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Nodal analysis currents and power loss

Short answer

Using nodal analysis, solve for node voltages (V1 = 6 V, V2 = 2 V, V3 = 6 V). Branch currents are 0.6 A, 0.4 A, 0.6 A, 0.2 A, 0.3 A. Total power loss in resistors is Σ I²R = 15 W.

Using Nodal analysis, determine currents in each branch of the network shown in below figure. Also find the total power loss in the network. [Figure: 1 A current source with 10 Ω in parallel on the left; 10 Ω and 20 Ω in the top branch; 20 Ω and 20 Ω with a 10 V source in the middle; 0.5 A current source on the right.]

Reading the figure

The bottom rail is ground (0 V). Node 1 has the 1 A source (into the node) and a 10 Ω to ground. Node 1 connects to node 2 through 10 Ω. Node 2 has 20 Ω in series with the 10 V source to ground (assumed + at the bottom, so the branch end sits at −10 V). Node 2 connects to node 3 through 20 Ω. Node 3 has 20 Ω to ground and the 0.5 A source (into the node).

Nodal equations (V₁, V₂, V₃)

Node 1: V₁/10 + (V₁ − V₂)/10 = 1 Node 2: (V₂ − V₁)/10 + (V₂ + 10)/20 + (V₂ − V₃)/20 = 0 Node 3: (V₃ − V₂)/20 + V₃/20 = 0.5

Solving: V₁ = 6 V, V₂ = 2 V, V₃ = 6 V

Branch currents

  • 10 Ω (node 1 to ground): 6/10 = 0.6 A
  • 10 Ω (node 1 to node 2): (6 − 2)/10 = 0.4 A
  • 20 Ω with 10 V (node 2 to ground): (2 + 10)/20 = 0.6 A
  • 20 Ω (node 3 to node 2): (6 − 2)/20 = 0.2 A (flows from 3 to 2)
  • 20 Ω (node 3 to ground): 6/20 = 0.3 A

Total power loss (resistors)

P = 0.6² × 10 + 0.4² × 10 + 0.6² × 20 + 0.2² × 20 + 0.3² × 20 P = 3.6 + 1.6 + 7.2 + 0.8 + 1.8 = 15 W

Answer: Total power loss = 15 W

Check: the 1 A source delivers 1 × 6 = 6 W and the 0.5 A source delivers 0.5 × 6 = 3 W. The 0.6 A in the 10 V branch enters the battery at its − terminal and leaves at its + terminal, so the 10 V source also delivers 10 × 0.6 = 6 W. Total supplied = 6 + 3 + 6 = 15 W, equal to the loss.

✓ Answer checked by Claude Opus 5.5 (high effort)

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