12 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 is connected in series with an ammeter and two similar cells . The current is 3A when the two cells aid the battery and is 2A when the cells and the battery oppose each other. How many cells in the battery are wrongly connected?

Explanation

IF n-number of cells are wrongly connected , then the number of cells correctly connected in the battery=12-n,emf of each cell=E.
`therefore` Emf of the battery=`(12-n)E-nE=(12-2n)E`
Again,emf of the additional two cells=2E. Let R be the resistance of the whole circuit.
In the first case,
`3=((12-2n)E+2E)/Ror,3/(14-2n)=E/R`
In the second case,
`2=((12-2n)E+2E)/Ror,32/(14-2n)=E/R`
`therefore3/(14-2n)=2/(10-2n)or,28-4n=30-6n`
or, 2n=2 or n=1
So, only one cell was wrongly connected.