In the circuit as shown in figure, the effective capacitance between A and B is

 

The capacitor C₄ and Cu are in parallel. Their equivalent capacitance C' = 2 +2 = 4 $\mathrm{\mu }$F. Now C' is in series with C₃. Their equivalent capacitance is given by
$\frac{1}{{\mathrm{C}}^{\prime \prime }}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\text{ or }{\mathrm{C}}^{\prime \prime }=2\mathrm{\mu F}$
Capacitance C₁ and C₂ are in series. Their equivalent capacitance C"' = 2 $\mathrm{\mu }$F
The capacitance a cross AB is given by $\mathrm{C}=2\mathrm{\mu F}+2\mathrm{\mu F}=4\mathrm{\mu F}$.