Question

A parallel plate capacitor of capacitance $20\mu\,F$ 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, would be, respectively :

• A. $60\mu\,A$, zero
• B. zero, zero
• C. zero, $60\mu\,A$
• D. $60\mu\,A$ $60\mu\,A$

Answer 1

The Correct Option is D

To solve this problem, we need to understand the concepts of conduction current and displacement current in a capacitor.

1. Conduction Current: The conduction current is the current flowing through the wires connecting the capacitor to the voltage source. For a capacitor, it is given by the formula:
   I = C \cdot \frac{dV}{dt}
   where I is the current, C is the capacitance, and \frac{dV}{dt} is the rate of change of voltage. Here, C = 20 \mu F and \frac{dV}{dt} = 3 V/s.
2. Substituting the given values:
   I = 20 \times 10^{-6} F \times 3 \, \text{V/s} = 60 \mu A
   So, the conduction current is 60 \mu A.
3. Displacement Current: Due to the time-varying electric field in the capacitor, the displacement current equals the conduction current flowing in the wires. According to the principle of continuity of currents, the displacement current (which flows across the capacitor plates through the dielectric medium) also equals 60 \mu A.

Thus, the conduction current through the connecting wires and the displacement current through the plates of the capacitor are both 60 \mu A.

Therefore, the correct answer is: 60\mu\,A, 60\mu\,A.