Question

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

• A. $(13/24)L$
• B. $(5/12)L$
• C. $(11/24)L$
• D. $(11/12)L$

Answer 1

The correct option is A

$(13/24)L$

Step 1: Given data

Resistance $=12r$

Internal resistance of cell $=3r$

Step 2: Find potential drop

According to ohm law, Potential, $V$ is directly proportional to the current, $I$

$$\begin{gathered} V=IR \\ I=\frac{V}{R} \end{gathered}$$ [$R$ is the resistance.]

Current in potentiometer wire, $i$

$$\begin{gathered} i=(\epsilon /12r+r) \\ i=(\epsilon /13r) \end{gathered}$$ [ $\epsilon$ is the EMF]

Voltage across the wire, $V_{AJ}$

$$\begin{gathered} V_{AJ}=i\times R_{AJ} \\ {V}_{AJ}=i\times \left(\frac{12r}{L}\right)x \end{gathered}$$ [$R_{AJ}$ resistance across the wire, $x$ is the length $AJ$, $L$ is the length of the potentiometer.]

As null point is at $J$, so potential drop across length is equal to EMF of the cell

$$\begin{gathered} V_{AJ}=\frac{\epsilon }{2} \\ \left[\right(\frac{\epsilon }{12r+r})\times (\frac{12r}{L})x]=\frac{\epsilon }{2} \\ \left(\frac{\epsilon }{13r}\right)\times \left(\frac{12r}{L}\right)x=\frac{\epsilon }{2} \\ \frac{12\epsilon }{13xL}=\frac{\epsilon }{2} \\ x=\frac{13L}{24} \end{gathered}$$

Hence, option $\left(A\right)$ is correct.