Find the minimum number of cells required to produce a current of `1.5A` through a resistance of `30Omega.` Given that the emf of each cell is `1.5V` and the internal resistance is `1Omega`.

Answer

`120`

Explanation

Here, `epsilon= 1.5 V , r = 1.0 Omega R = 30 Omega , 1 = 1.5 A`
As `r` comparable in `R`, so cells are to be connected in mixed grouping for maximum current .Let then be `n` cell in series in each row and `m` each rows in parallel
Current in the mixed grouping of cell is
`I = (mn epsilon)/(mR + nr) :. 1.5 = (mn xx 1.5)/(m xx 30 + n xx 1)`
or `45m + 1.5 n = 1.5 m n `.....(i)
For maximum current `R = (nr)/(m)`
so, `30 = (n xx 1)/(m) or n = 30m`.....(ii) we have
`45 m + 1.5 xx 30 m`
or `45 m + 45m = 45 m^(2)`
or `90 45 m or m = 2`
From (ii) `m n = 2 xx 60 = 120`