--- title: "Find RL for maximum power and Pmax" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:max-power-rl-8v-4v-6-ohm" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/max-power-rl-8v-4v-6-ohm site: BCT Engineering --- # Find RL for maximum power and Pmax Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2072 Chaitra, 8 marks - Video: [Maximum Power Transfer Theorem | Electric Circuits | Example 4.13 | Electrical Engineering](https://www.youtube.com/watch?v=x6b_DU7_8ro) — 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/max-power-rl-8v-4v-6-ohm - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question Find the value of RL for which the maximum power is transferred in the load resistance RL. Also find the maximum power that can be transferred to the load resistance RL. [Figure: 8 V source with 2 Ω shunt on the left; 4 Ω in the top branch; 6 Ω shunt; 4 Ω in series with a 4 V source; load RL on the right.] ## Short answer Thevenin voltage is 0.8 V and Thevenin resistance is 6.4 Ω. For maximum power, set RL = 6.4 Ω; the maximum power is 0.025 W. ## Answer #### Reading the figure Left: 8 V battery (+ on top) with a 2 Ω shunt at node P. 4 Ω from P to node M. 6 Ω shunt at M. Then 4 Ω in series with the 4 V battery to the load R_L. The 4 V battery has its + terminal on the 4 Ω (left) side (the longer plate is on the left). #### Step 1: Thevenin voltage (R_L removed) No current flows in the last 4 Ω or the 4 V battery. V_M = 8 × 6/(4 + 6) = 4.8 V (the 2 Ω is directly across the 8 V source and does not affect V_M) The 4 V battery opposes V_M (its + is on the M side), so V_th = 4.8 − 4 = **0.8 V** #### Step 2: Thevenin resistance (sources off) 8 V shorted (P to ground, the 2 Ω is then in parallel with the short), 4 V shorted: - At M: 6 Ω ∥ 4 Ω = 2.4 Ω - In series with the 4 Ω: R_th = 2.4 + 4 = **6.4 Ω** #### Step 3: Maximum power R_L = R_th = **6.4 Ω** P_max = V_th²/(4 R_th) = 0.8²/(4 × 6.4) = 0.64/25.6 = **0.025 W (25 mW)** **Answer: R_L = 6.4 Ω; P_max = 0.025 W** Note: if the 4 V battery polarity were reversed, V_th = 8.8 V and P_max = 3.03 W; the value above uses the polarity read from the figure. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is V_M? A. 8 x 6/(4+6) = 4.8 V; the 2 ohm doesn't affect it B. 8 V C. 4 V D. 3.2 V Answer: A. 2 ohm is directly across the source. (Key point: V_M) 2. How is V_th found? A. 4.8 + 4 = 8.8 V B. 4.8 - 4 = 0.8 V since the battery opposes V_M C. 4.8 V D. 4 V Answer: B. Battery + is on the M side so opposes. (Key point: V_th) 3. What is R_th? A. 6 + 4 + 4 = 14 ohm B. 2.4 ohm C. (6 || 4) + 4 = 6.4 ohm D. 4 ohm Answer: C. Sources shorted: 6 and 4 ohm in parallel, then the series 4 ohm. (Key point: R_th) 4. What is P_max? A. 0.1 W B. 3.03 W C. 0.64 W D. 0.8^2/(4 x 6.4) = 0.025 W at R_L = 6.4 ohm Answer: D. V_th^2/(4 R_th). 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