Question

In the circuit shown here, $C_{1}=6\, \mu\, F ,\, C_{2}=3\, \mu F$ and battery $B=20\, V$. 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. $20\, \mu C$

Answer 1

Answer: (C)

Common potential, $V=\frac{6 \times 20+3 \times 0}{(6+3)}$
$=\frac{120}{9}$ volt
So, charge on $3\, \mu F$ capacitor,
$Q_{2}=3 \times 10^{-6} \times \frac{120}{9}=40\, \mu C$