Nodal analysis, power in 5Ω

Reading the figure

Take the negative terminal of the 40 V source as the reference (0 V). Nodes:

• P: top of the 40 V source (+), so V_P = 40 V. Also the left end of the 6 Ω, the top of the 3 A source and the left end of the 5 A source.
• X: right end of the 6 Ω, right end of the 5 A source, top of the 3 Ω, and the − 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 (V_Z = V_X + 20), right end of the 5 Ω and of the 4 Ω. The 4 Ω returns to the reference.

Source directions: 5 A flows from X to P; 3 A flows from Y to P.

Nodal equations

X and Z are joined by the 20 V source, so use a supernode with V_Z = V_X + 20.

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

Supernode X + Z (currents leaving): (V_X − 40)/6 + 5 + (V_X − V_Y)/3 + (V_Z − V_Y)/5 + V_Z/4 = 0

Substituting V_Z = V_X + 20 and solving: V_X = −15.2 V, V_Y = −13.325 V, V_Z = 4.8 V

Current and power in the 5 Ω

I₅ = (V_Z − V_Y)/5 = (4.8 + 13.325)/5 = 3.625 A (from Z to Y)

P = I₅² × 5 = 3.625² × 5 = 65.7 W

Answer: P₅Ω ≈ 65.7 W

Figure notes: the 40 V source has its + terminal on top, the 20 V source has its + terminal on the right (Z side), the 5 A source points from X to P and the 3 A source points up (from Y to P). The vertical wire joining the 5 A and 3 A sources crosses the wire from the 40 V source to the 6 Ω; I took this crossing as a junction (otherwise the 5 A and 3 A sources would not satisfy KCL).