--- title: "Superposition theorem and maximum power to RL" subject: "Fundamental of Electrical and Electronics Engineering" chapter: 1 key: "semester_1/fundamental_of_electrical_and_electronics_engineering:superposition-max-power-transfer" source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/superposition-max-power-transfer site: BCT Engineering --- # Superposition theorem and maximum power to RL Fundamental of Electrical and Electronics Engineering · Chapter 1: Basic Circuits Concepts - Asked in: 2081 Baishakh (new course), 2+4 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/fundamental_of_electrical_and_electronics_engineering/superposition-max-power-transfer - Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-1 ## Question Explain superposition theorem with suitable example. Find the maximum power that can be delivered to the load resistor RL of the circuit shown in the figure below. [Figure: a 20 V source (+ at top) on the left; from its top terminal a 5 Ω series resistor leads to node A; a 10 Ω resistor connects node A to the bottom (reference) line; a 10 Ω series resistor connects node A to node B; a 4 A current source between the bottom line and node B, arrow pointing up (current into node B); the variable load resistor RL connects node B to the bottom line.] ## Short answer Superposition theorem states that in a linear bilateral network, the current or voltage in any element equals the algebraic sum of contributions from each independent source acting alone with all other independent sources replaced by their internal resistances (voltage sources shorted, current sources opened). For the given circuit, maximum power transfer occurs when RL = Rth = 13.33 Ω, delivering Pmax = 83.33 W. ## Answer #### Superposition theorem **Statement:** In a linear, bilateral network with more than one independent source, the current (or voltage) in any element equals the algebraic sum of the currents (or voltages) produced by each source acting alone, with all other independent sources replaced by their internal resistances. - An ideal voltage source is replaced by a **short circuit**. - An ideal current source is replaced by an **open circuit**. - Dependent sources are left in the circuit. - It applies to currents and voltages, not to power (power is not linear). **Example:** A 10 V source in series with 2 Ω feeds node X; a 2 A current source also feeds node X; a 3 Ω resistor connects X to ground. Find the current in 3 Ω. ```text 10 V alone (2 A open): I1 = 10/(2+3) = 2 A 2 A alone (10 V short): I2 = 2 × 2/(2+3) = 0.8 A Total: I = I1 + I2 = 2 + 0.8 = 2.8 A ``` #### Maximum power to RL Remove RL and find the Thevenin equivalent across node B and ground. **Thevenin resistance** (20 V source shorted, 4 A source opened): ```text 5 Ω || 10 Ω = (5×10)/(5+10) = 3.333 Ω Rth = 10 + 3.333 = 13.333 Ω ``` **Thevenin voltage** (RL open): the 4 A source current has no path except through the 10 Ω (A–B) into node A. KCL at node A (currents leaving A): ```text (VA − 20)/5 + VA/10 − 4 = 0 0.2VA − 4 + 0.1VA − 4 = 0 0.3VA = 8 VA = 26.667 V Vth = VB = VA + 4 × 10 = 26.667 + 40 = 66.667 V ``` (The 4 A flows from B to A through the 10 Ω, so B is 40 V above A.) **Maximum power transfer:** power is maximum when RL = Rth = 13.333 Ω. ```text Pmax = Vth² / (4 Rth) = (66.667)² / (4 × 13.333) = 4444.4 / 53.333 = 83.33 W ``` **Answer: RL = 13.33 Ω, Pmax = 83.33 W** (load current = 66.667/26.667 = 2.5 A). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When applying superposition, how are the ideal sources not under consideration replaced? A. Voltage source by open circuit, current source by short circuit B. Both voltage and current sources by short circuits C. Voltage source by short circuit, current source by open circuit D. Both are left in place but set to half value Answer: C. Independent voltage sources are shorted and independent current sources are opened, so only one source acts at a time. (Key point: Source replacement rules) 2. Superposition can be applied directly to which quantity? A. Power in a resistor B. Energy in a nonlinear element C. Power only if sources are DC D. Current or voltage in a linear network Answer: D. Superposition holds for linear relations (current, voltage); power depends on the square and is not linear. (Key point: Applicability of superposition) 3. For the max-power circuit, Rth seen from node B is 13.333 ohm. How is it obtained? A. 10 + (5 || 10) with sources deactivated B. 10 || (5 + 10) C. 5 + 10 + 10 D. (5 || 10) alone Answer: A. With the 20 V source shorted and the 4 A source opened, the 10 ohm (A-B) is in series with 5 ohm parallel 10 ohm = 3.333 ohm, giving 13.333 ohm. (Key point: Thevenin resistance of the circuit) 4. With Vth = 66.667 V and Rth = 13.333 ohm, what is the maximum power and the RL needed? A. RL = 13.33 ohm, Pmax = 166.7 W B. RL = 13.33 ohm, Pmax = 83.33 W C. RL = 6.67 ohm, Pmax = 83.33 W D. RL = 26.67 ohm, Pmax = 41.7 W Answer: B. Max power occurs at RL = Rth and Pmax = Vth^2/(4Rth) = 4444.4/53.33 = 83.33 W. 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