Question

The voltage between the plates of a parallel plate capacitor of capacity \(1 \mu \mathrm{~F}\) is changing at the rate of \(4 \mathrm{~V} / \mathrm{s}\). the displacement current in the capacitor is

• A. \(4\mu \text{ }\mathrm{A}\)
• B. \(3\mu \text{ }\mathrm{A}\)
• C. \(1\mu \text{ }\mathrm{A}\)
• D. \(6\mu \text{ }\mathrm{A}\)

Hint:

Displacement current in a capacitor equals capacitance times the rate of change of voltage.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Displacement current equals the conduction current in the circuit: \(I_d=\dfrac{dq}{dt}=C\dfrac{dV}{dt}\).

Step 2: Substitute
\(C=1\ \mu\text{F}=10^{-6}\text{ F}\) and \(\dfrac{dV}{dt}=4\text{ V/s}\).

Step 3: Result
\(I_d=10^{-6}\times4=4\times10^{-6}\text{ A}=4\ \mu\text{A}\).

Final Answer:
The displacement current is \(4\ \mu\text{A}\). \[ \boxed{\text{(A) }4\mu \text{ A}} \]