In the figure,the potentiometer wire AB of le | vedclass

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(B) The current $I$ flowing through the potentiometer wire $AB$ is given by $I = \frac{\varepsilon}{R_{wire} + r} = \frac{\varepsilon}{9r + r} = \frac

Vedclass · Accepted Answer

$\frac{5L}{9}$

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(B) The current $I$ flowing through the potentiometer wire $AB$ is given by $I = \frac{\varepsilon}{R_{wire} + r} = \frac{\varepsilon}{9r + r} = \frac{\varepsilon}{10r}$.The potential difference across the wire $AB$ is $V_{AB} = I 	imes R_{wire} = \frac{\varepsilon}{10r} 	imes 9r = \frac{9\varepsilon}{10}$.The potential gradient $x$ along the wire $AB$ is $x = \frac{V_{AB}}{L} = \frac{9\varepsilon}{10L}$.For the galvanometer $G$ to show no deflection,the potential difference across the length $AJ$ must be equal to the $emf$ of cell $C$. Let the length $AJ = \ell$.Thus,$x 	imes \ell = \frac{\varepsilon}{2}$.Substituting the value of $x$,we get $\frac{9\varepsilon}{10L} 	imes \ell = \frac{\varepsilon}{2}$.Solving for $\ell$,we get $\ell = \frac{\varepsilon}{2} 	imes \frac{10L}{9\varepsilon} = \frac{5L}{9}$.

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