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AC applied to a capacitor.

  1. Aug 31, 2014 #1
    In a purely capacitive ac circuit, we get,
    Im = Vm*ω*C, ....(1)
    Where, Im = Amplitude of the current
    Vm = Amplitude of the voltage

    Now, what I think is,
    We know that in a purely capacitive circuit, voltage lags behind current by a phase difference of ∏/2 rad. So, at any time 't',
    I = Im sin(ωt+∏/2)
    V = Vm sinωt

    Using Kirchhoff's Loop Rule,

    V = Vmsinωt = q/C

    Where q = charge on the capacitor at time 't',

    To find the current, I = dq/dt,
    dq = Idt,

    q = ∫Idt

    q = ∫Imcosωt dt
    q = Im∫cosωt dt
    q = Im*ω*sinωt

    So, Vmsinωt = Im*ω*sinωt /C
    Vm = Im*ω/C

    Im = Vm*C/ω, which is apparently, not equal to equation (1).

    Am I wrong in my approach?
     
  2. jcsd
  3. Aug 31, 2014 #2

    Simon Bridge

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    Have you applied a result to a situation where it does not occur.
    If there is no resistance in the circuit, then the voltage across the capacitor is going to be the same as the applied voltage - instantly. Draw the circuit diagram and see.
     
  4. Aug 31, 2014 #3
    this is wrong..

    q = (Im/ω)*sinωt.. You are integrating. remember not differentiating.
    So this becomes

    Vm = Im*1/ωC

    compare this to V = IR

    so 1/ωC plays the role of impeding the AC.
     
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