Superposition Theorem

Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman's certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.

The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all “superimposed” on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active. Let's look at our example circuit again and apply Superposition Theorem to it:

Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in effect. . .

            

                 . . and one for the circuit with only the 7 volt battery in effect:
              

                         

 The theorem states like this: In the network of resistors that is energized by two or more sources of emf, (a) the current in any resistor or (b) the voltage across any resistor is equal to: (a) the algebraic sum of the separate currents in the resistor or (b) the voltages across the resistor, assuming that each source of emf, acting independently of the others, is applied separately in turn while the others are replaced by their respective internal values of resistance.

This theorem is illustrated in the given circuit below:

                      

 The original circuit above ( left part ) have one emf source and a current source. If you like to obtain the current I which is equal to the sum of I' + I"using the superposition theorem, we need to do the following steps:

a. Replace the current source I_o by an open circuit. Therefore, an emf source v_o will act independently having a current I' as the first value obtained when the circuit computed.

b. Replace emf source v_o by a short circuit. This time I_o will act independently and I" now will be obtained when the circuit computed.

c. The two values obtained ( I' and I") with emf and current source acting independently will be added to get I = I' + I"