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Mesh analysis: voltage across 1 Ω resistor

Short answer

Using mesh analysis with supermesh from the 2 A source, set I1−I2 = 2 A, V1 = I2. Solve KVL equations to find I2 = 4.432 A. Voltage across the 1 Ω resistor is V1 = 4.43 V (+ at top).

In the given circuit determine the voltage across the 1 Ω resistor using the mesh analysis method. [Figure: 20 V battery (positive terminal at top) connected to the top node. From the top node, a 4 Ω resistor runs down to a middle node M and a 2 Ω resistor runs down (right side) in series with a dependent voltage source 2V1 (+ at top) to a right node N. A 3 Ω resistor connects M and N. A 2 A current source (arrow pointing down) is connected from M down, in series with a 5 Ω resistor to the bottom wire. A 1 Ω resistor connects N to the bottom wire, with V1 across it (+ at top). The battery negative terminal is on the bottom wire.]

Assumptions from the figure: the 20 V source feeds node T. 4 Ω connects T to M; (2 Ω + dependent source 2V₁, + at top) connects T to N; 3 Ω connects M to N; the (2 A source + 5 Ω) branch connects M to ground; 1 Ω connects N to ground and V₁ is the voltage across this 1 Ω resistor. The dependent source value is 2V₁ volts.

Mesh currents (all clockwise)

  • I₁: left mesh (20 V, 4 Ω, the 2 A branch).
  • I₂: lower mesh (3 Ω, 1 Ω, the 2 A branch).
  • I₃: upper triangle (4 Ω, 2 Ω + dependent source, 3 Ω).

The 2 A source (pointing down) is shared by I₁ and I₂, so I₁ and I₂ form a supermesh:

Constraint:  I₁ − I₂ = 2                         ... (1)
V₁ = 1 × I₂ = I₂                                 ... (2)

Supermesh KVL (loop through 20 V, 4 Ω, 3 Ω and 1 Ω; the 2 A source and its 5 Ω are left out)

−20 + 4(I₁ − I₃) + 3(I₂ − I₃) + 1·I₂ = 0
4I₁ + 4I₂ − 7I₃ = 20                             ... (3)

Mesh 3 KVL (4 Ω, 2 Ω, dependent source, 3 Ω)

2I₃ + 2V₁ + 3(I₃ − I₂) + 4(I₃ − I₁) = 0
−4I₁ − I₂ + 9I₃ = 0                              ... (4)   (using V₁ = I₂)

Solve

From (1), I₁ = I₂ + 2. Put in (3) and (4):

8I₂ − 7I₃ = 12
−5I₂ + 9I₃ = 8

I₂ = (12×9 + 7×8)/(8×9 − 7×5) = 164/37 = 4.432 A
I₃ = 3.351 A
I₁ = 6.432 A

Check by nodal analysis: V_M = 7.676 V, V_N = 4.432 V, which agrees with V₁ = I₂ × 1 Ω.

Answer: voltage across the 1 Ω resistor V₁ = I₂ × 1 = 4.43 V (+ at top).

✓ Answer checked by Claude Opus 5.5 (high effort)

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