Infinite number of cells having emfs and internal resistances
$(E, r), (\frac{E}{n}, \frac{r}{n}), (\frac{E}{n^{2}}, \frac{r}{n^{2}}), (\frac{E}{n^{3}}, \frac{r}{n^{3}}), . . . .$ ara connected in series in same manner across an external resistance of $\frac{n r}{n - 1}$ . Current flowing through the external resistor is:

\frac{E}{2 r}

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\frac{E}{(n + 1) r}

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\frac{(n - 1) E}{r}

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\frac{(n - 1) E}{2 r}

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Solution

The correct option is B $\frac{E}{(n + 1) r}$
$I = \frac{E_{1} + E_{2} + E_{3} + . . .}{r_{1} + r_{2} + r_{3} + . . . . + R_{l o a d}} E_{1} = E E_{2} = \frac{E}{n} E_{3} = \frac{E}{n^{2}} . . E_{m} = \frac{E}{n^{m - 1}} \sum_{m = 1}^{\infty} E_{\infty} = \sum_{m = 1}^{\infty} \frac{E}{n^{m - 1}} = \frac{E}{n - 1} r_{1} = r r_{2} = \frac{r}{n} r_{3} = \frac{r}{n^{2}} . . r_{m} = \frac{r}{n^{m - 1}} \sum_{m = 1}^{\infty} r_{\infty} = \frac{r}{n - 1}$
$∴ \frac{\frac{E}{n - 1}}{\frac{r}{n - 1} + \frac{n r}{n - 1}} = \frac{E}{n - 1} \frac{n - 1}{(n + 1) r} = \frac{E}{(n + 1) r}$

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