Question

A parallel plate capacitor of capacitance \(20 μF\) 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. \(zего, 60 μА \)
• B. \(60 μА, 60 μA \)
• C. 60 μA, zero
• D. zero,zero

Answer 1

The Correct Option is B

To solve the problem, we need to calculate both the conduction current through the connecting wires and the displacement current through the plates of the capacitor.

The given parameters are:

• Capacitance, \(C = 20 \, \mu F = 20 \times 10^{-6} \, F\)
• Rate of change of voltage, \(\frac{dV}{dt} = 3 \, V/s\)

The formula for the current through the capacitor is given by:

\(i = C \cdot \frac{dV}{dt}\)

Substituting the given values:

\(i = 20 \times 10^{-6} \times 3 = 60 \times 10^{-6} \, A\)

This calculation gives us the conduction current, which is \(60 \, \mu A\).

Understanding Displacement Current:

Displacement current is the concept used to describe the apparent current that flows in the dielectric of the capacitor due to the changing electric field. It can be calculated using the same formula as the conduction current, because both are equal in a capacitor with a changing electric field.

Thus, the displacement current is also \(60 \, \mu A\).

Answer and Conclusion:

The conduction current through the connecting wires and the displacement current through the plates of the capacitor are both found to be \(60 \, \mu A\). Therefore, the correct option is \(60 \, \mu A, 60 \, \mu A\).