Q94 / 125 · BITSAT 2018 · Physics

In the circuit given below, the value of resistance $X$ , when the potential difference between the points $B$ and $D$ is zero, will be

Solution

$P = 15 + 6 = 21 Ω, Q = \frac{8 X}{8 + X} + 3$

$R = 15 + \frac{6 \times 6}{6 + 6} = 18 Ω, S = 4 + \frac{4 \times 4}{4 + 4} = 6 Ω$

Since potential difference across $B D$ is zero hence it satisfies Wheatstone's Bridge condition,

$∴ \frac{P}{Q} = \frac{R}{S},$ so $Q = \frac{P S}{R} = \frac{21 \times 6}{18} = 7 Ω$

$3 + \frac{8 x}{8 + x} = 7 \Rightarrow x = 8 Ω$

Hence, the correct answer is option: 'B'.