
really important. You will find we
have power factor, apparent power, reactive
power, active power. All of these concepts are
really, really important.
116. Solved Examples 4: So let's have a solvent example on the apparent power
and the power factor. We have this current
and this voltage, supply voltage and the current
supplied by this voltage. Now, we would like to find the apparent power and the
power factor of solute. So first, what is
the apparent power? S is equal to V RMS
multiplied by RMS. So we have two
cosine waves so that V RMS is equal to V
max over root two. And I are mass is equal
to Imax over root two. So V max, which is 120, and iMacs which is four. So will be like this. Okay? And you can see it's a
unit is volt ampere for S, or the apparent power
is the unit is volt and paid for the active power, or the average power, or the real power. It is what? Because it is a consumed
electrical power. Now what we would like
to get is up our effect. So as we remember that power
factor assembly equal to cosine sita V minus C to C to V, which is negative 20, and C2, which is ten degrees. So you can see like this, cosine Theta v minus e to the negative 20 minus
ten gives us 0.866. And you will notice here
something important, which is we are
typing year leading. What does this mean? When Zack current leads? Voltage? We have a leading power factor
when the vault or when, let's say in the same sentence, if the current is lagging, legs v, it means that we have a power factor
which is lagging. So when we are saying
leading or lagging, we are talking
about the relation between the current with
respect to voltage. So here we have a power
factor which is leading. It means that the current
is leading, the voltage. If this power factor is lagging, it means that the current is
behind or lags the voltage. Now how can we know
if the current leads or lags from the angles? You can see here
is that the angle theta is equal to ten
degrees plus ten degrees. And the voltage
Sita V negative 20. So you can see is that
the current ten degrees and Theta v is negative two. So if you look at the difference between these two angles, you will find that the
current is leading, Paul is salty degrees. The difference between these
two numbers then minus -20. So it will give
us salty degrees. So it means that the current is leading by sorted degrees. The voltage. That's why we say a
power factor is leading. And what we can
notice here also, when the current is
leading the voltage, it means that we have
capacitive circuit. Add capacitive circuit. We have a capacitor. Capacitors, the value or the reactants of
the capacitor is much higher than the
reactants of the inductor. So you can see here,
current leads the voltage. Now let's have another one. We would like to get
the power factor of the entire circuit as
seen by the source, defined as the average power
delivered by the source. Okay? So first you have to
understand that we have a power factor for the supply itself and
the for each loop. Remember this? So here we are talking
about the power factor of the entire circuit as
seen by the source. So it means the power factor, it means cosine Theta
v minus Theta I. What voltage? The voltage of the supply. What current is the current
coming out from the supply? Okay? So first we have here is a voltage source
E and angle zero. Now what I would
like to get is the current coming out from it. So the equivalent
of this circuit is that these two are
parallel to each other. And series with a six ohm. And the current
will be the voltage divided by the equivalent
of this circuit. So you can see the total
impedance six ohm series with the parallel equivalent
for parallel to negative j to give
us this final value. Okay, now I need that current. It will be the
voltage divided by the impedance like this. The voltage wide wise, I'm Venus two will give us
the value of the current. Okay? Now, you can get from
here is the value of the current and
its angle thens r cosine Theta v minus
Theta I will give us the value of the perfect. However, however, if you look at and what we learned
before is that this angle of z is equal
to V minus theta, right? So we can say is that
the power factor is cosine this angle, which is 0.9 734. And if you look at this circuit, we have a six ohm 4
ω and a capacitor. So we don't have
any inductor here. So it means that we have
at leading the current. The current will
lead the voltage. That's why we say here leading. Okay. Now we need the average power delivered by the
source of the power. Average power delivered
by the source itself. So simply the power, since you are talking about
the average part will be S, or the apparent power
multiplied by the power factor, which is V RMS multiplied by R MS RMS multiplied by power
factor which we obtained. So first, as we
have said before, we get the current by dividing our voltage source and the angle zero divided bys I'm Baden seven and the
angle negative 13.24. So we will have
this final value. Now in order to get
the average of power, it will be S, which is V RMS, or you're a mess. So t multiplied by 4.286
multiplied by the power factor, which is 0.9, 734. So it will give us
finally 125 watt. Now, remember something
here which is important. Now, you can see that usually when we said
three and angle zero, we think is that this
value is max, V max. However, you can see
that in this problem, you can see V RMS. So t here representing
the root mean square. So when we divide
these two together, we get our Root mean
square knot I max. And as you can see, we can do it using
another method is that we can say
that the power is equal to root mean square multiplied by the
resistor, right? If we take the square
of the current, The multiplied by
the resistance, which is equivalent resistance. So we will get their power. So someone will ask me, where did we get this resistor? So you can see we have
all the root mean square. Root mean square is 4.286. Where did we get the value of
the resistor, which is 6.8. Now you can see that
this circuit equivalent of this part is, which is this angle seven
and the angle negative 0.24. So it is consisting of
R plus j Excel, right? Or J xl minus xc,
whatever it is. Okay? So in order to get
the resistance, it will be like this. Resistance will be seven
cosine angle negative 13.24. So seven multiplied by this
cosine will give us 6.8. Okay? So in this lesson we had
some soul with examples on the apparent power
and the power factor.
117. Complex Power and Power Triangle: Hi and welcome everyone to this lesson in our course
for electric circuits. In this lesson we will talk
about the complex power. So you have to understand that the complex power
is similar to what? Similar to the
apparent power is. The complex power is equal to the apparent power S. But the difference is that
the complex power here we write in the
form of phasor form or the form of real
plus imaginary part. Okay? So if you remember
that we said that S not S, Let's start with B, or the power is equal
to the voltage V max, or Emacs, cosine C
times V minus Theta I. And this part alone, this sport can be
equal to V RMS, RMS. We said that the apparent
power is equal to V RMS, RMS. So the apparent
power V RMS, RMS. Okay? But remember this rubbers
and thinking not phase, but as a magnitude, the value of the
apparent power is the magnitude of V RMS, RMS. But if we write it in their face or fall
in the phasor form. So it will be S equal to
V RMS multiplied by RMS. Conjugate, conjugates. Now, why conjugate? Because if you look at the
angles here for the power e.g. you will see that see TV, which is angle of the
voltage and negative c two, which is the conjugate
of the current. That's why when I write
S or the apparent power, we write it in this
form or the conjugate. And we know that
V RMS is equal to this value and I RMS
equal to this value. So from here we can conclude
that the apparent power S is equal to V RMS
multiplied by RMS. Okay? As the magnitude and
the angle C times V minus C. Okay? Okay, so let's throw
it all of this again. So here is our complex
power is really important in the power
analysis because it contains all of the
information related to the power absorbed
by a given load. What I mean by this, you will understand that we
have two types of power. We have the active power
and reactive power. The complex power or
the apparent power, will help us know how
much real power and how much reactive power
consumed or restore the boy, this loot, all
provided by supply. So as we have said that S or the apparent power
half VI conjugate, remember, this is the
maximum value V max or Emacs conjugate. Now, which is similar to V
RMS, RMS conjugate, right? And we said that V RMS is
this value and I RMS is Ali, RMS negative c two. So we will have this final
form which I exhaust, which I have written. So you can see that
we are a mess, all your mess and the
angle Theta v minus e toy. Now since this is
a phasor, we can, we can convert it into
the rectangular form to Israel plus j imaginary real
part and imaginary part. The real part will be V RMS, RMS cosine Theta v minus C2. And the imaginary
part will J V RMS, RMS sine Theta V minus C. Okay? And we also know that a voltage V RMS is equal to
z multiplied by the current. Okay? So we can take this one
and substituted here. So if you take this
one substituting here we have z, okay? Then I RMS multiplied
by I RMS conjugate. So what does this mean? I RMS multiplied by
I RMS conjugate. It will be equal to
the RMS is the force, the magnitude multiplied
by the second magnitude, which means the square. And the angle will
be the first one, which is c to the angle of
second one is negative c dy. So this angle is zero. So it will be I RMS square only. You can see all your
RMS square and the z, which is similar to also the V RMS square
divided by z, the conjugate. Now where did we get this? Simply, if you look
at this equation, we have V RMS as it is. And I RMS. So we will put like this conjugate I RMS
itself is the voltage divided by V RMS
divided by, Is it. Okay? So we will have V RMS
multiplied by V RMS conjugate, which is V RMS square divided
by z conjugate. Like this. You can see as we
have just obtained. Okay? So this are different
form is to obtain the apparent power or
the complex power. Now, if you look at this
equation for the power, for the power S, here, you can see it's consisting of a real part plus j imaginary. And we know that z is
equal to R plus j X, X here is XL minus
accessing, right? So if you look at this equation, this one here and
substituted here, this part. So S equal to I RMS
square multiplied by z, which is R plus j X. Now, this will give us two components like
this will be equal to the RMS square
multiplied by R plus j, or RMS square multiplied by x. So what we can see here
is that we have S, which is apparent power, S, which is apparent power equal to a real part plus
an imaginary part. Here we have a real part
plus an imaginary part, if you will remember, if you remember that I RMS square multiplied by
or is the real power. And V RMS, RMS
cosine Theta v minus Theta I is also the real power. Now what about the imaginary? If you multiply the current
square multiplied by x, you get the reactive power Q, which is similar to V RMS, RMS sine Theta v minus c. So this is also our queue. You will find that
our apparent power, which is produced by our supply, is, provides or gives
two types of power. It gives us the real power, which is a power
which is consumed by the resistor and real
power consumed and gives us as a
reactive power that is stored and return it back. Boy, Zach capacitor
or the inductor, which is reactive power. Now we have to remember
that reactive power are important in the electrical machines
because they are related to the magnetization for
the electrical machines. You will understand this in our course for
electrical machines. So here we'll find that
power is the real part of S, which is I RMS
square multiplied by r. And q is the
imaginary part of S, which is I RMS square
multiplied by x. So you will find that B is the
average or the real power, and it depends on
the load resistance. Q depends on the load, the reactants and the cold, very active, and sometimes we call it the quadrature power. But the most famous name
is that reactive power. We call queue as
the reactive power. As I've said before, that the power V RMS, RMS cosine Theta v minus Theta. And the Q is the imaginary
part, which is this part. We are a mess I error message
sine Theta v minus Theta. So you will find
that real power is the average power and
the measured in watts, which is delivered
the two lead to dilute it rubbers and things are useful consumed electrical
power, the reactive power, rubbers and things
up our xy change it between or energy exchange between the inductor or
capacitor and the supply itself. So you will find that S, or the apparent power is
measured in volts and bear. The real power is
measured in watts. The reactive power
is measured in var. Var, which is volt
and bear reactive. Okay? Okay. Now we will find that Q itself
has three different types. Q can be equal to zero or less than zero
or greater than zero. Okay? So what does this mean? When Q is equal to zero? So if you return it
back to the equation. Here we are a mess I RMS
sine Theta v minus Theta I. So if we write like this, Q equal to V RMS, RMS sine Theta v minus Theta. So first the case is that we
will have c v equal to C i. So when these two angles
are equal to each other, if you remember from
our previous lesson, we said we have a pure
resistive circuit, right? A pure resistive circuit. So in this case, when zeta V equals C2, this will be equal to zero. And the sine zero is zero, so Q, or the reactive
power is zero. When we have a resistive circuit
or a unity power factor. If you don't remember, the power factor is
cosine C v minus Theta. So in this case, cosine
Theta v minus e to i, difference between them is zero, so we have a unity power factor. Okay? Okay, so that's just really
the sport like this. Okay? Second case is that we
have a Q less than zero, or that Q is negative. Q is a negative value. When does this value is negative when sine is a negative angle. So when c is greater
than Theta v, which means that the current
is leading the voltage. You can see leading
power factor when the current leading voltage sign will be equal to negative. So q will be negative. So we have a Q negative. Now, when does this happen? When we have a capacitive
load when the x is c is greater than Excel. So as the current will
be leading the voltage. Same idea, when does, when we have Q
greater than zero, q becomes a positive. It means is that c, v greater than c. So this angle will be positive
and the Q will be posted. Okay? Now, what does Sita
v greater than zero? It means that the current
is lagging the voltage. That's why we say
lagging power factor. Karen lagging the voltage. So in this case we have
an inductive load and x is greater than accessing. It makes the current
lagging from the voltage. Okay? Okay. So in general, complex power, which is measured in volts and bear all the apparent power is a product of the root
mean square voltage. Phasor is a complex conjugate of the root mean square current. And it is a complex quantity as a complex quantity
consisting of two parts. Real part, which is B, or the power consumed, and is the imaginary part, which is the reactive power
or the stored energy, or the energy exchange
between the energy between the energy storage,
element and supply. In general, we have all of
these equations that will help us understand
the complex power. Complex power S equal p plus q, or voltage multiplied by the
conjugate of the current, which is V, and
the angle Theta v minus e to the apparent power. What does the
apparent power mean? It is magnitude, magnitude of S. So the magnitude of
S is b squared plus Q square root b squared
plus q squared, or the voltage
multiplied by current. That real part, or
the real power is B, which is real part of S. And the reactive power
is the imaginary part of Q, of imaginary part of
S, as you can see. So it will be V RMS, RMS cosine Theta
v minus Theta I. And, and Q will be as sine Theta v minus Theta because it is
the imaginary part, ends up our vector, as we'll learn it is B over S, which is cosine
Theta v minus c2. This will lead us to
represent this power as a power train or in
the form of a triangle. So it will be like this. So if we look at
the power triangle and the impedance string, we know that our z is
equal to R plus j X, or it is equal to
zero as a magnitude. The angle C, V minus
C are right angles, it a magnitude of z and the
angle Theta v minus c die. So r is our rubbers into
cosine Theta v minus Theta, cosine Theta v minus Theta and x representing that the multiplied
by sine Theta V minus C. Now, if I would like
to represent this in a real and imaginary axes, you will find that we
have the real part, which is all like
this real part. Here we have the real part exists and the vertical
is the imaginary part. The real part is
our imaginary part, which is x like this. Adding R plus j X, we will have our z. And the angle is sita, sita hero presenting
zeta V minus C. Okay? So if you look at this triangle, cosine theta multiplied
by z gives us, are the signs that
gives us sine c, that gives us sine C. Tomato
Blood buys it, gives us x. Same idea for some power. Real part is our power, imaginary part is our q, imaginary part is our q and
deserves summation gives us S and the angle is seed. So if you take S cosine theta, you will get some power. If you get sine theta, you get q sita here representing
zeta V minus C, The boy. Okay? Now, if we take this one
triangle here, we have S, we have the real part and the Q, we have two types of Q. We said that we can have zero q. We have a lagging power
factor, leading power factor. We said that when q is positive, which means that x
greater than accessing, which means that the current
is lagging the voltage. So we will have lagging perfect. So when we draw Q in the positive direction,
and we have S, it means that we have this triangle which are representing lagging
power factor. If x is greater than x L, which means that the
current is leading, then Q will be negative. So we will draw
our triangle like this, this triangle downwards. So we draw upwards when we
have our post FQ downwards, when we have negative Q. And if Q is zero, then our power
will be like this, our power and our S will be equal to each
other when Q is equal to z. Okay? So in this lesson, we talked about with the power triangle, we talked about
the complex power, and we now understand
the relation between the three
different types of power.
118. Solved Examples 5: Now let's have some solved
examples on the complex power. We have here, the
voltage across a load and the current through the
load is given as following. So we have a Z or the solute has a voltage across
it of this value. And the current going
through it is this value. Fors two, we need to find the
complex and apparent power. So what's the difference
between complex and the apparent power z are
similar to each other. The difference is
that apparent power is the magnitude of
the complex power, or S as the magnitude. And the complex power is as, okay, which is in
the complex form. So first we need to
get the complex form. So we know that the
complex power is V RMS multiplied by RMS conjugate. So we have the value
of the voltage and the current as
a maximum value. So we will take this and
divide it by root two. Divide this by root two. And the angle will be c times v, which is negative ten, minus c2, which is 50 degrees. So we will have like this. You can see 62 over root two and the current 1.5 over root two. So you can see here we have 62 over root two and
0.5 over root two. And the angle will
be negative ten -50. Negative ten -50,
which is negative six, because we are dealing with
the conjugate of the current. And it is measured
in volts and bear. The apparent power
itself is of course 45. Okay, the magnitude. Then the second part, we need to find the real
and reactive power. So if you take this one and, and make it a real
and imaginary parts, you will get the reactive power. So you can see 45
cosine negative 60 plus j 45 sign negative 60. You will have a real part
and an imaginary part. And we know that the rectangular
form here is P plus j Q. So from here you can
get a czar power, which is 22.5 and Q, which is negative 78.97. Now with an extra
requirement is that we need the power factor and
the load impedance. So the power factor is
really, really easy. Power factor is
cosine C, V minus C. Okay? Or the cosine of the angle of the complex power
that we obtained. We have power factor
cosine negative six, which is Theta v minus Theta I, or the angle of
the complex power. Is it leading or lagging? You can see that the difference
between them is negative, which means the angle
of the current is much higher than the angle
of the voltage, which means that the current
is leading the voltage. Okay? Now what we need also, we need to find the
load impedance. So we know that z is equal to the voltage divided by current. So we can get it as V
RMS divided by I RMS. And the angle C V minus
C two as well on it. So you can see here a
voltage divided by current. So the voltage RMS
divided by current RMS, or V-max divided by Imax. Both of them are the same
as we learned before. And the angle Theta v minus c, which is negative 60 degrees. And you can see it is a
capacitive impedance. Now why is this capacity? Because of course, the
angle is negative, which means that the
current is leading. The current is
leading the voltage. You can see it is leading
because the reactive power is negative, q negative. Or because C2
greater than c to v. So in these two cases we
have a leading effect. Okay? Now let's
have another one. So I load draws 12 V kilovolt and bear at
a power factor of point, it five-six, lagging from 120
volt RMS sinusoidal source. Vine does the average and
reactive power delivered to the load current and
the load impedance. So first we have as absorbs
at well V kilo volt and Ben. What does this mean? It means the apparent power as a magnitude S is equal to
12 kilo volt and pain. And the power factor
cosine C minus C two. Is equal to this value and the supply V RMS
is equal 220 volt. So the first requirement
is that we need the average and reactive power. The average power or the real
power is simply equal to S multiplied by
support vector, right? So you can see here
that the real power, power is equal to
S cosine Theta, which is c2 minus c2. Or ZAP 12 we kilovolt and being multiplied
by some power factor. So we will get our power. Then we would like
is a reactive power. The reactive power Q is
equal to S sign Sita. Okay? So how can I get
the angle simply? You can see that
the power factor cosine z w minus zeta is 0.856. So cosine minus one, this value will give us the
angle goes IN minus one. This value will
give us the angle. Now we have to remember that there is an
important part here. You can see that the
power factor is lagging. So it means that c v
is greater than c two. Remember this, C v
greater than c two. Which means that the
power factor is lagging. Y lagging because the
current lags the voltage. So in this case, this angle will be both Theta v minus Theta I will be a positive value, as
you can see here. However, if this power
factor is leading, it means that c v minus
Theta I should be what should be negative if
the power factor is leading. So in this case, Sita
will be negative one. Because current is leading. However, in this problem we have a lagging power factor is just to say cosine minus
one, the value here. Then we will get q by taking S and multiplied
by sine theta. Like this, q will
be us sign Sita. Now we need to find
the peak current. So you can see that the
current itself is equal to S divided by the voltage, right? So we have two methods here. First, which is the
easiest method, is the method we
need be current. So first we will get the
root mean square value, I RMS s as a magnitude. Remember S as a magnitude
equal to IR mess as a magnitude divided
by the voltage V RMS. S is equal to voltage
multiplied by current. So we have a 12 volt and bear as a magnitude divided
by the voltage which is 120. So it will give us 100 and bear that root mean square
value is 100 and bears. Now, if I would like the peak, if you remember that I RMS is equal to Imax
divided by root two. So to get Imax, we will multiply root
mean square by root two. So Imax will be hundred
root two and back. So remember this, this is the easiest solution
and bikes use. Okay? Second solution is
that you can just say is that since there's a
power factor is lagging, the complex power or B, b plus j Q. So it will be 10.272 plus
j Q, which is 6.204. Then you will get the actual value using
the complex power. We know that S is
equal to V root mean square root mean
square conjugate. So I root-mean-squared
conjugate will be S over V RMS. Remember here we are talking
about the phasor form. So we took this power, which is this one here. And we are a mess is 120
and the angles here. So dividing this together, we have this value, this final value which
are representing Ireland, square root conjugate, root
mean square conjugate. So I wrote many
square itself will be the reverse of this
angle to be negative 71.13. We have here conjugate. So T1 without conjugate
will be negative one. Okay? So we have this value, which is all your root mean
square as a magnitude, right? So the peak will be this
value multiplied by root two. So we'll get this value. As you can see, this
is another muscle, but the first one was
much easier by taking this magnitude and what it buys this magnitude we will
get root mean square. And from the root mean square, we multiply by root two
to get the maximum value. Now as a final requirement
is that load impedance. So how can I get
the load impedance? It is really, really easy. Simply, you can just take a voltage and voltage
by the current. So if we take V
RMS hundred 2010, the angle zero divided by I RMS, we will get the impedance. You can see that equals V RMS, RMS hundred 21, angle zero hundred and
angle negative one. It will give us this value. And of course we know that this is an inductive impedance. Excel greater than X is C. Why is this? Because we have a
lagging effect. So Excel is greater than x c. So in this lesson, we
had another example, or some solvent examples
on the complex power.
119. Power Factor Correction: Hi, and welcome everyone to this lesson in our course
for electric circuits. In this lesson, we
will talk about with the power factor correction. So you will find that most
of our domestic loads, such as washing machines, air conditioners, refrigerators, and industrial loads such
as the induction motors, are inductive and operate at
a low lagging power factor. Also that inductive nature of solute cannot be changed it, we can enhance or increase
its power factor. The process of increasing
the power factor without, with altering the
voltage or the current, is known as the power
factor correction. Since most of the loads
are inductive loads, the power factor is improved by installing a capacitor in
parallel with the loop. So let's understand why is that low lagging power factor
is bad for our system. So first you will find that we have an electrical generator. So let's say we have our
electrical generator like this. You will have to understand
that power of the generator, power that is produced by the
generator is measured in S or involved and bear
is apparent power or the vault and bear how many, how much or how many? Kilo volt and bear or
mega volt and bear. Okay. So when we say we
have a generator, we don't say that this generator, this
electrical generator, and the electrical substation or the electrical
generating station, we don't say it is
producing how many? What we say it produces how many volt and
bear. Why is this? Because we have different
loads such as e.g. a. Resistive load or load. Okay. So let's talk about e.g. a. Washing machine or
air conditioner or any industrial load. So we have our n. Since we have our resistance
and an inductive, it means that the volt ampere generate the Welby
devolved into parts. It will provide to
this load b plus j Q. So we have in the transmission
lines that Transmit electrical power from the
generator to our load, we will have P and Q. We have an active
power that will be consumed inside
the resistor, e.g. it will provide
mechanical power. Mechanical power,
such as inside, inside an induction motor. We have an inductor which
will consume or not consume, but we'll take a cue from the generator in order or
a store electrical energy, which is reactive power. So this inductor takes reactive
power, reactive power Q. Now, why does it take Q? Because it needs
magnetization or it needs magnetic field for the operation of the induction motor itself, as we will learn in the
electrical machines course. So anyway, this is P and the Q representing
assault and current. So we have a
transmission line that will take this power and
transmit it to our load. Okay? That this large amount of power, which is p and the q, b and q, this large amount of power is equivalent to a
certain amount of current. So you will see that we are overloading our
transmission line. We are providing more current
on the transmission line, which means it is heavily
loaded because it takes active power from the generator and it takes the reactive
power from the generator. Okay? How can I read the
use this amount of power or reduce this
current assembly? If I connect here a
capacitor like this, then what will happen is
that this capacitor will provide Q2 to the lute. The queue require the boys are load will be taken
from this capacitor. So that current that will take this amount of power
will be reduced. And the amount of Q taken from the supply will be also reduced. Okay? So again, the capacitor here connected
parallel to our motor. It is used to enhance
the power fact. How does it enhance
about vector by reducing the amount of Q taken
from the supply itself. Okay? So how does it occur? We will see right now. So as you can see, we
have the original load, we have our generator
that will have a certain voltage and current to an inductive
load such as e.g. an induction motor. This is the original case. And this case is with having a capacitor installed
parallel to our load. So let's see how
this will it change. So we have first is our voltage, which is applied here
and here, same voltage. Then we have the
original current, which is i l. This is the original current i l with a certain angle Sita one. So because we have
an inductive load, the current legs voltage, right? So how much Eight leg, It's like by a certain
angle, theta one. Now when we connected
a capacitor, we still have a
current going into the inductive load and another current going
into the capacitor. So what will happen is that
we have I l and I see. So you know, that that
current of the capacitor with respect to the voltage is
leading by 90 degrees. So you can see that we have the voltage and the
capacitor current. So it is leading by 90 degrees. Okay? Because the current
through a capacitor is leading the voltage applied
to it by 90 degrees. So we have now our current, our current here is
equal to the summation of I l plus r is c. So if you look at here we
have i L. Then we added IC, which is leading by 90
degrees from the voltage. We take this vector
and the boat it here. So when we add this vector
with the second vector, we will have a final current, I. In the second case, you will see that this
current has a lower angle, smaller angle, which means a higher power factor because the power factor is
cosine theta one, the original power
factor enzyme. After adding the capacitor, it will be cosine of theta2. You will see that theta
one is greater than C2, which means the
cosine theta one is less than cosine theta two. So by adding a capacitor, we enhance it the power
factor by reducing the ink. Now one important thing here is that you will find that
we have the original current was equal to I L. Now the new current is
equal to IL lost. I see, someone will tell me that the total
current increased. However, this is wrong. Now why is this? Because Zach current I L
is opposite to the IC. Okay. They are not saw
summation of them. They are opposite to each us. Okay? So if you remember
similar to XL minus XC, similar, Here's the total
current i l minus IC. They are not similar
to each other, zeros or phase
shift between them. Okay, so the total current, you can see that the total cost, this is the original current. And after adding the capacitor, you can see the
current is reduced. Smaller vector. So the current coming from
the supply is now reduced. Okay? So how does this, how can we translate this
to a power triangle? You can see that we had
an original power P, which is the active power
absorbed by solute. This power does noted change. It is the same power. If you look at this figure, you can see is our resistor here is similar to
the resistor here. So the same amount of power. Okay? So the same amount of power. That first case we had Sita one, which had a Q, L, this big Q1, representing the original case, all of this q. Now in the second case, we added a queue
of the capacitor. So instead of having this big, big line, we now will reduce
the zach you to this amount. Okay? Now why is this? Because if you remember that QL, the total Q is equal to at
the beginning of the Q L. After adding the capacitor, it will reduce the queue. It will be QC minus QC minus QC. It will drop it by Q
C. So we will have Q2, which is the final amount
of Q or the reactive power. Now, if you look at
this figure also, so we have Sita one, which is the original triangle, original triangle, original
power factor triangle. And we have C2. After
enhancing the power factor. In these two cases, we have the same power, but the amount of reactive
power is reduced. Also finds that the
power taken from the electrical supply, S1. Now the new power is s C2. And you'll find that
the power taken from the supply as c2 is
now less than S1. Power taken from the
apparent power taking from, taken from the
supply itself is now reduced the due to the
enhancement of the power factor. So it means that
we now will reduce these are overloading on
the transmission line. We reduce the Zak
current because the current is equal to
S over V, right? Or I conjugate is
equal to S over V. So when we reduce this S, we reduce the current flowing through the
transmission lines. Okay? So I hope the idea
why we improve the power factor
is clear for you. So we'll see that here. Let's define this by equations. So far as to we had the
original power factor. We had p and big Q1 and S1
with an angle theta one. So the first power, or zero power is equal to S
one cosine theta one ends. The original Q is equal
to S1 sine theta one. Okay? So we know that S
multiplied by cosine z, that gives us power
S multiplied by sine theta gives us
the reactive power. Now something important here, if you look at this figure
as this power triangle, look at angle theta. So tan theta one. Then see the one density, one is equal to what
equal to opposite, which is Q1 divided by the
adjacent, which is power. So you will find
that Q1 is equal to the power multiplied
by ten Sita one. Now, if we enhance
our power factor, if we enhanced it up, perfect. And do we have now Q2? We have the same power, but a new angle. So we can say is that Q2 is
equal to c2 sine C that, or we can say it is
equal to the same power. Simple row two doesn't change. Then see the two from
the second triangle, then see the two. Okay? So now we have the original Q and we have the new queue after enhancing
the power factor. So if I would like
to get the value of capacitor itself or the value
of the Q of the capacitor. We can say that Q C, which is this amount, is equal to Q1 minus Q2, which is B tan theta one minus, then see the two. Okay? So from this equation we can get Zara reactive power require
the boys are shunt, shunt capacitor means
the parallel capacitor. We need this amount of Q. And we know that q, which is reactive power, is equal to v squared
over ecstasy, right? So you can see that here. You can see QC is equal to v squared divided by x
c n. What voltage? The RMS effective value. And the one over x c is omega C. So from here we can
get that the value of the capacitor is equal to Q C divided by
omega V RMS squared. And the QC itself is B tan
theta one minus theta two. Now, so you usually, usually, when we enhances
our power factor, we are talking about adding a capacitor because it
is a dominant case. However, let's say
we have the reverse. We have e.g. a,
leading power factor. Okay? So you will have
to understand that leading or lagging
is not a good thing. The best thing is going to
near unity power factor. Okay? So becoming unity power factor is the best-case because we don't take any q
from the supply. Okay? So if we have leading lagging ball factory or a leading power factor,
it is not good. We need to reduce the leading
or reduce the lagging. Okay? So let's say we have
the reverse case, we have a capacitive load. And I would like to add
an inductor to reduce the power factor or to reduce the leading power factor or enhances at
eating Bohr effect. So we will add an inductor.
What is the value? It will be similar as before. So QL will be V RMS
squared divided by x l. And x here is equal
to omega n. From here we can get the inductor
required inductance circuit is equal to this value. And the QL itself is equal to the difference
between Q1 minus Q2, which is representing
the enhanced cement ends up our fact. As you can see here. So in this lesson, we talked about the
power factor correction using capacitor and inductor.
120. Solved Example 6: Now in this lesson we will have an example on the power
factor correction. You will find the more
practical examples about power factor correction. In our course for
electrical design. We will make it more
practical by giving you that tables and selecting from them. Okay? So we have this power
triangle, as we said before. And then we have our
supply, 120 volt RMS. So V RMS of the
supply is 120 volt. We have the frequency 60 horses. We have that our load will
absorb before kilowatt. So this is the active
power or the real power consumed at a lagging
power factor of 0.8. So this is our co-design. See that one finds the value of the capacitance and necessary to raise the power factor to 0.95. So this representing 0.295, representing cosine C tattoo
or the new power effect. Okay? So what do we need
here is that I need to find the value
of the capacitance. So in order to do this, first, we need to find the Q, right? So in order to find
the capacitance, so we need that Q are
required to reduce the queue or the
total reactive power and enhances the power factor. So we need Q C, okay? So QC is equal to
Q1 minus Q2, right? So we need Q1 and Q2. So the first step, Q1 has a two equations. Either to use Q1 and
Q2 has two equations, is or to use S sine theta. Or we can use B, then C then of course
Q1 sine theta 1.10, C21, Q2 will be signed
C2 and then Sita two. Okay? So we will start e.g. by using the S or
the apparent power. So you can see that the
first one or factor 0.8, then the angle cosine theta 1.8. So see Taiwan would be 36.87. Okay? Because IN minus
one off point date, the forest power factor, as you can see here. Then from here we
have our Power BI. So you can say is
that Q1 is equal to Q1 is equal to some
power which does not change equal to four. Then C21, which
is sold, is 6.87. Okay? This is a violet solution, k by putting the power
multiplied by ten. The other solution is that we
can say is that we get S or the apparent power by
dividing four by eight. Power BI divided by the
power factor gives us S, the blood by science it. You can see here, the power divided by the power
factor gives us, is the apparent power. So from here we can multiply by sine theta one to get 3,000 v. This will be similar to four. Then 6.87. Remember, this is four kilos, so it will be here, ten to the power three. So this will give us
the same as here. Okay? Now it's the same idea
for the second part. After enhancing the pole
vector, we have 0.95. So the new angle will be
smaller or equal to 18.19. So from here we can get the
S or the apparent power, apparent power 4,210 by dividing power divided
by cosine theta two. Then we take this multiplied by sine C two to get
the amount of var, or you can simply say the
power multiplied by ten, see that two, it
will give us Q2. Then we will subtract Q1
minus Q2 to get Zach QC. Okay? So we will have this amount
of words that representing the reactive power
supplied by the capacitor. Then we will equate it
with V RMS square omega C or Q C capacitor will be equal to Q C over omega V RMS squared. V RMS is 120, and omega is two
pi multiplied by the frequency to polymath
about my frequency, which is 60 outs. From here we can get the amount
of capacitance required. Okay. Now why did I mention here
is that as S1 and S2, because I would like to show
you that you can see is that the new power EC2 is 4,210
and the original is 5,000. So this one is 5,000
and this one is 4,210. So what we can learn from this is that by adding a capacitor, we reduce that amount of reactive power require
the firm's supply. Which means that the total S or total apparent
power is reduced, which means that the
current will produce and we are not overloading
our transmission system. Okay? So in this lesson
we talked about assault with the example on
the power factor correction.
121. Introduction to Resonance in Electric Systems: Hi and welcome everyone to our course for resonance
in electrical systems, or resonance appearing
in electric circuits. So in this course, we
are going to learn what is the meaning
of resonance and what is its effect on our
electrical system and why it is important to
understand resonance. So first, we need to have an introduction about resonance. Resonance in this course. This course introduces a very important resonant
or T owned circuit. So there is a circuit called a resonant circuit or resonance
in an electric circuit, which is our fundamental
to the operation of a wide variety of electrical and electronic
systems in today. So the resonant circuit, or the resonant
circuit is really important in many applications, as we will see in
the next two slides, will have to understand that
the resonant circuit, e.g. you have here, see it
as resonant circuit. And this one is a
parallel resonant circuit which you are going to
discuss inside our course. So the series resonant circuit, as you can see, it
is formed of R, L and C, three basic elements, the resistance, the
inductance, and capacitance. So R, L and C, as
you can see, R, L and C in that series
resonant circuit, we have all the
elements are in series. The supply is a voltage
source in series with a resistor in series
with an inductance, series with a capacitance or capacitor ends up
parallel circuit. We have a current source
battery to our resistor, better to an inductor
or capacitor. Okay? So you can see that
the resonant circuit is formed from R, L, and C. But what is the difference? What's the difference
in this circuit? The difference is, is that
we have a certain frequency, one circuit, certain
frequency at which we will have resonance response. Okay? So at a certain frequency, at a certain frequency, or which is called the resonance frequency or
resonant frequency. This frequency, e.g. in that series circuit, when we reach at certain
resonant circuit, you will find the
response like this. Okay, Let's see
the response e.g. in the series resonant circuit
at a certain frequency, at a certain frequency, at this frequency will find that x will be equal to accessing. So we'll find that in
our circuit we will have the minimum impedance. Okay? So as you can see in
this circuit, e.g. in this circuit, we have z, the total or the impedance
of this circuit is R plus j XL minus XC. You can see R plus j
XL minus XC, okay? At a certain frequency, at a certain frequency, which is a resonance frequency, you have to know that here. That Excel is equal to two pi multiplied by
a certain frequency. The blood boys or inductance. And xy as 1/2 pi multiplied by a frequency
multiplied by a capacitance, which is one over omega C. This one Excel is omega L. Now you will find that at a certain frequency when we
change as this frequency, when we change this frequency, you will find that the total
impedance also changes. But at a certain frequency,
at one frequency, you will find that x
will be equal to xy. And what will happen
in this case, we will have total or the total impedance
of the circuit will be p or resistant. Funds that, that impedance
is minimum in this circuit. So since it is the minimum, you will find that e.g. the current is equal to E
over sensors it is minimum, then the current
will be very high. Okay? That's why in the
resonant circuit, you will find that when we reach that resonance frequency, you will find that the value of current will be very high, which is a resonance
resonant case. You will find also that census
I current is very high. You will find that the voltage across the capacitor
or the voltage across the inductance
will be also very high. Okay? So the, most of the voltage and the
current will be very high. We'll find is that
since these are voltage here will be very high. That's why you see
here a circuit act as, as a voltage amplifier. It amplifies voltage source. So e.g. you will
find that E here, e.g. is our supply. You can find here that
the voltage across the capacitor in
resonance can be e.g. ten e, okay, ten
times or supply. Okay? Why? Because it is in resonance case or in the resonance formula. Now if we get back here, you will find also that
the electrical resonance, electric horizontal curves
in an electric circuit at a certain resonant frequency. One certain frequency. In e.g. the Sierras x equal to x cosine, the impedance of these two
elements cancel each other. Or e.g. in here, XX and XY or the
admittance cancel each us. Okay? So we will have in this case, or in this case, a pure resistive circuit. Okay? This is a case in
which we say we have a resonant frequency or we
have resonance formula, or resonance case at
a certain frequency. Now, in this case, you will find that we will have a pure resistive system because Excel goes with ecstasy
and Excel goes with xc, most of them cancel each answer. So we will have only a resistor. We will have here
only a resistor. So we will have a pure
resistive system. Now in this case, you will
find that when we have a B, your resistance, your
resistive circuit. What will happen in
this case is that the current will be in
phase with a supply. So we'll have this figure. The voltage is in phase
with the current. Okay? So what are the
applications of resonance? Why do we need to use resonance? E.g. resonant circuits, including the series
or parallel circuits, are used in many
applications such as e.g. selecting the desired station
in radio and TV receivers. Okay, switching
between channels, e.g. and you'll understand
that now how can we do this, e.g. in radio? How can we use resonance
to choose our channel? A series resonant circuit, and be used as a
voltage amplifier. You will find that the
output voltage across the capacitor will be
multiple of the input supply. In saboteur resonant circuit, you will find it is acting
as a current amplifier. It amplifies the current. And you'll find that also that resonant circuit can
be used as a filter. Now, e.g. if we have a
radio like this one, and when we switch
between the channels, how can we switch
between channels? Now we will find
that in real life, we have different frequencies. We have this frequency, we have this one, this one, each frequency
representing certain channel. Okay? So when we, when we turn
up the radio, this one, when we turn this it, you will find that
what we are doing is that we are tuning our radio. So when we turn this knob, what will happen is that
you will find that, that capacitance of the
capacitor is it changing? So finds that e.g. we have here, this is our circuit. We have an inductance, we have a capacitor,
we have a resistor. Ok? Now in this circuit, you will find that when we, when we rotate this, now b will find that the
capacitor, is it changing? When we change the capacitor, what will happen is that we are changing that
resonant frequency. We are changing the
resonant frequency. And they will understand in the course what is
the relation between the resonant frequency and that elements such as
L and C and supply. Okay, you will understand
this inside the course. But anyway, when we
change that capacitor, which changes the
resonant frequency, at which we will have very high value of
current or voltage. We change that
resonant frequency f. Okay? So you will find that when
we change it, if we e.g. it changes this capacitor. And Richard for exam 2.6, 0.296, 5 mhz, e.g. then what will happen
is that our radio, what will receive a
triangle number one, okay? Because it's a resonance, resonant frequency is equal to the frequency of a
channel number one. Okay? Now, if we change again
and we reach e.g. doing 7.075 mhz us. It means that we are now on a channel ten because the
resonant frequency of this circuit is equal to the frequency of the
receiving signal, or does that channel itself. So here you understand that
by changing the capacitance, which changes at
resonant frequency, which means that we are
selecting our channel. So as you can see here, FOR is a change by changing the capacitor of Zara h.
So by controlling it, we can control FOR when
we have this value e.g. FR equal to 26.2 9625. Then it means that we have the resonance Ads
channel number one, which means we are receiving
channel number one. Okay? So now we had an introduction
about resonance. Now we would like to get to understand more in
this course about that series resonant circuit ends up parallel
resonant circuit. Okay?
122. Definition and Equations of a Series Resonant Circuit: Hi and welcome everyone to this lesson in our
course for resonance. In this lesson we
are going to discuss a series resonance circuit. And what is the value of the frequency or the
resonant frequency? And what odds are power,
the power factor, the power consumed
in the resistor, the reactive power, and so on. Firstly, let's start. So what does that see it
as a resonant circuit. We said that the series
resonant circuit must have an inductive
and capacitive element. We said in the previous lesson, we need a resistor, we need an inductor, and we need a capacitor. So we need these three elements. You have to know that
resistive element is always present. Why? Because it is as a result of the internal resistance of
the supply voltage source, the internal resistance
of the inductor, RL, and any added resistance which control the shape of
the response curve. So let's just see this. If you look at this circuit, we have our supply source, E S, This is our voltage supply. Now you have to know
that any voltage source, any voltage source, if you are talking
about ideal case, then we have a voltage
source without any resistor. However, in practice, we have a resistor in
series with a supply. So any voltage source
into a voltage source. Any supply in a voltage
source will have a resistor in series R, S, okay? Which is the supply resistor, which is the internal
resistance of the source itself,
which is RA's. Okay? Now if you look at
any current source, in a current source, you will find that the
current source have what, have our resistor
in parallel to it. So that voltage source has
a resistance in series. And the current source have
a resistor in parallel. So that is a forest
to resist supply. R is in series with it. Then we have something which
is called the, our design. Leave it for now. We have
also that coil or inductor. And we have here capacitor, ok. Now the inductor itself have also add resistors in series, which is RL, which is the internal resistance
of the inductor. Why? Because the inductor
itself is a wire, which of course
have a resistance. Okay? Now, if you look at this curve, you will see that
this is a curve which representing the
variation of the current, the current going out from the supply with respect to two, that frequency, or the frequency at
which we are operating. Okay? Now as you can see or as you
will remember, that here, when we change the frequency
X L and X C changes, which means that
the total changes or the impedance changes. So the current here it changes, adds a resonant frequency. Resonant frequency, we will have a very large amount of current. Here, as you can see, this part, when we
have resonant circuit, or the frequency at which
we have resonance funds, or the current is
very, very high. Okay? Now, as you
can see that the current at resonance will be the supply voltage in the circuit divided by
the total resistance. So y divided by total resistance because
in Zaire resonant, we have or in the resonance. So we have excess
liquid ecstasy, so they cancel each other. And we have only resistance
inside the circuit. So if you increase
this resistor, when you change this resistor, as you increase it, the current will increase. If you increase the resistor. Since the current will decrease. So as you can see
here, when we have small value of resistance, we have larger curve. When the resistance is
medium goes down because the resistance is more than
this one. So it goes down. When we have very high resistor, that curve goes down. You can see the
effect of resistance. Now, RL and RS, we cannot control them. So we can control the
shape of this figure by adding an additional resistance. By changing this resistance, we can define our
shape of response. One shape of response, which is this shape, by changing this resistor. Okay? Now, let's find that
equivalent circuit. So we have this equivalent
circuit which is supply or S, R, D or L, L and C. We
need to simplify this. So as you can see
from the circuit, we can simply bought
it like this. We have inductance L, we have capacitances C, and we have r, which
is the total resistor, which is our L plus R, D plus our essences. They are all in series. So now we have one
circuit which is RLC. Now, the total impedance
of this network at any frequency modes
that this one is frequency at any
frequency in general. So total of this
circuit is equal to the resistance plus j
x over l minus j xc. So as a capacitance
is represented by negative j ecstasy and inductance represented
by plus j XL. And the resistance does
not have any angular. Okay? So in general you
will find that we have our plus j, xl minus xc. This is the impedance of this
circuit at any frequency. Okay? Now, when we have resonance, when we have resonance, what will happen is
that that circuit here, remember, resonance mean
unity power factor. Unity power factor. Okay? And at the same time, build resistive only
resistance inside the circuit. This is what does she mean? Unity power factor B0 resistive. So in order to have
the T equal to only r, it means that in
resonance condition, we will have x L
equal to z legs us. Why? Because XL minus
XC Z are equal, then this part will
be equal to zero. So we will have only
the total equal to R, as you can see, which is in
that resonance condition. Now, can we find that
resonance frequency? Yes. How can we get the
resonance frequency? Somebody we have when resonance. So when we are in
resonant condition, you will find that exhale
equal ecstasy, as we said. So Excel liquid ecstasy, which is XL is omega L. And X C is one over omega C. As you can see it, x L equal
to Omega L and z equal to one over omega C. So if you take this omega to the
other side and L2 this side, you will have omega square
equal to one over LC. Resonant frequency in radian
will be one over root LC. Or we can say that the
frequency not in radian, but as a frequency in hertz, this will be F equal to omega S divided by
two pi, like this. Okay? So this is in hortus, similar to 50 hz or
60 hz and so on. 1/2 pi root LC. Now what is the relation
between these two? Remember that that omega omega is equal to bar multiplied
by the frequency. Fs is equal to omega
S divided by two pi. Omega is divided by
two pi gives us F S. Remember this equation because
this is really important. This is what, this is
that resonant frequency, resonant frequency in
the series circuit. That's why e.g. in the radio, by changing the capacitance, we can change the frequency, which until we reach
that channel frequency. Okay? Now we have the total equal to x L equal XOR C and R
plus j X L minus X c. Now what is the value of
current at resonance? So first, let's
type it in general. So you have this
circuit which are LLC, and we need the current. So the current in any
electrical circuit is equal to the supply divided by the total. Okay? So let's apply E, which is E with an angle zero. We say, usually we say is that the angle of the supply
is equal to zero. Okay? Now what about the total? Total is equal to r plus
j x l minus x is c. Now what we need
here is the value of current at resonance, at resonance, at resonance
Excel equal ecstasy. So this part is equal to zero. So total will be equal to the resistance which
does not have any angle. So it will be r
with an angle zero. R does not have any angle. You can see j here, we're representing
the 90 degree, j representing 90 degrees. So we have total equal to r. So as you can see from
this equation is that the current will be E over
R as a magnitude. And the angle is
also equal to zero because they are in phase. So what we learn is that the current in resonance
is equal to E over R, The supply divided
by the resistance. And the angle of current
is similar to the angle also supply because
they are in phase, which means that we have
a unity power factor. Okay? Of course this is a maximum
current because the resistance R when we have
resonance Excel equal xy. So this part equal to zero. So in this case, total will be minimum, minimum value of the total, which means the maximum current, because I equal to E over R. So when that becomes minimum, very low, then the current
will be very high. That's why when you draw
the characteristics between the current
and the frequency, you will find that at resonance frequency we
have the maximum current, because we have the
minimum impedance. Now before it. If we increase the
frequency or decrease the frequency or funds as
the current goes down. Because in this
case we will have an additional term
which is x l minus x. Okay? Now, what is the value
of the voltages? The voltages of that inductor, and the voltage of
the capacitor V L is equal to what is the voltage
drop across an inductor? Voltage across an
inductor is equal to the current multiplied by x. So the current
multiplied by x l. And for the capacitor, it will be current, which is I multiplied
by Ecstasy. Okay, Very easy. I excel or accessing. But we have to remember that we have here an additional term, which is the angle. The angle. You can see that here
we have J Excel. And do we have
here negative J x, z, we have J x l. And we have here negative j. So j is translated
to 90 degrees. Negative j as translated
to negative 90 degrees. Okay? So we have J Excel. So Excel angle 9092
because we have j and X, C angle negative mine too, because we have negative j. So in total we will have
VL equal IXL angle. The angle is really,
really important. Voltage across the capacitor
is our angle negative mind. So as you can see, they are
out of phase by 180 degrees. Where did we get this value? Assembly? This angle minus
this angle gives us a phase shift of
hundred and 80 degrees. Okay? Okay. So as you can see, I excel and I ecstasy current is
equal to this current, equal to this current, and exhale equal
xC and resonance. So we'll find that
the magnitude, the magnitude of the
voltage is VL equal to VC, but they are shifted
by 180 degrees. Okay? Now, if we draw the
failures or diagram, what does this represent? This representing
is our voltages, current inside our circuit. So as you can see,
we have first E, which is our supply. Our supply have an
angular equal to zero. That's why it is
parallel to the x-axis. This is y-axis. So E is parallel to x
axis because it is our, because it's angular
equal to zero, angle equal to zero. What about the current? Current is equal to E
over R angles here. So what's the difference? It will be similar
to the voltage, but reduce the buyer
value of the resistance. That's all you will find
that the current I, this is our eye. You will find it is
a shorter vector, smaller vector, because it's, the magnitude is
smaller than E and parallel to it because it has
the same angle, angle zero. Okay? Now what about VR? Vr, which is the voltage
across the resistor, is equal to the current. I. Multiply it by the resistor, current I multiplied
by the resistor. So we have all multiply
this all by a resistor. So we will increase
its magnitude a little bit and reaching here. So we have now VR. What is VR? Vr is voltage across
the resistor. Now we need VL and VC. You can see that v l is equal
to x L with an angle 90. Okay? So you can
see it is leading the current and delete things or supply voltage by 90 degree. So as you can see,
this is our E. As you can see as
if it is here e. So it is leading by 90 degree. 90 degrees. In phasor diagram,
this means leading. So VL minus two degrees, so it is leading by 90. Then we draw our vector v L VL, which is I Xa. What about Vc? Vc is lagging
by 90 degrees negative 90. This one is zeros, this
one negative nine. That's why we draw
our exist negative. This is 90 degree also about
in negative direction. And we have VC. So from this figure, you will see that the
total voltages is VR, because VL going with VC
always going goals VC. So we have only one voltage. Here. If we're talking about the VR, you will find that
we have only vr v, l exhaust VC exist about z
are always opposing HR's. Okay. Now you can find this phasor
diagram indicates that the voltage across
the resistor at resonance is the
input voltage E. Okay? So we wouldn't multiply
that E equal to, remember that the current
is equal to E over our current equal to e over r. What is the voltage
drop across this one? Vr is equal to the current
multiplied by the resistor. So if you look at the current, current is equal to e over r, e over r multiplied
by the resistance. So this resistor goes
with this resistor, so we will have equal to E. So find that the voltage across that resistor is equal
to the supply voltage. What about with
the voltages here? What are the z value? You will find this in the
slides or next lessons. Okay? Now we take this phasor diagram
and we can convert it to a power triangle from phasor
diagram to up our training, converting the current
voltages into power. Okay? So first step, what is the power of the resistor or
the power consumed? And solid resistor assembly
is the power consumed in any resistor is equal to
I squared multiplied by r, i square multiplied by r. Now what about words
that inductor? What about inductor,
inductor assembly? Ql, which is a reactive
power of the inductor, is equal to the square of the current multiplied
by also Excel. Similar here, I square R. This one will be R-squared excel and QC will be I squared C. Okay? Now, you will see all
those in this Faisal diagrams as they are always
opposing each other, Q C and Q L. So the summation of
the reactive power at any time will always
be equal to zero. So the supply of
apparent power S, which is apparent power equal to the supply multiplied
by its account. So all of this is apparent power is equal to
going all to the resistor. So again, the apparent
power representing what S, Again, if you don't know as, as equal to b plus
j Q L minus Q C. So as you can see
from this figure, that QL is always equal to QC. So this part is equal to zero. So the parent of power coming
from the supply is equal to all of that active power
consumed in our resistor. Okay? Now, as you can see from
the power triangle, that at resonance, total power, apparent power is equal to the average power
dissipated by the resistor, since the QL equal
to c at any instant. Now what is the
power factor we said before without calculating it, we said that since we
have a pure resistive, your resistive
system in resonance. And we said also before that the current is in
phase with the voltage. So the phase shift between
them is equal to zero. So the power factor
will be unity. Power factor is equal to cosine. Theta is the phase
shift between E and I. Supply current and
supply voltage. You can see they
have the same angle. So it will be cosine V Theta. V minus Theta is the
angle of the voltage, minus the angle of the current. So this one is zero,
this one is zero, so it will be cosine zero, which means it will be unity. Or you will know that the
power factor is the ratio between the act of
power divided by, is that apparent power. Now we said before in the
previous slides that S is equal to p as equal to B. So the same factor dividing each other
gives us also unit. Okay? Now let's upload the
resistance power curve. So we have here three
curves which will help us understand
more about the power, that resistance power curve, the inductance power
curve and capacitance. Then combining all
of this together. So as you can see, we have
the current and voltage. Voltage here is a voltage
across the resistor. The current is the current
flowing through that resistor. So we know that both
of them are in phase. The voltage and current
are both in phase. So the power across the resistor
can be multiplied by e, which is the voltage
across the resistor, which is in this case
and resonance equal to the supply voltage. So we have all multiplied
by e. They are in phase, as you can see, as you
are following each other but with a
different magnitude. Now, I multiplied by
E when you multiply, we have here e and we have here. So this one is positive and
this one is also positive. So multiplication of two poles the values gives us
our balls stiff curve. Okay? Now let's see the other side. We have here current, negative current and
negative voltage. So negative current multiplied
by negative voltage, negative multiplied
by a negative gives us some balls the value. So as you can see, we
have our bowls difficult. You will find that in z
and our power consumed, the borrower will be like this. Like this. Okay? So this is the resistance or the power consumed
inside the resistor. It is with respect
to time like this. There is no negative part. What about the inductance? Inductance, we know
that we have E, which is a voltage
across the inductance. Remember here, E does not
representing the supply. It representing the voltage
across the inductance, is the current flowing
through the inductance. Okay? So as you know, is that the inductor mix as the
current is lagging, the voltage or VL, voltage across the inductor
is leading, leading, leads. Zavala has a value of
the current or the current by 90 degree. So when we block them, we have here our voltage, and we here have our current. You will find that here
between this point, any two similar point here, from here to here, or e.g. from here to here, or any other innocent. You will find that
that angle here, and the angle here is 90 degree. You will find that
the voltage here is leading the current
by 90 degrees. Okay? So this is a voltage, is this one is the current. Now what will happen when we
multiply them by each other? You will find that we have some times this
one ball stuff. As you can see here, the
current is positive, but the voltage is negative. So finds that their
multiplication is negative. As our time, both of
them are negative, so we have a bolster value, and so on are ones
that the inductor, inductor reactive power is in the form of a
sine wave like this. Like this. Part positive, part negative. Okay? Now let's see the capacitor. For the capacitor,
you will find that we have in the capacitor. The current leads. Voltage by 90 degrees is the
reverse of the inductor. Finds that the current
here, as you can see here, this current, and
this is our voltage. So you can see that the current
is leading by 90 degrees. From here to here, 90 degrees, or from here to here, 90 degrees and so on. So finds that the current
is leading in this case. Okay? So this is a current, this is a voltage across
what, across that capacitor. If you multiply these
two at any instant, you will also find we
have our sine wave. Sine wave. Okay? Similar to the inductance. But one important note, one important note,
you have to know that you will see that e.g. e.g. if we look at the E
or the induced EMF here, E here, and compare it
with e in the inductor, E and Z inductor e.g. here, or e.g.
starting from here, you will find that the
R naught starting at the same instant will
find that that curve is a power curve
of the inductance is lagging or leading
that capacity. So they are not
above each other. Okay, As you can see now, when we combine all
of this together, you will find that
the reactive power at any less than t is zero. Energy absorb it and released by inductor and capacitor
at resonance. You will find that this one is storing as this one
gives us energy, is this one stores it. Then this one provides energy
and this one is towards it. The energy reactive power
going from inductor, capacitor, from capacitor
or inductor and so on. Now, if we combine the
three previous covers, we will have this power curve. So this power curve, what does, it teaches us? You can see here this
PL is reactive power. Reactive power of
the inductor is, you can see the
power of the induct. On the other side, we have BC, which is Q C, or the power, reactive
power of the capacitor. So it will be like this. So you can see that B L and B C, B, C p, l, always opposite to each of us, will find that the summation of these two curves, or the PEL, PC or QL and QC, There's some mission at any
instant is equal to zero. So we'll find that we will have only this curve which
is resistive power. You can see this whole
region, this one, this yellow region, dark
halo region, this one. And this one is our
consumed resistance, consume the PowerPoint,
the resistor. Okay? So what do we learn from here? Is that what does
this mean? Also? Before we finish this lesson, we have to know that B L, which is a power. What does this represent? Power supply to the element? Power supply to the element. So how it is a power supply
to the element assembly, you can see B, L, or QL. All stuff means that the
inductor is absorbing power. Here QC is negative. It means it is returned
by this element. The combustor is giving it to, giving power to the inductor. Similar to here, you can
see this part is positive. So it means that the capacitor is absorbing reactive power. And here's the
inductor is negative. It means it is supplying
reactive power. At any instant you
have in general, you can have some knowledge. Capacitor giving two inductor, dr, giving to a
capacitor and so on. So it's n cycle, cycle goals on. Okay. So I hope this lesson
was helpful for you in learning about resonance.
123. Quality Factor of a Series Resonant Circuit: Hey everyone. In this lesson
we are going to discuss an important term in
the resonant circuits, which is Zach quality factor Q. So what does the quality factor? So the quality factor Q of
a series resonant circuit, as defined as the ratio
between the reactive power of the inductor or the reactive
power of the capacitor. Does the average power of the results resistor
at resonance, or the average power dissipated in the
resistor at resonance. So simply, Q is the ratio between reactive power
with respect to two, the power dissipated
in a resistor. So it can be written like this. Qs or the quality factor in
the series resonant circuit, is equal to the act of power, the capacitor or the
inductor with respect to the average power
dissipated in the resistor. Why it is important? Because it gives
us an indication of how much energy is placed in storage with respect to that dissipated
inside that resistor. So the higher the
quality factor, it means that we
have higher energy stored with respect
to z dissipate power. Okay? So the more que, more quality factor
of the inductor, e.g. it means it stores the energy
more than it dissipates it. Okay, It's really
important factor. Also will find that the
lower level of dissipation, lower dissipation means
that we will have higher Q. Lower dissipation means
higher q because we are having more energy stored. Okay? So it is a more concentrated
and intense the region of resonance. Now, how can we, or what's the
equation of Zach us? Reactive power over
the average power. The reactive power,
which is power e.g. inside the inductor.
The inductor, so power is equal to I
squared multiplied by Excel, the square of the current, multiplied by the reactants. And for the resistor I
square multiplied by the resistance R. So this
is the reactive power. This one is the power
dissipated in the resistor. Now, as you remember
that our circuit is a series resonant circuit. So since we have
a series circuit, inductance in series
with a resistor, in series with a capacitor. Then what does this mean? It means that the
current flowing through the inductor
is similar to the current flowing
through the resistor. So I square is similar
to this I squared, so we can cancel
them with each ours. Okay? So we'll have finally x L over R. And Excel is omega
S multiplied by L, which is angular frequency
at resonance omega S. Now, it was a
resonance resistance of this system is only the
resistance of the coil. We can say is that Q S is equal to Q L equal to x L over R n. So let's understand
the difference between these two photos that we have that quality factor of
the system, whole system. Okay? So as you remember, we have a voltage source with a series or S. And we have the coil, or L, which is the resistance of the coil and the coil
and the capacity. Okay? If we have here on our design or any
additional resistor. Anyway, if we're talking about with the quality
factor of the whole circuit, quality factor of
the whole circuit. Then it will be x, l or
ecstasy, which is reactants, or the reactants, or the active power with respect
to the total resistance, which is RS plus RL. Here, the QS, which is a quality factor of
the whole circuit. Now, if we are talking about
with the inductor alone, the quality factor of
the inductor alone. Then we will say x L over R L, which is the resistance
of the inductor only. So this is related
to the coil, only. This one related to
the whole circuit. Now, if we have, if we have only if
we don't have our S. We have only R L. It means that the total resistance inside
the circuit is also R L. So it means that Q
S will be equal to q L when we have only R L. Okay? So as you can see here, Q of the coil, q l is equal to x L over R L. Now we know that
XL is equal to omegas and omega S is equal to two pi multiplied by the
frequency at resonance. Frequency of resonance, as we obtained in the
previous lesson, equal to 1/2 pi root LC. Okay? So now what we are
going to do assembly, we can take this equation
and substitute it here as this one here.
And this one here. Omegas, okay? Or inside the omega S itself. Anyway, you can simply
write like this. We have Q S equal
to omega S over r. And I already know that Omega S by directly
without any simplification. But at the omega S is
equal to one over root LC. One over root LC. So you can see one over root. See, okay. Now, if you simplify this one, we have one over r, l divided by root L
gives us root l over c, root l over c. Okay? Rootsy. As you can see, your routine and LOG is to the power one and
this one to the power half. So one minus half gives us half, so it becomes root over c. So
this gives us that Q as Q, l to be more specific, the QL in general. So let's delete this.
You can see like this. So you have Q as
in the case of if, if q l is equal to Q S, okay? Or in general, it
does not matter. In general, this equation representing that
quality factor. If we would like you, then R will be, what will be RL. That's all. Okay. So as you can see here, q is
equal to omega S L over R. Omega is two pi F S and F S
is equal to 1/2 pi root LC. So as you can see, the
simplification gives us finally root l over
c multiplied by one over r. Now we would like to see what will happen of
the quality factor to the quality factor with
respect to the frequency. So as you can see,
that Zach q of the coin quality
factor of the coin Q L is equal to x L over
R L as we just obtained. So as you can see this
one x L is equal to two pi multiplied
by the frequency, multiplied by their inductance, by F S multiplied by
the inductance, okay? Or omega S multiplied
by the inductance. Now as you can see
that when f s, when the frequency increases, when frequency increases, what will happen to
the quality factor? Quality factor will increase
because excel increases. So the quality factor increases. Again, x L equal to two pi fs l. So as the frequency
increases, excel increases. So the q of the coin increase. So as you can see, ads
at lower frequency, you can see that as
frequency increases, Zach QL, what will happen to it? It starts to increase
like this. Okay. Now, until a certain point, until a certain point when it starts to Windsor
frequency increases, it starts to decay. Starts to decay. Okay? Now when does this happen? So let's see. So first, as you can see, the first region
here and this part, you can see it increases
linearly with the frequency. As the frequency increases, that QL increase, the
quality factor increase. Now as you can see, it is
true for the low range to the mid-range of
frequency, 5-50 hz, e.g. you, 50 khz, e.g. and as you can see,
unfortunately, as frequency
increases here, e.g. from 50, as we increase, what will happen when the frequency inside
the coil increases or the current
flowing it increases. What will happen is that. The effective
resistance of the coil increases y due to something which is called
the inside AC circuits, the skin effect. Okay? So thus can affect the
leads to the increase. Or to be more specific, r equal to root rho l over area. So the higher is the area of the conductor itself
or the cable, the lower the resistance. So as area increases, the
resistance decreases. Okay? Now what will
happen is that when the frequency starts increasing, the effective area of the conductor or the cable
itself starts to decay. The array itself
starts to decay. So the total resistance
increase when does this happen in the skin effect
in the AC systems. Okay. So when the frequency
becomes a very high, the area of the effective area of the conductor or the cable. In this case, our coil is
considered as a conductor. In this case, the area
starts to decrease. So the effective resistance
or the resistance itself starts to increase,
leading to what? As resistance increases,
that q decreases. Okay? As you can see that
when the frequency from here starts increasing,
excel increases. But at the same time due to the skin effect, our L increase. But you will see that
our L is much higher. The increase in RL is
much higher than Excel. That's why is that Q of the coil starts to
decay in general. So as you can see, it starts to decay due to the skin effect. Okay. Now also another effect which
also decreases that coil is Zach capacitive effect between the windings of the conductor,
the capacitance increases. Okay, due to the
increasing of what? Due to the increasing
inside their frequency. So it reduces the coil QL, or the quality
factor of the coil. Why? If you see
from this equation, Q S is equal to one
over r root l over c. So as the capacitance increases, when Zack capacitance increases, the quality factor
will start decay. Okay? So this is another factor. So we have two
factors leading to the King of q that
all factors are, number one, the skin effect which leads to increase
in the resistance. Number two is the
capacitive effect between the windings
also increases. So in this two factors helps indicating of the
quality factor. You will find that for the
same coil, for the same type, QL drops off more quickly for higher
levels of inductance. So as you can see a linearly, all of them, one Henry, hundred millihenries and
millihenry, one millihenry, all of them increases at the same time or
the same values. But you will find that
one Henry decay is much faster than 100 mainly is on ten millihenries
and one millihenry y. Due to the effect of
capacitance and effect of skin effect is much higher
in the higher inductance. This effect is more visible in the higher inductance values. Okay? Now, you will find that in the series resonant circuit used in communication systems, QoS is usually greater than one. By applying the
voltage divider rule to the circuit we
obtain like this. You will find is this circuit. We have here at supply
voltage, an AC supply. We have our resistance. We have x l and x c. So this x l and x z are
equal to each other, so they are in resonance. And they will understand now why is the quality
factor is important? Or what is the effect of
quality vector on our circuit? So let's say I would
like to find the voltage across C or the
voltage across XML. So what I'm going
to do is like this. First you can see that the
circuit is equal to E, r x l, and x z. Okay? Now the first step is
that we need the voltage across the XML, e.g. okay? So vl is equal to what? The voltage across the inductance
simply equal to L to x. Multiply it by the current. Current flowing
through the circuit. The voltage equal to the
reactants multiplied by current. Now, Excel, okay,
leave it as it is. Now, what is the value of
current, current in resonance? We are talking about the resonance circuit,
resonant circuit. So in resonance, the
current is equal to what, as we learned before, or equal to the supply
divided by resistance, ie over r. As you can see here, E over R X L, E over R X L
E over R. Now we know that the quality factor Q is
equal to x L over R. Okay? So X L over R can be
replaced by q s. So we can say QS multiplied by E, like this, Q S
multiplied by e. Now, what you will notice
here is that, but before we not something
we have to contain. Now, you can see that also the voltage
across that capacitor is equal to x is c
multiplied by the current, which is e over r. So x is c, e over r, which is the current. Now also xc over R gives
us the quality factor. Why? Because we said before that the quality factor is the
ratio between reactive power, all of the inductor
or the capacitor. The inductor or capacitor. So XL or ecstasy, because they are
similar to each other, XL equal to x cosine. So we'll find that the
voltage also gives us USE. So the voltage across excel is equal to the
voltage across C. In the resonance case. The R equal to Q S multiplied by e. So what you
will learn from here, you will learn is that the
voltage of the inductor or capacitor are multiple
of the supply. So you can see is if
QS greater than one, which is dominant case. If it is greater
than the, than one, it means that the voltage here, here or here is higher
than the supply. As an example, in this circuit, the Q S is equal to x L over R, X L over R, which means it will
be equal to 80. Okay? So the quality
factor equal to 80. So as you can see,
the supply ten volt. Okay? So what about the voltage of the inductor
or the capacitor? Most of them are equal
to QoS, which is 18. Multiply it by x multiplied by the supply voltage E.
E is equal to 10 v. So it gives us 100
volt as a magnitude. So as you can see
that the voltage, output voltage across
that capacitor, or the voltage
across the inductor, is now amplified equal to
voltage V is equal to 800 volt. So you have an input
supply of ten volt. But because we are in resonance, resonance formula or
in the resonance case, you will find that the
outward increase the by 80 times the output
voltage is 800 v, 800 v. Okay? So as you can see, that's why when we, in the introduction
of this course, so we said that see it as
resonant circuit. Series. Resonant circuit operates
as a voltage amplifier. Voltage amplifier because it
is at amplifies the voltage from ten volt, 200 volt. That's why it is an amplifier. As you can see, is
the quality factor determines how much our
output voltage will increase. So if we have Q S equal to, it means that voltage across the capacitor will be
two times the supply. Okay? So this is the effect of the quality factor
on our voltage. So as you can see, the L
equal vc equal to 800 v.
124. Total Impedance VS Frequency in a Series Resonant Circuit: Hey everyone, In this
lesson we are going to blot that oath and impedance
versus frequency. Would like to see how does the impedance a change
versus the frequency. And from the impedance, so we will get the current. So let's start first the total impedance of
a series RLC circuit, which we are discussing in
this part of this course. It's given by the total equal
to R plus j X L minus X c, or the total equal R plus j X L minus X c. This is what we learned in this
course because all of these elements are in
series with each other. Now, if I would like
to get the magnitude, we have the magnitude, this total have a
component plus j, another component, real
and imaginary component. So in order to get the
magnitude of F impedance, it will be the square
of the first root, R squared plus the square of x, l minus x is z. Like this. So the total, or the
total impedance equal to the square root
of r squared plus x minus x c all
squared is this is the magnitude of the impedance. Now what I would like to
get is that I would like to blow this against the frequency. Okay? So as you know that
here we have the resistance, which is, of course here
in our case does not, is not function in frequency. It is a constant value. The Excel function in frequency and XC function in frequency. So in order to block
this magnitude, we need to see the r with
respect to the frequency, X, L with respect to frequency, and xy with respect
to frequency. We need to find these
three values in curves, the curve of each
of these values. So first, for the resistance, we know that the resistance
is not function in frequency. So that resistance as a
function of the frequency is a constant value or constant
value with respect to time. Now, for the XL
excel, as we know, the inductance is equal to two pi multiplied by the
frequency multiplied by n or omega L. Omega L omega is two pi multiplied
by the frequency. So as you can see that
x equal to two pi f L x L equal to two pi
multiplied by the frequency. So as you can see, that two pi l multiplied by f, This
representing what? This can represent,
y equal to m x. So y is our y-axis,
which is Excel. M is the slope of the
line which is two pi L, the slope of the line two by l. And x is the frequency,
as you can see. Okay? So as you can see, Excel is directly proportional
with the frequency. As you can see as the frequency
increase, Excel increase. Now what about xc? If we look at accessing, is inversely proportional
with the frequency y, because x is c equal
to one over omega C. Okay? And omega is two pi multiplied by the
frequency multiplied by c. So as you can see, ecstasy is inversely proportional
with capacitance. So as capacitance, inversely proportional
with the frequency, the frequency here,
we would like to bloat extra z with
respect to frequency. As the frequency
increases or decreases. As you can see here, it
is a decaying curve. So now we have these
three curves and the way we would like to
combine them together. So first we will combine
excel and accessing. So we have x l, and x is z. We will combine them
into one curve because I would like to show you something which is really important. So as you can see here, we have accessing the
skirt and XL this curve. Now, if you look at this curve
with respect to frequency, you can see xc, very high value. Then as the frequency increases, it starts decaying with time. Excel starts from a small value, then it starts increasing. Now, you can find that there
is a certain point here, this point in which we
have x L equal to xy. Now this point is that
resonance state, okay? It's a resonance state. Here is, this is a
resonant frequency at which X equal to X c. Now before resonance, before
resonance, before this one, this part, you can
find that x is c value is higher than Excel. Okay? So x is greater than Excel. So we are in a
capacitive is state, our circuit is capacitive. Now, after that
resonant frequency, you will find that x l value
is higher than accessing. Okay? So we are in an
inductive estate. Okay? Someone ask is, why does ecstasy
here greater than Excel? Simply because you
know that x is c is equal to 1/2 pi FC. So let's say e.g. at this point, when the frequency is 0.1. Okay? So this is, frequency
is very small. So Xc becomes a very high. However, if we use 0.14
is the inductance 0.1, you will find that Excel
becomes a very, very small. Okay? So before resonance,
ecstasy has a small, there is a small frequency. Z is very high and they
excel is very low. After frequency, after resonant frequency
will find that here, F is very large. So ecstasy is decaying, as you can see here,
very small value. And at this region, F is very high, so Excel becomes a
really, really high. That's why we have two reasons, before resonance and
after resonance. Okay? So as you can see
is a condition of resonance is now
clearly defined, adds a point of intersection. So when Excel is
intersected with C at this point that we have
that resonant frequency. For frequencies less than fs. Here in this region, we find that this network
is capacity greater than Excel and after
resonance frequency greater than resonant condition, Excel is greater than c, So the network is inductive. Now we have these two curves. We have that resistance, we have that
capacitance inductance. Now we would like to combine all of this into one big figure. So when you combine all of this, you will find that the total
or the total impedance with respect to the frequency
is bloated like this. Okay? You will find that this region, which is this point, which is the minimum,
minimum impedance. Minimum impedance occurs at
a value of R B or resistive. At resonance. At resonance state, we
have x is equal to x L, x equal to x L. So
this part becomes zero and then we have
only resistance. So that's why at resonance, when f equal to f
resonant frequency, the total becomes
only pure resistive, which is the value equal to R. Okay? Now, before resonance
and after resonant, total increases both sides. Why? Because in this case, we have x l and x z component z on don't
they cancel each other? So they exist in this
region and in this region. Now one thing to notice here is that you will
find that this region, this region after resonance, and before resonance,
you will find that these two regions are
not equal to h awesome, they are not symmetric. Okay? Now, if we would like
to get the angle, so as we know that
angle of the total, the total is equal to again
r plus j x over L minus X c. Okay? So tan theta, which
is the phase angle, phase angle equal
to the j component, or the imaginary part, excel minus x is c divided by the resistance
or the real part. So Sita will be equal
to tan minus one. This part, as you can see, c t equals ten minus one
X L minus X c over r. Okay? So this is a phase angle, okay? And they will not understand
now the face angular presenting what
representing how much, how much the voltage is
leading the current. So what does this mean? You will find that
E, the supply, is equal to Ymax, maximum value and
angular equal to Sita. And the current is
equal to maximum angle. So this angle, CDA, representing how
much is the voltage, is leading the current. How much it is
faster than the car. Okay. So you will see that here. You will see that Sita, which are representing
how much is the voltage leading the current. Okay? So let's upload the relation
between C and the frequency. So first, as you know that when the frequency equal to
resonant frequency, what will happen is
that x is equal to xy. So this part is equal to zero. So ten minus 10 gives us what
gives us c t equal to zero. Okay? As you can see, at
resonant frequency, the angle theta equal to zero, this point is
corresponding to zero. Okay? So that is the first part. Second part is that you will see here that after
resonant frequency, if you remember from the curve, the curve here, this one
which is the impedance curve, you will find that
after resonant here, this case XL is greater
than accessing. Before resonance, x is
z greater than Excel. So what do we learn from this? When x becomes a
greater than ecstasy, It means that as x increases, then minus one, the
angle increases. Okay? So Sita will also increase. Again as x l becoming
greater than x is c, It means that we will have
more positive and that's why you will find that when
we increase the frequency, you will find that the
angular starts to increase, as you can see here,
until of course, the maximum value of 90 degree. Okay? As you can see here, when it becomes a
pure inductive. So as you can see in this case, when the XL increases,
what does inductance? Inductance, the
current lagging from the voltage makes as a current
lagging from the voltage. So here as you can see, is that in this case, when the frequency increases beyond that resonant frequency
XL becoming greater than c. So the effect of
inductance becoming greater than the effect of
capacitance, the effect, okay? So the inductance matrix is a current lagging
from capacitance. From voltage, makes the current
lagging from the voltage. That's why in this case, we say that the circuit is
inductive because Excel greater than ecstasy and
lagging power factor. Lagging power factor
means that the current is lagging the voltage
behind the voltage. Now before the
resonant frequency, the circuit is capacitive. Reverse happened because x
is c is greater than Excel. So when Xc becomes a
greater than Excel, Sita becomes more negative, becomes a negative
value, more negative. As x increases, it
becomes more negative. So as you can see
here, this is a 0.0. So as the frequency decreases, the angle becomes decreases in towards the negative side until the maximum value
of negative nine, which is pure
capacitive circuit. Okay? So what does that capacitance
you do when the effect of capacitance greater
than inductance, it will lead to the voltage
becomes lagging from current. So remember, voltage
lags is a current due to a new case in which we have large capacitance or large on the circuit,
becomes a capacitor. When the inductance effect becomes greater than
the capacitance, then the current
will be lagging. So we will learn into effect of capacitance or the frequency on the circuit becoming
capacitive or becoming inductive and its effect
on the power factor. So as you can see in general, when frequency is lower than
that and two frequency, then it will keep
becomes capacitive. And the current
leads the voltage. When the frequency greater
than the resonant frequency, it becomes inductive as the
voltage leads, the current. Ones, they are equal
to each other. It becomes pure
resistive circuit. So the voltage and the
current are now in phase. So I hope this lesson
was helpful for you in understanding the bloating of the total and the effect of frequency on the power
factor of the circuit.
125. Bandwidth and Selectivity Curve of a Series Resonant Circuit: Now let's understand
a concept in, as a series resonant circuit,
which is bandwidths. What is the meaning
of bandwidths? So as you can see, if we plot the total, which is a total impedance of our circuit as we have done
in some previous lesson. With respect to the frequency, you will find this
curve and you'll find that this off
towards our residents. Or after that
resonant frequency. Resonant frequency is zero
or not equal to each other. They are not symmetrical. Now, what if I would
like to blow that current with respect
to the frequency? And instead of that. So as you can see that
the current is equal to E over the total
divided by the total. Now e, which is our supply, is a fixed value, fixed supply. So let's say e is
a constant value. So we can say is that all is directly proportional
to one over z. Like this. Or to be more specific, it is the inverse of this curve. Okay? Okay, So let's see now. So as you can see
when we plot Zach current with respect to
the frequency or find it sells the same curve but
with reverse direction. So when the total becomes
very large at infinity, the current will be
very small, zero. Why? Because I is equal
to E over that. So when that becomes infinity, very large value here or here, you will find that one over
infinity gives us zero. That's why in this 0.0
and this point is zero. Now at resonant frequency, the total is minimum or
the impedance is minimum. So in this case, you will
find that the current is maximum at resonance. And its value at resonant
state is equal to E over R. Supply voltage divided
by resistance because we have only resistance or a
pure resistive circuit. So as you can see,
it rises from zero to a maximum value of e over r, in which is impedance
is minimum. Then it drops back to zero as the total increase,
as you can see here. And this part starts
to increase again, so the current starts
to drop again. Okay? Now we will find
that that curve, this curve is of course
is the inverse of the impedance versus frequency
curve, which is the scarf. Senses as a total curve is not symmetrical about the
resonant frequency. The curve of the current
versus frequency has the same property.
What does this mean? You can see that this part
is not equal to this part. The part of the
resonant frequency and support before resonant
frequency. Same here. You will find that
this part from here to here as not equal
from this part to here, as they are not symmetrical. So what we learn from this,
you will endorse now. Now, why do we need this? Here? You will find that there is a definite range of frequency at which the current is
near its maximum value. Enzyme maintenance is
also at a minimum veil. Those frequency
corresponding to 0.707 of the maximum current are
called Z band frequencies, cut off frequencies, half power frequencies,
or corner frequencies, all of them representing
the same sink, okay? They are indicated by F1 and F2. Now, the range of frequencies
between the two is referred to as the bandwidth
of the resonant frequency. So let's understand
what results to set. First, you will find that here, this region from here
to here, or this part. And this part, you
will find that that is still a small value. Still a small value, and the current is
still a large event. Okay? Now, this frequency,
what are the range, the art from a
frequency called F1, our frequency called F2. This tool frequency are
known as the band frequency, the cutoff frequency, how power frequency
or corner frequency. These two frequencies
form a bandwidths. From here to here. The bandwidth is a
representation on the curve that's representing
a group of frequencies. Group of frequencies in which current is
still a larger value, but the impedance is
minimum from F1 to F2. That's why it's a
bandwidths is equal to f1. F2 minus f1. This is a bandwidths
in which we have frequencies that produces
a larger current, still a larger current, close to the maximum value. Okay? Now what does F1
and F2 represent? Z represents something
which is called half power frequencies at this two
frequencies here and here, their power will be equal
to half the power maximum. Okay? It's up our maximum, which is I square maximum multiplied
by the resistance R. Okay? So at this frequency
F2 and frequency F1, we have a power equal
to half maximum power. Any value after here, from here, here or here is greater than, of course have the power. So anyway, F1 and F2, which are presenting
the bandwidths, are located where located
in the power equal to half the maximum power at the current equal
to 0.707 imax. And you will understand what is the relation between this value and the half power
frequency the next slide. So the half power frequencies
are Zoe's frequency, which is F1 and F2 at, which is a power delivered is the one-half that delivered
at the resonant frequency. So the power at half power frequencies
equal to half V-max. Now, the above condition is derived using the
fact that B max, or the maximum power at resonant state is equal
to our Emax is square, the square of the current
multiplied by the resistance. At resonant, we have
the maximum current, so we have the maximum power. Now, the half power frequency is equal to I squared
multiplied by R, The current here, and
the current here. If you take this value, 0.707, maximum and square
that we will have half maximum square r
I maximum squared R. So don't give us half b max. So again, this one or 1707 representing
one over root two. So when you take this current, which is one over root two, our Emax, this is the value
of the current at what? At the half power
frequencies. Okay? If you take this
current and squared it, so we say one over root two. Our Emax all squared
multiplied by r. You will find that it gives us half one over root two squared gives us half I omega
squared r squared, R is P max, so it will be half V-max. So the value of the
current one over root two or 0.707 imax gives us half of the
power at the resonance. Okay? So sensors are resonant, circuit is adjusted to select
a band of frequencies, which is called the
selectivity curve. The term is derived
from the fact that it's a one must be selective in choosing their frequency to ensure that it is
in the bandwidths. The smaller the brand with, the higher their selectivity. The shape of the
curve depends on each of the elements of the
series RLC circuit. So what does this all mean? So as you can see,
selectivity curve, it means that we have a curve
in which we select from. So as you can see here, we have in this figure, e.g. is our bandwidths. Okay? We have, we can select and solids is bandwidth frequencies
inside the bandwidth. This one, this one, this one, this one, as we would like, any values, any frequency
values and solidus curve. Now what will happen if that, if this bandwidths becomes
smaller like this? So the half frequency, e.g. it will be, the bandwidth
will be smaller. The bandwidth will be smaller. So the number of
frequencies which we can select a form
is much smaller. It means that we
should be higher. We should have
higher selectivity in a chosing our frequency. So as an example, if we are
having this bandwidths, hundred values of
frequencies, hundred values, then if this bandwidth
becomes smaller like this, it means that we will have e.g. a. T values. So it means we
need to be more selective. In choosing our frequency in order to be located
inside the bandwidths. Now is the shape of this
curve or the bandwidths. And all of this depends
on the RLC circuit, the elements of the RLC circuit. So e.g. if the resistance is smaller with a fixed
inductance and capacitance, the bandwidth will decrease
and selectivity increases. So what does this mean? As you can see, we have
three carbons, 12.3. This recovers have R1, R2, R3. R3 is greater than
or greater than R1. So as you can see, the
higher the resistance, look at our three
curve, this curve. And look at R1. R1. You will find that the
bandwidths of R1 at a smaller resistance formula, that bandwidth is very small. However, in a very
large resistance, you will find that the
bandwidth is much greater. So that's why the resistance
can shape our bandwidths. So as you can see here, Windsor, resistance is a smaller, such as R1 is the
bandwidth is smaller. Bandwidth decreases and it
means that we should be more selective in
our circuit, okay? Or in our frequency. Also, if the ratio of L over C increases with a
fixed resistance, the bandwidth will also
decrease with an increase in selectivity. So
as you can see, e.g. this circuit
representing L1, L2, L1 over C1 and C2, C2 LLC versus three. So as you can see, if this
ratio with a fixed resistance, it changes, the bandwidth,
will it change? So as you can see, e.g. in L3 and L3, you
will see that here. This one becomes more tight. Okay? So as this ratio increases, a bandwidths will
become smaller, which means that we need, we will have an increase
in selectivity. So as you can see,
bandwidth is three, band width to two. And bandwidth one here,
from here to here. Okay? Okay. So as you can see in terms
of QS or the quality factor, if you remember that the Q S is equal to x L over
R. So as you are, if you look at
this circuit, e.g. when we have larger resistance, R3 increases, or the
resistance increases, higher bandwidths,
higher bandwidths. And at same time
when r increases, the quality factor decreases. Okay? So as you can see,
r is larger for the same excel is NQS is less. Okay? Now, small q s, small QS is associated
with a resonant can have a large bandwidth
and small selectivity, while a larger Q
indicates the opposite y. As you can see here. That e.g. high resistance, lower QS, but higher resistance here, e.g. it gives us large bandwidths. Okay? So as you can see,
our increases, the quality factor
Q S decreases, the bandwidth in this figure,
bandwidths increases. So will find that resistance. And the bandwidths iron directly proportional to
charge more resistance, more bandwidth, which
means small selectivity. However, as the
resistance increase, the quality factor decrease. Okay, so small QS means that
we have large bandwidths. Q as a small, larger bandwidths. Similar idea, if
you have high QS, then you will have
low bandwidths. Okay? So as an example, high QS means that r is small. High QS means that
resistance is small. So it means that
resistance is small. It means that
bandwidth is small. Okay? I hope this idea is clear.
126. Derivation of Cutoff Frequencies: Now, how can we get the
cut off frequencies? We would like to
find the equation which are presenting that
cut off frequencies. So as you can see
in this figure, we said before that cut
off frequencies F1 and F2, existence at a current
equal to 0.707 or Emax. So as you can see here, that the current in this
point and this point drops to 0.707 of its Arizona. And a value, which means an increase in
impedance equal to 1/0, 0.707 equals root two primers
are resonant frequency. So what does this even mean? You will see that our Emax
is equal to E over R. And the value of the current
at F1 and F2 is equal to 0.707 Imax. Let's see. So the current here
at F1 and F2 is equal to 0.707 max. Now what is 0.707? 0 point is one over
root 21 over root two, corresponding to
0.707 Imax. Imax. What is the value of Imax? Is E over R, E over R. So we can say, is this equal to E over
root two multiplied by r? So the new current at F1 and F2 is equal to
E over root two r. Now, let's see, compare between these two equations are
Emacs and E are Emacs. Ea over R resistance R I here is equal to
E over root two. So what does the difference? You will find that in the
second equation at F1 and F2, the impedance becomes,
increases by root two. So that resistance was
only impedances only are. Now it becomes a root two. So it increases by root two. That's why is that
dropping the current is corresponding to an increase in Arizona and the frequency
and probably root two. So this is our
impedance at F1 and F2. So root two R will be equal
to all of this equation. So as you can see here, root two are equal two roots, r squared plus X L
minus X c all squared. Now, let's simplify this by taking the square
of the two sides. Squaring square
of root two gives us two square root of r squared. Square root of the square
root gives R outer squared plus x minus
c all squared. Now it takes us one
to the other side, so we'll have r squared
equal to x L minus X c. Okay? Now, taking the
square root of Z, two sides, square root of this, square root of that, we will have our
equal XL minus XC or r minus x l plus
x equals zero. This one is equal to this one. But we have to know something
which is really important, which will help us to
find F1 and F2. Okay? Now, if you remember
from mathematics, when you have r
square equal to x, l minus x all squared, or you have an x-squared
equals to y square. Now, very important thing
is that when you take the square root of both sides, you will have to
remember that you should add plus, minus. Okay? So it will be x equal
to plus minus y. So x can be all Steve, why? Or can be negative y. Why? Because if you
get the square, X square, if it is positive y, it will be y squared. If it is negative y, it will be also Y squared. Okay? So what do we learn
from this is that x equal to positive negative Y. So it means that
our resistance from this equation will be R will be equal to the negative x minus z. Okay? Now remember this because
we are going to use it now. So as we said now r
equal to plus minus x minus x sin k. So we
have two probabilities. We can have both are
equal to x minus x is c or r equal to
negative XL minus XC. So these two values are and
all should be positive. Of course, there is no
negative resistance. Resistance would
be bolster value. Now, in your opinion, which one of these
is the correct one? The first one or the second one? If you think about it, you can say, Okay, we have here a negative. It means that the resistance
will be negative. So this one is refused and
this one is a correct. Now, let me tell you that this one and this one
are both correct. But in uncertain conditions. Now what a member that we need, we need to get frequency
F2 and frequency f1. Now, remember from
the previous lessons, we said that after the
resonant frequency, we said that what
we said that x is greater than c. We said before, the resonant frequency,
we have x is c greater than Excel. Okay? So if we need frequency F2, it means our x l is
greater than accessing. And if we need F1, it means x is greater than XOR. So let's look at F2. F2, in this case, Excel greater than ecstasy. So if we choose this
equation or equal to x, L minus x is x0, is, will this be correct? Yes. Because x l
greater than ecstasy, so their difference will
give us a positive value, which means our
resistance is posted. If we look at the
second equation, negative x l minus x is c. Now x greater than z. So this part will be positive, but you will find
here a negative sign. So our resistance by using this equation
would be negative. So this equation is not correct. Okay? So for F2, we will use
the first equation. Okay? Now, what about
F1? What about F1? So an F1, we have x is
c greater than x L. So can we use the
first equation? X is z greater than Excel. So their difference will give us negative XL minus XC will
give us a negative value. So this equation is
not applicable for F1. What about this one? Xl minus xc gives us
a negative value. So negative multiplied
by negative gives us a positive value. It means this one can be used. So what do we learn from here? If I would like, if I would like to get F2 or omega two frequency
to after resonant, then I'm going to use
are equal xl minus xc. If I would like to get F1, then I will use r equal
a negative XL minus XC. Remember this? Now, let's see what will happen. So we need now F1 x L greater than c. So we are
talking about F2, okay? So XL greater than Xs and
we are going to use are equal x l minus accessing, okay? Because it will give
us a positive value, which is a positive resistance, which is related to
F2 or omega two. So you can see are
equal xl minus xc. Take x or minus x plus x. See Excel is omega
n omega L plus one over omega sine omega two and omega two because
we are talking about F2. Now multiplies this to
equate these two sides, the pi omega two. So it will be R omega two minus omega two square l
plus one over C. So we multiplied by omega two to eliminate
the omega two here. Now we have here our omega two minus
omega two square l plus one over c.
Now what we need is that we need to write
in this form a x squared plus bx plus
c equal to zero. Okay? So you can say negative
L omega two squared. You can say negative two, negative L omega
two square plus B, which is Omega to R, omega r omega two Zynga c is
one over c equal to zero. So you can say that a
is equal to negative l, b is equal to r, and c is equal to one over sin. Or you can simply take
this equation and divide by negative l so we can
eliminate the spot. Okay? So what we did is
that we divide it by negative l. So when we
divide by negative L, we will have omega two
square minus R over L, omega two minus one
over LC equal to zero. So we have now a quadratic
equation of the second degree. Quadratic is a second degree. So how can we solve this? We know that in a
second the degree e.g. x will be negative b plus minus root b squared minus
four ac divided by two, ac divided by two. Okay? Not to AC to a only. So as you can see, omega two, which is our variable, will be negative b. B is negative R over L
plus minus root b squared. B is negative R over L
from the mathematics, minus four ac minus
four multiplied by a, which is one, multiplied by c, which is negative one over LC, negative one over lc divided
by two multiplied by a. Okay? So when we have this
equation and simplify it, you will have this
final omega two. Omega two is equal to R over two L plus minus half root r squared over L squared
plus four over LC. Now, as you can see, we have out over two
L omega should be, what should be bolstered. Let's see z negative one
as the negative side. You can see all over two
L Now, plus, minus half. Now we can see r-squared over L squared
plus four over LC. Now, for now, in order to understand
which one is correct, Let's neglect to this part. Okay? So you can see root r squared over L squared
gives us root R over L. So you can see that this
term is equal to this term. Now, what happens if
we add four over LC? If we add it for over lc, It means that this term will
be higher than the stone. Okay? So what I mean is that this square root
is always greater than all over two n.
So we cannot have a negative sign because it will give us
negative frequency. Again, this part
greater than this part. So this part minus this part will give us
a negative frequency. So we cannot use
a negative sign. So we'll have R over two l plus half root r squared over L squared plus four
over n z, like this. So we cannot have a
negative sign, okay? So the frequency will be
omega divided by 1/2 pi, okay? So this is F2. Now what we can do to get
F1 the same procedure, but what is the difference? Remember, to get F1. R will be equal to
NOT x l minus x, it will be negative XL minus XC. Where did we get the spot? Remember from our
square equal to x L minus X c all squared. So we'll have F1 will be
equal to this final equation. So it will be similar to f two. But the difference is that
you will find that here. We have a negative sign here. Okay? Is that as the only
difference between f one and f two. Okay? Now we would like to
get the bandwidth. Now as we see from this figure, bandwidth is the
frequency f2 minus f1. F2 minus f1. So if you subtract
these two equations, we will have all over two pi L. If you subtract this
one from this one, you will have all
of our two by N. Okay? Now, you can relate this one to Zach quality
factor like this. How did we get this? It is really, really easy. So as you can see
bandwidths, Okay? You can see this derivation
or you can look at me and L. Bandwidth is equal
to R over two by L. Okay? So let's say I multiplied
here by the frequency, resonant frequency, and the multiplied here by the
resonant frequency. So we will have or multiplied
by the resonant frequency. And the two pi FSL
gives us excel. At resonance. We have our Excel. So we remember that
the quality factor is equal to x L over R. So out of our excel
gives us one over q S. Okay? Do we have here F S. So the bandwidth
equal to fs over QS. As you can see. Now, we have now a relation
between bandwidth and QS. And as we remember from
the previous slides, we said that as
the quality factor increases the bandwidth
to decrease the. Now we prove this using the Duration of F1 and F2. Now we can have
another equation which is the fractional bandwidth, which is f2 minus f1 over F s. So if we have the bandwidth
equal to f2 minus f1 equal to f2 minus f1. F2 minus f1 equal
to fs over q s. So from this equation, you can find that Q S is
equal to fs over F2 minus F1, or one over Q. One over Q S is equal
to f2 minus f1 over fs, which is called the
fractional bandwidth. Rewind the indication
of the width of the bandwidth compared to
the resonant frequency. So it provides a
width of this part. How, how much C02 hurt us? Or megahertz this with respect to a czar
resonant frequency. Okay? Now it's a financing in this lesson is a
voltage of plotting. Now remember that the voltage VL equal to x L
multiplied by current. And VC is equal to z
multiplied by the current. And R. The R is equal to the resistance multiplied
by the current. Each reactants and multiply by current and the resistance
multiplied by the current. So now we have
first this curve of the voltage current
and the meaning of the properties with
respect to the frequency. So you can see this current, this is the curve
of the current. The first one is the
curve of the current. So in order to get VR, we multiply this curve. By what way does this
dance are constant value. So we will have this
other curve, VR. Okay? Now second part, we have V L and V c. Now at
resonance, at resonance, you will find that this value, these two curves of VC and Vn
intersect with each other, which means they are equal
to each other at resonance. Now, another property,
another property we'll find is that here, this is vc and this one is v. Now before resonance
here we have our resonance. Before resonance, we said that
x is c greater than Excel. And after resonance,
x greater than c. So what does this mean? And this ends the forest, the case when x is
greater than x l, it means that V c
greater than Va. So if you look at this figure, you will find that V
C is higher than Vl. This one is V L, and this one is Vc. So at any frequency, you will find that in
VC is always greater than V. After resonance, Excel greater than x, z, so that v l is always
greater than VC. And at resonance, the anther sector because they are
equal to each. Awesome. Okay? So this is what does
this curve amine. So add a zero frequency. What is the value of the
voltage at zero frequency? At zero frequency, X l
will be equal to zero. So V L will be equal to zero. So as you can see, the curve
of v l gives us here zero. Okay? And what about VC? X is z is equal to
one over omega C. When omega equal to zero, it means that this
part will be infinity. So Vc will be infinity. Theoretically. Theory. However, you have to know that, of course we can not have
an infinity voltage. The maximum voltage is the
supply voltage, which is e. So add a resonance at zero frequency and resonance
at zero frequency. Vc will be equal to the supply. All the voltages go to here because we don't have any cards. Now, the same figures
when the quality becomes, quality factors becomes equal
to or greater than ten. Equal to or greater than ten. In this case, you
will find that V R and the current I
almost the same. If you look at VC and v l, you will find that it becomes
much closer to each other. Starting from F1 and F2. They are becoming very, very close to each Rs converge. They are as if they
are becoming very, very close to each other
and Z becomes higher. Okay? So in this lesson, we learned about bandwidths, the selectivity curve
out Select F1 and F2, and more properties apart with the bandwidth and the
cutoff frequency.
127. Example 1 on Series Resonant Circuit: So let's say I have
some examples on the series resonant circuit. In order to understand how can we get the values
of the voltage, bandwidths, the half power
frequencies, and so on. So let's just start
with the first example. So the first example is that we have a series
resonant circuit. As you can see here, we need to find the current. The voltage across the resistor is the voltage
across the inductor. The voltage across the
capacitor at resonance. We need also to find the
quality factor of the circuit. We have. Given also we have the
resonant frequency, which is 5,000 turtles. We need to find the bandwidths. And we need to find
the power dissipated in the circuit at the
half power frequencies. Okay? It seems that we have
a lots of requirement, but it is really, really easy. Easier than using forests. You have this circuit,
we have a supply voltage equal ten volt and angle zero. We have a resistance equal to x L equal ten ohm and
x equal turned on. You can see in the circuits
at x L equal to Ecstasy. Okay? Both of these reactants
are equal to each other, which means we are in the resin and restaurants or
in czarist state. Okay, So let's just start. So first we need to find
the current at resonance. Okay? So we know that the
current at resonance, at resonance in the
series resonant circuit, we know that our circuit
will be a pure resistor. We know that the
current is equal to the supply divided by the resistance because the two reactants
cancel each other. So the first
requirement is that the total at resonance is
equal to resistance, only pure resistive circuit equal to the two ohm resistor. The current will be E, which is our supply ten
volt and angle zero divided by the total which
are two and angle zero. So it will give us five
and there and angle z. This is the current
flowing in our circuit. Now second requirement is
that we need the voltage VR, the voltage across
that resistor. Okay? So as we know is
that the voltage across the resistor
is a restore, is equal to the current
flowing through it multiplied by the resistance. So as you can see, is that you will find that the
voltage across the resistor at resonance
is equal to the supply. Okay? So as you can see
here, in this equation, I equals E over R. And the voltage across
that resistor is equal to i multiplied
by the resistor. So it gives us E,
which is our supply. So as you can see here, the voltage across the resistor
at resonance is equal to the supply voltage equal to ten volt and angle zero
similar to the supply. Now so the requirement is the voltage across
the inductance. Okay? So the voltage
across the inductance, we have x L equal ten on, okay? So the voltage across
the inductance is equal to the current. Multiply it by the current
flowing through the inductor, which is I multiplied by one, multiplied by x L,
which is ten on. So as you can see here,
we have the current vl is equal to the current
multiplied by Excel. Okay? So the current is equal
to I and angle zero, XL is x l. And the angle mighty
because it is J Excel, if you remember, it is j X L. That's why we have
an angle mind. So X L is ten. Zac current is equal
to five and bear. So five multiplied
by ten gives us 50 volt and the angle 90 degree. Okay? Now what is the voltage
across that capacitor? Okay, it's a voltage
across the capacitor is the current flowing
through the capacitor multiplied by x is c. So the
voltage is c is equal to I, multiply it by ecstasy. Ecstasy is negative wget
accessing in a resistor. The impedance of the circuit. Negative z means
negative 90 degree. So we will have lights this
current oil and angle zero. X is c angle equaled
negative 90 degree. So it will give us 50 volt and the angle negative
nine degrees. So as you can see here,
it is pretty obvious that Vl at resonance is equal to VC because they have
the same impedance, same current flowing through
them, and same impedance. So they have the same
magnitude of the voltage, 50 volt, 54, 50 v. But the difference
is that we have a phase shift between them. Now was as sort of
the requirement or the second requirement is Q as the quality
factor of circuit. Now we know that the
quality factor is equal to x l over r. So x l is equal to ten ohm. Resistance is equal to
two, as you can see here. So it will give us five. Now, the requirement is that if the resonant frequencies
5,000 tortoise find ZAB bandwidths and
the power dissipation at hover half power
frequencies. So forth. Start with the bandwidths, Okay? We know that the band
width is equal to that fs over Q S equation which we obtained in the lesson. The resonant frequency,
which is 5,000, divided by Q, S, which is five. So it will give us
1,000. So let's see. As you can see, f2 minus
f1 equals F S over Q S, which is 5000/5,
gives us 1,000 hz. Now finally, we need the power dissipated at half
power frequency. So we know that the power at how far frequency you
can see half power. So it means that P, I'll talk about frequencies, is equal to half V-max. How V-max? That's why they
are called half power. So as the hall B max, the maximum power is
power at resonance, which is I max at resonance, square multiplied by a resistor. The current squared times the current at resonance
is five and there, and the resistance
is equal to two. So five squared gives us 25 multiplied by
two gives us 50. I think it will give us 25, 25%. So as you can see,
half V-max have I maximum squared r. This
is a power at resonance. So the square of the current, our current is five and bear. So the square of this current multiplied by the resistance, which is 2 ω, gives us
the power at resonance. Our at resonance
multiplied by half, half this power gives us the
power at all frequencies. Remember the how frequencies are the frequencies
of the bandwidths. Here. Okay? Here we have the bandwidth e.g. then this from here, while we're at this point
and this point is add 0.707. Next, if you remember, at F1 and F2, F2 and F1. Okay? Now one thing to
notice here is that we said before is that the, one of the equations
which you can learn is that the voltage
across the inductor, the voltage across
the capacitor at resonance is equal to V, L equal vc as a magnitude equal to the quality factor
multiplied bys or supply. If you remember this. So the quality factor is given
as what we obtained it as five multiplied by the
supply voltage is ten volt. So it will give us 50 volt. So as you can see here,
we have 50 volt, a volt. So our solution is correct. Okay?
128. Example 2 on Series Resonant Circuit: Now let's have another example on the series resonant circuit. So we have here is
our bandwidth of the series resonant
circuit is 400. So this is our bandwidths. Number one, we have also
the resonant frequency for yourselves on this, we need to find the Q, S or the quality factor. Second requirement, if that
resistance equal to ten oh, we forgot to ohm here. What is the value of
X L at resonance? That certain requirement if
the resonant frequency is 5,000 tortoise,
find the bandwidth. Then the force requirement finds the inductance L and
C of the circuit. Let's first, by our
requirements are givens. And the requirements was
that given is that we have bandwidths equal to 400 hz. We have resonant frequency. So this one is bandwidths. This one is F S. And we
have resistance than on. Now, the first requirement is the quality factor Q S. If you remember that the quality
factor Q S or not, the quality factor
first is a bandwidth. Bandwidth is equal to
f s over QS, right? That resonant frequency
over Zach quality factor. So from this equation,
the quality factor, which is needed equal to the frequency divided
by bandwidths. Resonant frequency is
given as 4,000. Tortoise. Bandwidth is given as 400 times. So it will be like this. It will give us, then. Second requirement is that
we need excel at resonance. Now, again, since
we obtained Q S, which is equal to ten, now we know that Q S itself
is equal to x L over R. Okay? X L over R equal to ten. Now, the resistance
is given as ten also. Then, from here
we can get Excel. So we have Q S equals x, L over R equal to ten, and the resistance
is given as ten. So from we have only one
unknown, which is Excel. So Excel will be 100. Okay? As you can see here, 100 ω. So the requirement is
that the bandwidth, the bandwidth is already
given as 400 hz, okay? As I said of the requirement is inductance and capacitance
of the circuit. So first, we have Excel. We have Excel. So Excel is equal to two multiplied by the
frequency at resonance, multiplied by the inductance. So frequency is 4,000 turtles. And the only unknown here
is L, the inductance. So as you can see here,
x L equal to pi FSL. So the L would be X
L over two pi f s. So XL is hundred frequencies
for ourselves on tortoise. So we will have
3.298 milli Henry. Okay? This is our inductance. Now the final requirement
is that capacitance. So it is really, really easy. Remember that frequency is
equal to 1/2 pi root LC. So the frequency
is 4,000 tortoise inductance obtained
3.98 millihenry. So we have only one unknown, which is the capacitance. Like this. We have this first equation, which is f x equal
1/2 pi root LC. Or you can obtain the same idea. You can obtain it from excess. See, you all know that
x L is equal to Xc. Excel liquid ecstasy. Ecstasy will be
equal to hundred, hundred all equal
to 1/2 pi f S C. And FCFS is 4,000 turtles. So from here, we can
get the capacitor. So this equation is correct and this equation
is also correct. Both of these solutions, solutions are correct
for this example, okay?
129. Example 3 on Series Resonant Circuit: Now let's have another example. We have a series RLC circuit, which have a resonant
frequency of 12,000 hz. Number one, we need to
find the resistance equal 5 ω and X l equal zero
hundred ohms at resonance, we need to find that band width. Second requirement, we need to find that cutoff frequency. Okay? Forest, as you can see here, that requirement we
have in this equation that will hurt us 12,000 thirds, which is F S. That
is 1.2 frequency. Resistance equal five
ohm and x equals 300. Now, the bandwidth, how
can we get the bandwidth? Remember that the bandwidth
equal to f s over q s. So that frequency itself
is equal to 12,000. This one is 12,000. As you can see here, the
quality factor assembly equal to x over r. So x l is zero hundred
and the resistance equal five or so. From this we can get the
quality factor and from which we can get the bandwidths.
As you can see here. So far as the quality
factor equal x over r x l zero hundred divided by
five, as you can see here. So the quality factor is 60. So the bandwidth is equal to the frequency 12-hour
does divided by 60. Like this. Now we need to find
the cutoff frequency. Now we remember that
we have two equations. We can get F1 and F2, which is the cut off
frequency by using Zoloft equation which we
have discussed, the pure. If you don't remember it. Let's get back quickly here. If you remember here. This part, you can see here f one and f two, these
two equations. We have R over L, we have that resistance, we have the inductance. We can get the capacitance since Excel equal ecstasy
at resonance. Then from here we can
get the F1 and F2. Okay? This is the direct
method, is a method. There is another method which
is much easier, which is, and here if you look
at this example, you will notice something which is really, really important. You will notice that
the quality factor, quality factor is equal to 60. S t is greater than ten. Remember that we
have a special case. When the quality factor
becomes greater than ten, we have this curve. Now, this is why this
example is important. Z will find that
here is that F1 and F2 are symmetric around
the resonant frequency. We said before that the distance
from here and this one, so from here at the curve of the current or the curve of the impedance
or not equal to each other. It is not symmetric curve. When does the quality
factor becomes equal to ten or
greater than ten? You will find that it becomes
more symmetric around the resonant frequency.
So notice e.g. is that this here, from resonant frequency to F2, this distance is equal to the distance from resonant
frequency to F1 here. So this distance equal
to this distance. Now, as you know that from F1, F2, all of this is
called the bandwidths. Okay? So all of this region
is bandwidths. So this part alone is
bandwidths over to this part is bandwidths over two because it is symmetric around the
resonant frequency. Now what is the value F2? So we have frequency
fs and we have this distance which is
bandwidth over two. So F2 will be equal to that resonant frequency plus the small distance
bandwidth over two. And this part will
be F1 will be equal to resonant frequency minus. This is small distance
bandwidth over two. So the distance here is
equal to the distance here, bandwidth, bandwidth over two. So from resonant
frequency to get F2, add bandwidths over two to get F1, subtract bandwidth overtone. So as you can see, is a
bandwidth is bisected by fs. Therefore, you will find
that F1 and F2, F2, f's frequency bandwidth
over to F1 is resonant frequency
minus bandwidths over two. As you can see here. So we'll have F2
equal to 812,100 hz and if one equals 11,900 hz. So I hope this example was
also helpful for you to understand more about the
series resonant circuit.
130. Example 4 on Series Resonant Circuit: Now, let's have a
Mozart example. Now what determines Q, S and the bandwidth
for that response? So we have this curve, response curve representing
the relation between the current and the frequency. Okay? Now, in this one we need to
find the quality factor. We need to find the bandwidths. We have the capacitance
equal 100,101.5 nanofarad. We need to find the L
and all of that circuit. And also finally, we need to
find the applied voltage. Okay? So here we have the
given capacitance equal 101.5 nanofarads. Leave this part for now. Now photos, do we need that quality factor
and the bandwidth? So if you look at this figure, we have some important
points which we can get. Now we know that the current, current is maximum at which
he stayed at resonance. So as you can see, this point is the maximum current, okay? This point which is corresponding
to this frequency here. Okay? So this, this point
at which we have maximum current or emax
equal to 100 million, bear 200 milli and beer. We have that resonant frequency. Resonant frequency
at this point. Now what does the value of this point assembly we have here 12 to 12,000. We have 3,000. Okay. And how many lines? 123-45-6789 and finally ten. So you will find the ten spaces. So this base between
these two or 3,000 -2000/10 gives us what? Oneself divided by
ten gives us hmm. Okay? So each one of these, so this 1200021002200 and so on. So here's 3,000, 2000, 902,800. So this part is 2,800 hz. This is a resonant frequency, the spot at which we
have maximum current. So that's the first
thing we need. Second part we need is that
we can get the bandwidth. The bandwidth is
equal to f2 minus f1. And F2 and F1 z are the
half power frequencies, cut off frequencies or half power frequencies,
half power. So remember that we
said before that Zack, current at half
power frequencies is equal to current at
all frequencies at F2 and F1 is equal to 0.707
multiplied by our Emax. So imax here is 200 milli and bear Here we'll be having a milliampere,
okay, milliampere. So 0.707 multiplied by 200
equal to 141 milli and bear. Okay? So, uh, why? Because
we said before in the curve at 0.707 is
one over root two. Okay? So when we square this, we will have half power. Now, in order to get F2 and F1, we will go to this curve at 141. So 104 to one, we can say is
at here at this point, e.g. if you go to the curve, you will find this point. And you'll find this point. This two points, this one
and this one representing, one representing is a
half cut frequencies. Okay? So if we go down here, you will find this point. And this point. Okay? So this one is 2,800, this one is 2000, 2,900, and this one is 2,700. Okay? So what does that bandwidth? Bandwidth is the difference
between these 22900 -2,700 or the gap between them, which will be 200 hz. Okay? So now we have bandwidths. We have our Emax, we have resonant frequency. So can we get que es, que es assembly equal to or to make it more
specific bandwidth equal to the resonant
frequency divided by Q. So Q S will be F S
divided by the bandwidth. Okay? So F S is 2,800
bandwidths, 200 hz. So from here, we can get our quality factor,
fs over bandwidth. Bandwidth 200 hz as we
get F S is 2,800, okay? Okay, as you can see here, now we need to find the
inductance and the resistance. So we know that at
resonance, at resonance, we know that x L equal to z or the resonant frequency
equal to 1/2 pi root LC. Resonant frequency is 2,800. Capacitance is 101.5 nanofarad. We have only one unknown, which is the
inductance, like this. So the inductance will be
equal to 31.83, mainly Henry. Now that the requirement
or the force requirement, whatever it is, it
is a resistance. So how can we get the
resistance really easy? We know that the quality
factor Q S equal to x L over R. Now x L is equal to two pi multiplied by zero is
one and the frequency multiplied by the inductance
which we just obtained. And the quality factor equal 14. So from here we can get the resistor q as
equal x L over R, or equal to x L over
QS from this equation. So we'll finally have
our resistance of 40. Now as a final requirement
is the applied voltage. We need to find the value
of the applied voltage. We know that again, at the resonance, at resonance, the current which is 200
million there at resonance, is equal to what we
said that our circuit is a resistive circuit. So it will be E over R. Resistance is equal to 40
own current 200 million bed. So e, which is our supply, will be equal to
200 million bear multiplied bys or for t. So it will give us eight
volt, I think, involved. Yeah. Okay. So as you can see, all what
we are going to do is that we are using the equations
of the bandwidth, the quality factor, the
resonant frequency, all of this to obtain the requirements
inside our problem.
131. Example 5 on Series Resonant Circuit: Now let's have another example. In this example, we have a series RLC circuit
designed to resonate. Resonate at omega S equals ten to the power five
radian per second. It has a bandwidth of 0.15
omega S. And draw with 16 from 120 volt supply at resonance, finds
the resistance. The bandwidth in hardest, L and C is the quality factor
as a fractional bandwidth. So let's first see how can
we solve this problem. As you can see, we have
a series RLC circuit. We have omega S equal ten to the power five
radian per second. The bandwidths 0.15,
16 what from 120 volt. Now, let's start
step-by-step resistance. How can we get the resistance? Resistance is
really, really easy. You'll notice that
we have our power, our power consumed by the
resistance, 16 watt, okay? Because at resonance, all
the power is voltage. In normal state. It goes to the, to
the resistance. So 16 what is consumed? Consumed from 120
volt at resonance. The power consumed inside
the resistor is equal to I squared multiplied
by the resistance. All right? Or we can say is that
same equation which is v squared over r. So we
have a resistance here. We have a current
flowing through it. And voltage between Act V. The power consumed inside
the resistor is equal to the square of the current
multiplied by the resistance, or the square of the voltage divided by R. They are
the same equation. Now, PR, the power consumed
by the resistor 16 what? Now? The voltage
across the resistor. Now remember that we are
in the resonance estate. So the voltage across the
resistor is equal to E, which is the supply. Since we are in resonance. So Zara supply is 120 v. So V square will be 120 square. So we have 16 watt equal to 120 square divided
by the resistance R. So from this equation, we can get that resistance
equal to 900 ω. Okay? That's the first requirement. Second requirement
is the bandwidth. Bandwidth z given
as 0.15 omega S. So simply omega S here is the resonant
frequency in radian, radian per second, or the angular frequency to
be more specific. So in order to convert
this to fs assembly, we have omega S equal to two multiplied by
the frequency fs. Fs will be equal to omega
S divided by two pi. So that resonant frequency is this value divided by two pi. From here we have bandwidths, bandwidths equal to 0.15 F s. If you would like it in radian, it will be 0.15 omega S, If you would like it in hertz. This as required in the problem. Remember here, we need bandwidth in hertz does not
radian per second. So you need to convert this
to hertz, as you can see. Okay? So as you can see here,
F S is equal to omegas over two pi and bandwidth equal 0.15 frequency
at resonance. Okay? Okay, so now we have
our bandwidths. Now is we have, as you can see, f, s
and the bandwidth. What does this helps you to get, it helps you to get QS. So bandwidths equal
to fs over us. So we have F S and
we have bandwidths. So you can get the
quality factor. Okay, so this is an
easy requirement. Now, L and C, you have the
quality factor Q S equal to x over r, x over r. So the quality factor I now
obtained and we have Excel. Do we have Excel? No,
we don't have Excel. It is two Pi f S multiplied by inductance
divided by the resistance. Resistance is. 900 ω. So we can get
the inductance L. And we know that at
resonance XL equal x is c. So we can get that capacitance. Or the frequency at resonance
is equal to 1/2 pi root LC. All of them will lead
to the same way. Another equation, and instead of getting quality factor first, what we can do assembly that. We can say that we can
get the inductance first. How? Remember that the
bandwidths is equal to fs over Q and Q S is
equal to f s here, the quality factor is x over r, X L over R. So we can put the R here. And x L is equal to two pi multiplied by the
frequency multiplied by l. So the frequency of goes
with the frequency we can have all over two, all over two by n, which is the original equation,
if you remember. So we can use this
equation like this. Bandwidth equal r over two pi L. The inductance,
we have bandwidth, we have resistance,
so we can get the inductance and
the capacitance. It can be obtained
from the frequency equal 1/2 pi root LC. We have L, we have
the frequency, so we can get the capacitance. Quality factor is x L over R. As you can see, x L over R, We have Excel, we
have resistance, so we can get, the
final requirement is a fractional bandwidth. If you don't remember, it, is F2 minus F1 divided by
the resonant frequency. So f2 minus f1 is what is
the bandwidth divided by S? Now, if you remember
what this even, I will tell you right now. Remember that bandwidth
is equal to fs over Q. The quality factor Q is
equal to F S over bandwidth. Bandwidth over F s. You know that quality
factor F S over bandwidth, bandwidth over F s is
the reverse one over Q. So we'll have legs. So F2 minus F1 divided by
that resonant frequency, which is a fractional bandwidth, equal bandwidths over fs, which is one over QS. One over Q SQS is
given as here, 6.67. So we're going to get 0.15 as
a fraction and bandwidths. Okay? So what do we learn from here? There are different methods to obtain all of these values. Okay? This is one of them
methods you can use other muscle that I just explained in order
to obtain the same idea. In the end, all of these laws will lead you to
the same solution. Okay?
132. Parallel Resonant Circuit: Hey everyone, In the
previous lessons, we discussed a series
resonant circuit and we had several examples on it. Now in this lesson
and the next lesson, we are going to discuss resonance circuit and we are going to have
some examples on it. So first, let's remember
that series circuit. We know that in the series
resonant circuit we have R, L, and C series with
unemployed voltage source, as you can see here. This is a series
resonant circuit. We have a supply source, we have a resistance
inductance stance all in series or parallel
resonant circuit. It is a same configuration. We have the interpreter
or in the C combination, z are parallel with an
applied current source. So as you can see here, we have a voltage source in
series with R, L, and C. Now, in that parallel
resonant circuit, we have a current
source and we have resistance battery to an
inductance, better to capacity. Okay? So in this case, we have four elements parallel to each other with a current source. Here we have four elements, citizens Chaucer In
a voltage source in the sea it as a resonant
circuit, okay, at resonance. So we have what maximum? We have a maximum current. Maximum. If you remember, we have a voltage source and the current becomes a
maximum at resonance. Here, we would like to
make the voltage maximum. So at resonance, the
voltage is maximum. Okay? So first, let's discuss
two important circuits, which is the ideal and the non-ideal parallel
resonant circuit. And we will understand
which one should we use. So for us that we have an ideal parallel resonant circuit. And what does this mean? First, you will see that
we have a current source. We have a resistance
parallel to it. We have an inductance, we have a capacitor. So what does it mean? It means that the inductance does not have any
resistance in series. So if you remember that
the inductor itself is a coil which has
format of wires. Therefore, it should
have a resistance. But in the ideal parallel
resonant circuit, we neglect this
resistance. Okay? So the ideal circuit
does not have a resistor in series
with the inductor. The mom ideal, which
is a practical case, has a resistor in series
with the inductor. So it has a resistance that practical one has
a resistance R, L in series with the inductor. Even though there
are L is pretty small compared to
other resistance R. However, this resistance has a great impact on the
parallel resonant condition. This resistance, we
cannot neglect it. Why? Because it has an impact on the conditions of the
resonant circuit. And as you will learn
in the next lessons, That's why we should
not use the circuit, but we have to use the
practical parallel RLC circuit. So we have a resistance
R, like this. Our battery to an inductor
with resistance or an inductor Excel with our resistance RL and the capacitance ecstasy
or the reactants x, z. So this circuit is the practical one and the
one which we should use. Now, as you can see, that due to the presence of this element,
Let's take it back. You can see that
here we have R, L, and C. So we can say
simply in the circuit, we can say that the
resonant condition occurs when x L
becomes equal to z, or the reactor
power becomes zero, or we have a unity power factor. However, due to the presence of an inductor resistance R
L inside the inductance. This effect, this
resonant condition. Now we cannot say
that the resonance occurs when x L
becomes equal to xy. Why? Because we have a resistance R, L and C it as with
the inductance. So this will affect the resonant condition
of our circuit. So what we have to do is that we would like to convert this for Excel to add two
parallel components, R and L, like this. So I would like to
convert our L series with Excel into our parallel, RB, which is our parallel
and XL are parallel. Okay, So we will
convert this for these two series components
and 22 parallel elements. So how can we do this? First? You can see that this element is
equal to RL plus j XL. This is a z of the
inductance with its own resistance,
RL Zhe Excel. Now, if you convert this to admittance Y, it
will be like this. Y. Or the admittance of this circuit is equal
to one over your nose. At the admittance is the
inverse of the impedance. So y equal to one over z equal
to one over r l plus j XL. Now one over r l plus j XL can be divided into these
two components. Okay? So how do we do this? Simply as you can see, we have one over r l, like this, one
over r l b plus j. Excel like this. Okay? Now you can multiply
numerator and denominator by the conjugate of
this complex number. So our logic cell is
a complex number. So we can convert this
into Another form, a by multiplying
by the conjugate, the conjugate assembly RL
minus j XL conjugate is the reverse of this sine
divided by R L minus J exon. So if I multiply here
by RL Manage Excel, and the multiply here by
R plus j xa minus j XL. We didn't need to do
anything because we can cancel this two with HR's. Now, if we multiply
them like it's rather our elbows x l
multiplied by RL Manage Excel. It will give us our L squared minus J x squared, x squared. However, since we
have J and J square, it will become plus. So this multiplication,
this part, with this part gives us R
squared plus XL square. And we have our l minus j XL. Have to know that this
form, or L minus x, L over R squared
plus XL squared is similar to one over
r l Jacque say, okay, it is a same formula. Now you can see that we have our Manage Excel
divided by this. So we can convert this into RL divided by RL squared plus j plus x L squared
minus j XL, okay? Divided by RL squared
plus x squared. So as you can see, RL divided by RL squared
plus x squared and the minus x l minus j X L divided by R
squared plus x squared. So simply we converted this for N22 elements and they will
understand why did we do this. So we're, as you can see, we have RL over
RL squared plus x squared minus j X L over R
n squared plus n squared. Now, we can do something else. What is it? You can
see that here we have RL over RL squared
plus x squared. So we can take this one here, now in here, like this. So we can put this down
here by making a division. Okay? So or, or to be more, much easier, you can divide here by RL and
divide here by RA. So we'll have one divided by RL squared plus
x squared over r. This formula similar
to this formula. Similar here we will divide by x l and divide here by XOR. So we will have RL squared plus x squared
divided by Excellent. Okay? And this is j. Negative j can be boating
down here as j plus j. Okay? Now why is this? Because if you
remember that one over j gives us negative g, The negative j here, it's converted into one over
j with a positive sign. So one over j,
which is this one, is originally negative j. Now I know that you will ask
me why we do all of this. You will understand that
now we have here one over something plus one
over j something. Okay? So we can say one over. This is a real part. This is an imaginary part. So you can say one over. So we have real part
and imaginary part. The real part is rp. So we can say one over r
p plus one over J x be. Okay, this is equal
to admittance. So from this we can
have RP equal to this element and XL be equal
to this element like this. So RP or the parallel resistance equal to RL squared
plus L squared over r l and x L p equals r l squared
plus x squared over x. And so as you can see here, what we did, why did
we do all of this? In order to convert
this to series element into two
parallel element. You'll notice that
here, y here, e.g. is one over z, one over z, which is RL plus j XL. For this, what is the
why of this system? Y equal to 1/1 plus
1/2, 1/1 plus 1/2. What does it do? One RP. What is Z2 is j x l. And this two circuits
are equal to each other. So why is this admittance
equal to this admittance? So 1/1 plus x n equals to one over RP
plus one over J x LLP. As you can see here, this
element equal to this one. And from that analysis
between them, we obtained RP and Excel will be equal to this two equations. What do you will notice
here is that our P, which is our, our battery. You can see it is a
function in what? Function in Excel. So the RP is function
in the frequency. So our p is function
in the frequency. Remember, because it is
really, really important. Now as you can see,
the same equations are x plus x L
squared over R L XL. This two equations here. Okay? Now, if we combine all of
this to the original circuit, we will have like this. We have a current source, that total, which is, we have first the resistance or supply, the
supply resistance. Okay? Do we have here
are parallel, okay, which is, which we just obtained
and XL P parallel ends. Our original x is c. Okay? So this was the original, what was this one? Or L plus J exon, we convert into two
parallel elements with ecstasy and parallel to r, s and the current source. Now as you can see, this circuit can be simplified like this. We have RS parallel
to RP like this. So we can say is that these two resistors that can be combined into one resistance, R, which is our supply
butter to RP, likes this. So as you can see, we have
this final circuit which is the resistance R
as patter two are parallel and x LP parallel. Remember Excel parallel nodes, the original Excel, but
Excel parallel, this one. Parallel to XY. So
this circuit is our final circuits that we are going to use in our analysis.
133. Unity Power Factor of a Parallel Resonant Circuit: Now let's discuss the
resonance conditions of a better resonance circuit. So after remember that in this circuit or the
pattern resin uncertainty in the new one. In Xenon idea, we
have a resistance R, which is equal to
RS battery to RP. And if you remember our B
as a function in Excel, which means its function
in the frequency, will find that here, something which is really,
really interesting. If we get back to the
series resonant circuit. If you remember that
the resonant frequency, resonant frequency
is a frequency at which we have impedance minimum. Guarantee becoming maximum. The input impedance becomes
a pure resistive circuit. X will cancel with ecstasy. Answer network will have
a unity power factor. Now, something which is really important is that
you will find that the circuit now in the
battle resonant circuit, you will find that
the condition, condition of that
unity power factor, condition of the
unity power factor, or the pure resistive, is different from the
condition at which we have maximum impedance or
the maximum voltage. So we have in this circuit two conditions, two
resonant frequency. So in the parallel circuit, in the butter second,
what do we need? The impedance, we
need the impedance to be maximum in the
parallel circuit. The total, we need it to
be maximum at resonance. And we need unity power factor. Unity power factor, and we need maximum voltage,
max voltage, Vc. Okay? That's all. Okay, unity power
factor equal to pure resistive, be resistive. Now what you will find
is that in this circuit, the condition at which we
have unity power factor. This is one condition which
is the resonant frequency f b is different from the conditions that
will make impedance maximum leading to maximum VCE is this is another
frequency called FM. So these two frequencies are
different from each other. They are not the same. Unlike the resonant frequency
of the series circuit, we have one frequency
which is F S, that satisfy all of
these conditions. Okay? We have one frequency in
the series RLC circuit. And in Paris on circuit
we have two frequencies. One which makes us have a unity power factor or a
pure resistive circuit. Another one which makes the impedance maximum
leading to maximum voltage. Okay? So funds that we need
to do to analysis in order to get this two values. The first one which
we would like to do is a unity power
factor condition. So the unity power
factor, if you remember, you only take factorials mean we have a pure resistive circuit. We have a circuit
with only resistance. We don't have any Excel or any excess see
in this circuit. So assembly lets the
first type our circuit, the admittance Y is equal to 1/1 plus one over R2
plus one over x0. X1 is one over r. One over x0, x1, x1 is our one over z. Two is one over J x b x L P, which is the Excel parallel
nodes, the original XL, remember this Excel,
but a real sense, we divide our Excel
into two components, plus one over negative J XC. Z. Number three is one
over negative j xc. Okay? Okay, now, we have
this three elements. So we can combine them together. We have one over R, and we have this
imaginary components. So we can say J, we
can put it here. It becomes negative j. One over j plus one over
z gives us negative Z. And one over negative
z gives us positive j. You can see it will
give us why total or the total admittance
is one over R plus j, one of our ecstasy
minus one over x LP. From what, from the admittance. This is the admittance at
any frequency in general. Okay? Now we need the
frequency at which we have unity power factor. So again, what does the
unity power factor means? It means the reactive
power component, or the imaginary component
is equal to zero. So we have only one over r. This component becomes
what becomes a zero. So as you can see from this
equation, you will find that One over x is c minus one over
x L becomes equal to zero. So x is c will be equal
to x L, like this. So in order to obtain
unity power factor, unity power factor,
the x L P parallel, Excel parallel should
be equal to xy x. Again, parallel node z x L x
L parallel equal to x is c. Now, then we are going to
get x LP parallel, right? Its equation. So we
know that XL parallel equal to RL squared plus
x squared over exon. So take this one and
substitute it here. So we'll have RL squared plus x squared over x L equal to x c. So from here, you can take
excel to the other side. So we will have our L plus X squared equal to z
multiplied by x. Now, x is one over omega C
and X L equal to Omega L. So we can take omega with omega, so we wouldn't have
L over C, Okay? So r l squared plus x squared
is equal to l over c. So x square can be equal to L
over C minus our L squared. Okay? Now what is the next step? The next step is that we
need is a frequency f p. Remember we have
F b and F M, F B, which is a resonance frequency
of the parallel circuit, which will provide this
unity power factor or a pure resistive circuit. Okay? This is a forest
resonant frequency. So Excel is omega L.
So as you can see, x L becomes omega L. This one. First you can say
x L squared equal to x L here from this
equation is equal to the square root of L
over C minus r square. You can see root l over c minus RL squared and x L is equal to two pi multiplied by the
frequency multiplied by L. And since we are talking about with that resonant condition, so it will be FB. So as you can see, it
will be two pi f b, which is this part, equal to the square root of
L over C minus RL squared. From here, take two by
l to the other side. So we will have F p equal 1/2 pi L root l over c
minus RL squared. Okay? So let's delete all of this. Okay? So we have this equation. You can do some simplification
to get finally, that FB is equal
to 1/2 pi root LC, the square root of one
minus r l square c over n. Okay? This equation, similar
to this equation, just the way that some
simplifications, Okay? Now someone will say, okay, why did we convert it from
this form to this form? You will see that here
we would like to reach a certain relation between the parallel resonant circuit and the series resonant circuit. So as you can see
when we converted this from this
form to this form, you will find something
which is really interesting. You can see this element. You will see 1/2 pi root LC. Does this remind
you of anything? Yes. This reminds me of the
series resonant frequency. Resonant frequency
is 1/2 pi root LC. So we can say is that our, our frequency in parallel
equal to the c. That's the resonant frequency
multiplied by the square root of one
minus r l squared, C over L, one minus RL
squared c over n. Now, as you can see from
this condition, you will find that one
minus RL squared c over l gives us something
less than zero. So square root of
something less than zero, we're reach in the end, less than z, less than one, sorry, less than one. So this square root is
always less than one. So anything multiplied
less than one multiplied by F S gives us a
frequency less than fs. So in the end from
this equation, we can note that f p is
always less than f S. Or Z. Apparel resonant
frequency is always less than the seal as
resonant frequency. So as you can see, again, f b is called the resonant frequency of
apparel resonant circuit. What frequency, what
does this frequency do? Resonance frequency. It does do what it makes
is a power factor, unity. We have a unity power factor
or a pure resistive circuit. And F s is the
resonant frequency of the series resonant circuit
when x equal to x is c. Now remember that f p, Where did we get FB when
x was equal to x is C. Okay? Now, unlike the seed is horizontal circuits are
resonant frequency. F v is a function of
the resistance RL. You can see FB is equal to
root one minus r l square. Function of resistance. However, F S was a
function in L and C only it was leading to
depend on the resistance. Now we will find that the
square root components, this component leads is at that particular frequency
becomes a less than that. See it as resonant frequency. Due to the basis of
this square root. Also finds that S or L
as r approaches zero. You will find that f
b becomes a pretty, pretty close to f s
or a process F S. Okay? Why? Because as you can see is that when our l is equal to zero, it means that we will
have brought one, which means that f b
will be equal to fs. If we neglect the resistance RL or the resistance
of the inductor. Okay? So now we discussed this, that unity power
factor condition of a parallel resonant circuit.
134. Maximum Impedance of a Parallel Resonant Circuit: Now let's discuss
another condition which is the maximum impedance of
a parallel resonant circuit. Now we have to know that we said in the previous lessons a unity power factor condition. This frequency at which we
have a unity power factor, is not the frequency at which we will have maximum impedance. It has a different frequency. Why? Because we have a resistance
in series with the inductor, which make the two frequencies
different from each other. So the frequency at
which we will have maximum impedance is called FM. Now, something which
is really important. Why is that maximum
impedance is important. Because the maximum
impedance is which, at which you will have voltage
of the parallel circuit, voltage across the
capacitor or across the inductor or resistance
or the current source. It will be maximum. Why? Because the voltage here, e.g. V B, VB is equal to the
total of this circuit, this part, the total multiplied
by the current source. Okay? So the current
source is constant. Now z is different. So as t increases,
the total increases, the voltage of the
parallel circuit increase. So we need to find
the maximum impedance to find the maximum
output voltage. Similar to in the series
resonant circuit. When we have Z T minimum, we had our maximum at
which we had resonance. That's why we need to find
the maximum impedance. Finds that at
frequency f equal f b, which is a unity power
factor frequency, you will find that, that
impedance is near its maximum, but not the maximum value. Why? Because our
parallel is dependent on the frequency equal to x L squared r l squared plus l
squared over r L squared. So it is dependent
on the exhale, which is dependent
on the frequency. Okay? So now we have another frequency at which
we have maximum impedance. This one is defined by F m n. It is slightly more than FP. So you will find that FFP, e.g. it can be here if m after it. Now, the frequency
fm is determined by differentiating the
general equation of z with respect to the frequency and equating this with
zero to get the frequency. So simply we have, we need that minimum. So we have z is equal
to this circuit, e.g. 1/1 over r plus one over x, LLP plus one over x is c, Okay? Okay, in addition
to the j, okay? So we have here j and do we
have here negative Zhe, okay? So we will get the
magnitude of this. So we have X12 equal
to the square root of one over r square plus
all of this, all squared. Again, as if we are
getting the magnitude. Okay? Then what we are going
to do is that we are going to take that
total equation, dy over d f, Since we are differentiating
with respect to the frequency and the
quizzes equation with z. After doing some analysis, you will get the frequency fm
from this differentiation. Now why did we differentiate
and equate to zero? Because at maximum value, at this maximum values or slope of the line is equal to zero. And the slope of any line is the differentiation
of the function with respect to the variable. So these are over d, f gives us the
slope of the line. And at maximum we
have zero slope. Okay? So of course you are not going
to do this large analysis because it is extensive and it will take
large amount of time. So in the end, we will
have F M equal to f s, the square root of one
-1/4 or L square c over l. So this is a frequency
at which we have maximum impedance leading
to maximum output voltage. So as you can see, if we combine the two
lessons, we have FB, we have f m function, of course, in fs, fs root one minus
r square c over l. You will find that if you
compare this two equations, find in the enzyme, the frequency of the
series RLC circuit is larger than the
frequency at which we have maximum mean that's larger
than the frequency at which we have resonant condition
of unity power factor. Now of course, as you can see, if our l becomes zero, if R L becomes zero, then F S will be equal to
f m will be equal to f t. So the only problem in Paris
on circuit being different from the series of
resonant circuit is the patience of RL or the
resistance of this coin. Now, you can see
that when we get m, the network can be used to get the magnitude and phase angle of the total impedance at resonance by substituting
at f equal FM. And performing the calculations, you will find that
Z2 at maximum or say total maximum equal
to R parallel to x L. Lateral to accessing. This three components are
parallel to each other and substitute with a
frequency equal to F n. E.g. z will be one
over omega c. So omega will be two pi
multiplied by f m. In the end. This will give us the maximum
impedance in our circuit. So now we learned about with
Jack maximum impedance, how can we get the frequency of the maximum impedance of a
parallel resonant circuit. So we have two frequencies. Again. If B at which we have
unity power factor and FM at which we have
maximum impedance.
135. Quality Factor of a Parallel Resonant Circuit: Hey everyone. In this lesson
we are going to discuss the quality factor inside a
parallel resonant circuit. So first, before we discuss
the quality factor we have the total
versus a frequency. So we would like to
see as uploading of the impedance inside the circuit with respect to the frequency. So if you remember from the
previous lesson, we had, or we learned that
the maximum impedance Z T maximum occurs at a
certain frequency called FM, which is a frequency at which
we have maximum impedance. That's why when you upload that total with respect
to the frequency, you will have this curve. Okay? So the total versus
frequency curve clearly reveals as a parallel
resonant circuit gives a maximum impedance
at resonance on like a series resonant circuit which experience minimum impedance
level at resonance. So if you remember the
series had, at resonance, it had minimum impedance in series ends up parallel
resonant circuit. We have maximum
impedance at resonance. You will also find
that z, the total, is approximately equal to r l as a frequency equals
zero, as you can see here, when the frequency
becomes zero or l, or does the total
is approximately r n. Now why is this okay? If you remember that at
frequency equal to zero, you will find that Excel, excel parallel, which it was
R squared plus XL square, as I remember, okay? Because I totally forgot. Here. As you can see,
Excel parallel equal to r inner squared
plus x squared over x. So when the frequency
becomes zero, okay? We have x L P equal to, here is this part will
be equal to zero. This one will be equal to
zero because it is omega L. And omega is equal to zero
since the frequency is zero. So this part is equal to zero. This part equals
zero, so it will be RL squared over zero, which means it will be
equal to infinity. Okay? So x l parallel to what? Infinity at frequency
equal to zero. So let's get back
to our lesson here. Okay? So here we have excel parallel equal to
infinity. What about Xc? Xc is equal to one over omega C. So when the frequency is zero, omega will be equal to zero. So 1/0 gives us also infinity. So this part will be infinity as this part
will be infinity. So infinity. What does this infinity mean? It means that we have an open
circuit and open circuit. So this part is an open circuit. This part is an open circuit. So we will have only R, okay? Now our itself is equal to
our supply parallel to r p. Okay? So Rp, which is the parallel component
of that inductance, and R S is a supply. Remember that the supply
itself is larger resistance, large resistance are parallel. We said before it's
equal to r l squared plus x squared divided
by R L squared. So x l is equal to zero when
the frequency becomes zero. So we have our L
squared over L squared. So this part is equal to this. There is no square here. So our L squared over
r l gives us our n. Okay? So are parallel at a frequency
equal to zero gives us our L. So we have our
supply parallel to RL. Now remember that our
supply is launched. Our L is small value. So when we have a parallel
or larger resistance, butter to a small resistance, Z equivalent will
be approximately equal to the small
resistance or n. That's why at zero frequency, we will have Z2 is
approximately equal to RA. So again, when the
frequency becomes zero, x z becomes infinity, x b becomes infinity. So this tool will
be an open circuit. So we have only resistance or resistance r is
equal to the supply. Parallel to our periphery
are parallel is equal to r l squared plus x
squared over r l. Now, Excel is equal to zero, so zero division gives us R
L. So R L is a small value, or L, which is the
resistance of the coil, is a small value and all
supply is larger value. So they are parallel, gives us approximately the smaller resistance,
which is RL. Okay? So I hope it's clear now. For the parallel circuit as Arizona and the resonance curve of interest is that of the voltage Vc across
the capacitor. Now why is the voltage here is the most important thing
across the capacitor. Because the capacitor
is usually becomes an input to another stage
of the network. Okay? So the capacitor is used as an intermediate stage
between two circuits. Okay? So the output voltage here is, can be taken to another circuit. Now as you can see,
the voltage across the capacitor is
equal to v parallel, equal to that current
of the supply. Supply multiplied by z
equivalent impedance of the total circuit. So as you can see,
we have the total is S curve representing
the total, okay? As you can see here.
And we have the current which is a fixed
it, current source. Okay? They're multiplication will
give us this final curve, which is voltage across the capacitor or across
the parallel components. So as you can see that
this curve similar to the curve of the impedance. But the difference is that is, it is multiplied
by a certain gain, which is a current source. Now, let's discuss
the quality factor. The quality factor Q of a parallel resonant circuit
is defined as the ratio of the reactive power
of the inductor or the capacitor to the real power of the resistor at resonance. Similar to what? Similar to
the series resonant circuit. Okay? So it will be like this. So we have reactive
power divided by power dissipated as a resistor or the real
power of the resistance. So it will be like this, q, p or q parallel. The quality factor is a
parallel resonant circuit equal to vb squared over r x l p divided
by v, v squared over r. Okay? So this part representing
the q or the reactive power, this part representing is the power dissipated
in a resistor, or the average power, which is P, or active power. Now someone will ask me, okay, you did in the previous lesson of the series resonant circuit, you said a Q S is equal to that, react the power
divided by real power. So we said before, I square
x divided by I square 0. So we had x over r, right? This is in the series
resonant circuit. So why didn't we use I
squared I squared R? Why did you use the voltage? Because in the series resonant
circuit we had the current was the current year
is a common factor. The current flowing through the inductance is similar
to the current of resistor. So we use this equation. And instead of v squared
over r over x L, or V squared over R, because we have common factor
I squared and I squared, which we can cancel with EHRs. We can have x L over R. So in the series circuit, we have the supply, supply E, and we have a
resistance inductance capacitor. So all of them have
the same current. So that's why I use this
relation to get x L over R. Now, in that parallel
resonant circuit, we have the voltage, same voltage across
them, which is VB. So we use the VB so that I can
cancel this with this one. And instead of the current,
if I use the current, then I need current flowing
through here or p squared. And I need all your R-square. If I use this relation, which will not help
the mean anything. However, as the
voltage is common, that's why I use
the v square over the reactants and
v squared over r. Okay? So as you can see, we can
cancel this with each other. So we will have all
of our XR like this, r over x l parallel,
which is our, is our supply parallel
to our parish. Okay? You can see
that q parallel is different from que
es, que es walls. X L over R. Over here, q is r over x, p. Remember p, Note XL only. Okay? Now why is this? Because we have this. If we take this with this one, we will have one over x l
b divided by one over r. One over r can be, they can award, and the one over x L can
be taken downward. So we'll have our over XL. Okay? Now, as you can see that x old parallel is equal
to xy at resonance. Therefore, we can
substitute with Excel parallel instead of
x l para we can have x z. Okay? So we have q equal to xy equivalent resistor divided by xy or divided
by x l parallel. Okay? Now, what if we have an ideal current source
if r is equal to infinity or if RS is very
large compared to RB, can make this following
approximation. So as you can see, when
we have two resistors, R, S and R are parallel. If this one is very large, which is usually,
usually the case, compared to our parallel. Then in this case, we can say it is approximately equal to the smaller resistance, which is R p. Okay? So we can type the
equation like this, Q equal to our
parallel divided by x. Okay? Very easy. Now, if you remember that our parallel equal to
R squared plus XL square divided by RL and XL a parallel or a low squared plus x squared divided by Excel. So we can take this
one with this one. So we will have x over r l. As you can see here, which is the quality factor
of zirconia UL, which is similar to
or similar to Q. Okay? Qs, if we have only one resistor inside
the circuit, okay? So as you can see,
q parallel will be X L over R L equal to q, q of the coil, or the quality
factor of the coil itself. When RS is very large
compared to our par. Okay? So you can see all of
these equations helps us to approximate the
quality factor, okay? Okay, so this is representing Zach quality factor
Q L of the query.
136. Bandwidth and Cutoff Frequencies of a Parallel Resonant Circuit: Now let us discuss how can
we obtain the bandwidth and the cutoff frequencies inside a parallel
resonant circuit. So first, as you know, is that the bandwidth
same equation is equal to F2 minus F1, which is the difference between the cut-off frequencies F2
and the cut off frequency F1. Or as the half power
frequencies F2 minus F1, which will be equal
to the frequency at resonance divided by q parallel. Now, which frequency
we are talking about? We're talking about the
frequency which gives us that unity power factor, or when we have a pure
resistive circuit. Okay? So similar to the series
resonance circuit, we have a big value, v-max. Here we will have not V-max, but the voltage will
be equal to the total. Multiply it by the
supply curve and the total multiplied by current and the
total in this case, in this resonant
frequency or F parallel, or the frequency at which we have unity power
factor will give us our multiplied
by the current. Okay? Now, the bandwidth
here at this point and this point at which we
have the voltage will be equal to 0.707 of this one. You can multiply it
by R multiplied by R. Okay? So it's equal to 0.707 of
the voltage at resonance. Remember that in the
series resonant circuit, we have our Emax at which
we have maximum power. And we have two
points at which we have the bandwidths, Okay? With that which we
have 0.707 imax, which will give us half
The power at resonance. Same case in the parallel
resonant circuit. And this point, we have
the voltage equal to the resistance multiplied
bys a current. Okay? Since we have a pure
resistive circuit, and at this point it will be 0.707 multiplied by r
multiplied by saccharin in order to y in order
to get off at resonance. Okay? It is not the maximum power, but it is half as a
power at resonance. Okay? So here you can see that the cutoff frequency
F1 and F2 can be determined as a unity
power condition. Remember we have two conditions, unity power factor and
the maximum impedance. Here we are talking about
unity power factor. So it means that
FOR will be F B, which is a frequency of apparel resonant circuit
laws that provide us with a pure resistive circuit or pure resistive circuit
or a unity power factor. The half power
frequency or defines that the condition at
which is a maximum, the output voltage is 0.707
times the maximum value. What maximum value
as a maximum value of the curve that
representing is a unity power factor H, a cut off frequency as the frequency at which
is the input impedance, is 0.707 times its
maximum value. And since there's
a maximum value is equivalent to resistance R. So it means that the resistance
will be equal to 0.707. Our adds a half god
frequency. Okay? So let's understand that. So as you can see,
this is a voltage At resonance equal to
R, the blood by eye. Now at the half got frequencies, as the voltage will
be equal to 0.707, this value, which is
r multiplied by a, so it will be 0.707
or multiplied by I. So what is the difference
between these two equations? You can find all the
points 707 hour. So it means that
we can say is that our resistance becomes a 0.7, 07.5 got frequencies,
as you can see here. Okay, So I'll want to use this in our analysis
to get F2 and F1. So as you can see in
this circuit first, we have y total, or the total admittance
of this circuit is 1/1 plus one over z, 21/31 over r, one over r j x p, one over negative
j x is c. Okay? So as you can see, one over r plus one over, let's type it. So you can understand it. One over x1 plus one
over R2 plus one over z, one over z, one, which
is our one over z2, which is J, x. Plus one over
negative J accessing. Okay? So one over j gives
us negative g. So as you can see, negative g goes one over j
gives us negative j. One over negative z gives
us bolster j one over x c. Okay? So here, this representing, representing our admittance
of this circuit. Now you can say is
that also Z2 is equal to one over y to one
over RC admittance. So you can take this
part and y equal to one over all of this. So we can say is that
total is equal to 1/1 over R plus j one over z minus one over XL department,
as you can see here. And remember that the
inductance here is L and capacitance is C. Okay? Now, here, this is our Z. Now since we are talking
about cut off frequencies, we said that the total
will be equal to 0.707 or so that
we can have 0.707. So the voltage will be 0.707 of V max of Arizona
and the frequency. Okay? Okay, so as you can see here, we will equate these two
equations with each other. So we can say, oh, 0.707 is equal to my exists. That is first. You can see here are over root 20.707 is one over root two, and r is r as it is. Now for this part you can, you can take one over r as
a common factor outside. So we have one over
r, two brackets, one plus j omega C
minus one over Omega. So if you take one over
R, multiplied it here, we will have one over r. Multiply this by
this gives us one. Now why did we do this? Simply? Because we can take
this to the other side, one over r, get it
to the other side. So we will have like this, one over root two equal one plus one plus j omega C minus one over omega
n. So as you can see, one over r, You can take
it to the other side as one over r. This r, We'll go with this r. So
we'll have one over root two. As you can see from
this equation, you can see one over this, equal to one over this. So it means that this part
is equal to root two. Or to be more specific, the magnitude of this
one equal to root two. So as you can see like this. Okay? Now why is this y r Omega c minus one over
omega L equal to one? Because assembly here
from this equation, one plus j omega C minus one over omega
L equal to root two. So the magnitude of this one is equal to the
magnitude of this one. So the magnitude of this is
equal to root one squared, which is a real part. Here, will be r. Here lets, it will be like this. R Omega c minus r over r Omega L squared
equal to root two. So we take the square of
the first plus square of the second one squared plus r omega c minus omega over omega l all squared equal to root two. Now, as you can see, is that all of the
squared root two. So it means that we have one plus one under the square
root gives us root two. So this part will
be equal to one. So as you can see, our
omega c or omega c minus R over omega L minus R over
omega L gives us one. You can also think about
this into another way. You can see that this magnitude
equal to this magnitude. So we can say e.g. equal to root k
equal to root two. The magnitude of this, now the magnitude of this
part is equal to root the real part squared plus
the imaginary part squared. So let's say real part squared. And the imaginary part
is this part squared. So let's say this part is B r Omega c minus
one over omega L. We would say it'd be, so it will be b squared. So wrote one a-squared plus
b-squared equals root two. So it means that one plus
d squared equal to two. So b squared would
be equal to one, or b would be equal to one. B is, what is our Omega
c minus one over Omega. So as you can see here, r Omega c minus one
over omega L equal to. Now, from this equation, we can get like this by substituting two pi
multiplied by the frequency. So we'll have f For square
minus f over two pi RC -1/4 pi squared
r l c equal to z. Then, by solving this equation, this is a second degree
or a quadratic equation, a fresco a x squared plus bx plus c. So it can be
solved with like this. We have a which is
the first component, which is the coefficient
of x squared. And this one's a
coefficient of b of x, which is negative 1/2 pi RC. And the coefficient
of the last element, c negative 1/4 pi squared LC. So the solution of
this equation is negative V plus minus root b squared minus four ac over
2a0, as you can see here. Okay, similar to what we did in the cut-off frequencies of
the series resonant circuit. So negative p plus minus b squared minus four ac over two. So by substituting,
you will have finally, F1 and F2 equal to
this two equations. Okay? So this two equations can
help you now to obtain the cut-off frequencies of the circuit in our
parallel resonant circuit. Now as you can see, one thing to notice here is
that F1 from this equation, you can find this is
always a negative value. Since we don't have a
frequency and negative. So F1 will be the
magnitude of F1. This one will take
its magnitude. Now we would like to finally
see the effect of R, L and L and C on the shape of
the petal resonance curve, we have the relation between
being and frequency. And the frequency. You can see is
that as we change, as we change our resistance, as our resistance increases, what will happen is that our
curve will start going down. You can see here, maximum value starts to decay as the
resistance increase. Okay, that's the first
thing you cannot. Now, if you increase the ratio, else L over C, if you increase this ratio, the curve will be going higher. It will go higher. And at the same time, if you look at this curve, you will find that as a
resistance increases, total starts to decay. And that same time the
band width will increase. The bandwidth increase. Here, the reverse, as
L over C increases, that increases, both, the
bandwidth becomes smaller. Okay? So again, if you
look at this for increasing or decreasing l, you will find that the
bandwidth increase. So as you can see, as
r increases or L3 e.g. you can see bandwidth
larger, okay? Or you decrease L
over C is our issue. Decrease ED such as Ilona for Q1 will find
that the bandwidth is also find that at
resonance and increase in RL, or decrease the ratio L over C results in a decrease
in Arizona and impedance. As you can see here. When RL increases,
such as oral history, you will find that the
total starts decaying. If you increase or decrease. If you decrease, it means that resonant frequency also be, will also start this to indicate that parallel
will starts to decay. Okay? Now, another thing here to
notice is that during Giza, before resonance and
after resonance, we are talking here about
the unity power factor. Remember unity power factor. So after resonance and before restaurants,
often resonance. You will find that here. That circuit at resonance
is a pure resistive. Pure resistive at
unity power factor or at resonance frequency, which gives us pure
resistive system. Okay? Now, before this, you will find that the
frequency is lower. Okay? So what will happen when
the frequency is low? Okay, let's look
at this circuit. So at low frequency, Excel part will be low value. Since we have a small
frequencies in this range. And ecstasy will be very high. Why? Because x is z is one over omega C. So at the small frequencies, we will have omega small. So accuracy becomes high. Now, since we have x l
parallel to z, okay? Have Excel small battery
to large existing. So what will happen here? What will happen is that since x is smaller than accessing, so XL, most of the current
will go through XR, okay? And small current
will go through Z. Okay? Or we can say z equivalent will be approximately close to Exxon because it is
a smaller impedance. Okay, That's why it's a circuit
becoming more inductive. Because Excel is smaller
than x and z are in battery. However, here in this region, x is c becomes a very low compared to Excel,
which becomes high. Why? Because larger frequency, when
frequency becomes a large, XL becomes high and
Xc becomes this mode. So x is c parallel to Excel, so exceeds the smaller
effect will be much higher. That's why a circuit
becomes more capacitive after the
resonant frequency. Okay? So as you can see
at low frequencies, the capacitive
reactants is high and the inductive reactance
is slow, is low. Sensors they are in parallel. The total impedance at
low frequency is zero. For inductor, at
high frequencies, the reverse is true and
the network is capacitor. At resonance FB,
the network will appear resistive or add
resonant frequency. Remember what resonant
frequency f be mult FM game. You will also that
it is the inverse of that appearing in
Sarasota on circuit. Because at low frequency,
if you remember, at low frequency
we have x is c Hi. And since it's the R in series, ecstasy effect will
be much higher. So the circuit walls capacity. And at high frequencies
XL beak was high. In the Sierras circuit is a
circuit boards inductive. So as you can see here, is that in series we had capacitive before frequency in parallel, we have inductor. In series after
resonant circuit, we have inductive and in the
parallel we have capacity. So it is the inverse is
the inverse of each. Awesome. Before we finish this lesson, I would like to mention a
very important thing here. Here you will find
when we obtain the equations for F1 and F2
will find here inductance. Now, one thing which
is really important, you can say, okay, is the inductance of
the parallel component, which is the most correct zinc. Okay? However, you will find
that in when we solve some examples in the
puddle resonant circuit, you will find that
we are using because the inductance L
and instead of LP. Now why is this? You
will find that in this example something
which is really important, the quality factor is larger
than or equal to ten. So there is an
approximation which we are going to discuss
in the next lesson. Is that wins, the quality
factor becomes pretty high, greater than ten funds that XML is Natalie equal
to XCP Excel P. So in this case we are using to say inductance L
will be equal to L. That's why we can use this
inside of our equations. Okay? And instead of L, okay? This is something
which I would like, which I would like to mention before we go to the next lesson.
137. Effect of High Quality Factor on the Parallel Resonant Circuit: Hey everyone. In this lesson we are going
to identify the effect of high quality factor
when the quality factor is equal to or greater than ten. So we would like to
simplify our equation based on high-quality vector. So as you can see in
the previous lesson, you can see that the analysis of the parallel resonant circuit
is much more complex. Zan is a series
resonant circuit. That's why you will find that in most pattern is one
on circuit Zach. Good thing is that
our quality factor is sufficiently large to permit
several approximations. These approximations
will help us simplify the required analysis. So first, let's see our
inductance x old battery. So we would like
to simplify this when we have high
quality factor. So first, we have x
parallel equal to RL squared plus x squared
over x L. Now, if we can say like this, r squared, this can be divided into RL squared over x l plus x squared over x squared over
x plus x L squared over x. Now, x L squared
over x L gives us x. As you can see here. Now for this part, for this part, you
can see RL squared. So if I multiply here by x
l and divide here by XR, what will happen is
that we would have r square over x square, multiply it by XL, RL squared over x
squared multiplied by x. Okay? Now, as you remember that Q L or the quality
factor of the coil, is equal to x L over R
L. So as you can see, x L over R L. However, this is a square. So we can say that
this part can be equal to one over Q square. So q L square, as you can see, is equal to x L squared over r squared
from this equation. So one over q n is RL
squared over x squared, R squared over x squared. So we will have x
over q squared, x over q squared. Okay? Now why did we do this? Because we need a
relation between the high-quality factor
and our element. So as you can see, when
Q is equal to ten, so it means that Q L
squared is equal to 100. Okay? So I value divided by hundred
plus the original value. What does this mean? It means that this part
will give us almost zero. So as you can see,
this can be written as one over Q squared plus one divided by x l. So as
you can see, 1/100 plus one. This can be approximated as one, because it is one plus
1/100, which is 1.01. So we can say it is
approximately equal to one. So it will be equal to XR. So as you can see, this part will be equal to zero. So x L a parallel will be
approximately equal to XR. Okay? That's why, if you remember
that in the equation of F1 and F2 or the cut off
frequencies, we wrote. Because we assume that we
have high quality factor. Now, as you remember that
X equal to X equal to x. So we can say in the
resonance condition or the resonant condition
is that x L is equal to x c. Now second part, second part is that we need to find that resonant frequency, FAP, which is representing
unity power factor. So FAB is equal to f s, the square root of one
minus r square c over n. I would like to use SQL to
simplify this equation. You can see is that this part, r squared, r squared c over l can be simplified
as one over x l. X is c over r l squared, which means it will be
approximately one over q squared. So all of this part can be simplified as one
over q squared. Now, let me write it for you. So we have our square, c over l, which is the spot. Now first, I will
multiply here by omega and the multiply
here by omega. Okay? So omega c is equal to what? Is equal to X c
and omega L is x. So now you can say that here, sorry, omega C is
one over omega C, one over omega C, which is Xc. Xc is one over omega C. So we have omega c here, so it will be equal to
one divided by x c. So we have our n square
over x l multiplied by x z. We can say one over x multiplied by accessing divided
by r square. As if we divide here by RL squared and y
here by RL squared. So dividing here gives us one dividing here it
gives us r squared. This equation is similar
to this equation. Okay? This is the first part. Second part here is
that we need QL. So you know that q l
is equal to x L over R n. Now remember that
in resonant condition, which results to set at quality
factor greater than ten, equal to or greater than ten, x is equal to x is equal
to x L the parallel, all of them are
equal to each other. So we can say is that since
x l multiplied by x c, So I can say x c becoming XOR. So it will be one over x
L squared over r squared. You know that the q of
the coil is x L over R L x L squared over r squared. So this part is
equal to Q square, as you can see here. Okay? So let's substitute with
this part in this equation. So we will have like this, f p is equal to f s root one
minus one over q squared. Now remember that here, quality factor greater than ten, let's say at least, then. It means that this part is 101 minus one over
hundred gives us 0.99, which is approximately
equal to one. So what is under the bracket is approximately equal to one. So this means that f b is
approximately equal to f s Windsor quality factor
becomes greater than ten, greater than or equal to ten. So as you can see,
a, B equal to F, S equals 1/2 pi root and C. Now let's see the second
resonant frequency, which is F M at which we have maximum voltage or
maximum impedance. So as you can see, f m
equal to f s root one -1/4 L squared c over n. Now, as you remember
from the previous slide, we said that this
part, or a square, c over l is equal to one
over Q square, like this. So as you can see, one -1/4
multiplied by this part. So this part alone is 1/4. Multiply it by at least
if q l equal ten, it means that one over hundred. So it is multiply 1/400. So this part under Zola root, it will be one -1/400. So this part is
really, really small. So we can say it is
approximately equal to one, similar to the previous slide. So you can see like this. So f m will be equal to f s because this part is really
small so we can neglect it. So as you can see, what
we learned is that at resonance or add, not resonance at QL or the quality factor of the
coil greater than ten. It means that equal to or
greater than ten it will, we will find that F M, F S, and F B are all
equal to each other. Now, let's apply the effect
of the quality factor on the RB parallel resistor. Remember that our
p is equal to RL squared plus L squared
over r l. So we can say it is equal to
our n square divided by R n plus z squared
divided by RL, RL squared divided by
RL gives us r n plus x squared over R n x L
squared over r. Now, if we multiply here by
R L and here by RL, then we wouldn't
have x L squared R L divided by R
squared x L squared, L over R squared. And we remember that this
part is equivalent to q squared or R L
plus q squared RL, which is one plus
q squared or n. Okay, So now as you can see, if our parallel equal to one plus q squared RL,
this is in general. Now what if q l
greater than ten? It means that this part, if it is at least, then it means that a
square is 100. Okay? So we have one plus
100 multiplied by RL. So as you can see, one compared to 100 is
really, really small. So we can neglect one. We can say this is
approximately equal to q squared multiplied by R. Okay? So as you can see,
for quality factor to greater than 101 plus q squared is approximately equal to Q. Square are parallel will
be approximately equal to q squared or ok. now, if we substitute
with x L over R L, we will have this
larger equation, which will give us in the end, are parallel equal
to L over R L C. Okay? So now there is another
thing which you can do, which is really, really simple. Remember that we said
in the previous slides, our square, c over n is
equal to one over q squared. So from this equation we have q squared equal to the
inverse of this term, which is L over R squared, q squared equal to
L over R square C. Okay? Now, we can take this one
and substitute it here. So our parallel will
be equal to L over R square C multiplied by RL. It means that this RL, we'll go with one of
these l. So we'll have our parallel equal
to L over R L C, L over R L C wins or quality
factor greater than, this is muscle is much easier than getting the
original equations. Okay? So if you don't remember, where did we get? Where did we get this equation? As this one? We obtained it from the previous analysis
in the previous slides. Okay. Now let's see. So final circuit. So as you can see, in the final circuit, we have our supply, current source or supply
then are parallel, is equal to what, q squared, q squared multiplied
by the resistance R L, which we just obtained
in the previous slide. And XL apparel is
approximately equal to x l. And we have xy. This is approximate circuit when Q is greater than
or equal to ten. So total is equal to RS
parallel to r p at resonance. So total impedance at
resonance, at resonance, what will happen is that
x will be equal to z, so z cancel each other. So we have only R
S parallel to rP, r s patterns to RB. So RS parallel to Q, L squared or n, which is r. Now, if our supply is, if we have an ideal current
source or R S is infinity, or R S is greater
than or be very, very large compared to RB. Then we can reduce
the equation to Z. Total will be equal
to q L square R. When in, when Q is greater than ten or
high-quality factor, and the supply
resistance is much greater than the
parallel resistor. So the quality factor is
now defined, the boy, that Q is equal to
resistance R divided by x, r divided by x, which is approximately
equal to x. What is x? And the
parent is equal to x l. And the resistance is
the total resistance, which is R S butter to RP. Or approximately, since we
have high quality factor, we can say it's equal
to l squared or n, like this, square R. Okay? Now, you'll notice that Q
or you know that X L over R L x L or R L over x
gives us one over q. So this part alone is
equal to one over QL. As you can see here. One over q and q L squared. Here we have Q L squared. So this will go with this
one. So we will have. Q parallel will be equal to
q. Q will be equal to Q L. When does this condition? When does the quality factor of the parallel resonant
circuit equal to the quality
factor of the coil. This will happen when the supply resistance
is very large, so we can neglect it. And the Windsor
quality factor of the coin is very large
or greater than ten. Okay? So in this case,
the quality factor Q P will be equal to q. Now remember that
what about bandwidth? Bandwidth is equal to f, f2 minus f1 equals
to f p over q b. Or the frequency at
resonance at which we have unity power factor and the quality factor of the
system, whole system. So first thing we
notice that when the Q is greater than ten, F B is equal to f s. So we can say this equal to
f b, which is f. And we know that at
high quality factor, high quality factor Q L, we can say is that QL QB is approximately equal to Q l when we have high
quality factor. And the same times
the resistance of sublime is very large. So we can say this Q P
will be equal to q L. Now, QL, QL is what? X L over R n. So we
can say x L over R l. And x L itself is equal to two pi multiplied by the
frequency at resonance, which is f b. Multiply it by is
the inductance L. Okay? But as you know that
F S is equal to F B, F S equal to LP, duality vector is very high. So we can cancel
this with this one. So we'll have our L divided by two pi n r l divided
by two by n. So as you can see here, our L over two. And this is when the
supply resistance is very large or very large
compared to our parallel. And at the same time the
quality factor is greater than, equal to or greater than. Now, what up our
voltages and current. So we learned that at resonance
and high quality factor. And third condition is that
S is equal to infinity. You will find that the total or the total impedance
at resonance is approximately equal to q
square, q square array. So the voltage across
the parallel circuit. So we have the total, which is q squared RL. And we need to find
the voltage here, which is equal to
the voltage here, equal to the voltage here, equal to the voltage here. So V equals V a local VR. What is its value? It is current source multiplied by the total
impedance of the circuit. The total, which is a supply, multiplied by z to the total at resonance equal
to q squared r. Okay? So we have here our voltage. Now, what about the
current? The current e.g. if we need IC, which is a current across, flowing inside a,
inside that capacitor. So we have total, but I will need now the current flowing
through the capacitor. The current through
a capacitor or any system from Ohm's law, it is the voltage across
it divided by reactants, VC divided by x c. Okay? So the current is equal to the voltage divided
by resistance, or in this case, reactants. So the voltage is
equal to I total Q square RL divided by x is c. Now, the next step
is that you will find that x is equal to x L. Okay? So you have total, you all squared RL divided by x. Now, you know that R L
divided by x l is one over Q. So we have a total q squared
I total Q L squared, r over x n is one over Q. So from this, you will find
that I total multiplied by one QL, like this. So the current
through the capacitor is a total multiplied by q. And so as you can see, is that the current
through the capacitor is amplified by the
quality factor Q L. Similar to what? Similar to the voltage in
the series resonant circuit. If you remember that the
voltage across the capacitor in series resonant
circuit was q. Multiply it by e, which was our supply. So our voltage was
amplified in our, see it as resonant circuit. Now, in parallel
resonant circuit, you will find that current
is amplified by q. Okay? So that's why it works as
that current amplifier. Now for inductance XML, it is a same idea as the current I l will
be equal to q l. Same value. Why is this? Because they have
the same voltage. They have the same reactants
XL equal to ecstasy. So the current
flowing here will be equal to the current
flowing here equal to Q. Total. Last thing about parallel
resonant circuit, this is a summary
which you can save on your laptop or PC
or whatever it is. In order to remind you of all of the equations of the
parallel resonant circuit, you can find that this
is the equation of the bandwidth Zach
currents at any cue. This one at q, or the quality factor is
high, and this one wins. The quality factor is high and our supply is very large
or equal to infinity. Combine the two q, l square or a, or compare the two are. So we'll find that
the values of all of the elements inside our
circuit from this table. Now what are we going to do? We are going to have
some examples on Zapata. There's one on circuit
to understand how can we solve equations.
138. Example 1 on Parallel Resonant Circuit: Hey everyone, In this lesson
and all the next lesson, we are going to
have some soul with examples on the parallel
resonant circuit. Okay? So in our first example here, we have a parallel network
consisting of ideal elements. So as you can see, we
have a current source with a resistance
RS parallel to it, which is ten kilo ohm, which is a source resistance. And we have an inductor
and capacitor. Now, this inductor is, since we have ideal elements, it means that this inductor does not have a
resistance in series, so RL is equal to zero. Okay? Now, first requirement is that we need to find the
resonant frequency f p. That is an anti frequency
of p. Secondary comment. We need a total
impedance at resonance. We need quality
factor bandwidths, cutoff frequencies as a
voltage Vc at resonance, the current I, l and I see at resonance and the
applied voltage. Okay? So let's start. So first
we need F parallel, not F. F better, yes, F parallel or the F parallel
resonant circuit, or the frequency of the
parallel resonant circuit. This can be obtained by the equation which
we have learned. This equation, f p
is equal to f s, the square root of one
minus r square sin over L. Okay? So first thing do we have our L? Do we have a resistance
for the inductor? No. Why? Because our inductor
is an ideal element. So our L would be equal to zero. So this part will
be equal to zero. So as you can see that F
P will be equal to f s. Okay? Do we know that F S, which is the frequency of
a series resonant circuit, is equal to 1/2 pi root LC. So we have the inductance,
one millihenry. We have the capacitance, one microfarad, so we
can get the frequency. Okay? So as you can see,
f b equal F as 1/2 pi root one millihenry,
one microfarad. Of course, remember
that one milli means ten to the
power negative three, and the micro means ten to
the power negative six. Okay? So this will give us
in the end the 5 khz, sort of presenting their
frequency of zeros, zeros on circuit and
the parallel circuit. And then now as you can see, since l is 0 ω, this results to a very
high quality factor for the coil QL, or the quality factor of the
coil is equal to x L over R L quality factor
of the coil and naught Q quality factor of
the query quality vectors, the color will be X L over R L. And since this one is zero, so Q L is infinity, of course not infinity,
but it is very, very large vein, okay? Okay? And we know that when the
quality factor is pretty high, it means that f b
will be equal to fs. Now second requirement
is the quality factor. Quality factor,
which is Q battery. Okay? Now sensors or quality
factor Q n is very large. It means that both of them
will be equal to each other. Okay? So as you can see, q parallel, as we know here. First, before the
quality factor, we need the total impedance, okay, the hotel impedance. So total impedance at resonance. Remember that total
At resonance, Excel, we'll go with ecstasy. Okay? We know that there
is no Excel or ecstasy or the cancel each
other to be more specific. So when we are in resonance, the total impedance will
be only our resistance, which is ten kilo-ohm. Or if you remember
that before we set our S, two are parallel. Okay? But are parallel does not exist because this
coil is ideal, so this does not exist. So we have only one
resistance which is RS. So if you look at this circuit, says it total or the input impedance at
resonance is equal to RS, which is ten kilo,
as you can see. Because that goes with that CC, cancel each other at resonance. Okay? Now, third requirement is
the quality factor Q b. Now let's remember what is the quality factor of the
parallel resonant circuit? We said q parallel
is equal to R S over R S. Or to
be more specific, queue parallel walls are. Total, the total
resistance divided by x. Okay? Since when we had here R L. So R L and X and inductance we convert into one resistance
parallel to an inductor. This inductor was parallel and this resistance
was air are parallel. Now we don't have
this resistance. So x l is equal to x L itself. And our total, there is
only one resistance, which is R S. So we have our S over two pi multiplied
by the frequency, multiplied by the
inductance itself. Now what frequencies? And so we are talking
about Q parallel. Then we are talking
about the frequency of the parallel
resonant circuit, which is 5.03 kw. As
you can see here. We finally obtain zero
hundred and 16.41 for the quality factor
is the bandwidth. We know that the
bandwidth is equal to F parallel divided bar is a
quality factor Q parallel. Now, F parallel is 5.03
khz and the q is 316. So as you can see
here, 5.03 divided by Serrano Augustine
gives us 15.29. Okay. Now, let's see, it's an
extra requirements for cut off frequencies F1
and F2 of the system. We need also the voltage
Vc at resonance and i, l and the IC at restaurants. So first is the cutoff
frequencies F1 and F2. Remember that we have
equations of F1 and F2, which we learned in the course. This two equations. So we are going to
substitute with capacitance resistance,
which is R, S and inductance, which is okay, because we don't have
accepted this one inductance. So by solving these
two equations, we will get finally that F1 and F2 are the
cutoff frequencies. Okay? Now, then augusta requirement
is a voltage Vc at rest, the voltage across this
system at resonance. Okay? Our ad resonant condition. Now at resonance
as a voltage here, which is the voltage
across resistance across inductance,
across capacitor, or VC, is equal to V0 of
the parallel circuit. The voltage across
parallel elements, which is the current supply
curve, multiplied by z2. At resonance. The current is done
manually and pair. What about the total? Is the total at resonance
is only pure resistive, which is ten kilo. So from here we can get V C, As you can see,
then Meli and bear, which is the current,
multiplied by ten kilo ohm, gives us 100 v. Okay? Then we need to find I L
and I see at resonance. You have to remember
that at resonance. At resonance, the voltage across this one is equal to the
voltage across this one. And that impedance of all reactants XL is equal
to the reactants xy. So it means that the
current I l will be equal to I c. Now i l and IC will be
equal to the voltage, which is the 100 volt, divided by the reactants
XL or ecstasy. Okay? So Excel will be two pi multiplied
by the frequency. Two pi multiplied
by the frequency, which is the frequency
of the parallel circuit, which we obtained in
the previous slide, multiplied by the inductance. Now remember that i l and the IC is the difference
between them is that the angle i l will be lagging by 90 degree and the IC will
be leading by 90 degree. So as you can see here, we have I l and I sing. So we said the voltage. You can see they have the
same magnitude as we said, because they have
the same reactants. You can see this reactance is equal to this
reactants are to 1.6, to 1.6, the same voltage,
hundred and hundred. So it gives us the same
amount of current. Okay? Now as you can see, also funds that the current
IC or current IL is equal to the queue parallel
multiplied by the current. Okay? Now where did we
get this equation? Simply as you remember
from here, we have VC. Vc is equal to the
current multiplied by z. So this one is a
current multiplied by z at the parallel resonant. In case of the
parallel resonance, the total in resonance is our S, which is the only resistance
inside the circuit. So as you can see, we have ecstasy
which is equal to x L R S over x L R S over XL is, what is our, this divided
by this one gives us q. So it will be the
current multiplied by q, the current multiplied by q. Okay? So all of this equation will lead to the same answer. So this was the first example on the parallel
resonant circuit.
139. Example 2 on Parallel Resonant Circuit: Now let's have another example. So in this example we have this parallel resonant circuit. We have our S equal to infinity. We need to find f S FM, FB and compares the levels, comparison between each other. Second, we need to find the maximum impedance and the
magnitude of the voltage. Vc at fm determines our quality factor Q
P, that bandwidth. And then we will do the
same all of this again. But when Q is greater than ten, with a simplification of
Q and greater than ten. And the compare between these
two, between these values. Okay? So the first requirement is this problem is that
we will get F S, F m, and F B. So for us, do we know that fs
is equal to 1/2 pi root LC? And we have equations of FM and FPU which we have learned
and inside the course, which are these equations. Okay? Now, as you can see, that fs equal to 29
point 06fm is equal to 28.58 and f b is
equal to 0.7, 0.06. So what we learned is that fs is greater than f m
greater than f b. That's what we learned
in the course. Now from this equation
you can see f as 29, f m 28.8, and f b is 27. So as expected, you can
see is that both F m and F p less than fs. In addition, f m is
much closer to f s Zen FB m is less than fs, but f b is also less than a, then f n. The difference between these frequencies
suggests that we have a low Q network
because if it is a high IQ, it means that F S will
be very close to FM. Fm will be very close to FP. Okay? We have some differences
between them, which means we have a
low quality factor. The second requirement is a maximum impedance and the magnitude of the
voltage Vc at fm, the quality factor
Q, B and bandwidths. So firstly, let's define
the maximum impedance. So maximum impedance
occurs at what curves? At a frequency f m
and voltage Vc at fm, which means we add the
same frequency fm, which is a maximum voltage. Okay? We need to find this too. So first, we know that our total in general is equal to from this
circuit is equal to, equal to x1 bar x2, x1 to x2. And x1 is L, L loss J x z. And z is equal to
negative j Omega C, or negative j, or negative
j XL XC, whatever it is. Okay? So anyway, we have our logic cell parallel
to negative j Omega C, which is equivalent
or the equivalent impedance at any frequency. Now we need maximum impedance, which means we will
substitute with frequency fm in this equation. So omega will be two pi fm
and Excel will be two pi f m. Okay? So as you can see, is
that total is equal to RL XL parallel to negative JSC, as you can see here,
at frequency f m excel equal to two pi f m l. Okay? So it will give us 53.87 ohm. Ecstasy is equal to one over omega C two pi multiplied
by the same frequency C. So we have 55.69 ω and we
have our L plus j XL RL. What is our L 20 0 j XL. What does Excel? Excel is to why Excel, which is 53.87 neutral
resource to calculate it here. So 20 0 plus GFF 3.87. If we would like to
write it in the form of phasor diagram or inside the
form of magnitude and phase, you can find 57.46 and
the angle 69 degree. Now for second
part, which is XC. Xc is this one with an
angle negative line. Since it is y negative 90
because it is negative g. Okay? So this is our x l,
this is our x is c, this is our x1, which is 57.46, and
this one is our z2. So we have two
parallel elements, capacitance and our LBJ Excel. So they're equivalent will
be a product that one. Multiplied by z2
divided by Z1 plus Z2. This is the equivalent of
two parallel elements. If you don't want to
know this gets back to our course for
electric circuits. So as you can see that
one multiplied by z2 divided by the summation. F1 is XL is our LBJ Excel, which is this equation, 57.46 ohm angle six to nine, as you can see here, multiplied by z2, which is 55 point 6.9 angle
negative 90, okay? And divided bys or summation. So you have 20 ohm plus j 53.87 on and this one which is ecstasy will be
negative j 55.609. Okay? So what this will give us, this will give us total ad, or the maximums that at f m is 159 and the
angle negative 15. Now we need the voltage VC. You know that the voltage
VC is the voltage here. The voltage we can say is that total multiplied bys account, which is to mainly Ambien. So I'll take this and multiply it by two milli and
bare legs this. So we will have to
mentally multiplied by 159 gives us 718 milli volt, which is the maximum voltage
at the maximum impedance. Okay? Now, the next requirement
is a quality factor Q P. Q P is equal to, equal to r over x
over x l parallel. Remember that q p is equal to our total was a
circuit divided by x. Okay? Now the resistance and South circuit will be
our parent only because we have an infinity supply or an infinity resistance
of the supply. So it will be our paradigm
of our laboratory. Now we remember that when
R S was equal to infinity, we had TOP equal to q. And if I remember correctly, okay, so as you can see here, Q will be equal to r over
x l parallel equal to q L, which is x L over R L. Okay? So it would be two
pi multiplied by the frequency multiplied
by inductance, which is 0.3 millihenry
divided by the resistance 21. Now what remember that
since we are talking about the quality factor of the
parallel resonant circuit. So that frequency here used
is f not f m, f p. Okay? So it will give us 2.55. Now, as you can see,
the quality factor of the circuit is 2.55, which is a low quality factor, which is corresponding with the differences between
the frequencies. F m is different from
the front from epi, which means that we have
a low quality factor. So as you can see, this confirms our conclusion of
the first two parts that we have a low
quality factor. As you can see, we have
a difference between FS, F m and F p. Now, as a final requirement
is the bandwidth. Bandwidth is equal to F
p over q p f parallel, which is resonant frequency
of the resonant circuit, 27 khz and q, p, which is 2.5 to five, which we just obtained. Okay? Now we need to do
the same or compare between them and the Windsor
quality factor of Zach. Remember Q L naught q p
q l is greater than ten. So QL was in that for
a circuit to 0.55. Now what about high
quality factor of the COI? High-quality markers? Then we have some
simplification. The first simplification
is that if you remember that when the quality is
greater than or equal to ten, F S, F, m, and b are all equal
to each other, equal to 1/2 pi root LC. So as you can see, all of the frequencies
equal to each other, equal to the resonant frequency
of the series circuit 29, which is 1/2 pi root. We'll see. Okay. Okay. Second requirement is
the maximum impedance. So what is the maximum emitters? A maximum impedance
occurs at fm. The remember that f m is
equal to F S equal to FB. Which means we have a
pure resistive system. Okay? So first, before this
is the quality factor. Quality factor Q V Q
P will be equal to Q. When Q is greater than ten, then the quality factor
Q will be equal to q L equal to x L over R l, which is 2.74 versus 2.55. Okay? Now, so what is the difference here between what is this
two values are different. Remember that Ql or equal to q parallel
when R S equal to infinity. This is in the first case, okay? Now, q power in the first
case was QL was x L over RL. And XL was two pi multiplied
by f frequency of F parallel inductance over
our resonant frequency. Now, this one was 27, okay? But in this case it was 29. Why? Because our FM equal to f b equal to f s When
quality factor is high. That's why these two values are different from each awesome. Now next requirement is
the maximum impedance. So we know that total
at resonance is equal to q squared, q squared RL. So quality factor squared, which is 2.74 squared
multiplied by RL, which is that when you own, it will give us 150
and angles here. Now, in the previous solution we had 159 and the
angle negative 15. Now someone will ask me, where did we get this equation? Remember, you can go back to the previous lessons or
you can somebody know that when the frequency becomes equal to the
parallel resonant circuit f, f b, it means that our
circuit is a pure resistive. And in this case we will
have one resistance, which is our parallel. Now with total at resonance
will be our parallel. When we are talking about
maximum impedance here, which is F m. And F
m is equal to F P, which means that we are in
a pure resistive estate, which is all parallel. Now, our pattern itself was
r squared plus x squared over R L. So if you remember that when
we simplify this equation, we have one plus q l
multiplied by our N q squared. Since we have here x L
squared square here. Then we said before that QL, when it is greater than
ten or equal to orbitals, and then this part is
much greater than one. So we can simplify this as
Q squared, as you can see. Okay? If you don't remember, this is a small reminder, remainder, reminder of this. Where did we get this equation? Okay. The next one is the
magnitude Vb, Vc. So the magnitude of the
voltage assemblies or current multiplied
by this impedance. So it will be too many
amperes multiplied by 150.15. So as you can see, this
is an older voltage. Then external comment is
the quality factor Q be. Okay, we obtained already
at here QB equal to 2.74. Finally is a bandwidth
is equal to f b over QB. So LP is equal to 2.74, sorry, q p is equal to 2.74
and f b is equal to 29. So you can obtain the
bandwidth as 10.61, similar to the one before. Now as you can see,
is that despite, despite that the
quality factor of q l was very small, it was 2.55. However, you will find that when we use this approximation
of q l greater than t, greater than or equal to ten. You'll find ones as the values
not far from each other. 150, 159 surrounded
it's 1,181,010.6. You can see that the
difference is not very large. Despite not being correct. However, the only problem was the frequencies
are frequencies where the difference between
them was pretty large. So the results
revealed that even for and low quality system, the approximate solution
I still close compared to those obtained using
the full equation. So this was another example on the parallel
resonant circuit.
140. Example 3 on Parallel Resonant Circuit: Hey everyone. In this
lesson we are going to have another example on
parallel resonant circuit. In this circuit we have FB provided the
resonance frequency, f b equal to 0.04 mhz us. We have QL, which is quality factor required
of the coil itself. We need are parallel. We need that total parallel. We need capacitance, we need QP bandwidths and
cutoff frequencies. So let's start. So we need forest Zach QL, or the quality
factor of the COI. Your nose at the quality
factor of the coil, q l is equal to x L over R. Okay? Now, excel here at
resonant frequency f p. Okay? So Q a parallel
circuit at a frequency f p q l in the series
resonant circuit at F S. Okay? So as a first step, which
will be two pi multiplied by the frequency of being
multiplied by the inductance, one millihenry divided by the resistance of the coil
itself, which is at ten. So it will be like this. Q l will be x over r l two
by FPL and to give us 25.12. Now as you can see, is that the quality factor of the
coil is larger than ten. So can use approximations
of Q and greater than, greater than ten. Okay? Second requirement is our parent or the parallel resistor. So if you remember
that's apparel resistor, when Q is greater than ten, it will be q squared
multiplied by the resistance. Or as you can see
as quality factor, quality factor is
greater than ten. Therefore, are parallel. Bq squared multiplied by RL
will give us 6.3 one-kilo. Now we need to add
parallel resonance. So we need the
total equivalent of this circuit at resonance. So we know that the total
at parallel resonance, x l will be equal to z. So we can cancel x
l and we can cancel x is c. Then we will have one resistance which is R S
bearer into another resistor, which is our parallel. This representing
the total resistance or the total impedance
at resonance. So it will be RS better to RP. So as you can see that total
at resonance or S butter to RP gives us 5.45 kilo ohms. Then another requirement
is capacitance C. So as you know is
that at resonance, at resonance x L is
equal to ecstasy. Or to be more specific, Xcel powder n equal
to accessing. However, remember
something which is really important that QL as greater than ten. In this case, x is
approximately equal to x-bar. So we can say that x
is equal to XR, okay? Using the frequency
f p, like this, or we can say is that the
frequency f p as equal to F S equal to 1/2 pi root LC. This is when the
quality factor is high. So this equation,
or this equation, they will give you
the same answer. So as you can see, a
pattern is equal to 1/2 pi root LC and the capacitance from
this equation will be equal to 15.83 nano. Tsar Nicholas, the requirement
is the Q part array, okay? So we know that q
parallel is equal to R equivalent of the circuit
divided by xA parallel. The equivalent resistance of that circuit divided
by Excel part. Now we know that XL power is
equal to x L itself, okay? Since the Q is greater than ten, the resistance is what is
RS parallel to our pattern. Okay? So odd is better to offer is
5.445 divided by x alpha, which we would like to obtain, which is this part. So as you can see,
Excel is two Pi f multiplied by 1 million or is this part is similar
to this part. And the equivalent resistance, which is 5.45 kilo ohm,
which is this one. Okay? So there the vision
will give us 21.68. Okay? Now, the next
requirement is that bandwidths and the
cutoff frequency, the bandwidth is equal
to what? Bandwidth? Equal to F p divided by q p, f p divided by q p
frequency at resonance, divided by quality
factor at resonance. Let's give us 1.85 kilo hertz. The cutoff frequencies.
As you remember, we have two equations. We have two equations for the cut-off
frequency, F1 and F2. So F1 and F2, it will be like this, this equation and
this equation and substituting with the
capacitance inductance, capacitance inductance
you will get finally, this two frequencies. Okay? Now you'll notice
something here. The f2 minus f1 is
this subtraction of these two frequencies gives
us the bandwidths 1.85, okay? Or close to it. Not exactly, but close to it. However, you will find
that the bandwidth is not symmetrical
about that as 1.2 frequency was one kilowatts
below and 840 hz above. So what does this mean? Okay, So this is our curve here. This is our resonant voltage. Here we have our 0.707 volt, which is the voltage at
which we have half power. Okay? Why exists by exist
and like this. So this is our
resonant frequency. You can see resonant
frequency equal to 0.04 mhz, which is 40 kilo hertz. And F2. F2 is the one which
after it is 4,840.84, 44. And F1 is equal to 39, 39. Okay? So we'll find that the
distance from here to here is our bandwidths. Okay? Now we will find
that the distance here is not equal to
the distance here. So the distance here is
40 minus certain time, which is 1 khz. And the distance
from here to here, which is 40.84 -40, which is 840 h. So you can see that the
resonant frequency is not symmetrical
around F1 and F2. Or the bandwidth is
not symmetrical. This part is not
equal to the spot. Okay?
141. Example 4 on Parallel Resonant Circuit: Now let's have another example. For the equivalent network
for the transistor. We have our transistor
here that provides us a collector of 2 million
bear current going out of it, 2 million, they're going
through this circuit. Now this transistor
circuit can be simplified as the circuit blue. We have a current source or S, R, L and inductance. And finally capacity. Now for the
equivalent to network of this transistor is a circuit. We need to find FP QB
bandwidths, vb at resonance. And finally, sketching
is a voltage across that is old
versus frequency. So let's start First. We need F, P and the Q
battery in this circuit. So we have our S, we
have our L, L, and C. Okay? So first, as you know that F parallel is equal to
the resonant frequency, multiply it by a
certain bracket, which is one minus one
minus r square c over l, something like this, okay, under the square root. So you can say, is that what is
the frequency F S? F S itself is equal
to 1/2 pi root LC. The inductance, which is 5
million henry capacitance, which is 15, be co-found. Okay? Okay, so now we have F S, But we need what under
the bracket here. Now, before we get to what this bracket we need to tap out. If F is approximately
equal to f s or not. How can we know this
quality factor Q, L greater than or equal to n? Okay? So the first step is that
we do are going to get q. And if this q l is
greater than tens and we are going to do lots of
simplification in our problem. In this case, b
will be equal to F, S will be equal
to this equation. Okay? So first exist, we are going
fs equal to 1/2 pi root LC equal to 1/2 pi root five
millihenry, 50 picofarads. This is a frequency fs, which we need five millihenry
and 50 be coherent. So this will give us
318.31 kilo hertz. Now we need to find the
quality factor Q L, QL itself as equal
to x L over R. Okay? L over R. Now exon, which is two pi multiplied by
the frequency at resonance, which is F p multiplied by
inductance L resistance RL, which is hundred ohms. Now, you have to know that
what are we going to do? We need to find the Q, okay? So QL can be obtained
using FP, okay? But we say, is that what
if L is greater than ten? Okay? If Q is greater
than tens and f b will be equal to F S. So we are going to use fs in this equation to see if the quality factor
is greater or not. So if we substitute like this, we obtained x l and we obtain
the quality factor Q L. You will find that tear
XL will be over r. L gives us ten kilo-ohm over
100 kiloohm gives us 100, which is greater than ten. Okay? So as you can see here, something which is
really important here. Here you will find one mistake, which is from France, which two of the lenses? We have XL, which
is ten kilo ohm. Rl is 100 ω hundred ohms. So there is no kilo here. So we'll find ten kilo, which is 10000/100,
gives us 100. Okay? So here we will find that
the quality factor Q L of a series resonant circuit
is greater than ten. So we are going to
make simplification. Simplification is that
f p is equal to F S. Okay? So I have p equal to f s. So from, in this case, what we can learn
is that we can get the quality factor Q P and the do lots of approximations
in our, properly. So Zach quality factor
Q b is equal to R, the resistance,
equivalent resistance of the circuit divided by the XR. Okay? So as you can see,
q parallel equal to the equivalent resistance
of a circuit divided by x L. Equivalent resistance
is R S and is a powder component
or p. And RB will be q squared multiplied by RL. As you can see, gives us one mega cube butter will
be equal to 50 kilo-ohm. Better to one mega ohm
divided by ten kiloohm. So it will give us 4.76, which is a quality
factor of the whole sec. Okay? Now you'll see that Zach Q drop the from q n equal 100 to Q will be equal to 4.7 60 due to the effect of the
resistance of the supply. So as you can see,
we had only are parallel if we substitute
it with our pattern, only if S does not exist, then we will have
q parallel equal to our parallel
divided by Excel. And the orbital is one
mega divided by X L, which is ten kilo-ohm, it will give us 100. So in this case, Q
will be equal to q p when we have
only one resistance, which is the are parallel. However, due to the presence of the supply air resistance, the quality factor drop
the form of 100 to 4.76. So this shows you the effect of that resistance also
supply on our circuit. Then we need bandwidth. Bandwidth is equal
to F P over Q P, or F B equal to fs divided
by q p, like this. Then we need the voltage of
the battery at resonance. The voltage at resonance, we know that at resonance. So we have a pure
resistive circuit. So our circuit
will be simplified to RS parallel to our pattern. So this z, which is at resonance and multiply
it by the current, which is a two milli and bear, it will give us what? It will give us the
voltage required. So as you can see, now, the last requirement is a curve
of VC versus a frequency. So we can draw it
like this, okay? So first, we draw Vc
versus frequency. We know that V maximum
is maximum value. Maximum at what value? It is maximum at
frequency equal to f m. Now sensors or quality
factor Q L is greater than ten. It means that f m is equal
to fs, equal to F p. Okay? So when we draw our circuit, you will find that at resonance, which at which we
have maximum voltage, it will be f m, which is F S, which is the voice frequency, zero hundred 18.3 1 kw. This frequency is
the frequency which we obtained adds the beginning, as you can see, f es
equal to this value, which is equal to
fm, equal to FB. And the bandwidths 66.87 s. This bandwidth
secretary 6.87. Okay? Okay. Now here,
if you would like the frequency F2 and
the frequency F1, approximately, approximately
it will be equal to F2, will be equal to that
resonant frequency plus bandwidth over two. And F1 is approximately equal to f t minus band width over two. As you can see here. And as you can see here. So it will give us
around 51.284, 0.9. Okay? Now this two frequencies
representing the half power frequencies,
half power frequencies. Now half power, it means that the voltage will
be equal to 0.707, the voltage at which
we have BMX, okay? Or not P max or at
unity power factor, at unity power factor. So 0.707 multiplied
by this voltage. This voltage is 85.24, which is a value here. This value at resonance, the voltage at restaurants, so it's 0.707, the
voltage at resonance. So it will be 0.707
multiplied by the voltage at resonance,
which is 95.24. So it will give us 67.34. So remember something
you here is that you will find that when quality factor Q greater
than or equal to ten, you will find that all
the different conditions combined with EHRs. So what does this mean? As you see that in the previous
or in the normal case, we had F S, we had FB, we had f and the frequency at which we have
maximum impedance, Z, T max or V max is different
from that frequency f b, which gives us that
unity power factor. Unity power factor, or
pure resistive system. Your resistance, pure
resistive system. So as you can see,
this condition was different from this one. Now we do to high-quality
factor resist to combine with each ours
as MCR one frequency. So the voltage here representing the maximum voltage and the voltage at which we
have maximum power unity, power factor unity,
pure resistive system. All of them combined
into one point. Okay? Okay. That's why this point is 0.707 or gives us half power, half of the power at resonance. Okay?
142. Example 5 on Parallel Resonant Circuit: Now in this example, we need to repeat
the same example, but ignore the effect of RS. So as IFRS does not exist to our circuit
will be like this. So remember, we have high
quality factor in this circuit. We have q greater
than ten and we have zero supply or R
S equal to infinity. So in this case we will have
several simplifications. So as you can see,
this is what we obtained in the
previous example. I have p equal to fs
equal to this value, Q equal to this, bandwidth equal to this. And VB at which we have maximum at resonance
equal to 95.24. Now is F B will be different. Now, when we neglect
as a supply, we still have the same condition
QL as greater than ten. Which means this
condition will be the same or the answer
will be the same. Our resistance will not
affect as a solution. So f v will be equal
to fs equal to 382nd problem queue parallel. So cute. Parallel. Butter is one
circuit equal 4.76. Okay? Now, if you remember
that we Q part is dependent on the supply
resistance R, s. So let's cancel the supply. So we will have q
parallel equal to one mega ohm divided
bys a ten kilowatt. So it will be like this. One Mega the world by ten
kiloohm gives us 100. Okay? So q parallel equal to q
when r is equal to infinity. This is also what we
learned in our lessons versus 4.76 here when
we had our supply, it is a very common bandwidth. Bandwidth will be F p
divided by q parallel. So it will be 0.8 in kilowatt-hours versus
the original one. When we had the q
power was lower. Vp. This means what? This shows us,
what this shows us that resistance itself
effect is the bandwidth. So it affected the bandwidth. And effect is Zach you. Okay? The presence of our S
or the resistance in general makes us
shape as a curve. We can change our curve. Then what is the value
of the voltage at resonance assembly
voltage will be the current milliampere
multiplied by the total impedance at
resonance, at resonance. So we have a pure
resistive circuit. So we have our S but R to R B. Now, we said before that
our S does not exist. So we have one
resistance which is R. Okay? Now, our parallel
is equal to what? Equal to q L squared
multiplied by RA. So it will give us here. As you can see, that
total equal to r, equal to one mega ohm. Since our pattern, if you
don't remember our pattern, we say equal to q squared
multiplied by RL. Now QL is 100, so 200 square multiplied
by RL, which is 100. So this product will
give us one mega ohm. So as you can see is that VB will be equal to the current
multiplied by the total. So two milli amps
multiplied bys or one mega gives us 2000 volt. So you can see that the R is affected the output
voltage at resonance. So before when we had our S, the voltage was only nine to 5.24 when we removed
the supply resistance. We have now 2000,
which is pretty, pretty much my Britain more
than the original value. Okay? So this results obtained
clearly reveals as a source, resistance can have a
significant impact on the response characteristics of a parallel resonant circuit.
143. Example 6 on Parallel Resonant Circuit: Now let's have one more example, owns a parallel
resonant circuit. So in this example, we need to design a parallel
resonance circuit to have the response curve in the following figure
using one millihenry, then ohm inductor and
a current source with an internal resistance
of 40 kilo-ohm. We have a current source with an internal resistance,
40 kilo-ohm. Our supply 40 kilo-ohm. We have an inductance,
one millihenry. We have our resistance of
the inductor that turn on and we have our capacitance
as you see, okay? Now, one thing which
is really important, you will find that here what we need to find
is the capacitance. We need to find the current
supply and we need to find if there is any
additional resistance. Okay? So originally we had
the current source are supply or l and inductance
and capacitance. Now we can add an
additional resistance. Why? Because as you remember, it can change our failure. It can change the bandwidth, it can change the maximum
voltage, and so on. Okay? So now we have bandwidths
2,500 hz is a frequency, f be resonant frequency 50 khz, and the voltage is ten volt. Okay? So let's start. So the first step is
that we have bandwidths. We have bandwidths. And can we get any thing using
this information? Yes, what a member that
bandwidths is equal to f be the rounded y, q, B. So from here we have bandwidths. We have the frequency. We can get the quality factor p, like this, bandwidth frequency
over the quality factor B. So we will have a quality
factor Q b equal to 20. Okay? So that's the first step. Second step is that here we need to find inductance capacitance. Capacitance can be
obtained from x L equal to x is c. We can
get this in the end. However, using the
quality factor, do you have any other information
for the quality factor? Yes, I know that q
p is equal to R, equivalent of the circuit
divided by, divided by XR. Okay? So do we have Excel? Yes, we have Excel two pi multiplied by
the frequency of p, which is given in our problem, multiplied by the inductance, which is also given. So we have Excel
and do we have Q? From here, we can get the equivalent
resistance of circuit. Okay? So if we get back here, you will see that this circuit can have how many
resistor we have, our S to R dash, which you may or may not exist. Parallel two are parallel, okay? Our l can be RL
and inductance can be exhaled the pattern
and are parallel. So the resistance R parallel, R dash and R S, this is at resonance. Or dash can be exist or not, depending on our calculations as you will see in
the next slide. So in general we have RS and RB. So we can say is that here that are equivalent
for now is our supply. Bilaterally, two are parallel. Okay? And since we have
high-quality vector, will find that there are
parallel is equal to Q square multiplied by RL. Rl is given Q L is equal
to the resistance of the x L over R L. Quality factor Q L
equal to x L over R XL two-parameter blood
buys a frequency Motorola inductance
whenever I l, which is given, okay? You will find it is high
power, high quality factor. So we can use this relation. So let's see. So first we need excel in all of our equation to y multiplied
by the frequency, multiplied by the inductance. Then we will get that QL. So QL is x over r l. Y starts to 1.4, which is larger than ten. So we can use simplifications. We can say is that our
parallel is equal to q alpha square RL parallel
equal to q squared r, which is the resistance
of the inductance. So it will give us 9,859. So this is our parish. Now, since we have QP, we have Excel, so we can get the equivalent resistance
of our circuit. So as you can see, here, we have our equivalent
divided by X L gives us 20, which is this value. And the resistance is the whole resistance
in our circuit. So we will assume that
there is no r dash. Okay? So we have our S parallel to RP. So our S parallel to
RP, which is 9,859. From this equation,
we will get that our supply should be
equal to 17.298 kiloohm. Okay? So the resistance of sublime
should be this value so that our quality factor of
the circuit becomes 20. Okay? Now, as you can see, as you can see is that our resistance of sub
y is 40 kilo Ohm. But our calculations
that gives us that our S should be 17. So we need to reduce
our resistance from 40 kilo-ohm to 17 kilo-ohm. So how can we do this by adding an additional resistance R dash. Okay? So in order to F,
we add our dash. It will reduce the 40 kilo-ohm
to 17.298, like this. So we will have our
supply parallel to an additional resistance
R dash. So let's type it. So we have our supply
equal to 40 kilo. However, the equivalent
is that we need is 17. So we need an additional
resistance or dash so that they're equivalent becomes
17.298 kilo ohms. Okay? So these two are
parallel to each other, so they're equivalent is multiplication
divided by summation. Multiplication
divided by summation gives us this equivalent value. So from here we can
have our dash which is needed certain 0.48 kilo ohm. However, in reality
or in commercial, we don't have any
certainty point to eight. We have exactly are the
closest value is 30 kilo ohm. So we are going to use this
as an additional resistance. Okay? So now we have our resistance. What does an extra step? We need to find the capacitance
and the current source. So the capacitance can be
obtained At resonance. We have x L equal to accessing. Excel equal to x L is
two point multiplied by the frequency to one multiplied by the frequency
multiplied by the inductance. Or we have X l equals
114 as we just obtained. So xc is 714 and x
is one over omega C. From here, we can
get the capacitance equal to this approximate value, which is a commercially
available one. Okay? Okay, so we have
our capacitance now and we have our resistance
which will be added. Now, the last thing we
need is the current, Okay? Current source. So we know that at resonance we have a voltage of ten volt. So this guy is equal
to the current multiplied by total
at resonance, equal to ten volt equal to the current source
which we need. We don't know it. Multiplied by that total. Total at resonance is the
whole resistors exist. So we have our S parallel
to r. So we have RS parallel to r dash
battery two are parallel. All of this is our
equivalent at resonance. So we will type like this. So that total at
resonance is our supply power to q squared or n. Okay? Now, you will tell
me all supply, but we have our dash here. When we say all supply, we mean that all of this, the spot or supply better to r dash is considered as
a supply resistor R S, which is the
equivalent of 17.298. So remember, our
current source has an internal
resistance of 40 kilo or by adding an additional one, we reduced it to 17.298. Okay? So this representing
RS parallel to attach better as
two are parallel. This is our R parallel. The equivalent that
will be 6.2 kilo. So we have current multiplied by 6.28 kiloohm gives us ten volt. So the current will be the
voltage divided by resistance. So our supply will be
approximately 1.6 milli and bear. So when we combine
all of our this, our knowledge in this problem, we have our final solution
which is 1.6 million pair. Our supply 40 kilo-ohm are
the additional resistance, 30 kilo ohm to reduce it to 17. And we have capacitance
or 0.01 microfarad. Now, before we end this course, I would like to say
thank you for learning or selecting our course in order to learn
about resonance. I hope this course
was helpful for you and wish you all the best. Thank you and see you
in another course.
Engr. Ahmed Mahdy/ Khadija Academy, Electrical Engineering Classes
