Q.
In the circuit shown here $C_{1} = 6\mu F,\ C_{2} = 3\mu F$ and battery $B = 20V.$ The switch $S_{1}$ is first closed. It is then opened and afterwards $S_{2}$ is closed. What is the charge finally on $C_{2}$
A) $120\mu C$
B) $80\mu C$
C) $40\mu C$
D) $20\mu C$
Explanation
(c) Common potential $V = \frac{6 \times 20 + 3 \times 0}{(6 + 3)} = \frac{120}{9}Volt$
So, charge on $3\ \mu F$ capacitor
$Q_{2} = 3 \times 10^{- 6} \times \frac{120}{9} = 40\mu C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.