Question

Potentiometer wire length is $10\text{m}$, having a total resistance of $75\Omega$ and a battery of emf $200\text{V}$ (of negligible internal resistance) connected to it as shown in figure. Find the value of $x$ if $\text{P}$ is a null point.

• A. $1\text{m}$
• B. $2\text{m}$
• C. $3\text{m}$
• D. $4\text{m}$

Answer 1

The correct option is B $2\text{m}$

Current in the main circuit is,

$200-25i-75i=0$

$i=2\text{A}$

The potential gradient across the potentiometer wire will be

$\varphi =\frac{V_{AB}}{L_{AB}}=\frac{2\times 75}{L_{AB}}=\frac{150}{10}=15{\text{Vm}}^{-1}$

For null point, potential drop across wire of $(10-x)\text{m}$ will be equal to emf $(=120\text{V})$ of secondary battery.

$\therefore \varphi (10-x)=120\text{V}$

$15(10-x)=120$

$\therefore x=2\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{B}\right)$ is the correct answer.