$\mathrm{I}_{1}$ is given by :

$I_{1}=\frac{n E}{R+n r}$   $. .(i)$

When the cells connected in parallel, current

$\mathrm{I}_{2}$ is given by :

$\mathrm{I}_{2}=\frac{\mathrm{E}}{\mathrm{R}+\frac{\mathrm{r}}{\mathrm{n}}}=\frac{\mathrm{nE}}{\mathrm{nR}+\mathrm{r}}$  $\ldots(\mathrm{ii})$

As $I_{1}=I_{2}$

So, $\frac{\mathrm{nE}}{\mathrm{R}+\mathrm{nr}}=\frac{\mathrm{nE}}{\mathrm{nR}+\mathrm{r}}$

$\therefore \quad \mathrm{R}+\mathrm{nr}=\mathrm{nR}+\mathrm{r}$

or $(n-1) r=(n-1) R$ or $r=R$