A potentiometer wire AB having length L and r | vedclass

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Question

(C) The current $I$ flowing through the potentiometer wire $AB$ is given by $I = \frac{\varepsilon}{R_{AB} + r} = \frac{\varepsilon}{12r + r} = \frac{

Vedclass · Accepted Answer

$\frac{13}{24}L$

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(C) The current $I$ flowing through the potentiometer wire $AB$ is given by $I = \frac{\varepsilon}{R_{AB} + r} = \frac{\varepsilon}{12r + r} = \frac{\varepsilon}{13r}$.The potential drop across the length $AJ$ (where $AJ = x$) is $V_{AJ} = I 	imes R_{AJ}$.Since the resistance of the wire is proportional to its length,$R_{AJ} = \frac{x}{L} 	imes 12r$.Thus,$V_{AJ} = \left(\frac{\varepsilon}{13r}ight) 	imes \left(\frac{x}{L} 	imes 12right) = \frac{12\varepsilon x}{13L}$.For the galvanometer to show no deflection,the potential drop across $AJ$ must equal the $emf$ of cell $C$,which is $\varepsilon/2$.So,$\frac{12\varepsilon x}{13L} = \frac{\varepsilon}{2}$.Solving for $x$:$\frac{12x}{13L} = \frac{1}{2}$$24x = 13L$$x = \frac{13}{24}L$.

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