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Nodal analysis current I

Short answer

Nodal analysis gives node voltages V₁ = 9 V, V₂ = 6 V. Current I through the 3 Ω resistor is (V₁ − V₂)/3 = 1 A.

Find the current I in the circuit of figure below by applying nodal voltage method. [Figure: 1 A current source across the top 3 Ω resistor carrying I; 6 Ω on the left in series with an 18 V source; 6 Ω shunt; 6 Ω shunt; 6 Ω on top right; 12 V source on the right.]

Reading the figure

Bottom rail = ground. Two unknown nodes:

  • V₁: junction of the left 6 Ω (to the 18 V battery, + on top), the 6 Ω to ground and the 3 Ω.
  • V₂: junction of the 3 Ω, the 6 Ω to ground and the right 6 Ω (to the 12 V battery, + on top).
  • The 1 A source is in parallel with the 3 Ω and pushes current from node 2 to node 1 (arrow to the left).
  • I is the current in the 3 Ω from node 1 to node 2.

Nodal equations (currents leaving the node)

Node 1: (V₁ − 18)/6 + V₁/6 + (V₁ − V₂)/3 − 1 = 0 Multiply by 6: V₁ − 18 + V₁ + 2 V₁ − 2 V₂ − 6 = 0 ⇒ 4 V₁ − 2 V₂ = 24 ... (1)

Node 2: (V₂ − 12)/6 + V₂/6 + (V₂ − V₁)/3 + 1 = 0 Multiply by 6: V₂ − 12 + V₂ + 2 V₂ − 2 V₁ + 6 = 0 ⇒ −2 V₁ + 4 V₂ = 6 ... (2)

Solution

From (1): V₂ = 2 V₁ − 12. In (2): −2 V₁ + 8 V₁ − 48 = 6, so V₁ = 9 V and V₂ = 6 V.

Current I

I = (V₁ − V₂)/3 = (9 − 6)/3 = 1 A

Answer: I = 1 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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