In the arrangement of resistors shown in the | vedclass

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(B) The circuit represents a Wheatstone bridge. For the potential difference between $B$ and $D$ to be zero,the bridge must be balanced.The resistance

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$2$

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

In the arrangement of resistors shown in the diagram,the potential difference between $B$ and $D$ will be zero when the unknown resistance $X$ is ............... $\Omega$.

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