### Combination of cell in the circuit

A.) Combination of cells when emf of cells are same: There are three types of combinations of cells in the circuit

1.) Series Combination of Cells

2.) Parallel Combination of Cells

3.) Mixed Combination of Cells

1.) Series Combination of Cells: Let us consider that the $n$ - cells having emf (electromotive force) $E$ and internal resistance $r$ are connected in series with external resistance $R$. Then from the figure given below
The total emf of the $n$ - cell = $nE$

The total internal resistance of the $n$ - cell = $nr$

The total resistance of the circuit = $nr+R$

The total current in the circuit

$i=\frac{Total \: emf \: of \: the \: n - series \: cell}{Total \: resistance \: of \: the \: circuit}$

$i=\frac{nE}{nr+R}$

2.) Parallel Combination of Cells: Let us consider that the $n$ - cells having emf (electromotive force) $E$ and internal resistance $r$ are connected in parallel with external resistance $R$. Then from the figure given below
The total emf of the $n$ - cell = $E$

The total internal resistance of the $n$ - cell

$\frac{1}{r_{eq}} = \frac{1}{r}+ \frac{1}{r}+.........n \: times$

$\frac{1}{r_{eq}}=\frac{n}{r}$

$r_{eq}=\frac{r}{n}$

The total resistance of the circuit = $\frac{r}{n}+R$

The total current in the circuit

$i=\frac{Total \: emf \: of \: the \: n - parallel \: cell}{Total \: resistance \: of \: the \: circuit}$

$i=\frac{E}{\frac{r}{n}+R}$

$i=\frac{E}{\frac{r+nR}{n}}$

$i=\frac{nE}{r+nR}$

3.) Mixed Combination of Cells: Let us consider that the $n$ - cells having emf (electromotive force) $E$ and internal resistance $r$ are connected in series in each row of $m$ parallel rows with external resistance $R$. Then from the figure given below
The total emf of the $n$ - cell in each row of $m$ parallel rows of the cells = $nE$

The internal resistance of the $n$ - cell in each row = $nr$

The total internal resistance of the $n$ - cell in each of $m$ parallel rows of the cells = $nr$

$\frac{1}{r_{eq}} = \frac{1}{nr}+ \frac{1}{nr}+.........m \: times$

$\frac{1}{r_{eq}}=\frac{m}{nr}$

$r_{eq}=\frac{nr}{m}$

The total resistance of the circuit = $\frac{nr}{m}+R$

The total current in the circuit

$i=\frac{Total \: emf \: of \: the \: cell}{Total \: resistance \: of \: the \: circuit}$

$i=\frac{nE}{\frac{nr}{m}+R}$

$i=\frac{nE}{\frac{nr+mR}{m}}$

$i=\frac{mnE}{nr+mR}$

It is clear from the above equation that for the value of $i$ to be maximum, the value of $(nr+mR)$ should be minimum. Now,

$nr+mR= \left[ \sqrt{nr}-\sqrt{mr} \right]^{2}+2 \sqrt{mnRr}$

Therefore, for $(nr+mR)$ to be minimum, the quantity $\left[ \sqrt{nr}-\sqrt{mr} \right]^{2}$ should be minimum. So

$\left[ \sqrt{nr}-\sqrt{mr} \right]^{2} = 0$

$\sqrt{nr}-\sqrt{mr} = 0$

$\sqrt{nr} = \sqrt{mr}$

$nr=mR$

$R=\frac{nr}{m}$

Here, $\frac{nr}{m}$ is the total resistance of the cells.

Thus, When the total internal resistance of the cells are equal to the external resistance then the total current in the external circuit will be maximum in the mixed combination of cells.