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  • Two cells of emf 2E and E with internal resistance respectively are connected in series to an external resistor R (see figure). The value

Two cells of emf 2E and E with internal resistance r_{1}\; \text{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
 
Option: 1 \frac{r_{1}}{2}+r_{2}
Option: 2 r_{1}-r_{2}
Option: 3 \frac{r_{1}}{2}-r_{2}  
Option: 4 r_{1}+r_{2}

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best_answer

\begin{array}{l} \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} \\ \\ \mathrm{R}=\frac{\mathrm{r}_{1}}{2}-\mathrm{r}_{2} \end{array}

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Deependra Verma

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