--- title: "Superposition theorem, current in RL" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:superposition-load-2-ohm-10v-5a" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-load-2-ohm-10v-5a site: BCT Engineering --- # Superposition theorem, current in RL Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2072 Chaitra, 6 marks - Video: [Superposition Theorem Solved Example Problem | Electrical Engineering](https://www.youtube.com/watch?v=4dsNAiZYwDw) — Electrical and Electronics Engineering - 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-load-2-ohm-10v-5a - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question Find current on load resistor RL, if its resistance is 2 Ω, using superposition theorem. [Figure: 2 Ω resistor along the top; 10 V source on the left with 2 Ω and 2 Ω resistors; load RL in the middle vertical branch; 5 A current source; 2 Ω resistor on the right.] ## Short answer With the 10 V source alone, current through the 2 Ω load is 2.5 A; with the 5 A source alone, it is 0.625 A. The total current using superposition is 3.125 A. ## Answer #### Reading the figure Top: 2 Ω from the left node L to the right node R. Middle row: 2 Ω from L to the middle node M, and 2 Ω from M to R. R_L = 2 Ω from M to the bottom rail. Left: 10 V battery (+ on top) between L and ground. Right: 5 A source (arrow up) into R, and a 2 Ω shunt from R to ground. #### Case 1: 10 V alone (5 A source opened) Node M: (V_M − 10)/2 + V_M/2 + (V_M − V_R)/2 = 0 → 3 V_M − V_R = 10 Node R: (V_R − 10)/2 + (V_R − V_M)/2 + V_R/2 = 0 → 3 V_R − V_M = 10 So V_M = V_R = 5 V and I′ = 5/2 = **2.5 A** (downwards through R_L). #### Case 2: 5 A alone (10 V source shorted) L is at 0 V. Node M: V_M/2 + V_M/2 + (V_M − V_R)/2 = 0 → 3 V_M = V_R Node R: V_R/2 + (V_R − V_M)/2 + V_R/2 = 5 → 3 V_R − V_M = 10 3 (3 V_M) − V_M = 10, so V_M = 1.25 V and I″ = 1.25/2 = **0.625 A** (downwards through R_L). #### Total Both currents flow in the same direction, so I_L = I′ + I″ = 2.5 + 0.625 = **3.125 A** **Answer: Current in R_L = 3.125 A** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When the 10 V acts alone, how is the 5 A source treated? A. Replaced by an open circuit B. A short circuit C. A 5 ohm resistor D. Left in place Answer: A. Current sources become open circuits. (Key point: Current source off) 2. When 5 A acts alone, what happens to the 10 V source? A. Opened B. Replaced by a short, so L is at 0 V C. Doubled D. Kept Answer: B. Voltage sources become shorts. (Key point: Voltage source off) 3. What are the two partial load currents? A. 5 A and 0.625 A B. 2.5 A and 5 A C. 2.5 A and 0.625 A D. 1.25 A and 1.25 A Answer: C. V_M = 5 V gives 2.5 A; V_M = 1.25 V gives 0.625 A. (Key point: Partial currents) 4. What is the final I_L and why? A. 1.875 A, as they subtract B. 7.5 A C. 2.5 A D. 3.125 A, as both flow downward and add Answer: D. Same direction, so algebraic sum = 3.125 A. 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