Question

In the circuit shown in figure, match the following two columns for flow of charge when switch is closed.

Answer 1

A) When S is open, total capacitance $C_i=\frac{3(4+2)}{3+(4+2)}=\frac{18}{9}=2\mu F$.
Equivalent charge is, $q_i=C_{i}V=2\left(30\right)=60\mu C$
when S is closed, $3\mu F$ is not working so equivalent capacitance $C_f=2+4=6\mu F$
and $q_f=C_{f}V=6\left(30\right)=180\mu C$
Thus charge flow from the battery $\Delta q=q_f-q_i=180-60=120\mu C$
$\therefore A\to 4$

B) As capacitor 4 and 2 are in parallel so charge distributes indirect ratio of capacity. Thus charge on 2 : $q_{2i}=\frac{2}{2+4}{q}_i=\frac{2}{6}\left(60\right)=20\mu C$ and $C_{2f}=\frac{2}{2+4}{q}_f=\frac{2}{6}\left(180\right)=60\mu C$
Thus flow from $2\mu F$ is $\Delta q_2=q_{2f}-q_{2i}=60-20=40\mu C$
$\therefore B\to 1$

C) Initial charge on $3\mu F$ is $q_{3i}=q_i=60\mu C$ .
when S is closed $3\mu F$ is not working so final charge $q_{3f}=0$
Thus flow from $3\mu F$ is $\Delta q_3=q_{3i}-q_{3f}=60-0=60\mu C$
$\therefore C\to 3$

D) Charge on 4 : $q_{4i}=\frac{4}{2+4}{q}_i=\frac{4}{6}\left(60\right)=40\mu C$ and $C_{4f}=\frac{4}{2+4}{q}_f=\frac{4}{6}\left(180\right)=120\mu C$
Thus flow from $4\mu F$ is $\Delta q_4=q_{4f}-q_{4i}=120-40=80\mu C$
$\therefore D\to 4$