AB is a wire of uniform resistance. The galva | vedclass

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(C) The given circuit represents a Wheatstone bridge configuration.According to the principle of a Wheatstone bridge,when the galvanometer shows no de

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$20$

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(C) The given circuit represents a Wheatstone bridge configuration.According to the principle of a Wheatstone bridge,when the galvanometer shows no deflection (null point),the ratio of resistances in the arms is equal to the ratio of the lengths of the wire segments.Let $R_{AC}$ and $R_{CB}$ be the resistances of the wire segments $AC$ and $CB$ respectively.Since the wire is uniform,the resistance is proportional to its length: $\frac{R_{AC}}{R_{CB}} = \frac{AC}{CB}$.For the balanced Wheatstone bridge,the condition is: $\frac{R}{80\,\Omega} = \frac{R_{AC}}{R_{CB}}$.Substituting the given values: $\frac{R}{80} = \frac{20}{80}$.Solving for $R$: $R = 20\,\Omega$.

$AB$ is a wire of uniform resistance. The galvanometer $G$ shows no current when the length $AC = 20\,cm$ and $CB = 80\,cm$. The resistance $R$ is equal to .............. $\Omega $

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