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12 cell each having the same e.m.f are cannected in series and are kept in a closed box. Some of the cell are wrongly connected. This battery is connected in series with an ammeter and the two cells identical with the other the current is 3A when the cells and battery aid each other and is 2A when the cells and battery oppose each other .How many cell are wrongly connected?
Let there be ‘n’ cells with opposite polarity inside the battery, so p.d. across the battery is (12-n)E-nE= (12-2n)E. A/Q, (12-2n)E+2E=3R ---(i) {R is the net resistance of the cuircuit} (12-2n)E-2E=2R --(ii) (i)/(ii) we get, (14-2n)/(10-2n)=3/2 =>28-4n=30-6n =>2n=2 =>n=1
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