For two identical capacitors in series, the equivalent capacitance is simply half the value of one ($2/2 = 1$). For non-identical ones, use "Product over Sum" ($18/9 = 2$).
Step 1: Reduce the upper branch (3 μF and 6 μF in series) using the reciprocal rule: \( \frac{1}{C_{up}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} \), so \( C_{up} = 2\ \mu\text{F} \).
Step 2: Reduce the lower branch (2 μF and 2 μF in series) the same way: \( \frac{1}{C_{low}} = \frac{1}{2} + \frac{1}{2} = 1 \), so \( C_{low} = 1\ \mu\text{F} \).
Step 3: Since the two reduced branches sit in parallel between A and B, add them directly: \( C_{AB} = 2 + 1 = 3\ \mu\text{F} \).
\[ \boxed{C_{AB} = 3\ \mu\text{F}} \]
