More on impedance and instant energy stored in capacitor

Discussion in 'General Electronics Chat' started by kiroma, Dec 22, 2014.

  1. kiroma

    Thread Starter Member

    Apr 30, 2014
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    Hello.
    I'd like to suggest and ask more about impedance, like
    What's the voltage across the capacitor in an AC series circuit?
    And what's the voltage across the resistor?
    And I'd like to ask the same for inductor, which is, probably, the inverse.

    And about the instant energy being stored in a capacitor (instantaneous voltage times the instantaneous current), I'd like to say that I've made a lot of progress about it, if you need my help, I'd be glad to help build this site.
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    All of this is first principles, and is already covered in the tutorials.

    I suggest you post the specific circuit you are having trouble analyzing.

    Energy stored on a capacitor
     
  3. kiroma

    Thread Starter Member

    Apr 30, 2014
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    You didn't mention about the voltage across the capacitor in an AC series circuit, so I think it's not all covered...

    About the energy stored in a capacitor is very common to think in that way, only in function of the capacitance and voltage, but I meant to publish a page about the energy BEING stored, I mean, in function of time, and by the way, in function of resistance (which is in series), capacitance and voltage.
    I'd like to talk about my achievements, but I don't know if this is the right place to do.
    But, just to let it be more clear, I got the equation Ipc=Ef²/R*e^(-2t/RC)
    Where, Ipc is the Instant energy being stored in the capacitor, Ef is the voltage of the source, t is for the time, R is resistance in series and C is the capacitance.
     
  4. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    That equation appears to be for a DC situation, since it has no sinusoidal component. I thought you were asking about an AC circuit?

    Here you go. The yellow trace and blue trace show the voltage across the cap and resistor respectively.
    ACvoltages.gif
     
    Last edited: Dec 23, 2014
  5. kiroma

    Thread Starter Member

    Apr 30, 2014
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    Well, you got me wrong.
    I was talking about 2 different things.
    One is the rms value of the voltage of a capacitor in a series AC circuit with resistance, because I didn't find it anywhere. I mean, the formula to find the rms voltage in function, I imagine, of the resistance, the frequency and the capacitance.

    Another thing is the instant energy being stored in a capacitor, which can be changed to AC, but my main purpose was the energy being stored in a DC circuit RC series.
    By the way, where do can I post a suggestion to make a page about this? I'm thinking I'm in the wrong session.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Shouldn't that be..

    \frac {E_f^2C}{2} \(1-e^{-\frac {t}{RC}}\)^2

    ....???

    For your equation consider two significant cases @ t=0 & infinity.

    At t=0 your equation yields a non-zero value which doesn't make sense - unless the capacitor is initially charged @ t=0. When t=infinity your equation yields zero energy, which again makes no sense.

    Also one would expect the energy to be directly dependent on the capacitance value - which your equation fails to show.

    A further comment about "instantaneous energy". The product of instantaneous voltage and instantaneous current isn't equivalent to instantaneous energy - rather to instantaneous power. So I suspect you have calculated the instantaneous power in the resistor.
     
    Last edited: Dec 24, 2014
  7. Wendy

    Moderator

    Mar 24, 2008
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    This appears to be asking more of questions than suggestions. The book has the info, but you have to read it and be clear about what it says. If you have a question the chat forum is the place to ask.
     
  8. kiroma

    Thread Starter Member

    Apr 30, 2014
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    You got me wrong again
    Power is dw/dt, where w is work

    I can say that power is the instant work being done to charge the capacitor

    And you're thinking that this is the energy stored in the capacitor, which isn't
    This is like the derivative of the (cv^2)/2 in respect to time
    So, to get the energy STORED in the capacitor, you'll have to integrate from 0 to infinity
    That will give you the widespread equation of the energy stored...

    And I ask you to have more attention when you look at my given formula
    It's not \frac {E_f^2C}{2} \(1-e^{-\frac {t}{RC}}\)^2
    It's \frac {E_f^2C}{R} \(e^{-\frac {2t}{RC}}\).
     
  9. WBahn

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    Mar 31, 2012
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    You mean the formula that you gave previously:

    Perhaps you need to pay more attention yourself, since you aren't being consistent.

    Check your units. Ef²/R has units of power, not energy. Ef²C/R has units of (A²·s), which is basically meaningless.

    The notion of "instant energy" is not defined. It sounds like you are talking about the rate at which energy is being stored in the capacitor, which is also known as the power being delivered to the capacitor (since the capacitor has no choice but to take the power delivered to it and store the resulting energy).

    Next, ask if the basic form of your formula makes sense. At t=0 there is no voltage across the capacitor, and hence there is no power delivered to it initially despite there being a current. The reason is basically that, initially, it takes no work to store the first bit of charge on the capacitor because there is no electric field that has to be worked against. So you expect a function that is going to start out zero, build to some maximum, and then decay back to zero.

    Two equivalent ways to get this are to differentiate the energy stored on the capacitor as a function of time or to multiply the instantaneous voltage across and the instantaneous current through the capacitor as a function of time. Both yield the same result.
     
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  10. kiroma

    Thread Starter Member

    Apr 30, 2014
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    Here's the proof
    Since always there's a capacitance, even if you have a wire
    And since always there's a resistance, even if you have a wire.

    You were telling me this is nonsense. How can you justify this? Is it luck? :p
     
  11. t_n_k

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    Mar 6, 2009
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    I believe all you've managed to "prove" (intentionally or otherwise) is the well-established condition that the energy delivered by the source is shared equally between (that lost as heat in) the resistor and (that stored as electric field in) the capacitor.
    Nothing particularly significant or novel in that.
     
    Last edited: Dec 25, 2014
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  12. kiroma

    Thread Starter Member

    Apr 30, 2014
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    Yes. But have you seen that you can set a period of time and get how much energy it has stored?

    By the way, I've made my calculus all over again and found that I was wrong.
    Yeah, that was the power lost in form of heat in the resistor.
    BUT, the result I found was the same, which is equal for the capacitor.
    The equation I found is V^2/R*(e^(-t/RC)-e^(-2t/RC)), which integrated from 0 to infinity gives the CV^2.
    This equation, yes, goes from 0, build some maximum, and goes back to zero. Sorry for the mistake.
    My suggestion keeps being the same, it's to show a new way to calculate the energy stored in the capacitor.
    Also, I calculated that the moment the capacitor is charging the fastest is at ln(2)RC. If you feel some interest, just tell me and I'll give all my calculations to help build this site.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    That's not a proof of anything related to what you are claiming. I can come up with an infinite number of integrands all of which integrate to that same result, would that mean that all of them represent the instantaneous power being delivered to the capacitor?

    And, yes, in your case it was pure luck. Your result relies on a coincidence of which you are completely unaware. If you determine the total energy delivered by the voltage source during the charging process you will discover that it is twice the total energy eventually stored on the capacitor. Since half of this ends up in the capacitor, that means that the other half must be dissipated as heat int he resistor. The end result is that the total energy dissipated in the resistor is equal to the total energy stored in the capacitor. So it is by that property that your integral works out to be what it is since your integrand is the power dissipated in the resistor and NOT the power delivered to the capacitor. Instead of blindly throwing equations at Wolfram Alpha and letting it do the thinking for you, you might consider what those equations represent in the first place (since Wolfram neither knows nor cares).

    Although the total energy in the resistor and the capacitor are the same, their time profiles are not. The power in the resistor starts high and monotonically decreases to zero, the power in the capacitor starts at zero, peaks at 1/4 of the peak power in the resistor at about 0.7RC (ln(2)RC, to be exact), and then decays to zero.
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    No, you can't. The only period of time for which the two are equal is t=∞. The bulk of the power dissipated in the resistor happens early on while the bulk of the energy stored in the capacitor happens much later. Plot the two and you will see.

    Good, you are learning.

    I hope you mean CV^2/2.

    And your suggestion keeps being wrong. Your "new way" is neither new nor is it a way to calculate anything other than the total energy stored in the capacitor. It also relies on a coincidence that works for a simple series RC circuit charged by a voltage source. It has little general application -- for instance, try applying it to a parallel RC circuit charged with a current source. By inspection you should be able to see that it doesn't work since the total energy dissipated in the resistor goes to infinity.

    Correct. You are definitely making progress.

    I vote that we pass. But keep trying. Achievement is almost always the result of lots of missteps along the way.
     
  15. kiroma

    Thread Starter Member

    Apr 30, 2014
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    I was talking about the energy being stored in the capacitor, not by the source. I know about that's half to the resistor and the other half to the capacitor, and my calculus does make this sense all the time.
    I think you're calling me an idiot. I don't throw blindly equations at Wolfram Alpha, I think first, and then use as input to see the graph. It's just a form to see complex equations working instead of making a hand-graph.
     
  16. kiroma

    Thread Starter Member

    Apr 30, 2014
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    Oh, yeah, sorry, It's CV²/2.

    Ok, I'll try to prove you I'm right with my equation.
    Let's take for random R=1, C=1, Ef=1 and t=1.
    It means that the RC constant is 1 and I'm going to calculate the energy stored from t=0 to t=1.
    Pc=V^2/R*(e^(-t/RC)-e^(-2t/RC)), where V is the same as Ef.
    Integrating Pc dt, from 0 to 1 (from the initial time to the time constant 1), you'll have 0,199788 Joules
    (the integral of the power is the work)
    This was the result given by my formula.

    Let's now use the CV²/2 equation.
    C=1 and V? Well, we have to discover by the equation of the charging capacitor Ec=Ef-Ef*e^(-t/RC)
    But I know you know that given the time constant RC=1, which is equal to the time we're looking for, you know the voltage in the capacitor will be ~63% of the source voltage.
    So, put this in the equation and you'll have 0.199788 Joules again.

    Am I this lucky or am I right with my "new" equation? hehe ^^
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    But now you are using the correct equation (the one that results from either approach I recommended in Post #9) and not your "new" equation.
     
  18. kiroma

    Thread Starter Member

    Apr 30, 2014
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    Ok, it's not mine, it's not of anyone...
    You just make me realise I was a bit wrong in my calculations. Thank you very much.
     
  19. WBahn

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    Mar 31, 2012
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    You're welcome.

    On a note that hopefully will be of value to you, saying things like R=1 or RC=1 are fundamentally wrong. If I were to tell you that the height of a certain athlete was 90, would I be more likely talking about a jockey or a basketball player? The simple fact is that 90 is not a height, it is a dimensionless number whereas height MUST have units of length to have any meaning. Similarly, resistance MUST have units of resistance and time constants MUST have units of time.
     
  20. kiroma

    Thread Starter Member

    Apr 30, 2014
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    Oh, yeah
    You're right. Thanks again.

    I didn't find anything in "the book" related to what we were talking about
    Could this subject go to the book?
     
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