COMEDK 2024 Shift 1 Paper

To determine the displacement current for a parallel plate capacitor connected to an AC source, we utilize the relationship between the capacitor's current, the voltage, and the given angular frequency.
The capacitive reactance, XC, of the capacitor can be calculated using the formula:
XC=‌

1

ωC

where:

‌ω‌ (omega) = angular frequency ‌=100‌rad‌ s−1
‌C=‌ capacitance ‌=400‌pF=400×10−12F
Substituting the given values:
‌XC=‌

1

100‌rad‌ s−1×400×10−12F

‌XC=‌

1

(100×400)×10−12

‌XC=‌

1

40000×10−12

‌XC=‌

1

4×10−8

‌XC=2.5×107Ω
Next, we need to find the RMS current. For a capacitor in an AC circuit, the RMS voltage, Vrms, and the RMS current, Irms, are related by:
Irms=‌

Vrss

XC

Given:
Vrms=100V
Substituting the values:

‌Irms=‌

100V

2.5×107Ω

‌Irms=4µA
This matches the given value of the RMS current. The displacement current in an AC circuit is equivalent to the RMS current of the capacitor. Therefore, the displacement current is:
Displacement current =4µA
The correct answer is therefore:
Option C: 4µA