# Superposition Theorem

For analysis of electrical circuits where we need to determine the current in a particular branch of the circuit containing several voltage and current sources we apply superposition theorem.

This can be applied in the circuits containing voltage sources, current sources or both.

## 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 short circuit and the current source is replaced by open circuit.”

## Application of Superposition Theorem

In general, the superposition theorem is telling that if there are several number of source acting simultaneously in an electric circuit then the effect of each source can be taken independently as if the other source do 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 are not considered.

The application of the theorem eliminates solving of simultaneous equations.

Why this theorem is not applicable in calculation of power?

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

## Procedure for Applying Superposition Theorem

Following are the procedure to be followed while applying superposition theorem in any electrical network or circuit.

1. Consider a source whose effect to be considered then replace all the other source by its internal resistance. If the internal resistance are negligible or not provided during the numerical problems then replace the voltage source by short circuit and replace the current source by open circuit.
2. Then compute the current in various branches of the circuit using Ohm’s law.
3. Repeat the process 1 and 2 for every source present in the circuit. Such as if the problem consists of 2 source then the step 1 and 2 has to be done for two times considering each source at a time.
4. 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 analysis of the electrical circuits or networks.

Example-1:

Here in the following electrical circuit we will find the current flowing through the 10 Ω resistor using 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 by short circuit as shown below.

Now the current through 10 Ω resistor is calculated as [The I1 is calculated using current divider.]

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

Then current through 10 Ω resistor is calculated as Finally the total current flowing through 10 Ω resistor is the algebraic sum of I1 and I2. Example-2:

Here in the following electrical circuit we will find the current flowing through the 100 Ω resistor using 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 by short circuit as shown below.

Now the current through 100 Ω resistor is calculated as Now let’s consider 5 V voltage source. So we will leave the 5 V voltage source as it is and replace the 10 A current source by open circuit s 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 the Figure: 2 is given by 