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Loop current method with 3 A source

Short answer

The 3 A source is across fixed battery voltages and does not affect the 2 Ω. Using loop currents with meshes, solve 3I1 − 2I2 = 9 and −2I1 + 5I2 = −12. Current through 2 Ω is I1 − I2 = 39/11 ≈ 3.55 A downward.

Use loop current method to calculate the current through the 2Ω resistance for the network shown below. [Figure: 3 A current source across the top; 9 V source with 1 Ω in the left branch; 2 Ω in the middle vertical branch; 3 Ω with a 12 V source in the right branch.]

Reading the figure

The 3 A current source is connected across the top between the left node L (+ of the 9 V battery) and the right node R (+ of the 12 V battery), direction left to right. Middle node M joins the 1 Ω (to L), 3 Ω (to R) and 2 Ω (to ground).

The 3 A source is in parallel with the (1 Ω + 3 Ω) path between L and R, but L and R are fixed at 9 V and 12 V by the batteries. So the source does not change the current in the 2 Ω.

Loop currents (clockwise)

  • I₁: left loop (9 V, 1 Ω, 2 Ω)
  • I₂: right loop (2 Ω, 3 Ω, 12 V)

Left loop: 9 = 1 × I₁ + 2 (I₁ − I₂) ⇒ 3 I₁ − 2 I₂ = 9

Right loop (going down through the 12 V battery from + to −): 0 = 3 I₂ + 12 + 2 (I₂ − I₁) ⇒ −2 I₁ + 5 I₂ = −12

Solving: I₁ = 1.909 A, I₂ = −1.636 A

Current through 2 Ω

I = I₁ − I₂ = 1.909 + 1.636 = 3.545 A (downwards)

Answer: I₂Ω = 3.55 A (39/11 A)

✓ Answer checked by Claude Opus 5.5 (high effort)

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