A cell on open circuit has e.m.f 2.0 V and in closed circuit having current of 0.05 A , the p.d is 1.5 V . Calculate internal resistance of the cell.

Explanation

e.m.f of cell E= 2V
Current in circuit I = 0.05 A
p.d across resistance V= 1.5 V
Let .r. be the internal resistance of the cell
E= (Ir+V)
`r= (E-V)/(I)`
`=(2-1.5)/(0.05) = (0.5)/(0.05) =10 Omega`
Internal resistance of the cell is `10 Omega`