To find the equivalent capacitance of the given circuit, we need to analyze the arrangement of capacitors.$\\$The circuit consists of:$\\$• A series combination of a $3 \mu F$ capacitor with a parallel combination of two $3 \mu F$ capacitors.$\\$Step 1: Calculate the equivalent capacitance of the parallel capacitors.$\\$For capacitors in parallel, the equivalent capacitance $C_{parallel}$ is given by:$\\ C_{parallel} = C_1 + C_2 \\ C_{parallel} = 3 \mu F + 3 \mu F \\ C_{parallel} = 6 \mu F \\$Step 2: Calculate the equivalent capacitance of the series combination.$\\$For capacitors in series, the equivalent capacitance $C_{series}$ is given by:$\\ \frac{1}{C_{series}} = \frac{1}{C_3} + \frac{1}{C_{parallel}} \\ \frac{1}{C_{series}} = \frac{1}{3 \mu F} + \frac{1}{6 \mu F} \\$Calculate the reciprocal:$\\ \frac{1}{C_{series}} = \frac{2}{6 \mu F} + \frac{1}{6 \mu F} \\ \frac{1}{C_{series}} = \frac{3}{6 \mu F} \\ \frac{1}{C_{series}} = \frac{1}{2 \mu F} \\ C_{series} = 2 \mu F \\$Therefore, the equivalent capacitance of the system is $2 \mu F$. This corresponds to Option 1.