Three capacitors of capacitance, 3 μF, 9 μF and 18 μF respectively are connected once in series and then in parallel. The ratio of equivalent capacitance in series to parallel combination will be:

1 : 15

The correct answer is Option (1) → 1 : 15

Case 1: When connected in parallel

$C_{eq}=3+9+18$

$=30μF$

Case 2: When connected in series

$\frac{1}{C_{eq}}=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}$

$C_{eq}'=\frac{18}{9}=2$

$\frac{C_{eq}'}{C_{eq}}=\frac{2}{30}=\frac{1}{15}$