If a variable frequency ac source connected to a capacitor then with decrease in frequency, the displacement current will

Current through capacitor,

$\mathrm{I}=\frac{\mathrm{E}}{{\mathrm{X}}_{\mathrm{C}}}=\frac{\mathrm{E}}{{\displaystyle \frac{1}{\mathrm{\omega C}}}}=\mathrm{\omega CE}=2\pi \nu \mathrm{CE}$or $\mathrm{I}\propto \nu$

$\therefore$decrease in frequency $\nu$ of ac source decreases the conduction current. As displacement current is equal to conduction current, decrease in $\nu$ decreases displacement current in circuit.