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(B) The circuit represents a Wheatstone bridge where the current between $B$ and $D$ is zero,implying the bridge is balanced.In the given circuit,the

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$2$

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(B) The circuit represents a Wheatstone bridge where the current between $B$ and $D$ is zero,implying the bridge is balanced.In the given circuit,the arms are:$AB = 12 \, \Omega + 4 \, \Omega = 16 \, \Omega$$BC = X \, \Omega$$AD = 1 \, \Omega + 3 \, \Omega = 4 \, \Omega$$DC = 1 \, \Omega \parallel 1 \, \Omega = \frac{1 	imes 1}{1 + 1} = 0.5 \, \Omega$For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal:$\frac{AB}{AD} = \frac{BC}{DC}$$\frac{16}{4} = \frac{X}{0.5}$$4 = \frac{X}{0.5}$$X = 4 	imes 0.5 = 2 \, \Omega$

In the circuit shown in the adjoining figure,the current between $B$ and $D$ is zero. The unknown resistance $X$ is ................ $\Omega$.

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