Output voltage Vo using Kirchhoff laws

Reading the figure

Bottom rail = reference. Two nodes:

• Node 1: top of the 6 A source (current into node 1), top of the left 1 Ω to ground, and the left ends of both 2 Ω resistors.
• Node 2: right ends of both 2 Ω resistors, top of the right 1 Ω to ground, top of the 3 A source (current into node 2), and the output terminal V_o (+).

The two 2 Ω resistors are in parallel between nodes 1 and 2, so they act as 1 Ω.

KCL equations

Let V₁ and V₂ = V_o be the node voltages.

Node 1: 6 = V₁/1 + (V₁ − V₂)/1 ⇒ 2 V₁ − V₂ = 6 ... (1)

Node 2: 3 + (V₁ − V₂)/1 = V₂/1 ⇒ V₁ − 2 V₂ = −3 ... (2)

Solution

From (1): V₂ = 2 V₁ − 6. In (2): V₁ − 4 V₁ + 12 = −3, so V₁ = 5 V and V₂ = 4 V.

Branch currents: left 1 Ω = 5 A, right 1 Ω = 4 A, the pair of 2 Ω carries 1 A from node 1 to node 2 (0.5 A each).

Answer: V_o = 4 V