The figure shows a 2.0 V potentiometer used for the determination of the internal resistance of a 1.5 V cell.

Question

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

Accepted Answer

The internal resistance $r$ of a cell determined using a potentiometer is given by the formula $r = R \left( \frac{l_1}{l_2} - 1 ight)$, where $l_1$ is the balancing length in open circuit, $l_2$ is the balancing length when external resistance $R$ is connected, and $R$ is the external resistance.

Given:
$l_1 = 76.3 	ext{ cm}$
$l_2 = 64.8 	ext{ cm}$
$R = 9.5 \ \Omega$

Substituting these values:
$r = 9.5 	imes \left( \frac{76.3}{64.8} - 1 ight)$
$r = 9.5 	imes (1.1775 - 1)$
$r = 9.5 	imes 0.1775 \approx 1.686 \ \Omega$.

Rounding to two decimal places gives 1.69 Ω, but 1.68 Ω is the closest valid option provided.

Step-by-Step Solution

Practice All PHYSICS Questions