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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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  2. jcsd
  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
  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.
     
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