Question:
Two cells of emf $2 \mathrm{E}$ and E with internal resistance $r_{1}$ and $r_{2}$ respectively are connected in series to an external resistor $R$ (see figure). The value of $R$, at which the potential difference across the terminals of the first cell becomes zero is
Correct Option: , 2
Solution:
$i=\frac{3 E}{R+r_{1}+r_{2}}$
$\mathrm{TPD}=2 \mathrm{E}-\mathrm{ir}_{1}=0$
$2 \mathrm{E}=\mathrm{ir}_{1}$
$2 \mathrm{E}=\frac{3 \mathrm{E} \times \mathrm{r}_{1}}{\mathrm{R}+\mathrm{r}_{1}+\mathrm{r}_{2}}$
$2 R+2 r_{1}+2 r_{2}=3 r_{1}$
$\mathrm{R}=\frac{\mathrm{r}_{1}}{2}-\mathrm{r}_{2}$
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