# Superposition Theorem

**Superposition theorem** states that in any linear, active, bilateral network having more than one source, the response across any element is the sum of the responses obtained from each source considered separately and all other sources are replaced by their internal resistance.The superposition theorem is used to solve the network where two or more sources are present and connected.

**Contents:**

In other words, it can be stated as if a number of voltage or current sources are acting in a linear network, the resulting current in any branch is the algebraic sum of all the currents that would be produced in it, when each source acts alone, all the other independent sources are replaced by their internal resistances. It is only applicable to the circuit which is valid for the ohm’s law (i.e., for the linear circuit).

## Explanation of Superposition Theorem

Let us understand the superposition theorem with the help of an example. The circuit diagram shown below consists of a two voltage sources V_{1 }and V_{2}.

First, take the source V_{1 }alone and short circuit the V_{2 }source as shown in the circuit diagram below

Here, the value of current flowing in each branch, i.e. i_{1}’, i_{2}’ and i_{3}’ is calculated by the following equations.

The difference between the above two equations gives the value of the current i3’

Now, activating the voltage source V_{2 }and deactivating the voltage source V_{1} by short circuiting it, find the various currents, i.e. i_{1}’’, i_{2}’’, i_{3}’’ flowing in the circuit diagram shown below

And the value of the current i_{3}’’ will be calculated by the equation shown below

As per the superposition theorem the value of current i_{1}, i_{2}, i_{3} is now calculated as

Direction of current should be taken care while finding the current in the various branches.

## Steps for Solving network by Superposition Theorem

Considering the circuit diagram A, let us see the various steps to solve the superposition theorem

**Step 1 –** Take only one independent source of voltage or current and deactivate the other source.

**Step 2 –** In the circuit diagram B shown above, consider the source E_{1} and replace the other source E_{2} by its internal resistance. If its internal resistance is not given, then it is taken as zero and the source is short circuited.

**Step 3 –** If there is a voltage source than short circuit it and if there is a current source than just open circuit it.

**Step 4 –** Thus, by activating one source and deactivating the other source find the current in each branch of the network. Taking above example find the current I_{1}’, I_{2}’and I_{3}’.

**Step 5 –** Now consider the other source E_{2} and replace the source E_{1} by its internal resistance r_{1 }as shown in the circuit diagram C.

**Step 6 –** Determine the current in various sections, I_{1}’’, I_{2}’’ and I_{3}’’.

**Step 7 –** Now to determine the net branch current utilizing the superposition theorem, add the currents obtained from each individual source for each branch.

**Step 8 –** If the current obtained by each branch is in the same direction than add them and if it is in the opposite direction, subtract them to obtain the net current in each branch.

The actual flow of current in the circuit C will be given by the equations shown below

## 7 Comments

fantastic…

hiii this is kartik, it benefits us to deep study of theorems

Hi,

Nice Tutorial. Keep it up.

very good

Hi, the last equation about I_3 isn’t wrong? It should have been:

I_3 = I’_3 + I”_3

i like the steps. It helped me a lot

Did extraordinary work.

many people make this complex.

making simple is special..