Obtain the condition fo rmaximum current through a resisitor, when a number of cells are connected in series.

Explanation

The cells are connected in series if they are connected one after the in cells in series other i.e., negative terminal of first cell is connected to the positive terminal of second cell, negative terminal of second : L cell is connected to the positive terminal : of third cell and so on.

Let n be the number of cells connected in series.
Let E be the e.m.f. and r be the internal resistance of each cell. This combination is further connected in series with an external resistance R.
As cells are connected in series, therefore, total e.m.f. of all the cells = n E.
The equivalent e.m.f. of a series combination of n cells is just the sum of their individual e.m.fs.
And Total internal resistance of n cells in series
`R_(i) = r+r+r+....n` times = nr
The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances.
Total resistance of circuit `= R +R_(i)= R + nr`
`:.` Current flowing through circuit is given by
`I= ("Total e.m.f.")/("Total resistance")= (nE)/(nr+R)`
If `R gtgt nr`
i.e., external resistance of the circuit is much greater than the internal resistance of the cell. Then we can neglect nr as compared to R.
From eqn. (1)
`I=n(E/R)`
I = n (current due to single cell).
So we can obtain maximum current through a resistor connecting the cells in series combination only when external resistance of the circuit-is much greater than the internal resistance of the cell.