The equivalent capacitance of the system shown in the following circuit is :

Two \( 3\mu\text{F} \) capacitors in parallel give \( 3+3 = 6\mu\text{F} \). This combination is in series with another \( 3\mu\text{F} \) capacitor: \( C_{eq} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\mu\text{F} \).