Explain the 'Mixed Connection' of cells and derive an expression for its equivalent emf and current.

(N/A) Consider $n$ identical cells,each of emf $\varepsilon$ and internal resistance $r$,connected in series to form a row. There are $m$ such rows connected in parallel,and this combination is connected to an external resistor $R$.
$1$. Total emf of one row: $\varepsilon_{eq} = n\varepsilon$.
$2$. Total internal resistance of one row: $r_{eq} = nr$.
$3$. Since there are $m$ such rows in parallel,the total equivalent emf of the combination is $\varepsilon_{total} = n\varepsilon$.
$4$. The total internal resistance $r_{total}$ of the $m$ parallel rows is given by $\frac{1}{r_{total}} = \frac{1}{nr} + \frac{1}{nr} + \dots (m \text{ times}) = \frac{m}{nr}$. Thus,$r_{total} = \frac{nr}{m}$.
$5$. The total resistance of the circuit is $R_{total} = R + \frac{nr}{m} = \frac{mR + nr}{m}$.
$6$. The current $I$ flowing through the external resistor $R$ is given by $I = \frac{\varepsilon_{total}}{R_{total}} = \frac{n\varepsilon}{\frac{mR + nr}{m}} = \frac{mn\varepsilon}{mR + nr}$.