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

Answer: (C)

Before opening the switch $S$, the two capacitors are connected to battery serially. The effective capacitance is
$C=\frac{C_1{C}_2}{C_1+C_2}=\frac{18}{9}\mu F$
The charge stored in each capacitor is equal because same current flows through each capacitor.
$Q_{tot}=CV=2\mu F\times 9V=18\mu F$
After opening the switch $S$, the voltage capacitors are same as that of voltage across resistors, that is, $3V$ and $6V$, respectively.
The charge stored in this situation in $3\mu F$ capacitor is
$3\mu F\times 3V=9\mu C$
and the charge in $6\mu F$ is
$6\mu F\times 6V=36\mu C$
Therefore, the total charge flow from $y$ to $x$ is
$9\mu C+18\mu C=27\mu C$