The combination of two identical cells, whether connected in series or parallel combination provides the…

When the cells are connected in series $i_s=\frac{2\epsilon }{2r+R}=\frac{2\epsilon }{2r+2}$

When they are connected in parallel, the emf is ${\epsilon }_p=\frac{{\displaystyle \frac{\epsilon }{r}}+{\displaystyle \frac{\epsilon }{r}}}{{\displaystyle \frac{1}{r}}+{\displaystyle \frac{1}{r}}}=\epsilon$ and the effective internal resistance will be $r_p=\frac{r\times r}{r+r}=\frac{r}{2}$. Therefore, the current will be,

$i_p=\frac{\epsilon }{R+{\displaystyle \frac{r}{2}}}=\frac{\epsilon }{2+{\displaystyle \frac{r}{2}}}=\frac{2\epsilon }{4+r}$

Since, $$\begin{gathered} i_s=i_p \\ \Rightarrow \frac{2\epsilon }{2r+2}=\frac{2\epsilon }{4+r}\Rightarrow r=2 \mathrm{\Omega } \end{gathered}$$