What does the superposition theorem state and how is it used with multiple sources?

In one line

In a linear circuit with several independent sources, the current or voltage in any element equals the algebraic sum of the values caused by each source acting alone, with the other sources replaced by their internal resistances.

Explained simply

Picture two friends pushing a stalled car from different sides. To see the effect of each, let one push alone and measure, then the other. The combined effect is just the sum (with direction) of the two separate pushes. Superposition does this with sources.

Technically, the theorem applies to circuits that are linear (element values do not change with voltage or current) and bilateral. Each independent source is taken one at a time; ideal voltage sources not in use are replaced by a short circuit and ideal current sources by an open circuit. Internal resistances stay in place.

How it works

1. Keep source 1; short other voltage sources, open other current sources.
2. Find the required current \( I' \) (with direction) through the element.
3. Repeat with each remaining source to get \( I'', I''', \dots \).
4. Add algebraically: same direction add, opposite directions subtract.
5. Power cannot be found by superposition; first find total current, then \( P = I^2R \).

Key formula

\( I = I' + I'' + \dots \) (algebraic sum)

\( I' \), \( I'' \) = currents (A) due to each source acting alone, measured in the same reference direction.

Quick example

A 12 V source in series with 4 \( \Omega \) feeds node A; from A to ground is a 4 \( \Omega \) resistor \( R_2 \); a 3 A current source also feeds into A. Find the current in \( R_2 \).

12 V alone (current source open): \( I' = 12/(4+4) = 1.5 \) A downward. 3 A alone (12 V shorted): the two 4 \( \Omega \) resistors split it equally, \( I'' = 1.5 \) A downward. Total \( I = 3 \) A. Check with nodal: \( (V-12)/4 + V/4 = 3 \Rightarrow V = 12 \) V, \( I = 12/4 = 3 \) A.

Common mistakes

• Adding powers from each source: \( (I'^2 + I''^2)R \neq (I' + I'')^2R \).
• Removing the internal resistance of a source along with the source.
• Opening a voltage source or shorting a current source (it is the other way round).

How MU asks it

Asked as "State and explain superposition theorem" (5 marks) or as a numerical (7–10 marks). For full marks: write the statement with "linear, bilateral", draw a separate circuit for each source acting alone, calculate each part current with its direction, then combine with a clear sign, and state the limitation (not valid for power or nonlinear elements).