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Loop current method, current in 1Ω

Short answer

With clockwise loop currents, I1 = 3.75 A, I2 = −0.75 A. The current through the 1 Ω resistor is I1 − I2 = 4.5 A, flowing downward.

Calculate the current flowing through 1 ohm resistor for network shown below using loop current method. [Figure: 4 A current source on top; 12 V source with 2 Ω on the left; 2 Ω in the upper middle; 1 Ω shunt in the middle; 6 V source on the right.]

Reading the figure

Both batteries have + on top. Left node L (top of 12 V) and right node R (top of 6 V). Between them: 2 Ω from L to the middle node M, 2 Ω from M to R, and 1 Ω from M to the bottom rail. The 4 A current source sits across the top between L and R.

The 4 A source is connected between two nodes whose voltages are fixed by the batteries (L = 12 V, R = 6 V), so it does not change the current in the 1 Ω.

Loop currents (clockwise)

  • I₁: left loop (12 V, 2 Ω, 1 Ω)
  • I₂: right loop (1 Ω, 2 Ω, 6 V)

Left loop: 12 = 2 I₁ + 1 (I₁ − I₂) ⇒ 3 I₁ − I₂ = 12

Right loop (going down through the 6 V from + to −): 0 = 2 I₂ + 6 + 1 (I₂ − I₁) ⇒ −I₁ + 3 I₂ = −6

Solution

From the first equation I₂ = 3 I₁ − 12. Then −I₁ + 9 I₁ − 36 = −6, so I₁ = 3.75 A and I₂ = −0.75 A.

Current in the 1 Ω

I = I₁ − I₂ = 3.75 + 0.75 = 4.5 A (downward)

Answer: I₁Ω = 4.5 A

Check by nodal method: V_M = (12/2 + 6/2)/(1/2 + 1/2 + 1) = 9/2 = 4.5 V, so I = 4.5 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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