A parallel plate capacitor is charged by an exteranl ac source straight the displacement current inside the capacitor is the same as the current charging the capacitor.

Explanation

Electric field between the capacitor plates
`E = (sigma)/(epsilon_(0)) = (q)/(epsilon_(0)A)`
Where q is the charge accumulated on the positive plate.
The electric flux through this plate
`phi_(E ) = EA = (q)/(epsilon_(A))cdot A = (q)/(epsilon_(0))`
`therefore` Dispacement current
`I_(d) = epsilon_(0)cdot (d phi )/( dt ) = epsilon_(0) (d)/(dt) [(q)/(epsilon_(0))] = (dq)/(dt)`
`(dq)/(dt)` is the rate at which charge flows to positive plate through the conducting wire.
`I_(d) = I_(C )`
i.e., displacement current between the capacitor plates = conduction current through wire.