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Power in 3Ω using Norton theorem (10 V, 10 A)

Short answer

Norton current is 7.5 A and Norton resistance is 2 Ω. Load current is 7.5 × 2/(2+3) = 3 A. The power dissipated in the 3 Ω resistor is 3² × 3 = 27 W.

Determine power dissipated in 3Ω resitor in the circuit shown in figure below using Norton's theorem. [Figure: 10 V source on the left; 2 Ω in the top series branch; 2 Ω shunt; 10 A current source; 1 Ω in series on top towards terminal a; load 3 Ω between terminals a and b.]

Reading the figure

Bottom rail = ground. 10 V source (+ on top) with 2 Ω in series feeds node N. At N: a 2 Ω shunt to ground and a 10 A source (arrow up, into N). From N, a 1 Ω in series goes to terminal a. The load 3 Ω is between a and b (b on the bottom rail).

Step 1: Norton current (short a-b)

With a shorted to ground, the 1 Ω is connected directly between N and ground.

Node N: (V_N − 10)/2 + V_N/2 + V_N/1 = 10 0.5 V_N − 5 + 0.5 V_N + V_N = 10 → 2 V_N = 15 → V_N = 7.5 V

I_N = V_N/1 = 7.5 A

Step 2: Norton resistance (sources off)

10 V shorted, 10 A opened: R_N = 1 + (2 ∥ 2) = 1 + 1 = 2 Ω

(Check: V_th = I_N R_N = 15 V, which is the open-circuit voltage of node N.)

Step 3: Load current and power

I_L = I_N × R_N/(R_N + R_L) = 7.5 × 2/(2 + 3) = 3 A

P = I_L² R = 3² × 3 = 27 W

Answer: Power in the 3 Ω = 27 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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