Mesh current method with current source

Reading the figure

Battery 50 V (+ at top) feeds the top rail T. Branches: 10 Ω from T to the middle node M, 5 Ω from M to the bottom rail, 2 Ω from T to node N, 1 Ω from N to the bottom rail. A 2 A source (with the 3 Ω in series) joins N to M, direction N to M. The 3 Ω carries 2 A whatever else happens, so it does not appear in the equations.

Mesh currents (all clockwise)

• I₁: left mesh (50 V, 10 Ω, 5 Ω)
• I₂: upper right mesh (10 Ω, 2 Ω, source branch)
• I₃: lower right mesh (5 Ω, source branch, 1 Ω)

Equations

Current source branch (flows N to M, i.e. I₂ minus I₃): I₂ − I₃ = 2

Mesh 1: 50 = 10 (I₁ − I₂) + 5 (I₁ − I₃) ⇒ 15 I₁ − 10 I₂ − 5 I₃ = 50

Supermesh (meshes 2 and 3 combined, because the source voltage is unknown): 2 I₂ + 1 I₃ + 10 (I₂ − I₁) + 5 (I₃ − I₁) = 0 ⇒ −15 I₁ + 12 I₂ + 6 I₃ = 0

Solution

Put I₃ = I₂ − 2: 15 I₁ − 15 I₂ = 40 and −15 I₁ + 18 I₂ = 12 Adding: 3 I₂ = 52, so I₂ = 17.33 A; I₁ = 20 A; I₃ = 15.33 A

Branch currents

• Battery: 20 A
• 10 Ω: I₁ − I₂ = 2.67 A (T to M)
• 5 Ω: I₁ − I₃ = 4.67 A (M to ground)
• 2 Ω: 17.33 A (T to N)
• 1 Ω: 15.33 A (N to ground)
• 3 Ω and current source: 2 A (N to M)

Answer: 20 A, 2.67 A, 4.67 A, 17.33 A, 15.33 A and 2 A as above.