In this lecture we are going to have discussion on superposition theorem. We will see what is superposition theorem and how to use this theorem to do the analysis of linear bidirectional networks and we will start with the statement of superposition theorem. Superposition theorem states that the voltage across or the current through an element in a linear circuit is the algebraic sum It is the algebraic sum of the voltages across or the currents through that element due to each independent source acting alone. This statement will make complete sense after solving one example. Now from this statement, we can say that in a linear bidirectional network containing more than one independent source, the response in any element is the sum of responses obtained.
with one source acting at a time and the other sources being turned off. Now, what do we mean by turned off? Turned off means all the independent sources are replaced by their internal resistances. That is, we replace every voltage source by 0 volt or we can say the short circuit and every current source by 0 ampere or we can say that open circuit. For example, here we have this voltage source and if we want to turn off this voltage source, then we will simply short circuit this voltage source.
Here we have short circuited this voltage source and if we have the current source and we want to turn it off, then we will open circuit this current source. Here we have open circuited this current source. So this is what do we mean by turning off the independent sources.
Now before we move on to the example to understand the superposition theorem in a better way, we will first note down two important points. According to the first point, we do not turn off the dependent sources. The dependent sources are left as they are. This is important point to remember.
In second point, we should note down that the superposition theorem is not valid in case of non-linear circuits. So this is all for the introduction. Now we will move on to the implementation.
And I have taken one question to understand how we can use superposition to find the voltage across an element. So here is our network and we need to find this voltage V by using the superposition theorem. I will give you few seconds to analyze the given network.
I hope you are done. Now we will move on to the solution. When you observe the given network, you will find we have one and two sources and we have two resistors out of which we need to find the voltage across four ohm resistor and for this purpose we will use superposition theorem.
And according to the superposition theorem, the voltage across an element is the algebraic sum of the voltages due to each independent source acting alone. We have two independent sources and we need to find out voltage across this resistor due to each source one by one. And then finally we will add the obtained voltages to have the net voltage V. This will be more clear once we start solving the question. so what i will do first is i will consider the voltage across 4 ohm resistor due to this source so first i will consider the 6 volt source and this means we need to turn off this 3 ampere source that means we will open circuit the source so let us copy the given network and After copying it, we will do some modifications and the modification is we will open circuit this source and we will write V as V sub 1 because this is not the voltage V. We have turned off this source and therefore the voltage obtained here is only due to this source. Now it is very easy to find out V1 if we can find out the current.
through this resistor we can have V1 and to find out the current in this loop we will first assume the current to be I1 and then we will apply KVL. After applying the KVL we will have plus 6 volts minus I1 multiplied to 8 minus I1 multiplied to 4 equal to 0. So we will have I1 equal to 6 divided by 8. plus 4 or we can say that current i1 is equal to 0.5 amperes now v1 it is equal to i1 multiplied to 4 we know i1 it is 0.5 therefore v1 is equal to 0.5 multiplied to 4 that is 2 volts So, in this way we have obtained V1 which is voltage across 4 ohm resistor due to 6 volt source alone. Now we will find out the voltage across 4 ohm resistor due to this source. So this time we will consider the 3 amperes source and this means we need to turn off the 6 volt source. So again we will do the modification in our network.
and this time we are going to short circuit the 6 volt source. So, I will remove this source and then we will have a short circuit plus I will change the name of V. We will call it V2. Now again we will focus on the calculation of voltage across this resistor but this time we will find it out due to this source here when this source is turned off. And to find out the voltage V2 we will first find out the current in this branch.
Let's say the current is I2 and once we have I2 we can have V2. To find out I2 we will use the current divider rule. According to the current divider rule we will have I2 equal to the total current entering multiplied to the other resistance which is 8 ohms divided by The sum of two resistances, 8 plus 4. So, we have I2 equal to 3 multiplied to 8 divided by 12. So, in this way we have obtained I2 equal to 2 amperes.
Now, we can find out V2. V2 is equal to I2 which is 2. multiplied to 4 ohms. So finally we are getting V2 equal to 8 volts.
So 8 volts is the voltage across 4 ohm resistor because of this 3 ampere source. Now the net voltage across this resistor which is V is equal to V1 plus V2 and this is from superposition theorem. we know v1 we know v2 we will have v equal to 2 plus 8 that is 10 volts so 10 volts is the answer