Superposition Theorem Explained (with Examples) – Howtoshtab – how to, lifehacks, tips and tricks

Hey friends, welcome to the YouTube channel 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. If you like this video, hit the like button and subscribe to the channel for more such videos.

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