When 10 identical cells in sereis are connected to the ends of a resistance of `59Omega`, the current is found to be 0.25A. But when the nsame cells, being connected in parallel, are joined to the ends of a resistance of 0.05Omega, the current is 25A. Calculate the internal resistance and the emf of each cell.

Explanation

`0.25 = (10E)/(59 +10r)` and `25 = (E )/(0.05 +(r )/(10))` On solving, we get
`E = 1.5 V` and `r = 0.1 Omega`