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

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Question

(B) The current $I$ flowing through the potentiometer wire $AB$ is given by the total $emf$ divided by the total resistance in the primary circuit:$I

Vedclass · Accepted Answer

$\frac{5L}{9}$

Vedclass · Answer

(B) The current $I$ flowing through the potentiometer wire $AB$ is given by the total $emf$ divided by the total resistance in the primary circuit:$I = \frac{\varepsilon}{r + 9r} = \frac{\varepsilon}{10r}$The potential difference $V_{AB}$ across the potentiometer wire $AB$ is:$V_{AB} = I 	imes 9r = \left( \frac{\varepsilon}{10r} ight) 	imes 9r = \frac{9\varepsilon}{10}$The potential gradient $K$ along the wire $AB$ is the potential drop per unit length:$K = \frac{V_{AB}}{L} = \frac{9\varepsilon}{10L}$For the galvanometer $G$ to show no deflection,the potential difference across the length $AJ$ must be equal to the $emf$ of cell $C$ $(\varepsilon / 2)$:$V_{AJ} = K 	imes AJ = \frac{\varepsilon}{2}$Substituting the value of $K$:$\left( \frac{9\varepsilon}{10L} ight) 	imes AJ = \frac{\varepsilon}{2}$Solving for $AJ$:$AJ = \frac{\varepsilon}{2} 	imes \frac{10L}{9\varepsilon} = \frac{5L}{9}$

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