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n identical cell, each of emf $ε$ and internal resistance $r_{'}$ are joined in series to form a closed circuit. One cell (A) is joined with reversed polarity. The potential difference across each cell, except A, is

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$\frac{n - 2}{n} ε$

$\frac{n - 1}{n} ε$

$\frac{2 ε}{n}$

$\frac{2 n}{n - 2} ε$

answer is A.

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Detailed Solution

$i = \frac{(n - 2) ε}{n r}$

All the cells except A will be discharging

$V = ε - i r$

$= ε - \frac{(n - 2) ε}{n r} r = ε [1 - \frac{n - 2}{n}] = \frac{2 ε}{n}$

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