A potentiometer wire AB having length L and r | vedclass

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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{

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$\frac{13}{24}L$

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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$.

Similar Questions

In a potentiometer circuit, when two cells of e.m.f. $1.5 \text{ V}$ and $1.2 \text{ V}$ are connected to assist each other, the balancing length is $270 \text{ cm}$. What will be the balancing length in $\text{cm}$ when these two cells are connected in opposition?

In a potentiometer experiment, a null point is obtained at a particular point for a cell on a potentiometer wire of length $L$. If the length of the potentiometer wire is increased by a few centimeters without changing the cell or the driving source, the balancing length will:

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