Two cells $\mathrm{E}_1$ and $\mathrm{E}_2$ having equal $\mathrm{EMF}$ ' $\mathrm{E}$ ' and internal…
Two cells $\mathrm{E}_1$ and $\mathrm{E}_2$ having equal $\mathrm{EMF}$ ' $\mathrm{E}$ ' and internal resistances $r_1$ and $r_2\left(r_1>r_2\right)$ respectively are connected in series. This combination is connected to an external resistance ' $R$ '. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of ' $R$ ' will be
$r_1-r_2$
$\mathrm{r}_1+\mathrm{r}_2$
$\frac{\mathrm{r}_1-\mathrm{r}_2}{2}$
$\frac{r_1+r_2}{2}$
Solution
The total current in the circuit is
$I=\frac{2 E}{r_1+r_2+R}$
....(Given cells are in series, $\mathrm{E}+\mathrm{E}=2 \mathrm{E}$ )
Now the potential drop across the first cell is
$\begin{array}{ll}
& \mathrm{V}_1=\mathrm{E}-\mathrm{Ir}_1=0 \\
\therefore \quad & \mathrm{E}-\left(\frac{2 \mathrm{E}}{\mathrm{r}_1+\mathrm{r}_2+\mathrm{R}}\right) \times \mathrm{r}_1=0 \\
& \frac{2 \mathrm{E}}{\mathrm{r}_1+\mathrm{r}_2+\mathrm{R}}=\frac{\mathrm{E}}{\mathrm{r}_1} \Rightarrow 2 \mathrm{r}_1=\mathrm{r}_1+\mathrm{r}_2+\mathrm{R} \\
\therefore \quad & \mathrm{R}=\mathrm{r}_1-\mathrm{r}_2
\end{array}$