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Power in 3Ω using Norton theorem

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.

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.]

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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