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?

Options

A. 1
B. 6
C. 4
D. 8

Explanation

Let E be the emf of each cell and let n be the number of cells connected wrongly. Then, the net emf of the battery is equal to the emf of correctly connected cells–emf of wrongly connected cells.
`(12-n)E-nE=(12-2n)E` ,br> When the two cells aid the battery, the net emf is `E_1=(12-2n)E+2E=(14-2n)E`
If R is the resistance in the circuit, the current is `i_1=E_1/R=((14-2n)E)/R`
When the two cells oppose the battery, the net emf is `E_2=(12-2n)E-2E=(10-2n)E`
Now, the current in the circuit is `i_2=((10-2n)E)/R`
Dividing Eqs. (i) by Eq(ii), we get `i_1/i_2=((14-2n))/((10-2n))=((7-n))/((5-n))`
Given, `i_1=3A and i_2=2A` Thus `3/2=((7-n))/((5-n))`
WHich gives n=1, Hence one cell in the battery is wrongly connected.