Q.
In given circuit when switch $S$ has been closed then charge on capacitor $A\ \&\ B$ respectively
A) $3\ q,\ 6\ q$
B) $6\ q,\ 3\ q$
C) $4.5\ q,\ 4.5\ q$
D) $5\ q,\ 4\ q$
Explanation
(b) $Q_{1} = \frac{C_{1}}{C_{1} + C_{2}}Q_{n}\ \ \ \ \&\ \ \ Q_{2} = \frac{C_{2}}{C_{1} + C_{2}}Q_{n}$
$\Rightarrow Q_{1} = \frac{6c}{6c + 3c} \times 9q\ \ \ \&\ \Rightarrow Q_{2} = \frac{3c}{6c + 3c} \times 9q$
$= 6q\ \ \ \ \ \&\ = 3q$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.