Question

In the circuit shown in the figure, $C_1=6\mu F,C_2=3\mu F$ and battery of EMF $E=20V$. The switch $S_1$ is first closed and $S_2$ is kept open. Then $S_1$ is opened , and $S_2$ is closed. What is the final charge on $C_2$?

• A. $120\mu C$
• B. $80\mu C$
• C. $40\mu C$
• D. $20\mu C$

Answer 1

Answer: (C)

$40\mu C$

After closing $S_1$, charge on $C_1$ is $q=6\times 20=120\mu C$.

Now $S_1$ is opened.

On closing $S_2$, charge $q$ will be distributed between $C_1$ and $C_2$ according to their capacitances.

So charge on $C_2$ is $q_2=\frac{C_{2}q}{C_1+C_2}=\frac{3\times 120}{3+6}=40\mu C$