Q. A $10\mu F$ capacitors and a 20$\mu F$ capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. The potential difference across each capacitor is
A) $\frac{400}{9}V$
B) $\frac{800}{3}V$
C) $400\ V$
D) $200\ V$
Explanation
(a) $\ C_{s} = \frac{10 \times 20}{10 + 20} = \frac{200}{30} = \frac{20}{3}\ \mu F$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q\ = C_{s}V$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q\ = \frac{20}{3}\mu F \times 200V$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q\ = \frac{4000}{3}\mu C$
Now, $\ \ \ \ \ V\ = \ \frac{4000\ \mu C}{3 \times 30\ \mu F} = \frac{4000}{90}V = \frac{400}{9}\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.