Two cells of emf 2 E


Two cells of emf $2 \mathrm{E}$ and $\mathrm{E}$ with internal resistance $\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ respectively are connected in series to an external resistor $R$ (see figure). The value of $\mathrm{R}$, at which the potential difference across the terminals of the first cell becomes zero is

  1. (1) $\mathrm{r}_{1}+\mathrm{r}_{2}$

  2. (2) $\frac{r_{1}}{2}-r_{2}$

  3. (3) $\frac{\mathrm{r}_{1}}{2}+\mathrm{r}_{2}$

  4. (4) $\mathrm{r}_{1}-\mathrm{r}_{2}$

Correct Option: 2,



$\mathrm{i}=\frac{3 \mathrm{E}}{\mathrm{R}+\mathrm{r}_{1}+\mathrm{r}_{2}}$

$\mathrm{TPD}=2 \mathrm{E}-\mathrm{ir}_{1}=0$

$2 \mathrm{E}=\mathrm{ir}_{1}$

$2 \mathrm{E}=\frac{3 \mathrm{E} \times \mathrm{r}_{1}}{\mathrm{R}+\mathrm{r}_{1}+\mathrm{r}_{2}}$

$2 \mathrm{R}+2 \mathrm{r}_{1}+2 \mathrm{r}_{2}=3 \mathrm{r}_{1}$


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