Find the minimum number of cells required to produce an electric current of `1.5` A through a resistance of `30 Omega`. Given that the emf of each cell is `1.5` V and internal resistance `1.0Omega`.

Explanation

`nm = ? , 30 = (nxx1)/(m) o rln = 30m`,
`1.5 = (nxx1.5)/(30+ (n)/(m))`
On solving we get n 60, m=2, total number nof batteries =
mm = 120.