Question

AC voltage V(t) = 20 sinωt of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m². The amplitude of the oscillating displacement current for the applied AC voltage is ______.

[take ε₀ = 8.85 × 10^-12 F/m]

पर्याय

• 21.14 µA

• 83.37 µA

• 55.58 µA

• 27.79 µA

Answer 1

AC voltage V(t) = 20 sinωt of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m². The amplitude of the oscillating displacement current for the applied AC voltage is 27.79 µA.

Explanation:

Frequency = 50 Hz

d = 2 mm, A = 1 m²

C = `(Aε_0)/d = (1 xx 8.85 xx 10^-12)/(2 xx 10^-3)`

= 4.425 × 10^-9 F

`X_c = 1/(omegaC) = 1/(2pi(50)(4.425 xx 10^-9))`

= `0.7193 xx 10^6 Ω`

displacement current (I_d) = `V_0/(X_C) = 20/(0.7193 xx 10^6)`

= 27.81 μA

= 27.79 μA