In the circuit diagram shown in the adjoining figure, the resultant capacitance between points P and Q is

In circuit, capacitors of capacitance 2 $\mathrm{\mu }$F and 3 $\mathrm{\mu }$F are in parallel. Their resultant capacitance is 5 $\mathrm{\mu }$F.

Now, capacitor 12 $\mathrm{\mu }$F, 5 $\mathrm{\mu }$F and 20 $\mathrm{\mu }$F are in series. So their resultant capacity,

                             $\frac{1}{\mathrm{C}}=\frac{1}{5}+\frac{1}{20}+\frac{1}{12}=\frac{1}{3}$

$\therefore \mathrm{C}=3\mathrm{\mu F}$