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Alternating Current question

2021 · 20 Jul · Shift 1 · Q60
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  5. /2021 · 20 Jul · Shift 1 · Q60

Alternating Current question

2021 · 20 Jul · Shift 1 · Q60

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
AC voltage V(t) = 20 sin ω\omegaω t 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 ‾\underline{\hspace{2cm}}​. [Take ε\varepsilonε 0 = 8.85 ×\times× 10 −-− 12 F/m]
  1. A
    55.58 μ\muμ A
  2. B
    21.14 μ\muμ A
  3. C
    27.79 μ\muμ A
  4. D
    83.37 μ\muμ A
View written solutionFree

Correct answer: C

  1. Given data
  • Applied voltage: V(t)=20sin⁡ωtV(t)=20\sin \omega tV(t)=20sinωt
  • So voltage amplitude, V0=20 VV_0=20\ \text{V}V0​=20 V
  • Frequency: f=50 Hzf=50\ \text{Hz}f=50 Hz
  • Plate separation: d=2 mm=2×10−3 md=2\ \text{mm}=2\times10^{-3}\ \text{m}d=2 mm=2×10−3 m
  • Plate area: A=1 m2A=1\ \text{m}^2A=1 m2
  • Permittivity of free space: ε0=8.85×10−12 F/m\varepsilon_0=8.85\times10^{-12}\ \text{F/m}ε0​=8.85×10−12 F/m
  1. Displacement current in a capacitor

For a capacitor,

id=CdVdti_d = C\frac{dV}{dt}id​=CdtdV​

If

V(t)=V0sin⁡ωt,V(t)=V_0\sin \omega t,V(t)=V0​sinωt,

then

dVdt=ωV0cos⁡ωt\frac{dV}{dt}=\omega V_0 \cos \omega tdtdV​=ωV0​cosωt

So the amplitude of displacement current is

I0=ωCV0I_0 = \omega C V_0I0​=ωCV0​
  1. Capacitance of the parallel plate capacitor
C=ε0AdC=\frac{\varepsilon_0 A}{d}C=dε0​A​

Substitute values:

C=8.85×10−12×12×10−3C=\frac{8.85\times10^{-12}\times1}{2\times10^{-3}}C=2×10−38.85×10−12×1​ C=4.425×10−9 FC=4.425\times10^{-9}\ \text{F}C=4.425×10−9 F
  1. Angular frequency
ω=2πf=2π(50)=100π rad/s\omega=2\pi f = 2\pi(50)=100\pi\ \text{rad/s}ω=2πf=2π(50)=100π rad/s
  1. Current amplitude
I0=ωCV0I_0=\omega C V_0I0​=ωCV0​ I0=(100π)(4.425×10−9)(20)I_0=(100\pi)(4.425\times10^{-9})(20)I0​=(100π)(4.425×10−9)(20) I0=8850π×10−9I_0=8850\pi\times10^{-9}I0​=8850π×10−9 I0≈2.779×10−5 AI_0\approx 2.779\times10^{-5}\ \text{A}I0​≈2.779×10−5 A I0=27.79 μAI_0=27.79\ \mu \text{A}I0​=27.79 μA
  1. Matching with options

Thus, the amplitude of oscillating displacement current is

27.79 μA27.79\ \mu \text{A}27.79 μA

So the correct option is C.

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