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Energy in a closed circuit

  1. Aug 27, 2011 #1
    Hi everyone,

    I was studying eletrodynamics and I've found the the following question


    In ideal circuit of the figure, initially opened, the capacitor of capacitance CX is initially loaded and stores a eletric potencial energy E. The capacitor of capacitance CY = 2CX is initially uncharged. After closing the circuit and get it to reach a new equilibrium, you can say that the sum of the energies stored in the 2 capacitors is :


    a) 0
    b) E/9
    c) E/3
    d) 4E/9
    e) E

    [PLAIN]http://img571.imageshack.us/img571/1196/imagemdnf.jpg [Broken]

    The resolution is like that:

    Initially the charge stored in capacitor Cx is Q, and the energy is E = Q²/2Cx

    When the circuitis closed, after reached the new equilibrum, the tension U in Cx and Cy are the same:
    Ux = Uy -> Qx/Cx = Qy/Cy -> Qy = 2Qx

    For the conservation of charges:
    Qx + Qy = Q

    We get Qx = Q/3 and Qy = 2Q/3
    So the energy stored after reached the equilibrum is:
    Ex + Ey = (Q/3)²/2Cx + (2Q/3)²/4Cx = Q²/6Cx = E/3

    My question is:
    For the law of energy conservation, the energy can't be created or destroyed, only transformed.
    If we consider the resistor R hasn't done anything (we can also take it out that the result'd be the same). Where's the other 2/3 of E?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 27, 2011 #2

    lewando

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    Why do you think the resistor hasn't done anything?
     
  4. Aug 27, 2011 #3
    I assume 2 reasons:

    First we don't use it in our resolution (so it makes no sense for it to be there)
    Second if the resistor is taken out of the circuit, wouldn't the answer be the same?
     
  5. Aug 27, 2011 #4

    vela

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    You have to consider what happens between the time t=0 and when the circuit finally reaches equilibrium. When the charge moves from the first capacitor to the second, there's a current, and a current through the resistor will dissipate energy.

    Of course, this doesn't explain your second question. In a practical sense, you can say that every wire will have some non-zero resistance, so saying R=0 is simply unrealistic. There are also other effects, which you can usually neglect in basic circuits; however, when R becomes very small, these effects become important. For example, the circuit forms a loop, which has an inductance, which will limit how the current flows and will cause the circuit to oscillate. (The inductance is always there, so you might wonder why the circuit doesn't always oscillate. It's because when R is relatively large, the oscillator is overdamped.) The important thing is when the charges move from one capacitor to the other, they have to accelerate, and accelerating charges emit electromagnetic waves, which carry off energy. That's the primary mechanism the system will use to get rid of the energy in cases where R becomes negligible.
     
    Last edited: Aug 27, 2011
  6. Aug 27, 2011 #5

    lewando

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    Sorry, not sure I understand "resolution". Whether or not it makes any sense for the resistor to be there is irrelevant-- it is part of the circuit in the given problem.

    No, you would get 2 different answers.
     
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