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(A) For a balanced Wheatstone bridge,the ratio of resistances in the arms must be equal,i.e.,$\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{CD}}$.Here,$R_{

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$25$

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(A) For a balanced Wheatstone bridge,the ratio of resistances in the arms must be equal,i.e.,$\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{CD}}$.Here,$R_{AB} = 100 \ \Omega$ and $R_{AD} = 200 \ \Omega$.The resistance between $B$ and $C$ is the parallel combination of $100 \ \Omega$ and $R$,which is $R_{BC} = \frac{100 	imes R}{100 + R}$.The resistance $R_{CD} = 40 \ \Omega$.Substituting these values into the balanced bridge condition:$\frac{100}{200} = \frac{\frac{100R}{100+R}}{40}$$\frac{1}{2} = \frac{100R}{40(100+R)}$$\frac{1}{2} = \frac{10R}{4(100+R)}$$4(100+R) = 20R$$400 + 4R = 20R$$16R = 400$$R = 25 \ \Omega$.

The given Wheatstone bridge shows no deflection in the galvanometer joined between the points $B$ and $D$ (see figure). Calculate the value of $R$ in $\Omega$.

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