Question

A parallel plate capacitor has a capacitance C = 200 pF. It is connected to 230 V ac supply with an angular frequency 300 rad/s. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

• A. 1.38 μA and 1.38 μA
• B. 14.3 μA and 143 μA
• C. 13.8 μA and 138 μA
• D. 13.8 μA and 13.8 μA

Answer 1

The Correct Option is D

The formula for current in a capacitor within an AC circuit is:

\[ I = V \frac{\omega C}{\sqrt{1 + (\omega C)^2}} \]

The parameters are defined as follows:

• \(V = 230 \, \text{V}\) represents the supply voltage.
• \(C = 200 \times 10^{-12} \, \text{F}\) denotes the capacitance.
• \(\omega = 300 \, \text{rad/s}\) is the angular frequency.

In a capacitor, the displacement current is equivalent to the conduction current. The current flowing through the capacitor is determined by:

\[ I = V \frac{\omega C}{X_C} \]

Here, \(X_C = \frac{1}{\omega C}\) signifies the capacitive reactance.

The calculation proceeds as follows:

\[ I = \frac{230 \times 300 \times 200 \times 10^{-12}}{X_C} = 13.8 \mu\text{A} \]

Consequently, the RMS current, comprising both conduction and displacement current, is 13.8 $\mu\text{A}$.