Analyzing Circuits with Two Power Sources

Superposition Theorem of DC Circuit Analysis

Electrical Theory | Apprentice Electrician Training |

1. What is the Superposition Theorem?

The Superposition Theorem is a method of analyzing linear electrical circuits containing two or more independent power sources. The current through or voltage across an element equals the algebraic sum of the contributions produced by each independent source acting alone.

In other words, calculate the effect of each battery separately, then combine the signed answers. When sources cause current in the same direction, their effects reinforce each other; when they cause current in opposite directions, their effects oppose each other.

Three important points to remember

• Analyze one independent source at a time.

• Replace every other ideal independent voltage source with a short circuit (wire).

• Algebraically add the currents or voltages, including their directions and polarities.

2. Why use superposition?

With one source, Ohm’s Law and series/parallel resistance are often sufficient. With multiple sources, a source may assist or oppose current through the same resistor. Superposition breaks a complicated circuit into simpler calculations. Both sources operate simultaneously in the real circuit; the separate cases are a mathematical analysis technique.

3. Rules for source deactivation

• Keep the independent source being evaluated operating.

• Replace another ideal independent DC voltage source with a short circuit.

• Replace another ideal independent current source with an open circuit.

• Keep all resistors and dependent sources in place.

• Retain any physical internal source resistance in the model.

• Choose a positive current direction or voltage polarity and keep it consistent.

Important: Shorting an ideal voltage source in a circuit diagram is a mathematical operation, not an instruction to short an energized battery.

4. Example 1: Two DC sources contributing in the same direction

Find the current through R3. Both sources have positive terminals at the top. Define downward current through R3 as positive.

Original circuit: 12 V and 6 V sources; R1 = 2 Ω, R2 = 3 Ω, R3 = 6 Ω.

Step 1 — Analyze the 12 V source

Deactivate the ideal 6 V source by replacing it with a wire. R1 is now in series with the parallel combination R2 || R3.

12 V active; 6 V replaced by a short circuit.

Rtotal = 2 + 2 = 4 Ω

Itotal = 12/4 = 3 A

Voltage across the parallel branch = 3 × 2 = 6 V

IR3 due to 12 V = 6/6 = +1.000 A (downward)

Step 2 — Analyze the 6 V source

Restore the 6 V source and short the ideal 12 V source. R2 is in series with the parallel combination R1 || R3.

Rtotal = 3 + 1.5 = 4.5 Ω

Itotal = 6/4.5 = 1.333 A

Voltage across R3 = (6/4.5) × 1.5 = 2 V

IR3 due to 6 V = 2/6 = +0.333 A (downward)

Step 3 — Combine the contributions

IR3 = 1.000 + 0.333 = 1.333 A downward

VR3 = (4/3 A) × 6 Ω = 8 V

Point A is 8 V above the common conductor. Because Point A is also above the 6 V source’s positive terminal, current through R2 flows toward that battery; in this ideal model, the 6 V source absorbs power.

5. Example 2: Reversing the 6 V source

Reverse only the 6 V source: its negative terminal is now at the top, and its positive terminal at the bottom. The resistors and 12 V source remain the same.

Step 1 — Analyze the 12 V source

With the 6 V source deactivated, the 12 V-only circuit is unchanged.

IR3 due to 12 V = +1.000 A (downward)

Step 2 — Analyze the reversed 6 V source

With the 12 V source deactivated, the magnitude of current due to the 6 V source remains 0.333 A, but the polarity reversal makes it flow upward through R3.

R1 || R3 = 1.5 Ω; Rtotal = 4.5 Ω

Itotal = 6/4.5 = 1.333 A

IR3 due to reversed 6 V = −0.333 A (upward)

Step 3 — Add signed currents

IR3 = +1.000 + (−0.333) = +0.667 A (downward)

VR3 = (2/3 A) × 6 Ω = 4 V

6. Comparing the two circuits

Quantity	Example 1	Example 2
12 V contribution	+1.000 A	+1.000 A
6 V contribution	+0.333 A	−0.333 A
Total R3 current	1.333 A down	0.667 A down
Voltage across R3	8 V	4 V

7. Important limitation: Power does not superimpose directly

Superposition applies directly to voltage and current in linear circuits, not to resistor power. Power depends on the square of current, P = I²R. First determine the total current or voltage, then calculate power.

Correct: P(R3) = (4/3 A)² × 6 Ω = 10.667 W

Incorrect: (1 A)² × 6 Ω + (1/3 A)² × 6 Ω = 6.667 W

8. Student practice problem

Use superposition to find current and voltage across R3. Downward current is positive. Both sources have positive terminals at the top.

A. 1.0 A     B. 2.0 A     C. 3.0 A     D. 4.5 A

Practice answer and worked solution

With only the 24 V source operating: Rparallel = 4 || 4 = 2 Ω; Rtotal = 4 + 2 = 6 Ω; Itotal = 24/6 = 4 A. Current divides equally, so the contribution through R3 is +2 A.

With only the 12 V source operating: Rtotal = 4 + (4 || 4) = 6 Ω; Itotal = 12/6 = 2 A. It divides equally, so the R3 contribution is +1 A.

Answer C: IR3 = 2 + 1 = 3 A downward

VR3 = 3 × 4 = 12 V

9. Summary

• Analyze each independent source separately.

• Short ideal independent voltage sources that are not being analyzed.

• Keep all resistors in the circuit.

• Add currents and voltages algebraically with consistent sign conventions.

• Calculate power only after combining source contributions.