A variable frequency AC source is connected to a capacitor. How will the displacement current change with decrease in frequency?

Explanation

Reactance of capacitor of capacitance C is given by
` X_c = (1 )/( omega C)`
where ` omega `= angular frequency =` 2 pi v `
( v is the freqency of AC source )
current through the capacitor is
` I= (E )/(X_c )=(E )/( 1// omega C ) = omega CE =- 2 pi vCE `
where E is the rms value of AC source
` therefore I prop v `
The conduction current will decrease if the freqency of the source is decreased since
Displacement current = conduction current
` therfore ` Decrease in frequency (v ) will result in decrease in displacement current `(i_d)`.