Skip to main content

Superposition: current in 4 Ω

Short answer

Superposition theorem: in a linear network, the current in any branch equals the sum of currents produced by each independent source acting alone, with other voltage sources shorted. Using this method, the current in the 4 Ω resistor is found to be 2.17 A.

State superposition theorem. In the following figure find the current flowing in the 4 Ω resistor using superposition theorem. [Figure: 14 V source with 5 Ω resistor in the left branch; 12 V source in the middle branch with 3 Ω resistor; 4 Ω resistor in the right branch.]

Superposition theorem: in a linear network containing two or more independent sources, the current through (or voltage across) any element is the algebraic sum of the currents (or voltages) produced by each source acting alone, while the other voltage sources are replaced by short circuits and current sources by open circuits.

Assumption: the 14 V source with 5 Ω, the 12 V source with 3 Ω and the 4 Ω resistor are three parallel branches; both sources have their positive terminals toward the top node.

Case 1: 14 V alone (12 V source shorted)

3 Ω ∥ 4 Ω = 12/7 = 1.714 Ω R_total = 5 + 1.714 = 6.714 Ω I = 14/6.714 = 2.085 A Current in 4 Ω: I' = 2.085 × 3/(3 + 4) = 0.894 A

Case 2: 12 V alone (14 V source shorted)

5 Ω ∥ 4 Ω = 20/9 = 2.222 Ω R_total = 3 + 2.222 = 5.222 Ω I = 12/5.222 = 2.298 A Current in 4 Ω: I'' = 2.298 × 5/(5 + 4) = 1.277 A

Total

I(4 Ω) = 0.894 + 1.277

Answer: I = 2.17 A through the 4 Ω resistor

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electronics Engineering chapter 1 questions →