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 battery are wrongly connected?

Explanation

Let m cells wrongly connected in the battery and `xi_("net") = (12-2m)xi`
When two cells aid the battery , then current.
`3=((12- 2m) xi+ 2xi)/(R) " "…..(i)`
where R is the total resistance of the circuit. When two cells oppose the battery , then
`2=((12-2m)xi - 2xi)/(R) " ".....(ii)`
Solving above equations , we get m = 1
Hence one cell is wrongly connected in the battery.