Question

Twelve cells, each having an e.m.f of E volt are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This 12 cell battery is connected in series with an ammeter, an external resistance R ohms and a two-cell battery (two cells of the same type used earlier, connected perfectly in series). The current in the circuit when the 12-cell battery and 2-cell battery aid each other is 3A and is 2A when they oppose each other.

Then the number of cells in 12-cells battery that are connected wrongly is:

• A. 14
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is D 1

Let $n$ be the number of wrongly connected cells.
Since each wrong cell has reverse connection of positive and negative terminal with neighboring cell, the total number of cells in a closed box is $(12-2m)$ and the e.m.f. is $(12-2m)E$.
Now, when the 12-cell battery and 2-cell battery aid each other, we get,
$I=\frac{\left(\right(12-2m)E+2E)}{R+14r}$
$\therefore 3=\frac{(14-2m)E}{R+14r}$
where, $r$ is the internal resistance of cells.
Also, when the 12-cell battery and 2-cell battery oppose each other, we get, $I=\frac{\left(\right(12-2m)E-2E)}{R+14r}$
$\therefore 2=\frac{(10-2m)E}{R+14r}$

Comparing the above equations, we get, $\frac{14-2m}{10-2m}=\frac{3}{2}$

$\therefore 2\times (14-2m)=3\times (10-2m)$

$\therefore 2m=2$

$\therefore m=1$