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Capacitors paradox

  1. Apr 25, 2013 #1
    if two equal capacitor one of then is charged and the other is not.then the switch is closed, half of the stored energy of the 1st capacitor will be lost.

    if there is a way to move electrons from one of the positive plates to the other positive of the other capacitor,will it require less energy than half energy stored in the beginning ?
     

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  3. Apr 25, 2013 #2

    mfb

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    Probably not like that, but there are ways to do the transfer more efficiently.
     
  4. Apr 25, 2013 #3
    with 2 identical capacitors as you have described you will always find that 1/2 of the energy is 'lost'.
    There will be lots of discussion about how it is lost but you cannot escape it. It is in the mathematics.
    You cannot make it 'more efficient'
     
  5. Apr 25, 2013 #4

    mfb

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    With a different setup, it is possible.
    If you just connect the two capacitors, you have to waste 50%, but that is not the only option.
     
  6. Apr 25, 2013 #5
    one capacitor: energy stored = 0.5QV
    2 identical capacitors in parallel Charge still = Q; V now V/2: energy stored = 0.5Q x 0.5V !!!!!
    what options are there?
     
  7. Apr 25, 2013 #6
    If the cap originally charged is large, then the switch closes allowing it to discharge into a smaller cap, the efficiency > 50%. I had it computed but I don't know where I put it. When I find it I'll post it. In other words, if the 1st cap is 1.0uf, and the 2nd cap is also 1.0uf, then the transfer efficiency is 1/(1+1)=0.50. But if the 1st cap is 2.2uf, 2nd cap is 1.0uf, efficiency is 2.2(1+2.2)= 0.688. If 1st cap is 4.7uf, 2nd cap is 1.0uf, efficiency is 4.7/(1+4.7)=0.825. Did I help?

    Claude
     
  8. Apr 25, 2013 #7
    the original post specified 2 equal capacitors and you have correctly calculated 50%
    It is a very straightforward capacitor calculation !!!!! turns up time and time again.
     
  9. Apr 25, 2013 #8
    effectively you are calculating the voltage across the parallel combination of these capacitors as a fraction of the original voltage across the single capacitor.
    One fact is undeniable: energy is lost when capacitors are connected together.
    It is 'over egging' it to call it 'transfer efficiency'. I have never met this term in the analysis of capacitor circuits
     
  10. Apr 25, 2013 #9
    Well I just introduced you to this term. BTW, what is "over egging"? I have never met this term in the analysis of cap circuits, nor elsewhere. Thanks in advance.

    Claude
     
  11. Apr 25, 2013 #10
    putting more eggs in a cake than are recommended in the recipe
     
  12. Apr 25, 2013 #11

    mfb

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    You can increase Q.

    I am sure there are better setups, but in doubt, use a DC-DC-converter. This will increase the charging efficiency, as charging current in the (initially) low-voltage capacitor starts larger than discharge current in the (initially) high-voltage capacitor.
    A coil and a quick switch could be even better.
     
    Last edited: Apr 25, 2013
  13. Apr 25, 2013 #12
    You cannot increase the charge (Q) if all you have is 2capacitors.
    What do you mean by 'better set ups'?
    What do you mean by ' increase the charging efficiency'
    What you have written is nonsense.
     
  14. Apr 25, 2013 #13

    mfb

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    I never questioned that.
    Setups with additional components.
    See my previous posts.
    It is not.
     
  15. Apr 25, 2013 #14
    'Additional components'. ,!!??
    Nonsense
    Where is the physics explanation?
     
  16. Apr 25, 2013 #15
    How?
     
  17. Apr 25, 2013 #16
    as you can see in the attached picture.how can you calculate the energy required to move half the electrons from the 1st positive plate to the 2nd positive plate,if we will assume the distance between the two positive plates is d and the starting voltage on the 1st capacitor is V

    attachment.php?attachmentid=58217&stc=1&d=1366932380.jpg
     

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  18. Apr 25, 2013 #17

    mfb

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    Read my posts, please, before you call them "Nonsense".
    I proposed two methods, a DC-DC-converter and a coil with a switch.

    @TAAREK: Multiply voltage difference with the charge you want to transfer.
     
  19. Apr 25, 2013 #18
  20. Apr 25, 2013 #19

    The Electrician

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  21. Apr 26, 2013 #20
    Electrician is correct. This question is standard text book physics.
    It is easy to calculate the energy 'lost' when capacitors are connected together. The references discussing circuits with inductors do not change the physics.
     
  22. Apr 26, 2013 #21
    energy to move charge Q from the positive plate to the 2nd positive plate = Q*V
    but if V is less that he voltage which was applied on the 1st capacitor then the power dissipated will be less than the half.

    so what is the potential defrance between the two positive plates ?
    the applied voltage on the 2nd plate equals the 1st plate
     
  23. Apr 26, 2013 #22
    It seems there are two discussions here - 1) The OP was about the Paradox, and 2) How to make this more efficient.

    As for the paradox it appears the energy is irradiated - in an ideal system, the current flow, would be extremely large - you would hear the "click" on a radio for example. An apparent mathematical proof is here : http://puhep1.princeton.edu/~kirkmcd/examples/twocaps.pdf

    The fact this was still being done in 2002 - seems remarkable to me.

    As for the 2nd "discussion" -- what is the actual question you are trying to settle?
     
  24. Apr 26, 2013 #23
    It seems there are two discussions here - 1) The OP was about the Paradox, and 2) How to make this more efficient.

    As for the paradox it appears the energy is irradiated - in an ideal system, the current flow, would be extremely large - you would hear the "click" on a radio for example. An apparent mathematical proof is here : http://puhep1.princeton.edu/~kirkmcd...es/twocaps.pdf [Broken]


    You are correct. This is standard A level text book stuff. If the 2 capacitors are equal then 1/2 the energy is 'lost' and in post #6 cabraham effectively gave more examples of energy lost for a range of different capacitors connected in parallel.
    The energy is lost by any resistance in the circuit, any sparking at the switch and ultimately by electromagnetic radiation caused by the changing current.
     
    Last edited by a moderator: May 6, 2017
  25. Apr 26, 2013 #24
    Maybe you simply misunderstood mfb? He was talking about an LC circuit with two capacitors in it. You do know how oscillations in an LC circuit work, do you? The energy is being transferred from C to L and then back again. Depending on the circumstances the energy lost in each cycle can be quite small. There is no law of physics that would stop you from transferring nearly 100% of the energy contained in a capacitor into another one.
     
  26. Apr 26, 2013 #25
    This thread is about the situation described in post one where two equal capacitors, one charged and one uncharged, are connected by a switch. In this case half of the charge will flow from the charged capacitor and to the second capacitor. Since energy stored equals Q squared/2C the final energy stored will be half of the initial energy stored.
    There is no paradox since the "lost" energy can be accounted for in terms of heating of the connecting wires, sparking and radiative losses.
     
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