Superposition Theorem states that in a linear network containing several independent sources, the overall response at any point in the network equals the sum of responses due to each independent source considered separately with all other independent sources made inoperative.

- This is only applicable to circuits with linear elements.
- If two or more than two independent sources (voltage or current) are operating in the circuit then the voltage across any element or current through any element is the sum of current and voltages due to individual sources.

In order to calculate the contribution made by each source in a circuit it is important to replace other sources without affecting the circuit. This theorem helps to solve complex circuits by converting them to Thevenin’s or Norton’s equivalent circuit.

The voltage source which we need to replace in the circuit is short circuited i.e set to zero. Similarly, if we want to replace the current source it should be open circuited i.e. set to zero.

## Points to consider

- The sign conventions are important while summing individual contributions from each source present in the circuit. The positive sign is assigned when the contribution from both sources is in the same direction. The negative sign is assigned when the contribution from both sources is in the opposite direction.
- The circuit components should be linear.
- Since power is a non-linear quantity this theorem is not applicable.

## Steps for applying Superposition Theorem

- Firstly, we need to select any one of the sources amongst all the sources present in the circuit.
- Secondly, we replace all the other sources in the circuit with their internal impedances.
- Now simplify the given network and calculate the value of current or voltage across the particular element in the network.
- We should repeat the same process for all the other sources as did for the first one.
- Finally, when we have the values against all the individual sources we sum up all the values of voltages and currents through the circuit element.

Question 1. Find the current trough 3Ω resistance.

Solution:

Firstly, we Short circuit 2V voltage source.

I_{1 }= 0

Secondly, we Open circuit 6A. Then the value of current through 3Ω is

I_{2 }= ⅔ A

## Special Case

Since two voltage sources with different magnitudes in parallel and we cannot connect them as in a single branch, two different currents are not possible (if 5V then I = zero).