--- title: "Superposition theorem with example" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:state-explain-superposition-theorem" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/state-explain-superposition-theorem site: BCT Engineering --- # Superposition theorem with example Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2072 Kartik, 4 marks; 2070 Chaitra, 5 marks (asked 2 times) - 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_1/basic_electrical_engineering/state-explain-superposition-theorem - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question State and explain Superposition theorem with an appropriate example. ## Short answer Superposition theorem: in a linear circuit, the current or voltage in any element equals the algebraic sum of contributions from each independent source acting alone, with other sources replaced by their internal resistances (voltage sources shorted, current sources opened). Example: with 10 V and 5 V sources, current through a 2 Ω resistor is 2.5 A. ## Answer #### Statement **Superposition theorem:** in a linear network containing more than one independent source, the current through (or voltage across) any element is the **algebraic sum** of the currents (or voltages) produced by each source acting alone, with all the other independent sources replaced by their internal resistances. #### Rules - Replace each unused **voltage source** by a short circuit. - Replace each unused **current source** by an open circuit. - Add the individual results with their signs (direction matters). - It applies only to **linear** circuits and to currents and voltages, **not to power** (power is not linear). #### Steps 1. Keep one source, set the others to zero. 2. Find the required current or voltage. 3. Repeat for every source. 4. Add the results algebraically. #### Example A 10 V source and a 5 V source feed a 2 Ω resistor R through resistors R₁ = 2 Ω (10 V side) and R₂ = 4 Ω (5 V side), all joined at the same node. ```text (10V)--[2]--+--[4]--(5V) | [R=2] | ----------------------- ``` - 10 V alone (5 V shorted): total R = 2 + (4 ∥ 2) = 3.333 Ω, I = 3 A; current in R = 3 × 4/(4 + 2) = 2 A. - 5 V alone (10 V shorted): total R = 4 + (2 ∥ 2) = 5 Ω, I = 1 A; current in R = 0.5 A. - Total current in R = 2 + 0.5 = **2.5 A**. Check by nodal analysis: V_R(1/2 + 1/4 + 1/2) = 10/2 + 5/4, so V_R = 6.25/1.25 = 5 V and I_R = 2.5 A. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. According to superposition, the response in an element equals what? A. The response of the largest source only B. The product of individual responses C. The algebraic sum of responses due to each source acting alone D. The average of the responses Answer: C. Contributions from each source are added with their signs. (Key point: Statement) 2. How are unused sources replaced? A. Voltage sources by open circuits, current sources by short circuits B. All sources by resistors C. Both by short circuits D. Voltage sources by short circuits, current sources by open circuits Answer: D. Ideal voltage source has zero internal resistance; current source infinite. (Key point: Turning off sources) 3. What cannot be found by superposition directly? A. Power, because it is not linear in current/voltage B. Currents C. Voltages D. Currents in linear circuits Answer: A. Power goes with I^2, so you must add currents first then compute power. (Key point: Limitation) 4. In the example the 10 V alone gives 2 A and 5 V alone gives 0.5 A in R. Total? A. 1.5 A B. 2.5 A, both in the same direction C. 2.0 A D. 1.0 A Answer: B. Add them algebraically: 2 + 0.5 = 2.5 A. 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