In an experiment with a potentiometer to measure the internal resistance of a cell. when the cell in the secondary circuit is by shounted by `5 Omega`, the null point is at `220 cm`. When the cell is shunted by `20 Omega ` the null point is at `300 cm`. Find the internal resistance of the cell.

Answer

`2.76 Omega`

Explanation

Let `epsilon` and `r` be the emf and internal resistance of the cell used in secondary circuit and `K` be the potential gradent along the potentiometer wire. According to equation
`((epsilon)/(r + 5)) 5 = K xx 220 `….(i)
and `(epsilon)/(r + 20)) 20 = K xx 300 `….(ii)
On solving (i) and (ii) we get
`r = 2.76 Omega`