A cell of emf `epsilon` and internal resistance r gives a current of 0.5 A with an external resistance of `12 Omega` and a current of 0.25 A with an external resistance of `25 Omega`. Calculate (a) internal resistance of the cell and (b) emf of the cell.

Answer

`1 Omega`, 6.5 V`

Explanation

EMF of the cell, `epsilon = I(R + r)`
In first case , `epsilon = 0.5(12 + r)`
In second case, `epsilon =0.25 (25 + r)`
`:. 0.5 (12 + r) = 0.25 (25 + r)`
On solving `r = 1 Omega`
so `epsilon = 0.5 (12 + 1) = 6.5 V`