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One capacitor charging another capacitor

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data
    You have one capacitor C1 (1uF) charged to 10V.
    Now the capacitor is switched to charge another capacitor C2 (0.25uF),
    Whats the voltage on C2?
    Everything's ideal. No loss caps, no wire resistance.


    2. Relevant equations



    3. The attempt at a solution
    I tried,
    Q1=C1V1
    Q2=C2V2

    total charge will be constant, so Qtotal, Qt=CtV
    Both caps are in series, so Qt= (C1+C2)V

    (C1+C2)V = C1V1+C2V2

    V2 is 0v

    V = C1V1/(C1+C2)

    So, V=8V.

    But when I do E = 0.5*C*V^2 for C1. (80uJ)
    and then for both the caps(32uJ and 8uJ), the numbers don't add up.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 10, 2011 #2

    gneill

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    Was there a question to go with this?
     
  4. Nov 10, 2011 #3
    Sorry, the question is to find the voltage at capacitor C2.
     
  5. Nov 10, 2011 #4

    gneill

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    It looks like you found it when you calculated the 8V.

    You could also have found it by interpreting the capacitors as being in parallel, and placing the total charge on the equivalent capacitor C1+C2.

    Don't fret about the loss of energy between before and after; When the charges were redistributed they had to move. Anytime something moves or changes spontaneously in nature (that is without some external energy source to drive it) it's because the system is seeking a lower energy state and one is available to be filled. If the components were truly ideal, with zero resistance, then the charge would continue sloshing back and forth between the capacitors (even a perfectly straight, 1cm length of wire has some inductance!). In real life the oscillations would eventually bleed away the energy via electromagnetic radiation and the system would settle down to the quiescent steady state.
     
  6. Nov 11, 2011 #5

    rude man

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    But when I do E = 0.5*C*V^2 for C1. (80uJ)
    and then for both the caps(32uJ and 8uJ), the numbers don't add up.

    Your capacitor leads have finite resistance. If you connect C1 to C2 with a resistor, no matter how small, you will find by the usual analysis that the missing energy is wholly disspiated by that resistor.
     
  7. Nov 11, 2011 #6

    NascentOxygen

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    And in the dielectric losses. (unless the dielectric is vacuum)
     
  8. Nov 12, 2011 #7
    I know the losses- ESR, dielectric loss, wire resistance.
    The Q=CV equation does not contain a loss term, but the result contains loss of energy. How is this possible?
    Let's assume ideal capacitor, idea wire (zero resistance).
     
  9. Nov 12, 2011 #8

    rude man

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    You have to accept that there is SOME resistance. It can be as small as you like, but not zero. ∫i^2*Rdt from 0 to ∞ will always be the difference between energies before & after connection is made, for ANY value of R.

    This is a well-known conundrum. It is always dangerous to deal with singularities. So just accept that R cannot be absolute zero.
     
  10. Nov 12, 2011 #9

    cmb

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    I would say that the reason you are getting caught in a quandry over how to answer this is because it cannot be answered. The question is, itself, 'wrong'.

    This, as posed, would lead to the condition well described in #4 above:

    That is to say, there would be no static value of voltage on C2 that could be defined. If everything was perfect and zero resistance with no inductance, the frequency of VACC2 would be infinite!

    Think of it as a Newton's cradle with two balls of different masses. The little one is stationary and the big one is dropped. There are no losses. How high is the smaller ball? You see, it is a time-dependent thing and the energy just swings one into the other, then back again.

    To answer the question, you must pragmatically include an assumption about energy losses, else the answers you get are meaninless. If you like, you could work out the max and min volts on each capacitor during you theoretically perfect oscillation, because those would be finite values. I'll start you off; the max volts on C1 is 10V, and the min on C2 is 0V. ;)
     
    Last edited: Nov 12, 2011
  11. Nov 12, 2011 #10

    NascentOxygen

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    The smaller the resistance, the greater the current surge, therefore the higher the i2R losses. With no wire resistance, the L-C ringing and associated radiated electromagnetic energy (radio waves) would be the principal energy loss from the system.
     
    Last edited: Nov 12, 2011
  12. Nov 12, 2011 #11

    NascentOxygen

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    If there was no inductance, then what mechanism would cause the charge redistribution to overshoot?
     
  13. Nov 12, 2011 #12

    rude man

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    Incorrect. The smaller the resistance, the greater the current surge, but the shorter the duration of the surge. So the integral of i^2*R is the same over t=0 to ∞.

    Work the math and forget radio waves.
     
  14. Nov 12, 2011 #13

    NascentOxygen

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    We can't simply "forget" EM radiation when in the limiting case it is the only existing loss mechanism.
     
  15. Nov 13, 2011 #14

    cmb

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    Bear in mind what I said about this being an unrealistic question in the first place.

    But we should still be able to theorise about idealised systems, and to address your question you should think again on the Newton's cradle example. You are asking why the falling ball does not instantly stop, but its energy 'overshoots' into the other smaller ball. (the analogy is - ball height is voltage).

    There is no 'overshoot', there is merely the bouncing of energy back and forth. There are two competing, opposed, osillatory drivers (as there usually are with oscillators!): You have the capacitors trying to even out the charge such that the smaller capacitor takes proportionately more voltage, whilst you have the system trying to even out the voltage.

    Nothing complicated in this, just think about the two germane elementary rules in electronics; the potential across serial capacitors will distribute itself proportionately to the capacitance, yet a differential voltage will cause a current.
     
  16. Nov 13, 2011 #15

    cmb

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    Yes we can. It is a theoretical device, and it is infinitely small. No radiating elements.
     
  17. Nov 13, 2011 #16

    NascentOxygen

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    If there is no overshoot, there is no oscillation--there can be no "bouncing back and forth".
     
  18. Nov 13, 2011 #17

    cmb

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    I am not describing inductive overshoot, per se.

    As I stated, this is not a real-world example for which there would always be some inductance. An infinite current (I) and a zero inductance (L)... what magnetic energy does the inductance formula [0.5xLxI^2] give you with those two values?

    Don't stress over it... the question is answered. It is either unrealistic question, or you make an assumption of some teeny amount of ESR so it damps promptly.
     
  19. Nov 15, 2011 #18

    NascentOxygen

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    Then what back and forth phenomenon are you talking about?
    If there is no inductance, there can be no stored magnetic energy. That's not too complicated, is it?
    Damps? :confused: What exactly do you imagine we need to damp? With no inductance, it is not a second order system, there is oscillation, so there is nothing to dampen.
     
  20. Nov 15, 2011 #19

    rude man

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    Good point! No antenna! Hertz would have understood.
     
  21. Nov 15, 2011 #20

    cmb

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    It is complicated by the fact that it is an impossible question. You say there is no magnetic energy because there is no inductance. But there is an infinite amount of current. In fact, there is an infinite amount of current squared!! I say my infinite current [squared] trumps your zero inductance! There WOULD BE magnetic energy there, in this dream-world fantasy!!!

    This is quite a simple problem to sort out, even if you do not have all the info for the losses. All you need to recognise is that there will be energy losses. Then you treat the two capacitors as parallel (hence you can deduce a sum capacitance, C1+C2). Then you calculate the amount of charge present before the switch was thrown. The charge cannot be lost, as the + and - terminals of each capacitor, so to speak, are mutually isolated. So we then know charge, and total capacitance - from whence the voltage is defined!

    Interestingly, you could stick a megohm or a milliohm resistance in the way, a Henry's worth of inductance, with a big Q or a little Q, or none at all, but the energy lost would be the same. Intuitively a bit wierd, but there it is!! The charge and capacitances are immutable values, so the final energy must also be immutable.

    (It is so counter-intuitive, I am not even sure of this!! Can anyone spot any flaws?)
     
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