The figure shows a 2.0V potentiometer used for the determination of internal resistance of a 1.5V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of `9.5Omega` 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.

Explanation

Given the balancing length when the cell is iri the open circit `l_(1) = 76.3cm`
The balancing length when the cell is in the closed circuit with a load resistor, `l_(2)=64.8` cm, and `R=9.5Omega`
We know that the internal resistance
`r=((l_(1))/(l_(2))-1)R`
i.e., `r=((76.3)/(64.8)-1)xx9.5=1.68Omega`
Hence internal resistance of the cell is `1.68Omega`