The current I in the given circuit will be :

Solution

Physics sir · 4 steps

Spot the balanced Wheatstone bridge: the 2 Ω resistor carries no current, so drop it.

1. Name the nodes. Call the left junction A, the left end of the 2 Ω resistor B, its right end C, and the right junction D. Then A–B is 4 Ω, A–C is 4 Ω (the top one), B–D is 4 Ω (the bottom one), C–D is 4 Ω, and the 2 Ω sits between B and C.
2. Check the balance. This is a Wheatstone bridge with the 2 Ω as its middle arm. The ratio test holds: R_(AB)/R_(BD) = 4/4 = R_(AC)/R_(CD). So B and C are at the same potential, and no current flows through the 2 Ω.
3. Equivalent resistance. Remove the 2 Ω. That leaves two paths from A to D, each 4 + 4 = 8 Ω, in parallel.

   R_(eq) = (8 × 8)/(8 + 8) = 4 Ω, I = (40 V)/(4 Ω) = 10 A

4. So. The battery supplies I = 10 A, option (a).

The idea to keep. Before writing Kirchhoff equations, look for a bridge. If the opposite arm ratios are equal, the middle resistor carries no current, so remove it and the circuit becomes plain series and parallel.

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