Question

In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 $\Omega$ is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

Answer 1

Balance point in open circuit, $l_1=350$ cm.
$R=9\Omega$ ,

New balance point $l_2$ = 300 cm.

Internal resistance of a cell is given by,

$r=R\left(\frac{l_1-l_2}{l_2}\right)=9\left(\frac{350-300}{300}\right)=\frac{9\times 50}{300}=1.5\Omega .$

Hence, the internal resistance of the cell is 1.5 ohm.