Superposition, current in branch AB

Reading the figure

Terminals A and B are joined by the 6 Ω resistor (R_L). The bottom-left corner is wired to G₁ (the bottom of the left 24 Ω resistor) and the bottom-right corner is wired to G₂; G₁ and G₂ are joined by the 2 Ω resistor. Take G₂ as the reference (0 V).

• A to G₁: 24 Ω, 24 Ω (diagonal) and the branch 6 Ω in series with the 6 V battery (+ on the 6 Ω side).
• B to G₂: 24 Ω, 24 Ω (diagonal) and the branch 6 Ω in series with the 12 V battery (+ on the 6 Ω side).
• 24 Ω ∥ 24 Ω = 12 Ω on each side.

Case 1: 6 V alone (12 V source shorted)

Right side from B to ground: 12 Ω ∥ 6 Ω = 4 Ω.

Node A: (V_A − V_G1 − 6)/6 + (V_A − V_G1)/12 + (V_A − V_B)/6 = 0 Node B: V_B/4 + (V_B − V_A)/6 = 0 Node G₁: V_G1/2 = (V_A − V_G1 − 6)/6 + (V_A − V_G1)/12

Solving: V_A = 2.5 V, V_B = 1.0 V, V_G1 = −0.5 V

I′(AB) = (2.5 − 1.0)/6 = 0.25 A from A to B.

Case 2: 12 V alone (6 V source shorted)

Left side from A to G₁: 12 Ω ∥ 6 Ω = 4 Ω.

Node A: (V_A − V_G1)/4 + (V_A − V_B)/6 = 0 Node B: (V_B − 12)/6 + V_B/12 + (V_B − V_A)/6 = 0 Node G₁: V_G1/2 = (V_A − V_G1)/4

Solving: V_A = 3.0 V, V_B = 6.0 V, V_G1 = 1.0 V

I″(AB) = (3 − 6)/6 = −0.5 A, that is 0.5 A from B to A.

Total

I(AB) = 0.25 − 0.5 = −0.25 A

The current in branch AB is 0.25 A flowing from B to A.

Check with both sources together: V_A = 5.5 V and V_B = 7.0 V, so I = (5.5 − 7.0)/6 = −0.25 A.

Answer: 0.25 A from B to A

Assumption: the polarities of both batteries are read from the figure with the + terminal on the 6 Ω side.