Question

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