# Superposition Theorem Explanation with Example

#### What is Superposition Theorem ?

Superposition theorem allows us to determine the effect of several energy sources (voltage and current sources) acting simultaneously in a circuit by considering the effect of each source acting alone and then combining or superposing these effects.

This theorem is applied to D.C. circuits.

The superposition theorem may be stated as :

In a linear, bilateral d.c. network containing more than one energy source, the resultant potential difference across any element or current through any element is equal to the algebraic sum of potential differences or currents for that element produced by each source acting alone with all other independent ideal voltage sources replaced by short circuit and all other independent current sources replaced by open circuits.

(Note : non-ideal sources are replaced by their internal resistance)

#### Procedure

The procedure for using the Superposition theorem  to solve d.c. network is as under :

##### Step 1 :

Select one source in the circuit and replace all other ideal voltage sources by short circuits and ideal current sources by open circuits.

##### Step 2 :

Determine the voltage across or current through the desired element or branch due to single source selected in step 1.

##### Step 3 :

Repeat the above two steps for each of the remaining sources.

##### Step 4 :

Algebraically add all the voltages across or currents through the element or branch under consideration. The sum is the actual voltage across or current through that element or branch when all the sources are acting simultaneously.

#### Prerequisites for Using Superposition Theorem

• Superposition theorem can only be used for the balanced bridge circuits, which are reducible to series/parallel combinations.
• All the components of the circuit must be linear and bilateral.
• Superposition theorem is only applicable for determining voltage and current, not power.

Fig.1

##### Step 1 :

Here we only have two voltage sources (batteries) in our example circuit in Fig.1.

Let’s consider the 40 V source alone first and short circuit the other 10 V voltage source.

Fig.2 : Second voltage source replaced with a short circuit

##### Step 2 :

Now, let’s determine the current through the 10 Ω resistor.

We can observe that in the above circuit, 10 Ω and 5 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 20 Ω resistor.

So, the equivalent or total resistance is obtained as below :

Total resistance RT = 20 + [(10 × 5)/15] = 23.33 Ω

By applying Ohm’s law,

IT = V/RT = 40/23.33 = 1.714 A

Now, the current through 10 Ω resistor can be determined by using current division method.

Hence I1 = IT × [5/(5+10)] = 1.714 × 0.333 = 0.570 A

##### Step 3 :

Now, consider the 10 V source alone and short circuit the other 40 V voltage source. The modified circuit is shown in Fig.3 below.

Fig.3 : First voltage source replaced with a short circuit

To find out the current through the 10Ω resistor, first we have to find out the total current IT .

We can observe in the above circuit, 10 Ω and 20 Ω resistors are in parallel with each other. This parallel combination is connected in series with a 5 Ω resistor. Hence the equivalent or total resistance is obtained as below,

Total resistance RT = 5 + [(10 × 20)/30] = 11.66 Ω

By applying Ohm’s law,

IT = V/RT = 10/11.66 = 0.857 A

Now, the current through 10 Ω resistor can be determined by using current division method.

Hence I3 = IT × [20/(20+10)] = 0.857 × 0.666 = 0.571 A

##### Step 4 :

By superposition theorem, the total current is determined by adding the individual currents produced by 40 V and 10 V.

Therefore, the current through 10 Ω resistor is :

I1 + I3 = 0.570+ 0.571 =  1.141 A