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Nodal analysis current through 10Ω

Short answer

Define node 1 (left) and node 2 (right) with bottom rail as ground. Nodal equations: V1/5 + (V1-10-V2)/10 = 5, (V2-V1+10)/10 + V2/3 + (V2+20)/2 = 0. Solving gives V1=16.67 V, V2=-10 V. Current through 10 Ω resistor = (V1-10-V2)/10 = 1.667 A.

Use the nodal voltage method to find the current flowing through 10Ω resistor for the network shown below. [Figure: 5 A current source on the left; 5 Ω shunt; 10 Ω in series with a 10 V source on top; 3 Ω shunt; 2 Ω in series with a 20 V source on the right.]

Reading the figure

Bottom rail = ground.

  • Node 1 (left): 5 A source into the node, 5 Ω to ground, and the branch 10 Ω in series with the 10 V battery (+ on the node 1 side).
  • Node 2: reached through that branch; 3 Ω to ground, and the 2 Ω branch in series with the 20 V battery (− on the 2 Ω side, + to ground).

Nodal equations (currents leaving the node)

Node 1: V₁/5 + (V₁ − 10 − V₂)/10 = 5 Node 2: (V₂ − V₁ + 10)/10 + V₂/3 + (V₂ + 20)/2 = 0

Simplify: (1) 3 V₁ − V₂ = 60 (2) −3 V₁ + 28 V₂ = −330 ... (multiply node 2 by 30: 3V₂ − 3V₁ + 30 + 10V₂ + 15V₂ + 300 = 0)

Adding: 27 V₂ = −270, so V₂ = −10 V; then V₁ = 50/3 = 16.67 V.

Current in the 10 Ω

I = (V₁ − 10 − V₂)/10 = (16.67 − 10 + 10)/10 = 1.667 A, from node 1 towards node 2.

Answer: I₁₀Ω = 1.667 A (5/3 A)

Assumption: battery polarities read from the plates and + marks in the figure (10 V: + on the node 1 side; 20 V: + on the ground side).

✓ Answer checked by Claude Opus 5.5 (high effort)

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