Basic Electrical Engineering Series

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Index

Definition

Superposition Theorem

“ An emf acting on any linear network produces the same effect whether it acts alone or in conjunction with other emfs ”…

Hence a network with many sources of emf may be analyzed by considering the currents due to each emf alone with all other emfs being replaced by their internal impedance’s – similar to what is done in Thevenins theorem.

Consider a two-mesh problem:

To find I, we can first replace E2/R2 with just R2 and calculate the current through RL due to E1. Then replace E1/R1 with just R1 and caculate the current due to E2. Then add (superimpose) them together. That’s it!

Example

1) E1=10, E2=12, R1=120, R2=120, R3=100 Find the current I_R3

Solution by superposition:

I_R3 due to E1, replacing E2

R2//R3 = (120*100)/(120+100) = 54.545 (resistors in parrallel)

V_R3 = (54.545 / 174.545)* 10 = 3.1249 (potential divider)

I_R3 = V_R3 / 100 = 0.03125. (Ohms law)

I_R3 due to E2, replacing E1

R1//R3 = (120*100)/(120+100) = 54.545 (resistors in parrallel)

V_R3 = (54.545 / 174.545)* 12 = 3.745 (potential divider)

I_R3 = V_R3 / 100 = 0.03745. (Ohms law)

Total I = 0.0687

Cross-Check using Millman’s theorem:

Vn = [(10/120) + (12/120)] / [(1/120) + (1/120) + (1/100)] = 6.875

Hence I = 6.875/100 = 0.0688 to 4 dp