A capacitor of capacitance \(C\) is connected across an AC source of voltage \(V\), given by;
\(V=V_0 \sin \omega t\)$\mathrm{}$
The displacement current between the plates of the capacitor would then be given by:
1. \( I_d=\frac{V_0}{\omega C} \sin \omega t \)
2. \( I_d=V_0 \omega C \sin \omega t \)
3. \( I_d=V_0 \omega C \cos \omega t \)
4. \( I_d=\frac{V_0}{\omega C} \cos \omega t\)

A parallel plate capacitor of capacitance \(20~\mu\text{F}\) is being charged by a voltage source whose potential is changing at the rate of \(3~\text{V/s}\). The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively:
1. zero, zero
2. zero, \(60~\mu\text{A}\)
3. \(60~\mu\text{A}\), \(60~\mu\text{A}\)
4. \(60~\mu\text{A}\), zero