Two identical cells, when connected either in parallel or in series...

Let $r$ be the internal resistance of each cell.

In above shown figure, first represents the series connection of cells. Second and third figures represents the parallel connection of cells.

Now, in case of series connection, ${ε}_{net} = 2εand{r}_{net} = 2r$. So, current $i = \dfrac{2ε}{5 + 2r}$ .....(1)

In case of parallel connection, ${ε}_{net} = \dfrac{εr + εr}{r + r} = εand{r}_{net} = \dfrac{r}{2}$. So, current $i = \dfrac{ε}{\dfrac{r}{2} + 5}$ .....(2)

Equating (1) and (2), we get

$\dfrac{2ε}{5 + 2r} = \dfrac{ε}{\dfrac{r}{2} + 5}$

$⇒ r + 10 = 5 + 2r$

$⇒ r = 5Ω$.