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

Explanation

Here `l_(1) = 76.3 cm, l_(2) = 64.8 cm`
`r = ? R = 9.5 Omega`
Now, `r = ((l_(1) - l_(2))/(l_(2))) R = ((76.3 - 64.8)/(6.8)) 9.5 = 1.68 Omega`