Consider N = n_1 n_2 identical cells, each having an emf E and an...

For a single line of $n_1$ cells connected in series, the equivalent emf is $n_1 \mathcal{E}$ and the equivalent internal resistance is $n_1 r$.

When $n_2$ such lines are connected in parallel, the equivalent emf of the entire combination remains equal to the emf of a single line, which is $\mathcal{E}_{eq} = n_1 \mathcal{E}$.

The equivalent internal resistance of $n_2$ parallel lines, each of resistance $n_1 r$, is $r_{eq} = \dfrac{n_1 r}{n_2}$.

The total current $I$ flowing through the external resistance $R$ is given by $I = \dfrac{\mathcal{E}_{eq}}{R + r_{eq}}$.

Substituting the values, we get $I = \dfrac{n_1 \mathcal{E}}{R + \dfrac{n_1 r}{n_2}}$.