Explain Superposition Theorem Example and Solution

Superposition theorem state that the voltage across (or current through) an element in a linear electric circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. The superposition theorem example can help understand the idea.

The above superposition theorem statement means that each independent source has its effect on each circuit’s passive elements. And the final effect can be found using the algebraic sum of all the above effects.

If an electrical circuit has multiple independent sources and the circuit is a linear circuit, so, the circuit can be analyzed using the superposition theorem.

Superposition Theorem Steps

To use the superposition theorem for electrical circuit analysis, follow the steps below.

  1. Replace all current sources with an open circuit and voltage sources with a closed circuit, expect, the source you are considering in the next steps.
  2. Find the current in each branch and voltage at each node, whatever technique you like.
  3. Repeat the above two steps for each independent source.
  4. Finally, find the algebraic sum of each branch current and node voltage from the above individual currents and voltages.

Consider the following superposition theorem example for explanation. 

Superposition Theorem Example

Find the value of voltage V2 across R2 using the superposition theorem in the following circuit.

Superposition Theorem example

Solution:

Short Circuit Sources:

Replace all independent current and voltage sources with short circuits expect we are considering in the next step. And I am going to consider only the voltage source V=6v in the next step.

Solve for single source:

To find the effect of Voltage Source, V= 6v, first of all, we open the circuit source, I, and the circuit will become something like this.

SuperPosition Theorem step1 Solved example

Now applying the applying Voltage Divider Rule to the above diagram will give us:

$v_{12}= \frac{V \times R_{2}}{R_{1}+R_{2}}$
$v_{12}= \frac{6 \times 4}{8+4}$
$v_{12}= \frac{24}{12}
v_{12}= 2$

Solve for another source:

To find the effect of Current Source, I= 3A, I short circuit the voltage source V=6v and the circuit will become something like this:

SuperPosition Theorem solved example

Now, apply the Current Divider Rule to the above circuit.
$i_{2}= \frac{I \times R_{2}||R_{1}}{R_{2}}$
$i_{2}= \frac{3 \times 4||8}{4}$
$i_{2}= \frac{3 \times 2.66}{4}$
$i_{2}=2 A$

Now finding the voltage across V$_{2}$, using Ohm’s law: 

$v_{22}=i_{2}R_{2}$
$v_{22}=2 \times 4$
$v_{22}=8$

Find Algebraic sum:

Finally, find the algebraic sum of the voltage from individual effects obtained in the above steps and we are done.

$v_{2}=v_{12}+v_{22}$
$v_{2}=2+8$
$v_{2}=10 v$

To wrap things up, the superposition theorem is the circuit analysis tool considering the individual effect of sources. And the theorem can only be applied to the linear circuits. Where the linear circuit is those which pose the property of homogeneity and additivity.   

7 thoughts on “Explain Superposition Theorem Example and Solution”

  1. Now, apply Current Divider Rule to the above circuit.
    i2=I×R2||R1R2
    i2=3×4||84
    i2=3×2.664
    i2=2A

    what about that parallel sign, isnt that division or something? please tell me, im stuck. i dont know what to google tho.

    Reply

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