Three capacitors of capacitances 3 $\mathrm{\mu }$F, 9 $\mathrm{\mu }$F and 18 $\mathrm{\mu }$F are connected once in series and another time in parallel. The ratio of equivalent capacitance in the two cases $\left(\frac{{\mathrm{C}}_{\mathrm{S}}}{{\mathrm{C}}_{\mathrm{P}}}\right)$will be

When capacitors are connected in series, then

               $\frac{1}{{\mathrm{C}}_{\mathrm{S}}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}+\frac{1}{{\mathrm{C}}_3}=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}$

                $\frac{1}{{\mathrm{C}}_{\mathrm{S}}}=\frac{1}{2}$

$\Rightarrow {\mathrm{C}}_{\mathrm{S}}=2\mathrm{\mu F}$

When capacitors are joined is parallel, then

                   ${\mathrm{C}}_{\mathrm{P}}=3+9+18=30\mathrm{\mu F}$

$\therefore \frac{{\mathrm{C}}_{\mathrm{S}}}{{\mathrm{C}}_{\mathrm{P}}}=\frac{2}{30}=\frac{1}{15}$