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

Answer: (D)

13.8 μA and 13.8 μA

Answer 2

The current through a capacitor in an AC circuit is given by the formula:

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

where:

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

The displacement current in the capacitor is the same as the conduction current in the circuit. The current through the capacitor is given by:

$I = V \frac{\omega C}{X_C}$

where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.

Now we calculate:

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

Thus, the rms value of the current is 13.8 $\mu\text{A}$ for both the conduction and displacement current.