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Nodal analysis branch currents

Short answer

Using nodal analysis, node voltages are V1 = 1.333 V, V2 = 5.667 V. Branch currents computed via Ohm’s law: 0.778 A (6 V branch), 0.444 A (3 Ω shunt), 0.333 A (5 V branch), 1.417 A (4 Ω shunt), 1.083 A (10 V branch).

Find the branch currents in the circuit of figure below by using nodal analysis. [Figure: 6 V source with 6 Ω on the left; 5 V source with 2 Ω in the middle top; 4 Ω on the right top; 3 Ω shunt; 4 Ω and 10 V source on the right side.]

Reading the figure

Bottom rail = ground (0 V). Two unknown nodes:

  • Node 1: top of the 3 Ω shunt. Left branch: 6 Ω in series with the 6 V battery (+ at top).
  • Node 2: top of the 4 Ω shunt. Right branch: 4 Ω in series with the 10 V battery (+ at top).
  • Between 1 and 2: 5 V battery (+ on the 2 Ω side) in series with 2 Ω.

Polarities are read from the plate lengths in the figure.

Nodal equations (currents leaving the node)

Node 1: (V₁ − 6)/6 + V₁/3 + (V₁ + 5 − V₂)/2 = 0 ⇒ 1.0 V₁ − 0.5 V₂ = 1 − 2.5 → V₁ − 0.5 V₂ = −1.5 ... (i)

Node 2: (V₂ − V₁ − 5)/2 + V₂/4 + (V₂ − 10)/4 = 0 ⇒ −0.5 V₁ + 1.0 V₂ = 2.5 + 2.5 = 5 ... (ii)

Check of (i): multiply node 1 by 6: (V₁ − 6) + 2V₁ + 3 (V₁ + 5 − V₂) = 0 → 6 V₁ − 3 V₂ = −9, same as V₁ − 0.5 V₂ = −1.5.

Solving (i) and (ii): V₁ = 1.333 V, V₂ = 5.667 V

Branch currents

BranchCurrent
6 V battery and 6 Ω (towards node 1)(6 − 1.333)/6 = 0.778 A
3 Ω shunt (node 1 down)1.333/3 = 0.444 A
5 V battery and 2 Ω (node 1 to node 2)(1.333 + 5 − 5.667)/2 = 0.333 A
4 Ω shunt (node 2 down)5.667/4 = 1.417 A
10 V battery and 4 Ω (towards node 2)(10 − 5.667)/4 = 1.083 A

KCL check at node 1: 0.778 = 0.444 + 0.333. At node 2: 0.333 + 1.083 = 1.417.

Answer: 0.778 A, 0.444 A, 0.333 A, 1.417 A and 1.083 A as tabulated.

✓ Answer checked by Claude Opus 5.5 (high effort)

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