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Twelve cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery of cells is connected in series with an ammeter and two cells identical with the others. The current is 3A when the cells and the battery aid each other
and is 2A when the cells and the battery oppose each other. How many cells in the batter are wrongly connected ?

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answer is A.

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Detailed Solution

$3 = \frac{(12 - 2 x) E + 2 E}{R_{eff}}$ …(1)
$2 = \frac{(12 - 2 x) E - 2 E}{R_{eff}}$ …(2)
Solvin (1) (2) x = 1

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