Mesh analysis, power in 15Ω

Reading the figure

Top branch (left to right): 20 Ω, 15 Ω, 10 Ω. Vertical branches: 400 V battery (left, + on top), 80 Ω, 90 Ω, 200 V battery (right, + on top). The bottom is a common wire.

Mesh currents (clockwise)

• I₁: left mesh (400 V, 20 Ω, 80 Ω)
• I₂: middle mesh (15 Ω, 90 Ω, 80 Ω)
• I₃: right mesh (10 Ω, 200 V, 90 Ω)

Equations

Mesh 1: 400 = 20 I₁ + 80 (I₁ − I₂) ⇒ 100 I₁ − 80 I₂ = 400

Mesh 2: 0 = 15 I₂ + 90 (I₂ − I₃) + 80 (I₂ − I₁) ⇒ −80 I₁ + 185 I₂ − 90 I₃ = 0

Mesh 3 (going down through the 200 V battery from + to −, a drop of 200 V): 0 = 10 I₃ + 200 + 90 (I₃ − I₂) ⇒ −90 I₂ + 100 I₃ = −200

Solution

I₁ = 6.8 A, I₂ = 3.5 A, I₃ = 1.15 A

The current in the 15 Ω is I₂ = 3.5 A.

Power in the 15 Ω

P = I₂² × 15 = 3.5² × 15 = 12.25 × 15 = 183.75 W

Answer: P₁₅Ω = 183.75 W

Check: the 80 Ω carries I₁ − I₂ = 3.3 A and the 90 Ω carries I₂ − I₃ = 2.35 A. The power from the 400 V source is 2720 W; the 200 V source absorbs 230 W; the resistors dissipate the rest.