There are n (n>2) cells, each of emf E and internal resistance r, connected in series with an external resistance R. One of the cells is wrongly connected, so that it sends current in the opposite direction. The current flowing in the circuit is

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$\frac{(n - 2) E}{(n - 2) r + R}$

$\frac{(n - 1) E}{n r + R}$

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

$\frac{(n - 2) E}{n r + R}$

answer is C.

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Detailed Solution

Out of n cells, two cells will cancel out each other's emf.
So, net emf = (n $-$ 2) E.
Total resistance = R + nr

Current, $i = \frac{(n - 2) E}{n r + R}$

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