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Capacitor's Reactance

  1. Dec 29, 2008 #1
    How do we derive the opposition to current flow (AC signal) due to capacitance without using complex numbers?

    Is the capacitor's reactance a constant or an average?

    If I were to measure the instantaneous voltage across a capacitor and the instantaneous current through it, shouldn't I get its reactance (according to Ohm's Law)?

    Why do we get the reactance of the capacitor dividing the peak voltage by the peak current if they are always 90 degrees out of phase?
     
  2. jcsd
  3. Dec 29, 2008 #2
    Xc (reactance) = 1/2 x pi x f x C IF the C is essentially lossless.

    R in series with C would give impedance Z = Root of the sum of the squares of Xc and R (vector sum)

    Xc is constant at given frequency f.

    I = V/Xc.


    Xc = Vrms/Irms and as Vrms is proportional Vpk and Irms is prop to Ipk. Xc = Vpk/Ipk
     
  4. Dec 29, 2008 #3
    I don't know if you've noticed but you didn't answer any of my questions. However, thanks for reply anyway :)
     
  5. Dec 29, 2008 #4

    berkeman

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    Staff: Mentor

    You generally use differential equations first, and then simplify to complex number notation.

    What are your thoughts on this? A capacitor is a physical thing, with the physical dimensions (generally) not changing with applied voltage....

    Can you show us the equation you are asking about?
     
  6. Dec 29, 2008 #5
    I know that but I'm asking how to fully derive it without any use of complex numbers.

    I'm sorry, my question didn't make sense. What I mean is; does reactance changes over time?

    What do you mean? What I was asking in here was if the reactance value would be the same as measuring the instantaneous values (voltage and current) at a given time and find their ratio (Ohm's Law).
     
  7. Dec 29, 2008 #6

    berkeman

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    I believe the correct answer is no, not for LTI (linear, time-invariant)systems. It does change over frequency, though. Usually when you are referring to a reactance, you are referring to a specific test frequency.

    Not sure I understand the question, but I think the answer is no. The reactance is the imaginary component of the complex impedance, not the instantaneous magnitude of the impedance.

    http://en.wikipedia.org/wiki/Reactance_(electronics)

    .
     
  8. Dec 30, 2008 #7
    I don't really know how to explain what I mean :S

    However, does Ohm's Law only work when both voltage and current are in phase?

    Imagine an AC circuit comprised of a resistor only. We both know that the voltage and current are both in phase. Now, if we want to know the R value we can just apply Ohm's Law at any given time. Obviously such value will also give you the circuit's opposition to current.

    If we now replace the resistor with a capacitor how can we manage to get the same current opposition as with the resistor?! Well, we just need to make X equal to R and we are done!
    Nonetheless, if we measure the instantaneous voltage (across it) and current (through it) at a given time and work out their ratio (Ohm's Law) we won't get the correct value of the opposition to current. Therefore Ohm's Law doesn't work with instantaneous values in AC circuits where there's a phase difference between the voltage and current.

    Lol... I think I've just answered my question although it seems a bit counter intuitive?!
     
  9. Dec 30, 2008 #8
    Intantaneous voltage or current is a difficult concept in AC circuits as it suggests a frozen moment in time and DC values which wont exist with a capacitor. It is a circuit element with energy storage and you have to analyse it over one or more AC cycles. The current flow depends on frequency. It has to be analysed dynamically.
     
  10. Dec 30, 2008 #9

    berkeman

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    Staff: Mentor

    Ohm's law works when you treat V, Z and I as complex numbers. There are Real and Imaginary components to each, in general.
     
  11. Dec 30, 2008 #10
    Fair enough, but could you tell me why do we get a capacitor's reactance working out the ratio between the Vp and Ip? Why the peak values if they are 90 degrees out of phase? If you apply Ohm's Law directly to instantaneous values and work out their ratio it will fail.
    Where does such ratio come from?
     
  12. Dec 30, 2008 #11
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