Question

A circuit is connected as shown in the figure with the switch open. When the switch is closed, what is the total amount of charge (in $μC$ ) that flows from $Y$ to $X$ ?

Answer 1

When switch is open capacitors are in series and net potential difference across them is $9v$

Charge on each capacitor before closing the switch is $q=C_{e q}V=\frac{3 \times 6}{3 + 6}\times 9=18μc$
After closing the switch current of $I=\frac{V}{3 + 6}=\frac{9}{9}=1A$ will flow across the resistors .potential difference across resistors and capacitors are
$V_{3 o h m}=IR=1\times 3=3VV_{3 ohm}=V_{3 μF}=3VV_{6 ohm}=IR=1\times 6=6VV_{6 ohm}=V_{6 μF}=6V$
Charge on each capacitor after closing the switch will be
$q_{3 \mu F}=CV=3\times 3=9\mu cq_{6 \mu F}=CV=6\times 6=36\mu c$
Hence, charge flown from $YtoX$ is
$q_{f l o w}=\left(36 - 18\right)+\left(18 - 9\right)=27μC$