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Kirchhoff law branch currents

Short answer

Apply KVL: loop 1 gives 3I1 - 7I2 - 5I3 = -12, loop 2 gives 7I1 - 13I2 + 18I3 = -20, loop 3 gives 18I1 - 10I2 + 7I3 = 10. Solving yields I1 = 2.00 A, I2 = 2.59 A, I3 ≈ -0.02 A.

Find I1, I2 and I3 in the circuit shown in the figure using Kirchhoff's law. [Figure: 12 V source on the left with 4 Ω on top and 3 Ω below; central 5 Ω resistor; 6 Ω on the top right with a 20 V source; 7 Ω on the right; bottom branch has 8 Ω with a 10 V source.]

Reading the figure

Nodes: Lw (left middle), P (top middle), M (centre) and Rw (right middle).

  • Lw to P: 12 V battery (+ up) then 4 Ω; current I₂ flows up from Lw to P.
  • P to M: 5 Ω, current I₃ downward.
  • P to Rw: 6 Ω then 20 V battery (+ at the Rw end).
  • Lw to M: 3 Ω. M to Rw: 7 Ω.
  • Rw to Lw: 10 V battery (+ on the 8 Ω side) then 8 Ω; current I₁ flows up into Lw.

Polarities are taken from the plate lengths in the scan.

Currents by KCL

  • Lw: I₁ enters, I₂ leaves, so the 3 Ω carries I₁ − I₂ (Lw to M).
  • M: 5 Ω brings I₃, 3 Ω brings I₁ − I₂, so the 7 Ω carries I₁ − I₂ + I₃ (M to Rw).
  • P: I₂ enters, I₃ leaves, so the 6 Ω carries I₂ − I₃ (P to Rw).

Loop equations by KVL

Loop 1 (Lw, 12 V, 4 Ω, 5 Ω, 3 Ω): 12 − 4 I₂ − 5 I₃ + 3 (I₁ − I₂) = 0 ⇒ 3 I₁ − 7 I₂ − 5 I₃ = −12

Loop 2 (P, 6 Ω, 20 V, 7 Ω, 5 Ω): −6 (I₂ − I₃) + 20 + 7 (I₁ − I₂ + I₃) + 5 I₃ = 0 ⇒ 7 I₁ − 13 I₂ + 18 I₃ = −20

Loop 3 (Lw, 3 Ω, 7 Ω, 10 V, 8 Ω): −3 (I₁ − I₂) − 7 (I₁ − I₂ + I₃) + 10 − 8 I₁ = 0 ⇒ 18 I₁ − 10 I₂ + 7 I₃ = 10

Solution

I₁ = 2.00 A, I₂ = 2.59 A, I₃ = −0.021 A

The negative sign means I₃ actually flows upward (from M to P), but it is very small.

Answer: I₁ ≈ 2.00 A, I₂ ≈ 2.59 A, I₃ ≈ −0.02 A (about zero)

✓ Answer checked by Claude Opus 5.5 (high effort)

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