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Nodal analysis, power in 5Ω

Short answer

Using a supernode for the 20 V source, node voltages are V_X = −15.2 V, V_Y = −13.325 V, V_Z = 4.8 V. Current through the 5 Ω resistor is 3.625 A, and power dissipated ≈ 65.7 W.

Determine the power dissipated by 5Ω resistor in the circuit shown in figure below by applying nodal voltage analysis. [Figure: 5 A current source on top; 6 Ω; 20 V source; 3 A current source; 3 Ω; 5 Ω; 40 V source on the left; 4 Ω in the bottom branch.]

This circuit is the same as the one in the 2071 Chaitra paper. Use the bottom node of the 40 V source as reference.

Reading the figure

  • P: top of the 40 V source (+), V_P = 40 V; left end of the 6 Ω; top of the 3 A source; left end of the 5 A source.
  • X: right end of the 6 Ω and of the 5 A source; top of the 3 Ω; − terminal of the 20 V source.
  • Y: bottom of the 3 Ω; bottom of the 3 A source; left end of the 5 Ω.
  • Z: + terminal of the 20 V source; right end of the 5 Ω and the 4 Ω (which returns to the reference).

The 5 A source takes current from X to P, and the 3 A source takes current from Y to P.

Nodal equations

Supernode (X and Z) with V_Z = V_X + 20: (V_X − 40)/6 + 5 + (V_X − V_Y)/3 + (V_Z − V_Y)/5 + V_Z/4 = 0

Node Y: 3 + (V_Y − V_X)/3 + (V_Y − V_Z)/5 = 0

Solution

V_X = −15.2 V, V_Y = −13.325 V, V_Z = 4.8 V

Current in the 5 Ω: (V_Z − V_Y)/5 = (4.8 + 13.325)/5 = 3.625 A

P = 3.625² × 5 = 65.7 W

Answer: Power in the 5 Ω resistor ≈ 65.7 W

✓ Answer checked by Claude Opus 5.5 (high effort)

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