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Alternating Current question

2021 · Q135
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Alternating Current question

2021 · Q135

NEETPhysicsAlternating CurrentMCQ+4 / −1
A capacitor of capacitance 'C', is connected across an ac source of voltage V, given by

V = V0sinω\omegaωt

The displacement current between the plates of the capacitor, would then be given by :
  1. A
    Id=V0ωCsin⁡ωt{I_d} = {V_0}\omega C\sin \omega tId​=V0​ωCsinωt
  2. B
    Id=V0ωCcos⁡ωt{I_d} = {V_0}\omega C\cos \omega tId​=V0​ωCcosωt
  3. C
    Id=V0ωCcos⁡ωt{I_d} = {{{V_0}} \over {\omega C}}\cos \omega tId​=ωCV0​​cosωt
  4. D
    Id=V0ωCsin⁡ωt{I_d} = {{{V_0}} \over {\omega C}}\sin \omega tId​=ωCV0​​sinωt
View written solutionFree

Correct answer: B

Given, V = V0sinω\omegaωt

We know, q=CVq = CVq=CV

Now displacement current Id is given by,

Id = dqdt=CdVdt{{dq} \over {dt}} = {{CdV} \over {dt}}dtdq​=dtCdV​

Id=C(V0ωcos⁡ωt){I_d} = C({V_0}\omega \cos \omega t)Id​=C(V0​ωcosωt)

=V0ωCcos⁡ωt = {V_0}\omega C\cos \omega t=V0​ωCcosωt

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