The current flowing through R₂ is :

Solution

Physics sir · 5 steps

Fold the network up from the far end at B, then unfold it back to find how the current splits.

1. Reduce from B. R₂ and R₁ are in series, giving 4 + 2 = 6 Ω from C to A. This is in parallel with R₃ = 6 Ω, so C to A is 3 Ω.
2. Reduce from D. Adding R₄ gives 3 + 3 = 6 Ω from D to A through C. In parallel with R₇ = 3 Ω this gives 2 Ω. Adding R₅ gives 4 Ω from E to A. R₆ sits directly across the battery, so it does not change the current in this branch.
3. Voltage at D. The current through R₅ is 8/4 = 2 A. The drop across the 2 Ω block from D to A is 2 × 2 = 4 V.

   I_(R₄) = V_(DA)/(R₄ + R_(CA)) = 4/6 = 2/3 A

4. Split at C. At C this current meets two equal 6 Ω paths, R₃ and R₂ + R₁, so it divides equally: 1/2 × 2/3 = 1/3 A.
5. So. The current through R₂ is 1/3 A, option (b).

The idea to keep. In a ladder network, reduce from the farthest end towards the battery, find the main current, then walk back and split it at each junction using voltage or equal-resistance division.

Answer from the NTA final answer key. Solution by PrepWiser.