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Nodal analysis, current through 15Ω

Short answer

Set up nodal equations at three nodes: 23V₁−15V₂=0, −3V₁+5V₂−2V₃=−192, −5V₂+22V₃=240. Solutions: V₁=−42.87 V. Current through 15 Ω = V₁/15 = −2.86 A (magnitude 2.86 A, flowing upward).

Use Nodal analysis method to calculate the current through the 15Ω resistor in the figure shown below. [Figure: 15 Ω on the left; 8 Ω, 12 Ω and 10 Ω along the top; 6 Ω in series with an 8 A current source in the second branch; 5 Ω shunt; 48 V source with 2 Ω on the right.]

Reading the figure

The bottom rail is ground. Nodes:

  • V₁: top of the 15 Ω (the 15 Ω hangs between V₁ and ground), joined by 8 Ω to V₂.
  • V₂: junction of the 8 Ω, the branch 6 Ω + 8 A source (current leaving V₂, downward) and the 12 Ω.
  • V₃: junction of the 12 Ω, the 5 Ω shunt and the 10 Ω going to the right.
  • Right branch: 10 Ω, then 2 Ω in series with the 48 V battery (+ on top), so it behaves as a 48 V source with 12 Ω in series.

The 6 Ω is in series with the current source, so it does not affect the equations.

Nodal equations (currents leaving each node)

Node 1: V₁/15 + (V₁ − V₂)/8 = 0 Node 2: (V₂ − V₁)/8 + (V₂ − V₃)/12 + 8 = 0 Node 3: (V₃ − V₂)/12 + V₃/5 + (V₃ − 48)/12 = 0

Simplified (multiply by 120, 24 and 60):

  • 23 V₁ − 15 V₂ = 0
  • −3 V₁ + 5 V₂ − 2 V₃ = −192
  • −5 V₂ + 22 V₃ = 240

From the first equation, V₁ = 15 V₂/23; from the third, V₃ = (240 + 5 V₂)/22. Substituting both into the second equation gives 655 V₂ = −43056, so

V₂ = −65.73 V, V₁ = −42.87 V, V₃ = −4.03 V

Current in the 15 Ω

I = V₁/15 = −42.87/15 = −2.858 A

The negative sign shows that the current flows from ground up through the 15 Ω towards node 1 (and on through the 8 Ω to the 8 A source).

Answer: I₁₅Ω ≈ 2.86 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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