--- title: "Find R for maximum power to load" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:max-power-load-4-ohm-find-r" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/max-power-load-4-ohm-find-r site: BCT Engineering --- # Find R for maximum power to load Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2073 Shrawan, 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-load-4-ohm-find-r - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question Find the value of Resistance R such that the load resistance RL which is equal to 4 Ω, will deliver maximum power. Also find that maximum power. [Figure: 3 A current source on the left; 6 Ω shunt; 20 V source in the top branch; R in series with a 4 Ω resistor in parallel; 3 Ω shunt; load RL = 4 Ω.] ## Short answer Thevenin resistance seen by RL is 2 + (R∥4). Set it equal to RL = 4 Ω, solving gives R = 4 Ω. Thevenin voltage is 12.67 V, so maximum power in RL is V_th²/(4·4) = 10.03 W. ## Answer #### Reading the figure Bottom rail = ground. Left node P: 3 A source (up) into P, 6 Ω to ground, and the 20 V battery (+ on the right) to node N. Node N: 3 Ω to ground, and two parallel resistors **R** and **4 Ω** from N to the top of the load R_L = 4 Ω. #### Step 1: Thevenin resistance seen by R_L Sources off (3 A opened, 20 V shorted): P is joined to N, so the 6 Ω and 3 Ω are in parallel at N: 6 ∥ 3 = 2 Ω Looking into R_L: R_th = 2 + (R ∥ 4) #### Step 2: Matching condition For maximum power in R_L, R_th = R_L = 4 Ω: 2 + R × 4/(R + 4) = 4 → 4R/(R + 4) = 2 → 4R = 2R + 8 → **R = 4 Ω** #### Step 3: Thevenin voltage (R_L removed) No current flows in R or the 4 Ω, so V_th = V_N. V_N = V_P + 20, and 3 = V_P/6 + V_N/3 3 = V_P/6 + (V_P + 20)/3 → V_P/2 = 3 − 6.667 → V_P = −7.33 V V_th = V_N = 12.67 V #### Step 4: Maximum power P_max = V_th²/(4 R_L) = 12.67²/(4 × 4) = 160.4/16 = **10.03 W** **Answer: R = 4 Ω; P_max ≈ 10.03 W** Note: this is the usual matching treatment (R_th seen by the load equals R_L). Strictly, with R_L fixed, the load power would be even larger for a smaller R, so the question is read as a matching problem. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What condition is used to find R? A. R_th seen by R_L must equal R_L = 4 ohm B. R equals the 6 ohm C. R equals 3 ohm D. R is infinite Answer: A. Maximum power at load needs R_th = R_L. (Key point: Matching condition) 2. What is the expression for R_th? A. 6 + 3 + R B. 2 + (R || 4), with 6 || 3 = 2 ohm C. (R + 4) || 2 D. R + 4 + 2 Answer: B. With sources off, 6 and 3 ohm are parallel and in series with R||4. (Key point: R_th) 3. What is R? A. 2 ohm B. 8 ohm C. 4 ohm D. 6 ohm Answer: C. 2 + 4R/(R+4) = 4 gives R = 4 ohm. (Key point: Value of R) 4. How is P_max found? A. V_th^2/R_L = 40 W B. V_th x 4 C. 3 A x 20 V = 60 W D. V_th^2/(4 R_L) = 10.03 W with V_th = 12.67 V Answer: D. No current flows in R and 4 ohm under open circuit, so V_th = V_N. 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