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Work done by load in charging a capacitor

  1. Apr 26, 2012 #1
    The energy stored on a charged capacitor is 0.5*Q*V or 0.5*C*V2
    BUT
    what is the electrical work done by the supply as it transfers the charge to the capacitor?

    I have heard two answers:
    W=QV and W=0.5*QV
    the first answer implies, i think, that energy is required to move charges against the already present charges on the capacitor plate. is this correct?
     
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  3. Apr 26, 2012 #2

    mfb

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    That depends a bit on your setup. If you connect both without any resistance in between, you get a short circuit and waste a lot of energy in cables/power supply/whatever.

    In general, if your power supply has the same voltage at all time, you need W=QV and waste 50% of this energy somewhere.
     
  4. Apr 26, 2012 #3
    What do you mean by 'waste energy somewhere'
    Is it exactly 50% that gets wasted every time
     
  5. Apr 27, 2012 #4

    mfb

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    The power supply supplies a charge of Q, with a voltage of V (unless it is a bad power supply or has a variable voltage or something else), which means that you take an energy of Q*V from it.
    The capacitor stores an energy of 1/2 Q*V.

    Both are simple equations, and if you divide both you get the result that only 50% of the energy is stored in the capacitor. You can use the other 50% for something else (heat a resistor, light a room, charge some other energy storage or whatever), but they won't get into your capacitor in this charging process.
     
  6. Apr 27, 2012 #5

    sophiecentaur

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    He wasn't being unspecific about the 50% loss. He was implying that the energy is lost "somewhere" and that can be in the connecting wires, the power supply internal resistance- in addition to the explicit resistance that is introduced.
    This may not apply where there is an Inductor involved in the circuit.
     
  7. Apr 28, 2012 #6
    OK, I have seen how the energy stored on a capacitor is 0.5QV because the graph of Q against V is a straight line and the energy is the area under the graph. (this is like the equation for energy stored in a stretched spring = 0.5Fx)
    Also I know that the charge coming from the battery is always at the voltage V and so the energy from the battery is QV. So 0.5QV is missing.
    I don't think the battery internal resistance comes into this because the V is at the terminals of the battery. If energy is missing it must appear as another kind of energy somewhere . What kind of energy is the missing energy?
     
  8. Apr 28, 2012 #7
    This is an old problem folks but it would appear the answer is not well known !

    Consider an idealised situation of a 'stiff' voltage source (a power supply which maintains voltage whatever the current drawn) connected via wires with no resistance to two parallel metal plates with a slab of dielectric between them.

    A charge Q is transferred to one plate and -Q to the other plate.

    But where is the energy? Remove dielectric slab and put it into identical set of plates connected not to a power supply but to a voltmeter. The voltmeter will resister a voltage and if replaced by say a lamp the lamp will glow.

    The energy you are missing was stored in the polarisation of the molecules in the dielectric.

    Regards

    Sam
     
  9. Apr 28, 2012 #8
    "The energy you are missing was stored in the polarisation of the molecules in the dielectric"
    Surely that cant be right !!!Isnt the dieelctric part of the operation of the capacitor, just like the plates? Isnt that where the 0.5QV is stored?
    And it is possible to have no dielectric in a capacitor, just 2 metal plates.
     
  10. Apr 28, 2012 #9
    In the original capacitor disconect the power supply and text the voltage between them - it will suprise you !
     
  11. Apr 28, 2012 #10
    we have done measurements on capacitors and the voltage is equal to the battery voltage when it is disconnected. We measured the voltage falling in an exam practical to measure time constant.
    It is the energy explanation that sounds wrong. I cannot see anything in my text book along those lines.
     
  12. Apr 28, 2012 #11

    sophiecentaur

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    Surely it would be better to discuss what happens in a vacuum capacitor first before getting bogged down with what happens inside a dielectric when it is introduced. The Energy Situation is the same and there is one less factor to consider.

    The fact is that supplying energy to a Capacitor, in a practical situation, always involves series resistance. If you really want to ignore any series resistance, you still can't ignore the inductance of connecting leads. If the series inductance dominates then you just get an oscillation, in which the energy is shared (over time) between the L and the C.
    One needs always to avoid the 'irresistable force and immoveable object' type of scenario.
     
  13. Apr 28, 2012 #12
    There is something wrong here....when the caoacitor is charged ther is no current so I do not see what part inductance would play
     
  14. Apr 28, 2012 #13

    sophiecentaur

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    Not wrong. Whilst charging (through the conductor) , energy is stored in the magnetic field associated with the Inductor. As the Capacitor volts increase, the current decreases, and the voltage on the inductor changes (change producing volts). Eventually, the volts at the capacitor get higher than battery volts and current is zero. Then the capacitor discharges through the L and volts drop. This process goes on for ever or until the resistive elements have dissipated the 50% of energy supplied.
     
  15. Apr 28, 2012 #14

    mfb

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    This voltage is the voltage at zero load - while you are charging the capacitor, you get a current and therefore a lower voltage in a realistic power supply.


    @sambristol: Your post explains how the 1/2 QV in the capacitor are stored, but that is the part which is not discussed here.
     
  16. Apr 28, 2012 #15
    There is an awful ot of confusing and misleading physics in these responses.
    There has been no mention of the fact that a changing electric current produces electromagnetic radiation and this amounts to an energy 'loss'.
    When the connection to the battery is made a current flows to charge the capacitor. This current will produce heat energy in any resistance and electromagnetic radiation from the wires. If the resistance is small then the current will be large and the radiation could be the major means of energy loss. If the resistance is large the current will be small and resistive heating could be the major energy loss mechanism.
    It is also possible that a spark is produced when the switch is closeds which is also an energy loss mechanism.
    The same processes also apply when charged capacitors are connected in parallel to uncharged capacitors.
     
  17. Apr 28, 2012 #16

    sophiecentaur

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    "Confusing"? I should say, rather, that we're probably being a bit too simplistic.
    Your EM idea is very fair comment and, as we try to get more and more 'ideal' in our thought experiment, it must be the limiting factor for energy loss, as it can't be eliminated, once you've chosen the dimensions of the set up. (Unless you do it all in a silver lined box, of course)

    Wouldn't the radiation resistance of a small structure be pretty low (arm waving values) compared with the likely internal resistance of power supply and wires?
    Would you really expect an arc when a switch closes? Breaking a circuit when current is flowing through an inductor would certainly be an issue but I think you are going another step deeper if you want to introduce sparks here.

    We'd have to be a bit more specific about the actual setup before taking this much further, I think.
     
  18. Apr 28, 2012 #17
    I typed 'energy loss charging capacitors' into Google and got a reference to physicsforums :Nov4-06 by jumanicus who also states energy loss by EM radiation....it has already been discussed here!!!!
    I hope (I am not certain how to do it) that I am attaching an anaysis of EM radiation loss in capacitors.
    In my humble opinion this is the FIRST loss mechanism that should be discussed when dealing with this topic....it is always there. It is a pure physics, fundamental explanation that does not rely on resistance, inductance, dieelectric constant etc.
    How to get sparks....Touch a charged capacitor onto the terminals of an uncharged capacitor (almost zero resistance)
    PS: I am a physics teacher and passionate that students get sensible explanations to their questions.
     

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  19. Apr 28, 2012 #18

    mfb

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    And this is reasonable, as the energy released EM radiation is really small, unless you add some fancy oscillators to the setup, charge it with another capacitor and superconducting wires or do something else unusual I don't think of at the moment.

    Resistance is always there, too, unless you use superconductors everywhere.

    Right.
     
  20. Apr 28, 2012 #19
    Would any one agree that the answer to the original post:
    "the first answer implies, i think, that energy is required to move charges against the already present charges on the capacitor plate. is this correct?"
    is incorrect and a good answer would be :
    The energy lost is due to
    a) Heat produced in any resistance
    b) electromagnetic radiation from the connecting wires
    c) sparks (if there are any)
     
  21. Apr 28, 2012 #20

    sophiecentaur

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    It's certainly always there but it can hardly be thought of as the major loss in most 'real' circumstances. In many ways, it can be looked upon as a bit of a distraction for a student who is probably struggling with the basics of circuits and Maths. It appears as a resistance in series with the Ohmic resistances in the circuit and can be included in the little rectangle that's always drawn in the 'charging a capacitor' circuit.
    'Sensible' explanations don't usually bring in yet another can of worms for a poor student to deal with. A level doesn't actually deal with launching em waves into space. What level do you teach?
     
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