Question

A potentiometer wire $AB$ having length $L$ and resistance $12r$ is joined to a cell $D$ of emf $\epsilon $ and internal resistance $r$ . A cell $C$ having EMF $\epsilon /2$ and internal resistance $3r$ is connected. The length $AJ$ , at which the galvanometer, as shown in the figure, shows no deflection is

• A. $\frac{5}{12}L$
• B. $\frac{11}{24}L$
• C. $\frac{11}{12}L$
• D. $\frac{13}{24}L$

Answer 1

Answer: (D)

Potential difference across $AJ=IR_{A J}$ ,
$=\frac{\epsilon }{13 r}\cdot \left(\frac{12 r}{L} \cdot x\right)$
For no current in galvanometer,
$\frac{\epsilon }{2}=$ potential difference across $AJ$
$\frac{\epsilon }{2}=\frac{12 \epsilon }{13 L}x$
$x=\frac{13L}{24}$