Study the following circuit diagram and mark the correct options.

When S is open $i=\frac{18}{9}=2A,\frac{v_1}{v_2}=\frac{C_2}{C_1}=\frac{3}{6}=\frac{1}{2}$

$V_A-V_B=6\times 2=12V$

$\Rightarrow v_1=\frac{1}{3}18=6V$

$v_2=\frac{2}{3}18=12V$

$V_a-V_c=3\times 2=6V$

$v_a=6(\because v_c=0)$

$v_1=v_B-v_a=6V$

$v_2=v_b-v_D=12V$

$v_{ab}=v_a-v_b=6-12=-6V$

$v_b=12v(v_D=0) Q_1=c_1{v}_1=6\times 6=36\mu c$

When S is closed $v_1^{\text{'}}=12V,\Rightarrow Q_1^{\text{'}}=C_1{v}_1^{\text{'}}=6\times 12=72\mu C$

$v_2^{\text{'}}=6V,\Rightarrow Q_2^{\text{'}}=C_2{v}_2^{\text{'}}=3\times 6=18\mu C$

Change from through $S=Q_1^1-Q_2^1=72-18=54\mu C$

$\Rightarrow$a and c are correct