Q. A $10\mu F$ capacitor 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. What is the potential difference across each capacitor
A) $\frac{400}{9}V$
B) $\frac{800}{9}V$
C) $400\ V$
D) $200\ V$
Explanation
(b) Initially potential difference across each capacitor
$V_{1} = \frac{20}{(10 + 20)} \times 200 = \frac{400}{3}V$
and $V_{2} = \frac{10}{(10 + 20)} \times 200 = \frac{200}{3}V$
finally common potential $V = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}}$
$V = \frac{10 \times \frac{400}{3} + 20 \times \frac{200}{3}}{(10 + 20)} = \frac{800}{9}V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.