Question

In an experiment with a potentiometer, the null point is obtained at a distance of $60\text{cm}$ along the wire from the common terminal with a Leclanche cell. When a shunt resistance of $1\Omega$ is connected across the cell, the null point shifts to a distance of $30\text{cm}$ from the common terminal. What is the internal resistance of the cell ?

• A. $1\Omega$
• B. $2\Omega$
• C. $3\Omega$
• D. $4\Omega$

Answer 1

The correct option is A $1\Omega$

Given,
$R=1\Omega ;l_1=60\text{cm};l_2=30\text{cm}$

Internal resistance of the cell is given by

$r=R\left(\frac{l_1-l_2}{l_2}\right)$

$r=1\times \left(\frac{60-30}{30}\right)=1\Omega$