--- title: "Power in 3Ω using Norton theorem" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:norton-power-3-ohm-20v-10a" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/norton-power-3-ohm-20v-10a site: BCT Engineering --- # Power in 3Ω using Norton theorem Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2074 Chaitra, 6 marks - Video: [Norton's theorem problem solution |2024](https://www.youtube.com/watch?v=DnUxSe9j6ls) — Engineers Around The World - 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-power-3-ohm-20v-10a - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question Find power dissipated in 3 Ω resistor using Norton's theorem. [Figure: 20 V source, 2 Ω in series, 3 Ω shunt, 10 A current source, 4 Ω in series towards terminal x, load 3 Ω between x and y with current IL.] ## Short answer Norton short-circuit current I_N = 0 A because the 20 V source and 10 A source cancel; Norton resistance R_N = 5.2 Ω. With zero current from the Norton source, power in the 3 Ω load is 0 W. ## Answer #### Reading the figure Bottom rail = ground. 20 V source (+ on top) in series with 2 Ω to node N. At N: 3 Ω shunt to ground, 10 A source (arrow down, taking 10 A out of N) and 4 Ω in series to terminal x. The load 3 Ω is between x and y (y is ground). #### Step 1: Norton current (short x-y) With x shorted to ground, the 4 Ω is directly across ground, so N has the 3 Ω and 4 Ω as shunts: (V_N − 20)/2 + V_N/3 + V_N/4 + 10 = 0 V_N (1/2 + 1/3 + 1/4) = 10 − 10 = 0 So V_N = 0 V and the short-circuit current is **I_N = V_N/4 = 0 A** (The 20 V source through 2 Ω supplies exactly 20/2 = 10 A, which is all taken by the 10 A source.) #### Step 2: Norton resistance (sources off) 20 V shorted, 10 A opened: R_N = 4 + (2 ∥ 3) = 4 + 1.2 = **5.2 Ω** #### Step 3: Power in the 3 Ω Load current: I_L = I_N × R_N/(R_N + 3) = 0 A P = I_L² × 3 = **0 W** **Answer: I_N = 0 A, R_N = 5.2 Ω, so the power in the 3 Ω is 0 W.** Note: with the polarities read from the figure the two sources exactly cancel. If the 10 A source were reversed, V_th would be 24 V and P = (24/8.2)² × 3 = 25.7 W. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is I_N = 0? A. The 3 ohm is open B. The 4 ohm is infinite C. With x shorted, V_N = 0 because the 20 V source supplies exactly the 10 A the current source takes D. The 10 A source is open Answer: C. 20 V / 2 ohm = 10 A, which equals the 10 A drawn out, leaving nothing for the short. (Key point: Short-circuit current) 2. What is R_N? A. 2 + 3 + 4 = 9 ohm B. 4 ohm C. 1.2 ohm D. 4 + (2 || 3) = 5.2 ohm Answer: D. With sources off, 2 and 3 ohm are in parallel and in series with 4 ohm. (Key point: R_N) 3. What is the power in the 3 ohm? A. 0 W, because I_L = I_N R_N/(R_N + 3) = 0 B. 25.7 W C. 3 W D. 12 W Answer: A. With I_N = 0 no current flows in the load. 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