--- title: "Norton, R for maximum power, 48 V and 5 A" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:norton-max-power-r-48v-12-ohm-5a" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/norton-max-power-r-48v-12-ohm-5a site: BCT Engineering --- # Norton, R for maximum power, 48 V and 5 A Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2069 Chaitra, 6 marks; 2068 Chaitra, 6 marks (asked 2 times) - 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/norton-max-power-r-48v-12-ohm-5a - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question Use Norton's theorem to calculate the value of R that will absorb maximum power from the circuit shown in the figure below. Also calculate the maximum power drawn by it. [Figure: 48 V source in series with 12 Ω on top; R in a vertical branch; 6 Ω shunt; 5 A current source.] ## Short answer Norton equivalent: I_N = 9 A, R_N = 4 Ω. For maximum power, R = R_N = 4 Ω. Maximum power P_max = I_N²·R_N/4 = 81 W. ## Answer #### Reading the figure From left to right: 48 V battery (+ on top) in series with 12 Ω; the variable resistor R; a 6 Ω shunt; a 5 A source (arrow up) feeding the top rail. R is the load. #### Step 1: Norton current (short R) With R replaced by a short circuit, the 6 Ω shunt is also shorted. - Current from the 48 V branch: 48/12 = 4 A (towards the short) - The 5 A source also feeds the short. I_N = 4 + 5 = **9 A** #### Step 2: Norton resistance (sources off) 48 V shorted and 5 A opened: R_N = 12 ∥ 6 = 12 × 6/18 = **4 Ω** #### Step 3: Value of R for maximum power R = R_N = **4 Ω** #### Step 4: Maximum power Current in R = I_N/2 = 4.5 A P_max = I² R = 4.5² × 4 = **81 W** (also P_max = I_N² R_N/4 = 81 × 4/4 = 81 W) Check with Thevenin: V_th = I_N R_N = 36 V, P_max = V_th²/(4 R_N) = 1296/16 = 81 W. **Answer: R = 4 Ω; maximum power = 81 W** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is I_N = 9 A? A. Because 48/6 = 8 A plus 1 A B. Because 5 A minus 4 A C. Shorting R shorts the 6 ohm, so 48/12 = 4 A plus the 5 A source flow into the short D. Because only the 5 A flows Answer: C. With the short, the 6 ohm carries nothing and both sources feed the short. (Key point: I_N) 2. What is R_N? A. 12 + 6 = 18 ohm B. 12 ohm C. 2 ohm D. 12 || 6 = 4 ohm Answer: D. 48 V is shorted and 5 A opened, leaving 12 and 6 in parallel. (Key point: R_N) 3. For maximum power, what is R? A. R = R_N = 4 ohm B. R = 9 ohm C. R = 2 ohm D. R = 12 ohm Answer: A. Load must equal the Norton resistance. (Key point: Value of R) 4. What is the maximum power? A. 162 W B. 81 W, since current is I_N/2 = 4.5 A C. 40.5 W D. 9 W Answer: B. I^2 R = 4.5^2 x 4 = 81 W. 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