A battery of 6 V is connected to the circuit as shown below. The current I drawn from the battery is :

Solution

Physics sir · 3 steps

Spot the balanced Wheatstone bridge first: the 5 Ω in the middle carries no current, so drop it.

1. Check the bridge. The four outer resistors form a Wheatstone bridge with the 5 Ω as the middle arm. The ratios match on both sides: 3/3 = 6/6, so the bridge is balanced.
2. Remove the middle arm. In a balanced bridge, both ends of the middle arm are at the same potential, so no current flows through the 5 Ω. Remove it. That leaves the top branch 3 + 3 = 6 Ω in parallel with the bottom branch 6 + 6 = 12 Ω.

   R_(eq) = (6 × 12)/(6 + 12) + 2 = 4 + 2 = 6 Ω

3. So. I = V/R_(eq) = (6 V)/(6 Ω) = 1 A, which is option (a).

The idea to keep. Whenever you see a resistor bridging two branches, test the Wheatstone ratio before doing any heavy calculation. If P/Q = R/S, the middle resistor drops out completely.

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