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A basic math question about the current in a capacitor

  1. Dec 4, 2013 #1
    The current in a capacitor is calculated from

    [itex]i_{C}(t)=C \dfrac{dv_{C}(t)}{dt}[/itex]​

    when [itex]i_{C}(t)[/itex] is the current in the capacitor, [itex]t[/itex] is the time variable, [itex]C[/itex] is the capacitance and [itex]v_{C}(t)[/itex] is the capacitor's voltage.
    This equation can also be expressed, using [itex]\omega t[/itex] as the variable, as

    [itex]i_{C}(\omega t)=\omega C \dfrac{dv_{C}(\omega t)}{d(\omega t)}[/itex]​

    Now my basic question is why there's a multiplication in [itex]\omega[/itex]? That is, why is it correct to multiply when changing in independent variable and multiplying in by frequency [itex]\omega[/itex]? Seemingly, the new variable, [itex]\omega t[/itex], is derived in the same manner as [itex]t[/itex] is derived when the function uses only [itex]t[/itex] as the independent variable.
    I don't quite understand this little issue. Hope someone could explain that in a sentence.

    Thanks in advance! :tongue:
     
  2. jcsd
  3. Dec 4, 2013 #2

    tiny-tim

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    Hi Adir_Sh! :smile:
    chain rule

    dvC/dt

    = dvC/d(ωt)*d(ωt)/dt

    = dvC/d(ωt)*ω :wink:
     
  4. Dec 5, 2013 #3

    rude man

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    Realize that d(wt) = w dt and you have your answer.
     
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