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Mesh analysis, current in 5Ω

Short answer

Assign clockwise mesh currents I₁, I₂, I₃. Solve: 22I₁−20I₂−2I₃=8, −20I₁+25I₂−2I₃=10, −2I₁−2I₂+9I₃=12. The 5 Ω resistor carries I₃ = 560/167 ≈ 3.35 A.

Determine current in 5Ω resistor by mesh analysis in figure below. [Figure: 8 V source on the left; 20 Ω shunt; 10 V source on the top right; 3 Ω on the right; 2 Ω and 2 Ω resistors in the middle; 12 V source with 5 Ω in the bottom branch.]

Reading the figure

  • Left branch: 8 V battery (+ on top).
  • Top middle: 20 Ω shunt between the top wire and the middle node M.
  • Top right: 10 V battery (+ on the right) and 3 Ω in the right vertical.
  • Middle row: 2 Ω (left) and 2 Ω (right) with the 20 Ω joining at M.
  • Bottom branch: 12 V battery (+ on the left) in series with the 5 Ω.

Battery polarities are read from the plate lengths in the figure.

 +--------+--[20]--+--(10V)--+
 |        |        |         |
(8V)     ...      ...       [3]
 |   I1   |   I2   |         |
 +--[2]---+---M----[2]-------+
 |            I3             |
 +--(12V)------[5]-----------+

Mesh currents (all clockwise)

  • I₁: top left mesh (8 V, 20 Ω, 2 Ω)
  • I₂: top right mesh (20 Ω, 10 V, 3 Ω, 2 Ω)
  • I₃: bottom mesh (2 Ω, 2 Ω, 5 Ω, 12 V)

Mesh equations

Mesh 1: 8 = 20 (I₁ − I₂) + 2 (I₁ − I₃) ⇒ 22 I₁ − 20 I₂ − 2 I₃ = 8 Mesh 2: 10 = 3 I₂ + 2 (I₂ − I₃) + 20 (I₂ − I₁) ⇒ −20 I₁ + 25 I₂ − 2 I₃ = 10 Mesh 3: 12 = 2 (I₃ − I₁) + 2 (I₃ − I₂) + 5 I₃ ⇒ −2 I₁ − 2 I₂ + 9 I₃ = 12

Solution

I₁ = 4.679 A, I₂ = 4.411 A, I₃ = 3.353 A (Cramer's rule, determinant 1002)

The 5 Ω carries the mesh 3 current only.

Answer: Current in the 5 Ω resistor = I₃ ≈ 3.35 A (560/167 A)

✓ Answer checked by Claude Opus 5.5 (high effort)

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