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Power in 5Ω using Norton theorem (2 A, 10 V)

Short answer

Norton current with 5Ω shorted is 7 A, Norton resistance is 1 Ω. Load current = 7 × 1/(1+5) = 1.167 A. Power dissipated in 5Ω = 1.167² × 5 = 6.81 W.

Find the power dissipated by 5 ohm resistor of the following circuit using Norton's theorem. [Figure: 2 A current source; 1 Ω in series on top; 10 V source in the top branch; 2 Ω shunt; 2 Ω shunt; 5 Ω load.]

Reading the figure

Bottom rail = ground. The 2 A source (arrow up) feeds terminal a, then a 1 Ω in series to node N₁. N₁ has a 2 Ω shunt. Between N₁ and N₂ is the 10 V battery (+ on the N₂ side). N₂ has a 2 Ω shunt and the 5 Ω load.

 (2A)--[1 Ω]--N1--(10V)--N2
               |          |   |
              [2]        [2] [5]
               |          |   |
 --------------+----------+---+--

Step 1: Norton current (5 Ω shorted)

With N₂ shorted to ground, the battery makes V_N1 = −10 V (V_N2 = V_N1 + 10 = 0). The 2 Ω shunt at N₂ carries nothing.

Current from ground into N₁ through its 2 Ω shunt: 10/2 = 5 A Current from the 2 A source: 2 A

I_N = 2 + 5 = 7 A (through the battery into the short)

Step 2: Norton resistance (sources off)

2 A opened, 10 V shorted: N₁ and N₂ are joined, so the two 2 Ω shunts are in parallel (the 1 Ω is in series with the open source and carries nothing):

R_N = 2 ∥ 2 = 1 Ω

Step 3: Current and power in the 5 Ω

I₅ = I_N × R_N/(R_N + 5) = 7 × 1/6 = 1.167 A

P = I₅² × 5 = 1.167² × 5 = 6.81 W

Answer: Power in the 5 Ω = 6.81 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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