Two cells, having the same e.m.f. are connected in series through an external resistance $R$. Cells have…
Two cells, having the same e.m.f. are connected in series through an external resistance $R$. Cells have internal resistances $r_1$ and $r_2\left(r_1 > r_2\right)$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is:
$r_1+r_2$
$r_1-r_2$
$\frac{r_1+r_2}{2}$
$\frac{r_1-r_2}{2}$
Solution
According to the question
$E-I r_1=0 \text { and } \mathrm{I}=\frac{E+E}{r_1+r_2+R}$
$\begin{array}{l}
\therefore \frac{E}{r_1} =\frac{2 E}{r_1+r_2+R} \\
\Rightarrow r_1+r_2+R =2 r_1-r_2 \\
\Rightarrow R =r_1-r_2
\end{array}$