**Superposition Theorem:** For analysis of electrical circuits where we need to determine the current in a particular branch of the circuit containing two or more voltage and current sources, we apply the superposition theorem.

We can apply this theorem in the circuits containing voltage sources, current sources, or both.

Contents

## Statement

Superposition theorem states that:

“For a linear resistive network containing two or more number of voltage-source, current source or both then the current through any element (resistance or source) can be determined by algebraically adding the currents produced by each source acting alone when all the other voltage source or the current source present in the network are replaced by their internal resistances. If the sources do not have their internal resistance then the voltage source is replaced by a short circuit and the current source is replaced by an open circuit.”

## Application of Superposition Theorem

In general, the superposition theorem is telling that if there are several numbers of sources acting simultaneously in an electric circuit then the effect of each source can be taken independently as if the other source does not exist in the circuit and the current in any branch of the circuit will be equal to the algebraic sum of the currents due to each source separately when the other source effect is not considered.

The application of the theorem eliminates solving of simultaneous equations.

**Why this theorem is not applicable in the calculation of power?**

*Since power is the square of current which is non-linear so power cannot be calculated using the superposition theorem. Also, we cannot apply this theorem to the circuit consisting of non-linear elements and the circuit consisting of less than two independent sources. *

## Procedure for Applying Superposition Theorem

We can apply the following procedure while applying the superposition theorem in any electrical network or circuit.

- Consider a source whose effect is to be considered then replace all the other sources with its internal resistance. If the internal resistance is negligible or not provided during the numerical problems then replace the voltage source with a short circuit and replace the current source with an open circuit.
- Then compute the current in various branches of the circuit using Ohm’s law.
- Repeat the process 1 and 2 for every source present in the circuit. Such as if the problem consists of 2 sources then the step 1 and 2 has to be done for two times considering each source at a time.
- Finally, compute the total current in the branch. This total current will be equal to the algebraic sum of the currents due to each source.

## Examples

We will be looking at some examples of this theorem. This will familiarize us with the theorem and its procedure for the analysis of electrical circuits or networks.

__Example-1:__

__Example-1:__

Here in the following electrical circuit, we will find the current flowing through the 10 Ω resistor using the superposition theorem.

Here at first let’s consider the 30 A current source. So we will leave the 30 A current source as it is in the circuit and replace the 60 V voltage source with the short circuit as shown below.

Now the current through 10 Ω resistor is calculated as

_{1}is calculated using the current divider.]

Now let’s consider a 60 V voltage source. So we will leave the 60 V voltage source as it is and replace the 30 A current source with the open circuits shown below.

Then current through 10 Ω resistor is calculated as

Finally, the total current flowing through the 10 Ω resistor is the algebraic sum of I_{1} and I_{2}.

__Example-2:__

__Example-2:__

Here in the following electrical circuit, we will find the current flowing through the 100 Ω resistor using the superposition theorem.

Here at first let’s consider the 10 A current source. So we will leave the 10 A current source as it is in the circuit and replace the 5 V voltage source with a short circuit as shown below.

Now the current through 100 Ω resistor is calculated as

Now let’s consider a 5 V voltage source. So we will leave the 5 V voltage source as it is and replace the 10 A current source with open circuits shown below.

Then current through 100 Ω resistor is calculated as

The total current through 100 Ω resistor with the reference direction as shown in Figure: 2 is given by