Superposition Theorem

This theorem states that:
“In a network of linear resistances containing more than one generator,the current which flows at any point is the sum of all the currents which would flow at that point if each generator where considered separately and all the other generators replaced for the time being by resistances equal to their internal resistances ”
Explanation
This theorem is only applicable to the linear networks where current is linearly related to voltage as per ohm’s law. This theorem states that if some emf’s act simultaneously in any linear bilateral network, than each emf will act freely. The amount of current in any conductor is the algebraic sum of the voltages that each emf would have produced while circulating freely.
In other words, any conductor of the network is obtained by superimposing the currents and a voltage due to each emf is present in the network. Figure (a) shows the conditions obtained when left hand side battery acts alone whereas the figure (b) shows that the right hand side battery acts alone in the circuit.
By combining the current values of Figure a and Figure b, we get

I1=I1’-I1’’; I2=I2’’-I2; I=I’+I’’



Example

  By using Superposition Theorem, find the current in resistance R shown in Figure.
R1 = 0.005 Ω, R2 = 0.004 Ω, R = 1 Ω, E1 = 2.05 V, E2 = 2.15 V
Internal resistances of cells are negligible.

Solution.

 In Fig. (b)  E2 has been removed.  Resistances of 1 Ω and 0.04 Ω are in parallel
across poins A and CRAC = 1 || 0.04 = 1 × 0.04/1.04 = 0.038 Ω.  This resistance is in series with
                = I1 + I2 = 0.896 + 1.16 = 2.056 A.


0.05 Ω.  Hence, total resistance offered to battery E1 = 0.05 + 0.038 = 0.088 ΩI = 2.05/0.088 = 23.3 A. 
Current through 1-Ω resistance, I1 = 23.3 × 0.04/1.04 = 0.896 A from C to A.
When E1 is removed, circuit becomes as shown in Fig.(c).  Combined resistance of paths CBA and CDA is = 1 || 0.05 = 1 × 0.05/1.05 = 0.048 Ω.  
Total resistance offered to E2 is = 0.04 + 0.048 = 0.088 Ω.  
Current I = 2.15/0.088 = 24.4 A.  
Again, I2 = 24.4 × 0.05/1.05 = 1.16 A.

To current through 1-Ω resistance when both batteries are present