Study the following circuit diagram and mark the correct options 

When ‘S’ is open across capacitors

$\mathrm{i}=\frac{18}{9}=2\mathrm{A}$

$\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}=\frac{3}{6}=\frac{1}{2}$

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{a}}=6\left(2\right)=12;{\mathrm{V}}_1=\frac{1}{3}\left(18\right)=6$

${\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{C}}=3\left(2\right)=6;{\mathrm{V}}_2=\frac{2}{3}\left(18\right)=12$

${\mathrm{V}}_{\mathrm{a}}=6(\because {\mathrm{V}}_{\mathrm{C}}=0);{\mathrm{V}}_1={\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{a}}=6$

${\mathrm{V}}_2={\mathrm{V}}_{\mathrm{b}}-{\mathrm{V}}_{\mathrm{D}}=12$

${\mathrm{V}}_{\mathrm{ab}}={\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}=6-12=-6$

${\mathrm{V}}_{\mathrm{b}}=12({\mathrm{V}}_{\mathrm{D}}=0)$

${\mathrm{Q}}_1={\mathrm{C}}_1{\mathrm{V}}_1=6\left(6\right)=36\mu \mathrm{C}\text{;}$

${\mathrm{Q}}_2={\mathrm{C}}_2{\mathrm{V}}_2=3\left(12\right)=36\mu \mathrm{C}$

When ‘S’ is closed

${\mathrm{V}}_1^1=12\mathrm{V}\therefore {\mathrm{Q}}_1^1={\mathrm{C}}_1{\mathrm{V}}_1^1=\left(6\right)12=72\mu \mathrm{C}$

${\mathrm{V}}_2^1=6\mathrm{V}\therefore {\mathrm{Q}}_2^1={\mathrm{C}}_2{\mathrm{V}}_2^1=3\left(6\right)=18\mu \mathrm{C}$

$\therefore$ Charge flown through ‘S’

$={\mathrm{Q}}_1^1-{\mathrm{Q}}_2^1=72-18=54\mu \mathrm{C}$