Question

The figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5$\Omega$ is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer. The internal resistance of the cell is

• A. 1.63$\Omega$
• B. 1.66$\Omega$
• C. 1.69$\Omega$
• D. 1.72$\Omega$

Answer 1

Answer: (C)

1.69$\Omega$

The expression of internal resistance of a cell in potentiometer is given by

$r=R[\frac{l_1}{l_2}-1]$

or $r=9.5[\frac{76.3}{64.8}-1]=1.69\Omega$