45 cells each of internal resistance `0.4Omega` and emf 2.0 V are used to send current through an external circuit of resistance `2.0 Omega`. What will the best mode of grouping them for maximum current? Find the currnet in the external circuit.

Explanation

Here, `r=0.4 Omega , epsilon = 2.0V , R= 2.0 Omega`
As internal resistance `r( =0.4 Omega)` of a cell is comparable of external resistance R`( = 2.0 Omega)`, so for maximum current, the cells are to be connected in mixed grouping.
Let there be n cells in a row and m rows of cells. Then
mn =45
For maximum current in the mixed grouping
`(nr)/(m)= R` or `(nxx0.4)/(m) = 2` or `n/m= 5` or n=5m
From (i) and (ii)
`m xx 5 m= 45` or `m^(2)=9` or m=3
`:. n = 5 xx 3=15`
Therefore for maximum current there should be 3 rows of cells and each row should have 15 cells.
`I_(max) = (m n epsilon)/(mR + n r) = (45xx2.0)/(3 xx2 +15xx0.4)=7.5A`