Question

12 cells each having same e.m.f 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 cells identical with the others. The current is 3A when the cells and the battery support each other and 2A when the cells and the battery oppose each other. The number of cells in the battery that are wrongly connected is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A

1

Let m cells be connected correctly and n cells be connected wrongly.

Then m +n = 12 …(1)

If E is the emf of each cell, then the net emf of the battery is (m - 2n) E.

When the cells and the battery aid each other, $\frac{(m-2n)E+2E}{R}=3$ …(2)

When the cells and the battery oppose each other, $\frac{(m-2n)E-2E}{R}=2$ …(3)

From eq’s (2) and (3); m - n = 10 …(4)

On solving eq’s (1) and (4), we get m = 11 and n = 1

Thus, one cell is wrongly connected.