Thirty six cells each of emf 1.5 V and internal resistance `0.5 Omega` are used to send current through an external resistor of resistance `2 Omega`. What is the best mode of grouping them and the current through the external resistor

Explanation

Here, `epsilon = 1.5 V, r=0.5 Omega. R=2 Omega`.
Since the value of r is comparable to R, for the best mode, the cells are connected in mixed grouping. Let there be m number of rows and n cellsin each row. Then mm =36
For maximum current in the mixed grouping
`nr/m = R or nxx0.5/m=2 or n=4m`
From (i), ` 4m xx m = 36 or m^(2) =9` or m=3
and `n =4 xx 3=12`
Thus for maximum current, there should be 3 rows and 12 cells in series in each row.
Maximum current,
`I_(max)= (m n epsilon)/(mR +nr) = (36 xx1.5/3 xx 2 + 12xx 0.5) = 4.5 A`