Q. In a potentiometer experiment, the balancing length of a cell is 560 cm. When an external resistance of $10\Omega$ is connected in parallel to the cell, the balancing length chages by 60 cm. The internal resistance of a cell is
NTA AbhyasNTA Abhyas 2022
Solution:
Consider a cell of emf E and balancing length $l_{1}$
$E=kl_{1}$
The potential difference is balanced by length $l_{2}$ .
$V=kl_{2}$
The internal resistance of the cell is given by
$r=\left(\frac{E - V}{V}\right)R=\left(\frac{E}{V} - 1\right)R=\left(\frac{l_{1}}{l_{2}} - 1\right)R$
$=\left(\frac{560}{560 - 60} - 1\right)10=\left(\frac{56}{50} - 1\right)10$
$=\frac{6}{5}=1.2\Omega$
$E=kl_{1}$
The potential difference is balanced by length $l_{2}$ .
$V=kl_{2}$
The internal resistance of the cell is given by
$r=\left(\frac{E - V}{V}\right)R=\left(\frac{E}{V} - 1\right)R=\left(\frac{l_{1}}{l_{2}} - 1\right)R$
$=\left(\frac{560}{560 - 60} - 1\right)10=\left(\frac{56}{50} - 1\right)10$
$=\frac{6}{5}=1.2\Omega$