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Mesh analysis, voltage across 20Ω

Short answer

With mesh currents I1 = 6 A, I2, I3 and supermesh constraint I3 − I2 = 9 A, KVL gives 12 = 20I3 + 10(I2 − I1). Solving yields I3 = 5.4 A, so voltage across the 20 Ω resistor is V = 20 × 5.4 = 108 V.

Using mesh analysis, calculate the voltage across the 20 Ω resistor shown in the figure below. [Figure: 6 A current source on the left; 12 V source on top; 10 Ω resistor; 9 A current source; 20 Ω resistor on the right.]

Reading the figure

The left branch holds the 6 A source (upward). The 10 Ω resistor (the "10" in the figure, with + at its top) is the second branch. The 12 V source (− on the left, + on the right) is in the top wire between the second and third branches. The third branch holds the 9 A source (upward) and the fourth branch holds the 20 Ω. The 10 Ω value is assumed from the label.

Mesh currents (clockwise)

  • I₁ = 6 A (left mesh, through the 6 A source, upward)
  • I₂: middle mesh (10 Ω, 12 V, 9 A source)
  • I₃: right mesh (9 A source, 20 Ω)

Source branch of 9 A (upward, shared by meshes 2 and 3): I₃ − I₂ = 9, so I₃ = I₂ + 9

Supermesh (meshes 2 and 3)

KVL around the outer loop of meshes 2 and 3: going through the 12 V from − to + gives a rise of 12 V.

12 = 20 I₃ + 10 (I₂ − I₁) 12 = 20 (I₂ + 9) + 10 (I₂ − 6) 12 = 30 I₂ + 180 − 60 I₂ = −3.6 A

I₃ = −3.6 + 9 = 5.4 A

Voltage across 20 Ω

V = 20 × I₃ = 20 × 5.4 = 108 V

Answer: V₂₀Ω = 108 V (check: 10 Ω carries 9.6 A, so 96 V; 96 + 12 = 108 V)

✓ Answer checked by Claude Opus 5.5 (high effort)

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