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 $\frac{\epsilon }{2}$ and internal resistance $3r$ is connected. The length AJ at which galvanometer as shown in figure shows no deflection is

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

Answer 1

Answer: (C)

Given, length of wire, $AB=L$
Resistance of wire, $AB=12r$
emf of cell $D=\epsilon$, internal resistance of $D=r$
emf of cell $C=\frac{\epsilon }{2}$, internal resistance of $C=3r$
Current in potentiometer wire $(i)=\frac{\text{ Total emf }}{\text{ Total resistance }}$
$i=\frac{\epsilon }{r+12r}=\frac{\epsilon }{13r}$
Potential drop across the balance length $AJ$ of potentiometer wire is
$V_{AJ}=i\times R_{AJ}$
$\Rightarrow V_{AJ}=i(\text{ resistance per unit length }\times \text{ length }AJ)$
$V_{AJ}=i\left(\frac{12r}{L}\times x\right)$
where, $x$ is the balance length $AJ$.
As null point occurs at $J$, so potential drop across balance length,
$AJ=$ emf of cell $C$
$V_{AJ}=\frac{\epsilon }{2}\Rightarrow i\left(\frac{12r}{L}\times x\right)=\frac{\epsilon }{2}$
$\Rightarrow \frac{\epsilon }{13r}\times \frac{12r}{L}\times x=\frac{\epsilon }{2}$
$\Rightarrow x=\frac{13}{24}L$