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(B) The circuit represents a Wheatstone bridge where the potential difference between points $B$ and $D$ is zero. This implies the bridge is balanced.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The circuit represents a Wheatstone bridge where the potential difference between points $B$ and $D$ is zero. This implies the bridge is balanced.For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal: $\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}}$.Here,$R_{AB} = 12 \, \Omega + 4 \, \Omega = 16 \, \Omega$.$R_{BC} = x$.$R_{AD} = 1 \, \Omega + 3 \, \Omega = 4 \, \Omega$.$R_{DC}$ consists of two $1 \, \Omega$ resistors in parallel,so $R_{DC} = \frac{1 	imes 1}{1 + 1} = 0.5 \, \Omega$.Applying the balance condition: $\frac{16}{x} = \frac{4}{0.5}$.$\frac{16}{x} = 8$.$x = \frac{16}{8} = 2 \, \Omega$.

In the given circuit,if the voltage between points $B$ and $D$ is zero,then what is the value of resistance $x$ in $\Omega$?

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