show 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

Here,
`l_(1) = 76.3 ` cm,
`l_(2) = 64.8` cm
R = 9.5 `Omega`
With the help of potentiometer , interna resistance of cell ,
` r = R ((l_(1) - l_(2))/(l_(2)) ) `
`therefore r = 9.5 (( 76.3 - 64.8)/(64.8)) = (9.5 xx 11.5 )/(64.8)`
= ` 1.68 Omega approx 1.7 Omega`