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AC Circuits in capacitor

  1. Jul 28, 2004 #1
    there is one question i need someone to help me to check and another question that i need someone to help with.

    an alternationg voltage, V = Vosin (omega)(t) is applied across a pure capacitor of capacitance C.

    derive an expression for the capacitive reactance.
    i do it this way..

    given V = Vosin (omega)(t)
    but Q = CV
    Q = CVosin (omega)(t)

    I = dQ/dt
    I = Io cos (omega)(t), where Io = CVo(omega)

    capacitive reactance = Vo / Io
    = 1 / C*(omega)

    this is the only way to do it right?

    the next question is

    what is the effect on current in the circuit if the frequency of the power supply is increased, but its capacitance C of the capacitor and rms value of the alternating voltage V remain constant? without referring to the reactance of the circuit, explain your answer.

    im not sure how to answer this question, and i dont answer for this question as well. what i can think of is... the value of the max current decreases because Io = CVo*omega

    please correct me. thanks
  2. jcsd
  3. Jul 28, 2004 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    First answer is correct.

    Second - approach is right. But, like you said [tex]I_o = CV_o~ \omega [/tex]. So, how does increasing the frequency ([tex] \omega[/tex]) cause the current to decrease ?
    Last edited: Jul 28, 2004
  4. Jul 28, 2004 #3
    when f increase, capacitor will be charged and discharged in a shorter time and more frequently. therefore, I increases.

    am i making sense?
  5. Jul 28, 2004 #4
    You need to look at the complete expression for the current and not just the amplitude of the function. Once you have fixed a particular omega, the amplitude is fixed really and all that you can hope to extract is from the argument of the sinusoidal function. The higher the omega is, the greater is the amplitude (as evident from your expressions denian) but greater is the ANGULAR FREQUENCY (radians per second) and so the function oscillates at a faster rate now.

    Now I am really wondering that since [tex]\omega = 2\pi f[/tex] and so increasing omega is equivalent to increasing f (or vice versa), why do you consider only the peak value in the answer? Or am I misunderstanding your statement? Please clarify...

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