Q. Twelve cells, each having emf E volts are connected in series and kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This 12-cell battery is connected with an ammeter, an external resistance R ohm 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 2A when they oppose each other. Then, the number of cell in 12-cell battery that are connected wrongly is
A) 4
B) 3
C) 2
D) 1
Explanation
(d) Let polarity of $m$ cells in a 12 cells battery is reversed, then equivalent emf of the battery $= (12 - 2m)E$
Now the circuit can be drawn as
When 12-cell battery and 2-cell battery aid each other, then current through the circuit
$i_{1} = \frac{(12 - 2m)E + 2E}{R}$
$or\ \ \ 3 = \frac{(14 - 2m)E}{R}\ \ \ldots.(i)$
When they oppose each other, the current through the circuit.
$i_{2} = \frac{(12 - 2m)E - 2E}{R}$
$or\ \ \ 2 = \frac{(10 - 2m)E}{R}\ \ \ \ \ \ldots(ii)$
Dividing Eq. (i) by Eq. (ii), we have
$\frac{3}{2} = \frac{14 - 2m}{10 - 2m}$
$or\ \ \ 30 - 6m = 28 - 4m\ $
$\ or\ \ \ \ \ \ \ \ \ \ \ \ 2m = 2$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ m = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.