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Charging two parallel capacitors one of them is %50 charged

  1. May 15, 2012 #1
    hi
    i need an equation which can describe the charging time of two parallel capacitors one of them is %50 charged


    and if i have a capacitor which is always charged from 50% to 100% and never fully discharged will it have a different impedance than the fully charged/discharged capacitor?
     
  2. jcsd
  3. May 15, 2012 #2

    sophiecentaur

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    The charging time will depend upon the resistance (and any stray inductance) involved.
    The 'Impedance' is not really a relevant factor when you are talking about charge / discharge - Impedance describes the effect of a circuit on alternating current signals. 'time constants' are more the relevant quantity in your problem.
     
  4. May 15, 2012 #3

    phinds

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    Having two caps in parallel with one of them 50% charged and the other not charged at all is a really good trick. How do you propose to accomplish it?
     
  5. May 15, 2012 #4

    sophiecentaur

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    Good point. I was assuming that he'd switch one in and out but perhaps not.
     
  6. May 15, 2012 #5
    yes it can be done by switching and assume that this capacitors are ideal and they are connected to a resistance R
     
  7. May 15, 2012 #6

    sophiecentaur

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    The time constant would probably be RC, where the C value is the two capacitors connected in series.
     
  8. May 15, 2012 #7

    dlgoff

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  9. May 15, 2012 #8
    "in serise"!! why they are connected in parallel as one capacitor
     
  10. May 15, 2012 #9

    sophiecentaur

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    But one must be discharging into the other one? Perhaps you need to draw us a picture with the details. It's not clear.
     
  11. May 15, 2012 #10
    What does it mean for a capacitor to be half charged?
     
  12. May 15, 2012 #11

    sophiecentaur

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    More optimistic than half empty? :tongue2:
     
  13. May 15, 2012 #12

    psparky

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    Hmmmm....seems to me "that the voltage across is the same".

    Since you are saying 50% charged I assume you are talking DC?

    If talking DC....the capacitors should act independently in regards to charging. The non charged capacitor doesn't care what the 1/2 charged one is doing or visa versa.

    Without some type of "resistance" value or "R"......you cannot calculate an RC constant.
    If you are talking ideal....then your e^-((t/(RC))....if R=0....then should be instantaneous.

    If you are talking real rather than ideal....you wires and caps have resistance built in.
    But still you need to take them individually. Maybe take an ohmeter to the non charged caps and circuit without power supply. Easier if you just go ideal and insert your own resistor.

    In regards to the cap you are interested in.......to find the RC constant.....if the resistor is in series with the power source......simply short the voltage source (on paper)....open the cap you dont want....and walla....RC constant is found. Or just go the easier route on that....you know. (Edit: prob not correct here.....you have to find the total resistance first to get the correct voltage drop across the resitor........then expand back out. This will then give the correct voltage across your parallel connection for your equation)

    If you put the resistors in series with each cap......That RC circuit is much easier. Just take the RC from the branch you are interested in.

    I believe their parallel nature makes them act indepedent of eachother if the resistor is NOT in series with the power source. Now if you start discharging one into the other or whatever, then yes the system must be combined as a whole.

    It's the "voltage across is the same" is what is important to remember.

    Ok.....I just fried my brain. I hope what most of what I wrote is correct.

    If not....Sophie will school me shortly as per usual......which I have learned to enjoy.
    And if nothing else, hopefully I have spurred further conversation.
     
    Last edited: May 15, 2012
  14. May 15, 2012 #13

    sophiecentaur

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    HAHA again. Just because you're paranoid doesn't actually mean I'm not out to get you.

    I don't think we can decide on anything definite until the problem has been specified better. At the mo, I think we all have different pictures in our minds.
     
  15. May 15, 2012 #14

    psparky

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    I have concluded you are much smarter, experienced and knowledgeable than me.....

    Therefore, like it or not....you are now my Mentor.

    Congratulations....lol.

    Ya.... a more definite circuit would be better....but I threw out some points to make everyone think....especially myself!
     
  16. May 15, 2012 #15

    sophiecentaur

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    Older and uglier, you mean!
     
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