12 cells, each having same emf, are connected in series and are kept in a closed box. Some of cells are wrongly connected. This battery is connected in series with an ammeter and two cells, identical with the others. The current is 3 A with the cells and battery aid with each other and 2 A when the cells and battery oppose each other. How many cells are wrongly connected.
Let each cell have emf E and internal resistance r.
For aiding: (12E + 2E) / R = 3A → 14E/R = 3.
For opposing: (12E – 2E) / R = 2A → 10E/R = 2.
Solving, x = 2 wrongly connected.
Answer: (b) 2.
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