In the circuit shown in figure. Find the equivalent capacitance between points a and b when switch is open and switch is closed.

When switch is open, the circuit can be re-drawn as

2 $\mathrm{\mu }$F & 4 $\mathrm{\mu }$F are in series, 4 $\mathrm{\mu }$F, 8 $\mathrm{\mu }$F are in series, 2 $\mathrm{\mu }$F, 4 $\mathrm{\mu }$F are in series and all are in parallel.

The equivalent capacitance is $\frac{16}{3}\mu F$ 

When the switch is closed the circuit can be re-drawn as

So the 8uF and 4uF will be in series with the parallel combination of 4uF and 2uF.

$C_{eq}=\frac{12}{3}+\frac{4}{3}=\frac{16}{3}\mu F$