--- title: "Superposition: current in 3 Ω" subject: "Basic Electronics Engineering" chapter: 1 key: "semester_2/basic_electronics_engineering:superposition-10v-12v-3ohm" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/superposition-10v-12v-3ohm site: BCT Engineering --- # Superposition: current in 3 Ω Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts - Asked in: 2069 Bhadra, 5 marks - Video: [Superposition Theorem](https://www.youtube.com/watch?v=ZJ8zD8m-B1Q) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/superposition-10v-12v-3ohm - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1 ## Question 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.] ## 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. ## Answer **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.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the current in the 3 Ω found by superposition? A. Sum of the currents due to each source acting alone B. Average of both C. Difference of both sources D. Use only the larger source Answer: A. Each source acts alone with the other shorted, and the results are added. (Key point: Method) 2. With 10 V alone, what is R_total? A. 5 + (4 ∥ 3) = 6.714 Ω B. 4 + (5 ∥ 3) = 5.875 Ω C. 5 + 4 + 3 = 12 Ω D. 5 ∥ 4 ∥ 3 = 1.28 Ω Answer: A. The 12 V branch is shorted; 4 Ω ∥ 3 Ω = 1.714 Ω in series with 5 Ω. (Key point: Case 1) 3. With 12 V alone, the 3 Ω current is 2.043 A × ? A. 4/7 B. 5/(5 + 3) C. 3/8 D. 1/3 Answer: B. The current divider gives the 3 Ω a fraction 5/(5+3), giving 1.277 A. (Key point: Case 2 divider) 4. Total current in the 3 Ω and the nodal check agree at A. 2.13 A B. 1.70 A C. 0.85 A D. 3.0 A Answer: A. 0.851 + 1.277 = 2.13 A; nodal V = 6.383 V gives 6.383/3 = 2.13 A. 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