# Verification of Superposition Theorem

## Aim

Verification of Superposition Theorem

## Apparatus

Breadboard, Batteries or DC regulated power supply, Resistors, Digital multimeter, Connecting wires, Alligator clips, Computer and Multisim software for simulation.

## Theory

Statement: If a number of voltage or current source are acting simultaneously in a linear network, the resultant current in any branch is the algebraic sum of the currents that would be produced in it, when each source acts alone replacing all other independent sources by their internal resistances.

### Steps to Solve Circuits using Superposition Theorem

1. To find the current/ voltage in any branch, take one source at a time and replace rest of the sources by their internal resistances (if given).
2. Calculate the current/voltage by any method (mesh/nodal/KVL/KCL).
3. Now calculate the current/ voltage in the same branch by taking the other source in the circuit and replacing rest of the sources by their internal resistances.
4. Repeat steps 1,2 till all the sources have been considered.
5. Total current/voltage in the given branch= algebraic sum of all the currents/voltages in the branch due to all the current sources.

Note:

• Replace Voltage sources by Short Circuit if their internal resistance is not given.
• Replace current sources by Open Circuit if their internal resistance is not given.

Consider the circuits shown in the figure 1,2 and 3. As shown in figure 1, total current flowing through resistor of 1Ω can be calculated by using Ohm’s Law. Hence total current flowing through resistor is Itotal=3A.

Now, Let’s find the current by Superposition theorem,

Step 1: As shown in figure 2

Only 1V source present (Replace 2V source by short circuit)

∴I’ = 1A

Step 2: As shown in figure 3

Only 2V source present (Replace 1V source by Short Circuit)

∴I” = 2A

Step 3: Sum of currents as per Superposition Theorem

Itotal = I’ + I”

= 1+2

= 3A

## Circuit Diagram Figure 5: V1 is acting alone and V2 is short circuited Figure 6: V2 is acting alone and V1 is short circuited

## Procedure

1. Connect the circuit as shown in the figure 4.
2. Measure and note down the current I1, I2 and I3flowing through resistors R1, R2 and R3
3. Connect the circuit as shown in the figure 5 that is replace V2 by short circuit.
4. Measure and note down the current I1’, I2’ and I3’flowing through resistors R1, R2 and R3
5. Connect the circuit as shown in the figure 6 replace V1 by short circuit and connect V2.
6. Measure and note down the current I1’’, I2’’ and I3’’flowing through resistors R1, R2 and R3

## Precautions

• All the connection should be tight.
• Ammeter must be connected in series while voltmeter must be connected in parallel to the components (resistors).
• Before circuit connection working condition of all the components must be checked.
• The electrical current should not flow the circuit for long time, otherwise its temperature will increase and the result will be affected.

R1=_____ Ω

R2=_____ Ω

R3=_____ Ω

For figure 4

I1 =

I2 =

I3 =

For figure 5

I1’=

I2’ =

I3’ =

For figure 6

I1’’=

I2’’ =

I3’’ =

## Calculations

For measured values:

I1 = I1’ + I1

=

I2 = I2’+ I2

=

I3 = I3’ + I3

=

[Calculate current flowing through each resistor by using any method, preferably mesh analysis]

## Result

 I1 (mA) I2 (mA) I3 (mA) I1’ (mA) I2’ (mA) I3’ (mA) I1” (mA) I2” (mA) I3” (mA) Theoretical Practical

## Conclusion

It is found that the current flowing through each branch (Resistor) when both voltage sources are acting together is equal to/ nearly equal to algebraic sum of the currents that would be produced in it, when each source acts alone replacing remaining voltage source by their internal resistances.