$I_{1}=\frac{\varepsilon_{2}-V}{r_{1}} \Rightarrow I_{2}=\frac{\varepsilon_{2}-V}{r_{2}}$

Combining the last three equations

$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}=\frac{\varepsilon_{1}-\mathrm{V}}{\mathrm{r}_{1}}+\frac{\varepsilon_{2}-\mathrm{V}}{\mathrm{r}_{2}}$

$=\left(\frac{\varepsilon_{1}}{\mathrm{r}_{1}}+\frac{\varepsilon_{2}}{\mathrm{r}_{2}}\right)-\mathrm{V}\left(\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}\right)$

Hence, $\mathrm{V}$ is given by, $\mathrm{V}=\frac{\varepsilon_{1} \mathrm{r}_{2}+\varepsilon_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}-\mathrm{I} \frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}$

If we want to replace the combination by a single cell, between $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ of emf $\varepsilon_{\mathrm{eq}}$ and internal resistance $\mathrm{r}_{\mathrm{eq}},$

we would have $\mathrm{V}=\varepsilon_{\mathrm{eq}}-\mathrm{Ir}_{\mathrm{eq}}$