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Loop current method, current in 10Ω

Short answer

Solving three mesh equations yields I₁ = 0.137 A, I₂ = −0.308 A. Current in the 10 Ω resistor is I₁ − I₂ = 0.445 A, flowing downward.

Find the current through the 10 Ω resistor using loop-current method? [Figure: 10 V source on the left; 4 Ω, 2 Ω and 5 Ω along the top; 10 Ω shunt in series with a 5 V source; 6 Ω shunt; 20 V source on the right.]

Reading the figure

Three meshes side by side. All three batteries have the + terminal on top.

  • Left mesh: 10 V, 4 Ω, then the shunt branch (10 Ω in series with the 5 V battery).
  • Middle mesh: the 2 Ω on top, the 6 Ω shunt on the right.
  • Right mesh: 5 Ω and the 20 V battery.

Loop currents (all clockwise)

I₁ (left), I₂ (middle), I₃ (right).

Loop 1: 10 = 4 I₁ + 10 (I₁ − I₂) + 5 ⇒ 14 I₁ − 10 I₂ = 5

Loop 2 (going up through the 5 V from − to +, a rise of 5 V): 5 = 2 I₂ + 6 (I₂ − I₃) + 10 (I₂ − I₁) ⇒ −10 I₁ + 18 I₂ − 6 I₃ = 5

Loop 3 (going down through the 20 V from + to −, a drop of 20 V): 0 = 5 I₃ + 20 + 6 (I₃ − I₂) ⇒ −6 I₂ + 11 I₃ = −20

Solution

I₁ = 0.137 A, I₂ = −0.308 A, I₃ = −1.986 A

Current in the 10 Ω

I = I₁ − I₂ = 0.137 + 0.308 = 0.445 A, flowing downward (top to bottom)

Answer: I₁₀Ω ≈ 0.445 A

✓ Answer checked by Claude Opus 5.5 (high effort)

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