In the given circuit,the galvanometer shows z | vedclass

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(D) For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal. The equivalent resistance of the parallel combination of

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$25$

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(D) For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal. The equivalent resistance of the parallel combination of $100 \ \Omega$ and $R$ is $R_{eq} = \frac{100 	imes R}{100 + R}$.The condition for a balanced bridge is:$\frac{100}{200} = \frac{R_{eq}}{40}$Substituting $R_{eq}$:$\frac{1}{2} = \frac{100R}{40(100 + R)}$$\frac{1}{2} = \frac{10R}{4(100 + R)}$$4(100 + R) = 20R$$400 + 4R = 20R$$16R = 400$$R = \frac{400}{16} = 25 \ \Omega$.

In the given circuit,the galvanometer shows zero deflection. What is the value of $R$ in $\Omega$?

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