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Voltage across a capacitor?

  1. May 8, 2013 #1
    Lets say if i got RC series circuit.

    If i increase the voltage supply frequency then what happen to the voltage across a capacitor ??

    I rekon the voltage gonna increases and the I going to decrease to negligible.
     
  2. jcsd
  3. May 8, 2013 #2

    psparky

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    Due to it's impedance of 1/jωc, increasing frequency (ω) will lower the impedance of the capacitor. As you raise the frequency, the resistor (R) will start to gain voltage and the capacitor (c) will start to drop in voltage.

    As far as negible voltage across the capacitor, this will be true if your frequency is super high. If the frequency is moderate, the capacitor may have a decent amount of voltage across it. It's just a math equation, or simple voltage division in this case.
     
  4. May 8, 2013 #3
    Thanks
     
  5. May 24, 2013 #4
    Hi,
    Do the formulas Zc = 1/jωc and ZL = jωL apply for other voltage shape like square, sawtooth waves?
     
  6. May 24, 2013 #5
    The formulas you show are the impedance's of a capacitor and an inductor. These formulas are always correct, and they can be used with square and sawtooth waves.

    HOWEVER, in order to use them, you will have to decompose the square wave or sawtooth wave (or any signal shape) into it's constituent components, and to do this, you need to know Fourier Analysis.

    Do you have any experience with Fourier or Laplace transforms?
     
  7. May 24, 2013 #6
    Thanks Runei,
    I know these transforms. And therefore, there are different frequencies and different impedance for each frequency.
    Ah, it seems to me that it is like a filter, right?
     
  8. May 24, 2013 #7
    You are completely right. And actually, it IS a filter :)
     
  9. May 24, 2013 #8

    psparky

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    Don't forget about your C*dv/dt=it and L*di/dt=vt

    They also always hold true. (ideal conditions of course)
     
  10. May 26, 2013 #9
    To help keep the logic straight - DC is F=0 - at this point capacitors are at 0 current, and inductors are pure conductors. This boundary condition is alwas a good test of your logic - since the f=0 - is easily dropped into the math as well.
     
  11. Aug 14, 2013 #10
    Hi,
    I got a bit confused, hope anyone can help!
    For example, I have a AC voltage source with various frequencies from very low frequency (DC) to ultra-high frequencies (UHF). The voltage source is connected in parallel with an ideal capacitor. At very high frequencies, the capacitor will be short because Zc = 1/ωC is very small, almost zero.
    Will this short damage the voltage source?
     
  12. Aug 14, 2013 #11
    As far as I know a short circuit is a short circuit regardless of situation. That being said, the supply would definitely be shorted - although the damage would be a function of its protective circuitry.
     
  13. Aug 15, 2013 #12
    To be precise you should refer to the 'reactance' of a capacitor and an inductor.
    Reactance is given as Xc = 1/ωC for a capacitor and XL = ωL
    Impedance is the combination of resistance and reactance.
    For R and C in series the impedance is given by Z2 = R2 + XC2
    I don't think you will find ZC used to represent the reactance of a capacitor
    You will not find Z2 = R2 + ZC2 in a text book

    Check the hyperphysics site for more detail
     
    Last edited: Aug 15, 2013
  14. Aug 15, 2013 #13
    Hi,
    For example, in this circuit there are two capacitors that are used to filter out low and high frequencies.
    attachment.php?attachmentid=60912&stc=1&d=1376547863.jpg
    I don't see protective circuitry anywhere. At very high frenquencies, the ceramic capacitor will be source and will this damage the bridge rectifier?

    I know these formulas but my assumption that the capacitor is ideal!
     

    Attached Files:

  15. Aug 15, 2013 #14
    If the capacitor is ideal, with zero resistance, then you have only capacitative reactance XC
     
  16. Aug 15, 2013 #15

    meBigGuy

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    I'd like to state this in a more accurate way.

    The formulas only apply to sine waves. In order to use them with complex waveforms you must de-compose the complex waveform into its sine waves (all infinitely repeating waveforms can be decomposed into sine waves).

    The responders seem to be making this much more confusing that it really is.

    It is technically incorrect to say a capacitor across an AC supply will "short" the supply. It will provide a complex load determined by its reactance and resistance and the supply will respond accordingly.

    The voltage across a series LC circuit becomes 0 at the series resonant frequency (for ideal components) even though the reactance of the L and C are non-zero at that frequency. In fact, at resonance they have equal reactance and opposite phase, which cancels. If you configure and analyze a divider using the formulas you wrote above, you can solve for the voltage across the capacitor.
     
    Last edited: Aug 15, 2013
  17. Aug 15, 2013 #16
    For ideal components, your circuit (without the diode bridge) would draw extremely large currents if you impressed a voltage on the transformer primary with a frequency in the upper RF bands. That won't happen, though, for components that are physically realizable. Foremost, the frequency response of the grid transformer will have low-pass characteristics and you'll generally have parasitic inductance just about everywhere. Maybe you've seen these types of charts before for the impedance of real capacitors as a function of frequency:

    http://www.analog.com/library/analogdialogue/archives/39-09/3909_01.gif

    You'll notice that they start to behave an awful lot like inductors when you pass their resonance frequency.

    Edit:
    I wanted to give you an idea of what would happen if you had a situation as per your post #10. If you include the diode bridge without actually loading the rectifier (assuming ideal components) it will just draw an extremely large inrush current until the smoothing capacitance is charged.

    There's nothing wrong or imprecise about referring to the impedance of an ideal capacitor and/or an inductor. There's a very significant difference between impedance and reactance, but it has to do with complex numbers.

    I have a friend who's a very good electrician. He talks about impedance and reactance in the same fashion you do since he was taught a version of AC analysis where only the magnitude of the impedance mattered, because that was mainly what was needed in sizing components and such. In this way, he didn't need to learn the theory of complex numbers, which really wouldn't benefit him a great deal in his work anyway.

    That version of AC analysis is not the whole story and it's not what you're taught as an electrical engineer. I can't rule out every university/college of course, but I highly doubt any would define impedance as you do.
     
    Last edited: Aug 15, 2013
  18. Aug 15, 2013 #17
    There is a difference between reactance and impedance and it has nothing to do with complex numbers!
    Complex numbers is a branch of mathematics that is very useful in many areas of physics. Complex numbers are not AC theory.
    Reactance and impedance are 2 different terms that have different meanings and their difference does not require a knowledge of complex numbers. ( if anything all you need to recognise is Pythagoras theorem).
    You should check some basic school and college text books (and the hyperphysics site) to gain greater insight into the meaning of these terms.
    The difficulty is on a par with referring to a weight of 1kg....I am sure this would be corrected here.

    Do you have access to suitable text books?
     
  19. Aug 15, 2013 #18
    Then I doubt you have studied electrical engineering.

    Here's two text books I often see recommended for freshman/sophomore EE courses:

    Electric Circuits, 9th Edition, Nilsson, Riedel, ISBN-13: 978-0-13-611499-4.
    Excerpt from p. 320 under the heading "Impedance and Reactance":
    Engineering Circuit Analysis, 7th Edition, Hayt, Kemmerly, Durbin, ISBN-13: 978-0-07-286611-7.
    Excerpt from p. 387 under the heading "Impedance":
    Here's a text book I used in my first year for my bachelor's degree in EE/mechatronics:

    Basic Engineering Circuit Analysis, 9th Edition, Irwin, Nelms, ISBN-13: 978-0470-23455-6.
    Excerpt from p. 389 under the heading "Impedance and Admittance":
    I don't claim that these are great, or even good, books. I do claim that they present the basics correctly.

    And with regards to the HyperPhysics site:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imped.html

    You'll notice, under the heading "Impedance":
    which is what you're taught in EE courses. You should also notice the abundance of complex numbers throughout the rest of that page.

    Maybe it's best if we just agree to disagree so we don't derail yet another thread.
     
    Last edited: Aug 15, 2013
  20. Aug 15, 2013 #19
    OK.
    I have studied electrical engineering and fully realise the value of complex numbers.
    I had no problem grasping AC theory before complex numbers were introduced. They are a mathematical tool...that is all

    Ps...the definition impedance = V/I I totally agree with
     
    Last edited: Aug 15, 2013
  21. Aug 15, 2013 #20
    That's good, but I didn't write what I did to discuss complex numbers. I wrote it because you're using a definition of impedance that discards the phase information that's included with the use of complex numbers and that is one which I highly doubt is part of any college/university EE curriculum.

    But you realize that V and I are phasors, i.e. they're not real numbers, as in they're not in ℝ? Thus impedance Z cannot, in general, be a real number. The magnitude of Z, which is what you find with |Z| = √(R2 + X2), is a real number.

    Edit:
    This PF library entry goes into greater detail:
    https://www.physicsforums.com/library.php?do=view_item&itemid=303
     
    Last edited: Aug 15, 2013
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