Consider N = n1n2 identical cells, each of emf ϵ and internal resistance r. Suppose n1 cell are joined in series to form a line and n2 such lines are connected in parallel. The combination drives a current in an external resistance R.
(a) Find the current in the external resistance.
(b) Assuming that n1 and n2 can be continuously varied, find the relation between n1, n2, R and r for which the current in R is maximum.
a) 
; b) n1r=n2R
Solution:
a)
When n1 cells each with emf ‘E’ are connected in series, the total emf, Enet in one branch is,
Since, n2 of such the branches are connected in parallel, the Total emf in every branches are the same, Enet.
The resistance of n1 cells each with resistance ‘r’ in series is,
The total resistance for such n2 number of branches, connected in parallel, is
It is given that the whole setup is connected to an external resistance R. Hence the total net resistance will be,
Hence the current through the external resistor is,
Or,
b)
We know that the relation connecting n1, n2, R and r is
To get the minimum current through the resistor R, the denominator in the above expression should be minimum.
In order to minimize the term ‘n1r+n2R’, we re-write it as,
SO, this term should be minimized. But, the above value is minimum only when the term in the bracket is zero.
So,
Or,
Hence i is maximum when n1r=n2R