Figure below 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 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Answer –

According to the question statement –

Internal resistance of the cell is r = 1.5 V cell

Balance point of the cell in open circuit is l = 76.3 cm

External resistance is R = 9.5 Ω

New balance point is l₁ = 64. 8 cm

Then, the expression for internal resistance is given by the relation –

\[r=R\left[ \frac{l}{{{l}_{1}}}-1 \right]r=9.5\left[ \frac{76.3}{64.8}-1 \right]\]

\[r=1.69\Omega \]