Question

The capacitance of a parallel plate capacitor is 400 pF. It is connected to an ac source of 100 V having an angular frequency 100 rad s$^{-1}$. If the rms value of the current is 4 μA, the displacement current is:

• A. \(4 \times 10^{-2}\) μA
• B. 0.4 μA
• C. 4 μA
• D. 4 A

Hint:

For displacement current, use the relation \( I_d = C \omega V_{\text{rms}} \). This is useful when dealing with ac circuits involving capacitors.

Answer 1

The Correct Option is C

The displacement current \( I_d \) in a capacitor is related to the capacitance \( C \), the rms voltage \( V_{\text{rms}} \), and the angular frequency \( \omega \) by the formula: \[ I_d = C \omega V_{\text{rms}} \] Given: - \( C = 400 \, \text{pF} = 400 \times 10^{-12} \, \text{F} \) - \( V_{\text{rms}} = 100 \, \text{V} \) - \( \omega = 100 \, \text{rad/s} \) Substitute these values into the formula: \[ I_d = 400 \times 10^{-12} \times 100 \times 100 \] \[ I_d = 4 \times 10^{-2} \, \text{A} = 4 \, \text{μA} \] Thus, the displacement current is 4 μA.