--- title: "Superposition, current in 8Ω" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:superposition-8-ohm-90v-60v-2a" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-8-ohm-90v-60v-2a site: BCT Engineering --- # Superposition, current in 8Ω Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2068 Shrawan, 8 marks - Video: [How to Use Superposition to Solve Circuits | Engineering Circuit Analysis | (Solved Examples)](https://www.youtube.com/watch?v=uoZ0E6zxoi0) — Question Solutions - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-8-ohm-90v-60v-2a - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question Using superposition theorem, find the current in 8Ω resistor of the network shown below. [Figure: 36 Ω in series with a 90 V source; 12 Ω in series with a 60 V source; 2 A current source; 8 Ω resistor, all in parallel.] ## Short answer By superposition: 90 V source gives 1.3235 A, 60 V gives 2.647 A, 2 A gives 1.059 A, all in the same direction. Total current in the 8 Ω resistor is 5.03 A. ## Answer #### Reading the figure Four branches are in parallel between the top rail and the bottom rail: 36 Ω in series with the 90 V battery (+ up), 12 Ω in series with the 60 V battery (+ up), the 2 A source (arrow up) and the 8 Ω resistor. All three sources push current into the top rail. #### Case 1: 90 V alone (60 V shorted, 2 A opened) Rest of the circuit across the 8 Ω: 36 Ω in series with the source; the 12 Ω is in parallel with the 8 Ω. 12 ∥ 8 = 4.8 Ω V = 90 × 4.8/(36 + 4.8) = 10.588 V I′ = 10.588/8 = **1.3235 A** #### Case 2: 60 V alone (90 V shorted, 2 A opened) 36 ∥ 8 = 6.545 Ω V = 60 × 6.545/(12 + 6.545) = 21.176 V I″ = 21.176/8 = **2.647 A** #### Case 3: 2 A alone (both batteries shorted) 36 ∥ 12 ∥ 8: 1/R = 0.02778 + 0.08333 + 0.125 = 0.23611, so R = 4.235 Ω V = 2 × 4.235 = 8.471 V I‴ = 8.471/8 = **1.059 A** #### Total (all in the same direction, top to bottom) I₈ = 1.3235 + 2.647 + 1.059 = **5.03 A** Check by nodal analysis: V = (90/36 + 60/12 + 2)/(1/36 + 1/12 + 1/8) = 9.5/0.2361 = 40.24 V, and I = 40.24/8 = 5.03 A. **Answer: Current in the 8 Ω = 5.03 A** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 60 V case, how are the other sources replaced? A. 90 V shorted and 2 A opened B. 90 V opened and 2 A shorted C. Both shorted D. Both opened Answer: A. Voltage sources become short circuits and current sources become open circuits. (Key point: Turning off sources) 2. With only the 90 V source, the 8 ohm is in parallel with which resistor, and what is V across the 8 ohm? A. 36 ohm, V = 90 V B. 12 ohm (4.8 ohm combined), V = 10.588 V C. 12 ohm in series, V = 60 V D. None, V = 0 Answer: B. 12 || 8 = 4.8 ohm; V = 90 x 4.8/(36 + 4.8) = 10.588 V. (Key point: 90 V case) 3. With only the 2 A source, what resistance does the source see? A. 36 || 12 || 8 = 4.235 ohm B. 12 || 8 = 4.8 ohm C. 8 ohm only D. 36 + 12 + 8 = 56 ohm Answer: A. Shorted batteries leave three resistors in parallel, so V = 2 x 4.235 = 8.47 V and I = 1.059 A. (Key point: 2 A case) 4. Why are the three currents simply added to get 5.03 A? A. Because the sources are identical B. They are multiplied C. The 2 A case is subtracted D. All three flow top to bottom in the 8 ohm, so they add in the same direction Answer: D. All three sources push current into the top rail, so each contribution is downward: 1.3235 + 2.647 + 1.059 = 5.03 A. (Key point: Adding results) ## Related in Basic Electrical Engineering - [Superposition theorem, 5 Ω current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-5-ohm-15v-20v) — Basic Electrical Engineering, chapter 3 - [Source current by series-parallel reduction](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aseries-parallel-current-30v-r1-r9) — Basic Electrical Engineering, chapter 2 - [Superposition theorem, current in RL](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-load-2-ohm-10v-5a) — Basic Electrical Engineering, chapter 3 - [Superposition, current in branch AB](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-branch-ab-6v-12v) — Basic Electrical Engineering, chapter 3 - [Superposition, current in 0.5Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-0-5-ohm-15v-20v) — Basic Electrical Engineering, chapter 3 - [Source current from parallel branch currents](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aparallel-branch-currents-time-expression) — Basic Electrical Engineering, chapter 5 ## Related in other subjects - [State model of parallel RLC with current source](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Astate-model-parallel-rlc-current-source) — Electrical Circuit Theory, chapter 4 - [Superposition theorem and maximum power to RL](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Asuperposition-max-power-transfer) — Fundamental of Electrical and Electronics Engineering, chapter 1 - [Superposition: branch currents, 6 V and 12 V](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Asuperposition-6v-12v) — Basic Electronics Engineering, chapter 1 - [Superposition: 90, 50, 130 ohm circuit](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Asuperposition-90-50-130) — Basic Electronics Engineering, chapter 1, asked 3×