Question

$12$ cells each having same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two identical cells. Current is $3A$ when cells and battery aid each other and is $2A$ when cells and battery oppose each other. The number of cells wrongly connected is :-

Answer 1

Assume E is emf of each cell.
and R = resistance of the circuit.
Since, $I=\frac{V}{R}$
When $x$ cells are wrongly connected then
$E_{net}=12E-2xE$
When cells and battery aid each other
Then, $3=\frac{12E-2xE+2E}{R}.....\left(1\right)$
When they oppose each other
$2=\frac{12E-2xE-2E}{R}........\left(2\right)$
Solving equation (1) and (2),

$\frac{3}{2}=\frac{14-2x}{10-2x}$
$x=1$