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ALL ABOUT ELECTRONICS. So, in this video,
we will learn about the Superposition theorem
in the electrical circuits. So, in this video
we will see some features of this superposition
theorem, like under which condition this superposition
theorem can be applied, what is the usefulness
of the superposition theorem, what are the
limitation of the superposition theorem, and
we will solve some problems based on this
superposition theorem. So we will understand
this superposition theorem by taking one example.
So, as you can see here, we have total two independent
voltage sources. And let us say we want to
find the current I that is flowing through
this resistor R1. So using this superposition
theorem what we can do, we can consider this
one particular voltage source or a one particular
independent voltage source at a time. And we
can find the current that is flowing through this
resistor R1. So, suppose if we consider
this voltage source V1 alone then we can
turn off this voltage source V2. And we can
replace this voltage source V2 by it's
internal impedance. Now here all the independent
voltage sources that we are considering our
ideal sources. So, the ideal voltage source
has zero series internal impedance. So we can
replace this voltage source V2 by short
circuit. So in this case, we can find the current
that is flowing through this resistor R1.
Let's say the current that is flowing
through this resistor R1 is a I1. Similarly,
in the second case what we can do, we can consider
this voltage source V2 alone and we can turn
off this voltage source V1. So we can replace
this voltage source V1 by a short circuit.
And in this scenario we can again find the
current that is flowing through this resistor
R1. Let us say, in this case the current
is flowing through this resistor R1 is
I2. Now, the total current that is flowing
through this resistor R1 will be nothing
but algebraic sum of this current I1 and
I2. So, the total current that is flowing through
this resistor R1 is nothing but a algebraic
sum of this current I1 and I2. So, according
to the superposition theorem in any linear
bilateral network containing more than one
independent sources the response in any particular
branch either voltage or current, that is equal
to the algebraic sum of the responses that
is caused by individual forces acting alone.
And at the same time all the other sources
are replaced by the internal impedance. Now
in the most of the cases, as we are considering
the ideal sources, so all the ideal voltage
sources can be replaced by the short circuit
and all the ideal current sources can be
replaced by the open circuit. So, now in this
statement if you see, we are considering linear
bilateral network. So, this superposition theorem
can be applied only for the linear and the
bilateral network. Now, if you don't know about
this linear and bilateral network, then you
can check my video on the Classification of
the Electrical Network. The link for the same I will provide in the description below.
So you may check that out. So now let us see some features of this superposition theorem.
So as I said earlier,
this superposition theorem is only applicable
for the linear and bilateral networks. Now this
particular superposition theorem is useful
when a circuit contains more than one sources
with different frequencies. Now, if your circuit
contains only DC sources, then there are many
ways you can find that solution. But if the circuit
contains AC sources and also with a different
frequencies then this superposition theorem
is quite useful. Now, apart from the independent
sources, the circuit contains dependent sources
as well, then those dependent sources will be
considered ON during the analysis. So, this
will get clear to you when will take one example
based on this dependent sources.
Now, this superposition theorem is not applicable for the power calculations.
So, we will understand more about it when 
 we will take one example based on this power calculations
So, now let's see examples based on this superposition
theorem. So, in the first example we have given
given this electrical circuit and we have been asked
to find the current that is flowing through
this 1 ohm resistor. So, we will solve this
example using this superposition theorem.
So, what you will do, we will consider only one particular
independent source at a time and because of
that independent sources, we will find the
current that is flowing through this resistor
R and the algebraic summation of those currents
will be the total current that is flowing through
this resistor R. So, first of all let us consider
this one ampere current source. Now suppose
if we consider only this one ampere current
source only then this voltage source will
get short circuited. And if you see the equivalent
circuit, the equal circuit will look like this.
So, now in the equivalent circuit if you see,
these two 1 ohm resistor are in parallel. Similarly,
these two 1 ohm resistors are also in a parallel.
S,o we can redraw this circuit like this. So,
now here, as these two 1 ohm resistors are in parallel, so
we can replace them by the equation parallel
resistance, that will be the 0.5 ohm. Similarly,
the equivalent parallel resistance of these two
two 1 ohm resistor will be 0.5 ohm. So, now the equivalent
circuit will look like this. So now, this one
ampere current will get divided in two branches.
Some portion of this current will flow through
this 0.5 ohm resistor and the remaining current will
flow through this series combination of this
0.5 ohm and  1 ohm resistor. So, the equivalent
circuit will look like this. So, the current
is flowing through this 1.5 ohm resistor is
nothing but 0.5*1/(0.5+1.5), which
will come out as a 0.25 ampere. This
is nothing but a current divider rule. So, using
this current divider rule the current that
is flowing through this series combination
of 1 ohm  and 0.5 ohm resistor is nothing but
0.25 ampere. So, we can say that current
I1 that is flowing for this resistor R is
nothing but 0.5 ampere. So, now let's
consider only this 1 volt voltage source.
And let us assume that the current that is
flowing through this resistor R is nothing
but I2. So, when we consider only this particular
voltage source, then this current source can be replaced
by the open circuit. And the equivalent circuit
will look like this. So, now let us redraw this
circuit. So, as you can see here, this is the
terminal A. So we can say that this is also terminal
A. And this is terminal B. so first of all
as you can see two 1 ohm resistors are
connected to this positive terminal of this
1 volt voltage source. So we can we draw this
circuit like this.So these are the two 1 ohm
resistors which are connected to the positive
terminal of this 1 volt voltage source.
And this terminal will be a thing but A terminal.
Similarly, at the negative terminal of this
1 volt voltage source, once again these two 1
ohm resistors are connected. So, we can redraw
these two 1 ohm resistors like this.So,  again here
this point will be nothing but a terminal
B and across this terminal A and B, one ohm resistor R
is connected. So, the equivalent circuit now we
can redraw like this. So, now if you see the
circuit, this circuit is nothing but a balanced
Wheatstone Bridge circuit. So as this Wheatstone Bridge is a balance Bridge, so the current
that will flow through this 1 ohm resistor
will be nothing button 0. So, we can say that
the value of this I2 is nothing but a zero
ampere. So in this way we found out that the
current I1 is 0.25 ampere, when this
one ampere current source is acting alone
and current I2 is zero when this only one
voltage source is acting alone. The total
current I that is flowing through this resistor R
 is nothing but I1+I2. That
will be nothing but 0.25 ampere. So, in this
way using this  superposition theorem we can find
the total current that is flowing through this
 resistor R. Now let's see the second example. So in this
example we have given one resistive network
and in this resistive network, three independent
voltage sources are connected. And we have
given that when this independent voltage sources
are acting alone then the power that is dissipated
across resistor R is 18 watt, 50 Watt, and  98 watt.
So, we have been asked to find the possible
maximum and minimum value of power that is
dissipated across resistor R when all the three
sources are acting simultaneously. So here
we have given that when voltage source E1
is acting alone then the power P1 is 18 watt.
Similarly when the voltage source E2  is acting
alone and the power that is dissipated across
the resistor R is 50 Watt. And similarly
when this voltage source E3 is acting alone,
then the power P3 that is disputed across the
resistor R is a 98 watt. Now we can write this
P1 as (I1^2*R). Let us say I1 
is the current that is flowing through this  resistor R
when this E1 Source is acting alone. Similarly,
we can write this power P2 as (I2^2*R).
and we can write this power P3 as
a (I3^2*R). Now, earlier we have discussed
that for the power calculations we cannot
apply this superposition theorem. So if 3 voltage
sources are acting simultaneously. then the total
power P1 +P2 +P3 will not
be equal to  (I1^2*R) + (I2^2*R)+ (I3^2*R). So
directly we cannot add this three powers to
find the total power that is dissipated across this
resistor R. But rather when 3 voltage sources
are acting alone and the total current that
is flowing through this resistor R, that is
I (total) will be nothing but I 1 + I2 + I3. So
the total power P that is dissipated across
this resistor R is nothing but [I(Total)]^2 *R.
So, this will be the total
power that is dissipated across this  resistor R.
Now here in this example we have been asked
to find the possible maximum and the minimum
value of this power. So, first of all, let's
find the value of this I1, I2 and I3.
So, as we know, P1 is equal to 18, is equal
to (I1^2)*R. So, we can say that I
is nothing but √(18/R). Similarly,
I2 will be nothing but √(54/R)
and I3 will be anything but √(98/R). So for maximum possible power Pmax
will be nothing but [(I1 +I2+I3 )^2]*R. That is nothing but
[√18 +√50+√98]^2. And if you
further simplify it then we will get
2*[√9 +√25+√49]^2. So, we'll get the maximum possible power
that is 450 Watt. So we will get the maximum
possible power as a 450 watt. So similarly let's
find out the minimum possible power. The minimum
possible power P min can be given as
[(I3-I2-I1)^2]*R. And if we further
simplify it then we can write it as
[√98 -√50-√18]^2 .
And if we further simplify it then we can
write it as 2*[√49 -√25+√9]^2. And
further if we simplify it then we will get 2*[7-5-3]^2. So the
value of minimum possible power will
come out as 2 watt. So in this way we found
out the maximum possible value as a 450 watt
and minimum possible power as 2 W. So,
here the point is that for the power calculation
we cannot directly use this superposition theorem.
So, for the power calculation, first of all
either we have to find the total current I(T)
or voltage across that particular element.
And from that we can find the total power
that is dissipated across that particular
element. So now let us see the third example.
So, in this example we have to find the power
that is dissipated across this 5 ohm resistor.
Now in the circuit as you can see, we have
one dependent source. So as we have discussed
earlier if the circuit contains a dependent
sources then for the supervision analysis we
will consider this voltage source ON during
the analysis. Now, here to find the power that
is dissipated across this 5 ohm resistor what
we will do, we will find the total current I that
that is flowing through this 5 ohm resistor. So
to find the total current I, we will use the Supervision
theorem and for that will consider only a
one particular source at a time. So first of
all, we will consider this 10 volt voltage
source is acting alone. So in the circuit if
only these 10 volt voltage source is acting
alone then we can replace this 2 ampere current
source by open circuit. And the equivalent
circuit will look like this. So now let us
assume that the current that is flowing through
this 5 ohm resistor is I1. And let us assume
that the voltage across this 1 ohm resistor
is V1' and hence the voltage of this
dependent source will be 4V1'. So we
can apply kvl in this loop. So now if we apply
kvl then we can write it as a 10 + V1'+ 4V1' - 5I1 =0
Now, closely if you look here, thisV1' is
nothing but -I1, that is a drop across
this 1 ohm resistor. So we can replace this
V1' by (-I1). So, we'll get 10 - (5*I1) - (5*I1)=0
Or we can write I1 as 1 ampere. So
in this way we found the current I1 that
is flowing through this 5 ohm resistor, while
this 10 volt voltage sources acting alone.
Now let us find out the current that is flowing
through this 5 ohm resistor when this 2 ampere
current source is acting alone. Let us assume
that the current that is flowing through this
5 ohm resistor is I2. And in that case we can
replace this 10 Volt voltage source by a short
circuit.In that case the equivalent circuit
will look like this. And let us assume that
the voltage across this 1 ohm resistor is nothing
but V1". And the voltage of
the dependent source is 4V1".
So, if you see this circuit, we have total two nodes.
This is the first node and this is the second node.
And let us assume that the voltage at this node
is V1", that is a voltage across
this 1 ohm resistor. And let us assume that
at this and the voltage is nothing but V2.
And the current that is flowing through
this 5 ohm resistor is nothing but I2. So, if
you see here between these two nodes, we have
a supernode. So we will consider these two
node as a super node. And let apply KCL
for this supernode. So applying KCL, we can
write V1"-2+ (V2/5) =0
Now here I2 is nothing
but V2/5, where I2 is the current that is
flowing through this 5 ohm resistor. Now for this
supernode, we can write V2-V1"= 4V1"
Or we can write V2=5V1"
So, now if we
put this value in this equation then we will
get I2= V1". And if we
put this value in this first equation then
we will get I2-2+I2=0
Or we can say that I2 is equal to 1
ampere. So in this way we have found the value
of current that is flowing through this 5
ohm resistor, when this 2 ampere current source
is acting alone. So in this way we found that
value of I1 as 1 ampere and value of I2
as  also 1 ampere. So, the total current I that
is flowing through this 5 ohm resistor is
nothing but the algebraic sum of these two
currents. That is 2 ampere. So in this way the
total current I that is flowing through this
5 ohm resistor is nothing but  2 ampere.
So, now the power that is dissipated across
this 5 ohm resistor is nothing but (I^2)*R.
That is [(2)^2]*5. Which will come
out as 20 W. So, in this way we can find the
power that is disputed across this 5 ohm resistor.
So, in this way using this superposition theorem
we can find the power that is dissipated across
this 5 ohm resistor. So, I hope in this video you
understood what is superposition theorem in
the electrical circuits and using this supervision
theorem how we can solve the circuit problems.
If you have any question or suggestion please
let me know in the comments section below.
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