Question

In a potentiometer arrangement for determining the internal resistance 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

$\begin{array}{c}{r}_{\circ }=\left(\frac{l_O-l_c}{l_c}\right)R\\ L_o=longlength\\ l_c=smalllength\\ r_{\circ }=\left(\frac{350-300}{300}\right)\times 9\\ =\frac{50}{300}\times 9\\ =\frac{3}{2}\\ Hence,\\ r=1.5\Omega \end{array}$