Twelve cells, each having emf E volts are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This 12 cell battery is connected in series with an ammeter, an external resistance R ohms and a two-cell battery (two cells of the same type used earlier, connected perfectly in series). The current in the circuit when the 12-cell battery and 2-cell battery aid each other is 3A and is 2 A when they oppose each other. Then, the number of cells in 12-cell battery that are connected wrongly is :

Question

Twelve cells, each having emf E volts are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This 12 cell battery is connected in series with an ammeter, an external resistance R ohms and a two-cell battery (two cells of the same type used earlier, connected perfectly in series). The current in the circuit when the 12-cell battery and 2-cell battery aid each other is 3A and is 2 A when they oppose each other. Then, the number of cells in 12-cell battery that are connected wrongly is :, 4, 3, 2, 1

MARKS App · Accepted Answer

Let polarity of m cells in a 12 cells battery is reversed, then equivalent emf of the battery = ( 12 − 2 m ) E Now the circuit can be drawn as : When 12-cell battery and 2-cell battery aid each other, then current through the circuit, i 1 ​ = R ( 12 − 2 m ) E + 2 E ​ or 3 = R ( 14 − 2 m ) E ​ ...(i) When they oppose each other, the current through the circuit. i 2 ​ = R ( 12 − 2 m ) E − 2 E ​ or 2 = R ( 10 − 2 m ) E ​ ...(ii) Dividing Eq. (i) by (ii), we have 2 3 ​ = 10 − 2 m 14 − 2 m ​ or 30 − 6 m = 28 − 4 m or 2 m = 2 ∴ m = 1

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