Twelve cells, each having emf $E$ volts are connected in series and are kept in a closed box. Some of these…
Twelve cells, each having emf $E$ volts 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 $2 \mathrm{~A}$ when they oppose each other. Then, the number of cells in 12-cell battery that are connected wrongly is :
4
3
2
1
Solution
Let polarity of $m$ cells in a 12 cells battery is reversed, then equivalent emf of the battery
$=(12-2 m) E$
Now the circuit can be drawn as :
When 12-cell battery and 2-cell battery aid each other, then current through the circuit,
$i_1=\frac{(12-2 m) E+2 E}{R}$
or $\quad 3=\frac{(14-2 m) E}{R}$ ...(i)
When they oppose each other, the current through the circuit.
$i_2=\frac{(12-2 m) E-2 E}{R}$
or $\quad 2=\frac{(10-2 m) E}{R}$ ...(ii)
Dividing Eq. (i) by (ii), we have
$\frac{3}{2}=\frac{14-2 m}{10-2 m}$
or $\quad 30-6 m=28-4 m$
or $\quad 2 m=2$
$\therefore \quad m=1$
