(i) Suppose n cells, each of emf `xi` volts and internal resistance r ohms are connected in series with an external resistance R. as shown in Figure.
(ii) The total emf of the battery `n xi`
The total resistance in the circuit `nr+R`
By Ohm's law, the current in the circuit is
`I=("total emf")/("total resistance") =(n xi)/(nr+R)`
Case (a) If `r lt lt R` then,
`I=(n xi)/(R)~~ nI_(1)`
where `I_(1)` is the current due to a single cell
`(I_(1)=(xi)/(R))`
(iii) Thus, if r is negligible when compared to R the current supplid by the battery is n times that supplied by a single cell.
Case (b) a single cell.
Case (b) If `r gt gt R, I=(n xi)/(nr)~~(xi)/(r)`
(iv) It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cell.
(v) Thus series connection of the cell is advantageous only when the effective internal resistance of the cells is negaligibly small compared with R.
Cells in parallel
(i) In parallel connection all the positive terminnals of the cells are connected toone point and all the negative terminals to a second point. These two points from the positive and negative terminals of the battery.
(ii) Let n cells be connected in parallel between the points A and B and a resistance R is connected between the point a and B as shown in figure. Let `xi` be the emf and r the internal resistance of each cell.
(iii) The equivalent internal resistance of the battey is `(1)/(r_(eq))=(1)/(r)+(1)/(r)+....(1)/(r)` (nterms) =`(n)/(r)`
So `(I)/(r_(eq))=(r)/(n)` and the total resistance in the circuit `R+(r)/(n)`. The total emf is the potential difference between the points A and B, which is equal to `xi` the current in the circuite is given by
`I=(xi)/((r)/(n)+R)`
`I=(n xi)/(r+nR)`
Case (a) If `r gt gt R, I=(n xi)/(r)=nI_(1)`
(iv) Where `I_(1)` is the current due to a single cell and is equal to `(xi)/(r)` when R is negligible. Thus, the current through the externla resistance due to the whole battery is n times the current due to a single cell.
Case (b) If `r lt lt R, I(xi)/(R)`
(v) The above equation implies that current due to the whole battery is the same as that due to a single cell. Hence it is advantageous to connecte cells in parallel when the external resistance is very small compared to the internal of the cells.
