Question

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $2\text{V}$ and a negligible internal resistance. The potentiometer wire is $4\text{m}$ long. When the resistance $R=9.5\Omega$ is connected across a secondary cell the balancing length on the potentiometer wire was found to be $2.85\text{m}$. If the balancing length for the emf of the secondary cell is $3\text{m}$, the value of internal resistance of the cell is

• A. $0.5\Omega$
• B. $0.95\Omega$
• C. $0.25\Omega$
• D. $0.75\Omega$

Answer 1

The correct option is A $0.5\Omega$

Given:
$l_1=3\text{m},l_2=2.85\text{m}$
$R=9.5\Omega$

Using the relation,

$r=(\frac{l_1}{l_2}-1)\times R$

$r=(\frac{3}{2.85}-1)\times 9.5$

$\therefore r=\frac{0.15}{2.85}\times 9.5=0.5\Omega$

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