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:

Displacement current is the same as:

The charge of a parallel plate capacitor is varying as; \(q = q_{0} \sin\omega t\). The magnitude of displacement current through the capacitor is:
(the plate Area = \(A\), separation of plates = \(d\))
1. \(q_{0}\cos \left(\omega t \right)\)
2. \(q_{0} \omega \sin\omega t\)
3. \(q_{0} \omega \cos \omega t\)
4. \(\frac{q_{0} A \omega}{d} \cos \omega t\)

The S.I. unit of displacement current is:
1. Henry
2. Coulomb
3. Ampere
4. Farad

A variable frequency AC source is connected to a capacitor. Then on increasing the frequency:

Instantaneous displacement current of \(2.0~\text A\) is set up in the space between two parallel plates of \(1~\mu \text{F}\)$$ capacitor. The rate of change in potential difference across the capacitor is:
1. \(3\times 10^{6}~\text{V/s}\)
2. \(4\times 10^{6}~\text{V/s}\)
3. \(2\times 10^{6}~\text{V/s}\)
4. None of these

A larger parallel plate capacitor, whose plates have an area of 1 $m^2$ are separated from each other by 1 mm, is being charged at a rate of 25.8 V/s. If the plates has dielectric constant 10, then the displacement current at this instant is:

1. 25 $\mu A$

2. 11 $\mu A$

3. 2.2 $\mu A$

4. 1.1 $\mu A$

A capacitor is made of two circular plates each of radius \(12~\text{cm}\) and separated by \(5.0~\text{cm}.\) The capacitor being charged by an external source. The charging current is constant and equal to \(0.15~\text{A}.\) The displacement current across the plates is:
1. \(0\)
2. \(0.14~\text{A}\)
3. \(0.16~\text{A}\)
4. \(0.15~\text{A}\)