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.
$Q_{ tot }= CV =2\, \mu F \times 9 \,V =18\, \mu F$
After opening the switch $S$, the voltage capacitors are same as that of voltage across resistors, that is, $3 \,V$ and $6\, V$, respectively.
The charge stored in this situation in $3\, \mu F$ capacitor is
$3\, \mu F \times 3 \,V =9\, \mu C$
and the charge in $6 \,\mu F$ is
$6 \,\mu F \times 6\, V =36 \,\mu C$
Therefore, the total charge flow from $y$ to $x$ is
$9 \,\mu C +18 \,\mu C =27\, \mu C$