Four capacitors are connected as shown below in the figure. Here C1 = 2 μF, C2 = 3 μF, C3 = 4 μF and C4 = 5 μF.

What is the equivalent capacitance between A and B?

3.27 μF

2 $\mu F$ is in parallel with the series combination of the other three $C_2 , C_3 , C_4$ .

Equivalent Capacitance = $\frac{C_2C_3C_4}{C_2C_3+C_3C_4+C_2C_4} = \frac{60\mu F}{47}$

Equivalent Capacitance = 2+1.27 = 3.27$\mu$ F