--- title: "Maximum power transfer, find R" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:max-power-transfer-6v-8v-2a" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/max-power-transfer-6v-8v-2a site: BCT Engineering --- # Maximum power transfer, find R Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2074 Asoj, 8 marks - Video: [Maximum Power Transfer Solved Example | Electrical Engineering](https://www.youtube.com/watch?v=xvGeIgLsxd4) — 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-transfer-6v-8v-2a - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question Find the value of Resistance R to have maximum power transfer in the circuit as shown in below figure. Also obtain the amount of maximum power. [Figure: 2 A current source on the left; 6 Ω in series with a 6 V battery on the top; 15 Ω, 3 Ω and R in vertical branches; 8 V battery in the bottom branch between the 3 Ω and R branches.] ## Short answer Thevenin voltage V_th = 11 V, Thevenin resistance R_th = 2.625 Ω. For maximum power transfer, R = R_th = 2.625 Ω. Maximum power P_max = V_th²/(4R_th) = 11.52 W. ## Answer #### Reading the figure Bottom rail = ground. Left node a: 2 A source (up) into a, 15 Ω from a to ground, and 6 Ω in series with the 6 V battery going to node b (the battery's + terminal is on the 6 Ω side). Node b: 3 Ω to ground. The load R is connected between b and the bottom of the 8 V battery (+ terminal on the ground side), so the 8 V battery is in series with R. The load R is taken between terminals **b** and **c**, where c is the lower end of the R branch (−8 V relative to ground). #### Step 1: Open-circuit voltage (R removed) Node a: 2 = V_a/15 + (V_a − 6)/(6 + 3) Multiply by 45: 90 = 3 V_a + 5 V_a − 30, so V_a = 15 V Current in the 6 Ω and 3 Ω = (15 − 6)/9 = 1 A V_b = 1 × 3 = 3 V V_c = −8 V (battery + is on the ground side), so V_th = V_b − V_c = 3 − (−8) = **11 V** #### Step 2: Thevenin resistance (sources off) 2 A opened, batteries shorted: from b, the 3 Ω is in parallel with (6 Ω + 15 Ω). R_th = 3 × 21/(3 + 21) = **2.625 Ω** #### Step 3: Maximum power R = R_th = **2.625 Ω** P_max = V_th²/(4 R_th) = 11²/(4 × 2.625) = 121/10.5 = **11.52 W** **Answer: R = 2.625 Ω; P_max ≈ 11.52 W** Assumption: plate polarities are read as 6 V (+ towards the 6 Ω resistor) and 8 V (+ on the ground side). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the approach for finding R? A. Set R = 3 ohm B. Short R and compute C. Find the Thevenin equivalent seen by R, then set R = R_th D. Use superposition only Answer: C. Maximum power transfer needs the Thevenin resistance seen by the load. (Key point: Which method) 2. How is R_th found? A. 3 + 6 + 15 = 24 ohm B. 6 || 15 = 4.29 ohm C. 3 || 6 = 2 ohm D. Open the 2 A source, short the batteries: 3 ohm || (6 + 15) = 2.625 ohm Answer: D. From b, the 3 ohm is in parallel with the 6 ohm in series with 15 ohm. (Key point: R_th) 3. What is V_th? A. 11 V = V_b - V_c = 3 - (-8) B. 3 V C. 5 V D. 8 V Answer: A. The 8 V battery adds to the 3 V at node b. (Key point: V_th) 4. What is P_max? A. V_th^2/R_th = 46 W B. V_th^2/(4 R_th) = 11.52 W C. V_th^2/(2 R_th) D. 23 W Answer: B. 121/(4 x 2.625) = 11.52 W. 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