Two cells with same e.m.f. ‘E’ and different internal resistances r1 and r2 are connected in series to an external resistance ‘R’. The value of R so that the p.d. across the first cell be zero is

# Two cells with same e.m.f. ‘E’ and different internal resistances r1 and r2 are connected in series to an external resistance ‘R’. The value of R so that the p.d. across the first cell be zero is

1. A

$\sqrt{{\mathrm{r}}_{1}{\mathrm{r}}_{2}}$

2. B

${\mathrm{r}}_{1}+{\mathrm{r}}_{2}$

3. C

${r}_{1}-{r}_{2}$

4. D

$\frac{{\mathrm{r}}_{1}+{\mathrm{r}}_{2}}{2}$

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### Solution:

$\begin{array}{l}\mathrm{i}=\frac{{\mathrm{E}}_{1}+{\mathrm{E}}_{2}}{\mathrm{R}+{\mathrm{r}}_{1}+{\mathrm{r}}_{2}}=\frac{2\mathrm{E}}{\mathrm{R}+{\mathrm{r}}_{1}+{\mathrm{r}}_{2}};{\mathrm{V}}_{1}={\mathrm{E}}_{1}-{\mathrm{ir}}_{1}=0\\ ⇒\mathrm{i}=\frac{\mathrm{E}}{{\mathrm{r}}_{1}}=\frac{2\mathrm{E}}{\mathrm{R}+{\mathrm{r}}_{1}+{\mathrm{r}}_{2}};\mathrm{R}+{\mathrm{r}}_{1}+{\mathrm{r}}_{2}=2{\mathrm{r}}_{1}\\ ⇒\mathrm{R}={\mathrm{r}}_{1}-{\mathrm{r}}_{2}\end{array}$  