JEE MainPhysicsAlternating Current
A parallel plate capacitor of capacitance 50 F is connected to an AC voltage source of RMS voltage 220 V . If the peak value of the displacement current in the capacitor is 1.1 2 A , the angular frequency of the source is :
Options
- A100 rad s ⁻¹
- B100 2 rad s ⁻¹
- C200 rad s ⁻¹
- D50 rad s ⁻¹
Correct answer
A. 100 rad s ⁻¹
Step-by-step solution
The peak value of the displacement current is equal to the peak value of the conduction current. Given peak current I₀ = 1.1 2 A . The RMS value of the current is: I_ rms = I₀ 2 = 1.1 2 2 = 1.1 A In a purely capacitive circuit, the RMS current is related to the RMS voltage by: I_ rms = V_ rms X_C = V_ rms C Rearranging to solve for the angular frequency : = I_ rms V_ rms C Substitute the given values: = 1.1 220 50 10⁻⁶ = 1.1 11000 10⁻⁶ = 1.1 1.1 10⁻² = 100 rad s ⁻¹ Answer: 100 rad s ⁻¹