Question

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 of the potentiometer wire. Determine the internal resistance of the cell.

Answer 1

Internal resistance of the cell $=r$

Balance point of the cell in open circuit, $l_1=76.3cm$

An external resistance (R) is connected to the circuit with $R=9.5\Omega$

New balance point of the circuit, $l_2=64.8cm$

Current flowing through the circuit $=l$

Using the relation connecting resistance and emf is,

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

$=\frac{76.3-64.8}{64.8}\times 9.5=1.68\Omega$

Therefore, the internal resistance of the cell is 1.68$\Omega$.