Call the top nodes 1 (after the 6 Ω) and 2 (after the first 2 Ω); the bottom wire is ground. Reduce from the right:
- Right branch: 2 + 4 = 6 Ω, in parallel with the 6 Ω at node 2: \(6 \parallel 6 = 3\ \Omega\)
- Add the 2 Ω to node 1: 5 Ω, in parallel with 3 Ω: \(3 \parallel 5 = \tfrac{15}{8} = 1.875\ \Omega\)
- Total: \(R_{eq} = 6 + 1.875 = 7.875\ \Omega\)
\(I_0 = \dfrac{15}{7.875} = \dfrac{40}{21} \approx 1.905\ \text{A}\), so \(V_1 = I_0 \times 1.875 = \tfrac{25}{7} \approx 3.571\ \text{V}\).
Current from node 1 towards node 2: \(\tfrac{V_1}{5} = \tfrac{5}{7}\ \text{A}\), which splits equally between the two 6 Ω paths.
1. \(I_5 = \tfrac{5}{14} \approx 0.357\ \text{A}\).
2. The circuit is linear, so the ratio \(\tfrac{I_5}{I_0} = \tfrac{5/14}{40/21} = \tfrac{3}{16}\) holds for any source. With I₅ = 7 A:
\(I_0 = 7 \times \tfrac{16}{3} = \tfrac{112}{3} \approx 37.33\ \text{A}\) (the source would then have to be 7.875 × 37.33 ≈ 294 V).
— rhs