The internal resistance of a cell is determined by using a potentiometer. In an experimetn, an external resistance of `60 Omega` is used across the given cell. When the key is closed, the balance length on the potentiometer decreases form 72 cm to 60 cm. calculate the internal resistance of the cell.

Explanation

According to equation `epsi_(0)=xl_(0)…………(i)`
`V=IR=xl.....................(ii)`
`I=epsi_(0)/(R+r)...............(iii)`
From equation (i),(ii)and (iii) we get
`r=R(l_(0)/l-1)`
here `l_(0)=72 cm , l = 60 cm , R=60 Omega`
`thereforer=(60)(72/60-1)or r=12Omega`