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Superposition: current in 3 Ω

Short answer

Applying superposition, short one source at a time: with 10 V alone, current is 0.851 A; with 12 V alone, it is 1.277 A. Both flow the same direction, so total current in the 3 Ω resistor is 2.13 A.

Find the current flow in the 3 Ω resistance. Use the superposition theorem to solve the problem. [Figure: 10 V source in series with 5 Ω in the left branch; 12 V source in series with 4 Ω in the middle branch; 3 Ω resistor in the right branch.]

Method: superposition says the current in the 3 Ω resistor is the sum of the currents due to each source acting alone (the other voltage source replaced by a short circuit).

Assumption: three parallel branches: 10 V with 5 Ω, 12 V with 4 Ω, and the 3 Ω load; both sources have their positive terminals toward the top node.

Case 1: 10 V alone (12 V shorted)

4 Ω ∥ 3 Ω = 12/7 = 1.714 Ω R_total = 5 + 1.714 = 6.714 Ω I = 10/6.714 = 1.489 A Current in 3 Ω: I' = 1.489 × 4/(4 + 3) = 0.851 A

Case 2: 12 V alone (10 V shorted)

5 Ω ∥ 3 Ω = 15/8 = 1.875 Ω R_total = 4 + 1.875 = 5.875 Ω I = 12/5.875 = 2.043 A Current in 3 Ω: I'' = 2.043 × 5/(5 + 3) = 1.277 A

Total

I = I' + I'' = 0.851 + 1.277

Answer: I = 2.13 A through the 3 Ω resistor

(Check by nodal analysis: V = (10/5 + 12/4)/(1/5 + 1/4 + 1/3) = 5/0.7833 = 6.383 V; I = 6.383/3 = 2.13 A.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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