The value of unknown resistance (x) for which the potential difference between B and D will be zero in the arrangement shown, is :

Solution

Physics sir · 4 steps

Shrink each arm to one resistor, then demand the ratio of the top arms equal the ratio of the bottom arms.

1. Reduce each arm. Arm AB: the two 12 Ω in series make 24 Ω, and that sits in parallel with 24 Ω, so R_(AB) = 12 Ω. Arm BC: two 1 Ω in parallel give R_(BC) = 0.5 Ω.
2. Lower arms. Arm AD: the two 12 Ω in parallel give 6 Ω, in series with x, so R_(AD) = x + 6. Arm DC is R_(DC) = 0.5 Ω.
3. Balance condition. B and D are at the same potential when the bridge is balanced, that is, when the ratios of resistances on the two sides match.

   R_(AB)/R_(BC) = R_(AD)/R_(DC) ⇒ 12/0.5 = (x+6)/0.5

4. So. x + 6 = 12, which gives x = 6 Ω, option (c). The 14.5 V battery does not affect the balance.

The idea to keep. For zero potential difference across a Wheatstone bridge, first reduce every arm to one equivalent resistance, then use P/Q = R/S. The battery voltage never matters for balance.

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