Question

$3$V potentiometer used for the determination of internal resistance of a $2.4$V cell. The balance point of the cell is open circuit is $75.8$cm. When a resistor of $10.2\Omega$ is used in the external circuit of the cell the balance point shifts to $68.3$cm length of the potentiometer wire. The internal resistance of the cell is?

• A. $2.5\Omega$
• B. $2.25\Omega$
• C. $1.12\Omega$
• D. $3.2\Omega$

Answer 1

Answer: (C)

$1.12\Omega$

Given:

The voltage of the potentiometer is $3V$.

The voltage of the battery whose internal resistance is to be determined is $2.4V$.

Balance point of cell in open circuit is $75.8cm$.

The new balance point of the wire is $68.3cm$.

The internal resistance of the cell is given by:

$r=R(\frac{l_1}{l_2}-1)$

$r=10.2(\frac{75.8}{68.3}-1)$

$r=1.12\Omega$