Two cells, having the same emf, are connected in series through an external resistance R. Cells have internal resistances r₁ and r₂ (r₁ > r₂), respectively. Which the circuit is closed, the potential difference across the first cell is zero. The value of R is

Current in the circuit is given by Ohm's law.

Net resistance of the circuit = r₁ + r₂ + R

Net emf in series - E + E = 2E

Therefore, from Ohm's law, current in the circuit

$\mathrm{i}=\frac{\mathrm{Net}\mathrm{emf}}{\mathrm{Net}\mathrm{resistance}}$

$\Rightarrow \mathrm{i}=\frac{2\mathrm{E}}{{\mathrm{r}}_1+{\mathrm{r}}_2+\mathrm{R}}$             …..(i)

It is given that, as circuit is closed, potential difference across the first cell is zero. That is

$\begin{array}{l}\mathrm{V}=\mathrm{E}-{\mathrm{ir}}_1=0\\ \Rightarrow \mathrm{i}=\frac{\mathrm{E}}{{\mathrm{r}}_1}\end{array}$              ……(ii)

Equating equations (i) and (ii), we get $\frac{\mathrm{E}}{{\mathrm{r}}_1}=\frac{2\mathrm{E}}{{\mathrm{r}}_1+{\mathrm{r}}_2+\mathrm{R}}$

$\begin{array}{l}2{\mathrm{r}}_1={\mathrm{r}}_1+{\mathrm{r}}_2+\mathrm{R}\\ \mathrm{R}=\mathrm{external}\mathrm{resistance}={\mathrm{r}}_1-{\mathrm{r}}_2\end{array}$