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Energy shared by capacitors

  1. Aug 5, 2009 #1
    hi, this is a basic conceptual question.
    Assume two identical capacitors of capacitance C without any leakage. potential difference across one capacitor V volts and the other capacitor is 0 volts. now the energy stored on the first cap is 1/2*C*V*V and there is no energy stored on other capacitor. Now if i connect these capacitors in parallel the overall cap is 2C and since there is no way for charge to escape the voltage across the parallel combination will be V/2. now the energy stored in the entire network is 1/2*(2C)*(V/2)^2=1/4*C*V*V. I couldn't figure there the pther half of the energy is gone.

    please help me understand where i am going wrong in understanding the concept here.
    thank you.
     
  2. jcsd
  3. Aug 5, 2009 #2

    berkeman

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    Staff: Mentor

    I guess I need to make this a stickie FAQ. Could somebody please search up some links to the solution to this problem, and I'll take care of the stickie. It's late, and I'm out for tonight. Thanks.
     
  4. Aug 5, 2009 #3
    Hello asifshaik,what do you see happening in the connecting wires when the capacitors are joined together?

    (Can someone please tell me what is meant by "sticky" and by "sticky FAQ".Yes,I am dopey.)
     
  5. Aug 5, 2009 #4
  6. Aug 5, 2009 #5

    berkeman

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    Staff: Mentor

    A stickie is a thread that get's pinned or stuck at the top of a forum, and doesn't get pushed off the bottom of the first page by newer posts. Look at the top of the EE forum to see some stickied threads.

    A FAQ is a frequently asked question.

    This is a handy website for decoding common acronyms, BTW:

    www.acronymfinder.com

    .
     
  7. Aug 5, 2009 #6
    Thank you berkeman.
     
  8. Aug 7, 2009 #7
    Hello asifshaik,what do you see happening in the connecting wires when the capacitors are joined together?


    I never tried it to see what happens when i physically connect these capacitors, its just a theoretical question popped up in some discussion. I came to some conclusions but wanted to confirm them.
    thats why i posted this question.

    thanks to TurtleMeister for the link. it was very useful.
     
  9. Aug 7, 2009 #8
    Hello again asifshaik,my question wasn't meant to be taken literally,it was meant to get you thinking about the problem in a particular way.My fault,I should have phrased it more carefully.Anyway, when the capacitors are joined together there is a momentary current flow as charge flows from one set of plates to the other through the connecting wires.This results in electrical heating losses.There can also be losses due to any spark that may be produced and any electromagnetic wave pulse that may be generated.Most of the energy is lost as heat in the connecting wires.
     
  10. Aug 7, 2009 #9
    I never tried it to see what happens when i physically connect these capacitors, its just a theoretical question popped up in some discussion. I came to some conclusions but wanted to confirm them.
    thats why i posted this question.

    thanks to TurtleMeister for the link. it was very useful.

    Hello again asifshaik,my question wasn't meant to be taken literally,it was meant to get you thinking about the problem in a particular way.My fault,I should have phrased it more carefully.Anyway, when the capacitors are joined together there is a momentary current flow as charge flows from one set of plates to the other through the connecting wires.This results in electrical heating losses.There can also be losses due to any spark that may be produced and any electromagnetic wave pulse that may be generated.Most of the energy is lost as heat in the connecting wires.


    My way of analysis is that, when i assumed a resistance R between the two caps, and calculate the energy dissipated on resistor it comes to be CV^2/4 which is independent of R. Hence when I connect those capacitors I think a infinitely large current flows for zero time like a delta signal.


    *************************************************************************************************************
    When this delta current signal is squared and multiplied by R and integrated overtime as R->0 the limit if that integration results in (CV^2)/4. In other words an infinite amount of current going through a zero resistance may result in finite energy dissipation.

    But I was not happy with the conclusion because i didn't understand what this energy is getting converted into, till I have gone through this site.
    http://www.smpstech.com/charge.htm
    which i mentioned in this post by TurtleMeister.
    There I came to know that It will result in some form of heat or radiation like an arcing.
     
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