Exams › NEET › Physics

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 V/s. The conduction current through the connecting wires and the displacement current through the plates of the capacitor are, respectively,

1. Zero, 60 \(\mu\text{A}\)
2. 60 \(\mu\text{A}\), 60 \(\mu\text{A}\)
3. 60 \(\mu\text{A}\), zero
4. Zero, zero

Correct answer: 60 \(\mu\text{A}\), 60 \(\mu\text{A}\)

Solution

The current is given by \(I = C\,\frac{dV}{dt}\). Substituting \(C = 20\,\mu\text{F}\) and \(\frac{dV}{dt} = 3\,\text{V/s}\), we get \(I = 60\,\mu\text{A}\). In a charging capacitor, the conduction current and displacement current have the same magnitude.

Related NEET Physics questions

• A capacitor is fully charged by a battery, and then the battery is disconnected. When a dielectric slab is inserted between the capacitor plates, what is the effect?
• In a charged capacitor, where is the energy stored?
• A capacitor of capacitance $C$ is charged to a potential $V$. The energy stored in it is:
• A capacitor of capacitance $50\,\mu\text{F}$ is charged to $10\,\text{V}$. What is the energy stored in it?
• A parallel-plate capacitor has air between its plates. When the air is replaced by a dielectric material of dielectric constant $K$, the capacitance:
• Eight identical small drops, each of radius $r$ and carrying charge $q$, coalesce to form one large drop. What is the ratio of the electric potential of the big drop to that of one small drop?

⚔️ Practice NEET Physics free + battle 1v1 →