Question
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 of previous cells. The current is 3 A when the cells and the battery add each other and is 2 A when the cells and the battery oppose each other. How many cells in the battery are wrongly connected?
Solution
Let us assume that, m cells are connected correctly and n cells are connected wrongly.
Then, we have
m + n = 12
If, E is the emf of each cell, the total emf of the battery is (m – n) E.
When the battery and the cells add each other, the net emf is = (m – n) E + 2 E
If R is the total resistance of the circuit, the current is given by
I = ...(1)
When the battery and the cells oppose each other, the net emf is, (m – n) E – 2 E.
Then, the current across them is given by,
I= ...(2)
Dividing equation (1) by equation (2) gives us,
m – n = 10
But, m + n = 12
Hence m = 11 and n = 1.
Thus, one cell is wrongly connected.
