Two identical cells, each of emf E and internal resistance r , are connected with a load resistance $\displaystyle \mathrm{R}$, first in series and then in parallel. Obtain the condition under which the current through $\displaystyle \mathrm{R}$ is same in both cases.

In series, \[\begin{aligned} & \varepsilon_{\mathrm{eq}}=\varepsilon+\varepsilon=2 \varepsilon \\ & \mathrm{I}_{\mathrm{s}}=\frac{2 \varepsilon}{2 \mathrm{r}+\mathrm{R}} \end{aligned} \] In parallel, \[\begin{aligned} & \varepsilon_{\mathrm{eq}}=\varepsilon \\ & \mathrm{r}_{\mathrm{eq}}=\frac{\mathrm{r}}{2} \\ & \mathrm{I}_{p}=\frac{\varepsilon}{\dfrac{\mathrm{r}}{2}+\mathrm{R}} \\ & \because \mathrm{I}_{p}=\mathrm{I}_{\mathrm{s}} \\ & \frac{2 \varepsilon}{2 \mathrm{r}+\mathrm{R}}=\frac{\varepsilon}{\dfrac{\mathrm{r}}{2}+\mathrm{R}} \\ & \frac{2}{2 \mathrm{r}+\mathrm{R}}=\frac{1}{\dfrac{\mathrm{r}}{2}+\mathrm{R}} \\ & 2 \frac{\mathrm{r}}{2}+2 \mathrm{R}=2 \mathrm{r}+\mathrm{R} \\ & \mathrm{r}=\mathrm{R} \end{aligned} \]