Question

A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is :

• A. $0$
• B. $54\mu C$
• C. $27\mu C$
• D. $81\mu C$

Answer 1

The correct option is C $27\mu C$

When steady state reached , the capacitors are fully charged and no current pass through the capacitor branch.
Thus, the current through the resistors is $I=\frac{9}{3+6}=1A$.
When S is closed, potential across $3\mu F=$ potential across $3\Omega =3I=3V$
and potential across $6\mu F=$ potential across $6\Omega =6I=6V$
Charge on $3\mu F$ is $Q_1=3\times 3=9\mu C$ and Charge on $6\mu F$ is $Q_1=6\times 6=36\mu C$
Here charge $-Q_1$ is shifted from the positive plate of $6\mu F$ capacitor. The remaining charge on the positive plate of $6\mu F$ capacitor is shifted through the switch.
Thus, charge passing through switch $=36-9=27\mu C$