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Lost energy when capacitor discharges

  1. Feb 15, 2015 #1
    1. The problem statement, all variables and given/known data

    I have two capacitors: 1-Farad capacitor A, and 9-Farad capacitor B.
    In state (a), I charged A with 1 Coulomb of charges, as depicted in figure (a).
    In state (b), I paralleled A with B, by doing this I discharged A, as depicted in figure (b).

    I try to find the total energy of state (a) and state (b). From what I have calculated, the system lost 90% of its energy when it goes from (a) to (b). Where does the energy go?

    2. Relevant equations

    [itex]Q=C\phi[/itex]
    [itex]U=(1/2)C\phi^2[/itex]

    3. The attempt at a solution

    Please refer to the image below.
    IMG_0050 copy.jpg
     
    Last edited: Feb 15, 2015
  2. jcsd
  3. Feb 15, 2015 #2

    mfb

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    Staff: Mentor

    You heat the wires and the capacitors from the resistances in the setup. A tiny bit is also lost to electromagnetic radiation.
     
  4. Feb 15, 2015 #3
    I think my approach to find energy assumes there is no resistance on wires. So I suspect that energy would not turn to heat on these wires.

    What I see during the discharge process is a flow of electrons. Can you explain how electromagnetic radiation is produced?
     
  5. Feb 15, 2015 #4

    mfb

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    Staff: Mentor

    Your approach is independent of the resistance.
    Connecting capacitors like that without any resistors is a very special case, it will give you an extremely good oscillator (as you always have a small inductance) and all the lost energy will be emitted as radiation over time.
    You have a variable current flow = accelerated charges. With small resistances, current will oscillate back and forth for a while.
     
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