The equivalent capacitance between $A$  and  $B$  in the circuit given below is  $in\mu F$ :

The simplified circuit of the circuit given in question as follows:

The equivalent capacitance between $C$  &  $D$  capacitors of  $2\mu F,5\mu F$  and  $5\mu F$ are in parallel.
$\therefore C_{CD}=2+5+5=12\mu F$   ( $\because$  In parallel grouping  $C_{eq}=C_1+C_2+\mathrm{....}+C_n$)
Similarly equivalent capacitance between  $E$  &  $B{C}_{EB}=4+2=6\mu F$
Now equivalent capacitance between   $A$  &  $B$
 $\frac{1}{C_{eq}}=\frac{1}{6}+\frac{1}{12}+\frac{1}{6}=\frac{5}{12}$
 $\Rightarrow C_{eq}=\frac{12}{5}=2.4\mu F$   ( $\because$  In series grouping, $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\mathrm{......}+\frac{1}{C_n}$ )