Question

A cell balances against a length of $200cm$ on a potentiometer wire, when it is shunted by a resistance of $8\Omega$. The balancing length reduces by $40cm$, when it is shunted by a resistance of $4\Omega$. Calculate the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.

Answer 1

Let balancing length be $l_1$ and internal resistance be '$r$'.
$r=R\left(\frac{l_1-l_2}{l_2}\right)$
$r=8\left(\frac{l_1-200}{200}\right)$ ....(1)
$r=4\left(\frac{l_1-160}{160}\right)$ ....(2)
From equations (1) and (2)
$8\left(\frac{l_1-200}{200}\right)=4\left(\frac{l_1-160}{160}\right)$
$\frac{l_1-200}{10}=\frac{l_1-160}{16}$
$16l_1=3200+10l_1-1600$
$l_1=266.66cm$
$r=8\left(\frac{266.66-200}{200}\right)$
$r=2.66\Omega$