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I Capacitor energy

  1. May 30, 2016 #1
    All, there have been many threads regarding where the 1/2 factor comes in to the energy storage of a capacitor, but I am at a loss as to why the capacitor voltage can be equal to the battery voltage...if energy is 1/2, shouldn't voltage be 1/2? I feel like I'm missing something fundamental and obvious here. Thanks--
     
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  3. May 30, 2016 #2

    David Lewis

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    Forget about capacitance for a moment. If you attach two very long wires to a battery, and measure the voltage at the far end of the wires, you are just going to read the battery voltage, right? Well, a capacitor works the same way except, since the plates have a lot of area, and are close together, it may take some significant time for the voltage to reach its final value. But if you wait long enough, eventually the voltage on the capacitor will reach battery voltage.
     
  4. May 31, 2016 #3

    sophiecentaur

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    In order to charge a capacitor, you need an Energy source (battery etc). All sources of Electrical Power have an internal resistance which is in series with the output. Every Coulomb of charge that goes into the Capacitor will require Energy to get through the resistor. This amount of energy is the same as the energy that's stored, however low the series resistor happens to be; it will just happen quicker for a low resistor value. Whilst the Voltage across the Capacitor is less than V, current will keep flowing from battery to Capacitor. The value of the current flowing will reduce exponentially with time until equilibrium is reached. The final value must consequently end up the same as the battery volts. It's an exponential change, so it never 'really' gets there - as with heating and cooling curves etc.. But it's 'near as dammit' within a few RC 'time constants' (look it up).
     
  5. May 31, 2016 #4
    Thanks for the replies. How can voltage be equal on both, but the energy stored be only half? Let's say a 12 V battery charges a capacitor. Why is the available energy not 6 V now instead of 12? I can't reconcile the energy loss in terms of voltage...especially if voltage is a form of potential energy.
     
  6. May 31, 2016 #5

    anorlunda

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    Voltage is not energy. Power (which is rate of flow of energy) is voltage times current.

    Does that make it clearer?
     
  7. May 31, 2016 #6

    sophiecentaur

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    You cannot always expect to get an intuitive answer to this sort of question. There are a number of 'proofs' for it and they are based on Calculus or a simpler, verbal version of calculus. This link gives a fairly rigorous argument for you.
    One reason why you wouldn't expect to get the 'whole 6V' worth of energy is that it is easy to put in the first bit of charge into the Capacitor and it gets harder to increase the charge as the capacitor volts get higher. Likewise, the energy available from the first few charges you take from a fully charged Capacitor is high because the Voltage is the full charged value. When the capacitor is nearly discharges, the volts across it are low so the energy available for the remaining charges is much lower. (Voltage is Energy per unit charge). If all the charge (Q) were delivered at V volts, you would get QV but the average energy, over the range of Voltages as V decays to 0, is only half that amount. (=QV/2, or, alternatively CV2/2)
    This factor of 1/2 occurs all over the place, in Physics. Energy in a stretched spring is kx2/2 where k is the stiffness of the spring and x is the extension.
     
  8. May 31, 2016 #7
    Part of my problem is also trying to reconcile these graphs, which do make sense based on sophiecentaur's explanation that energy is required to place (especially) the last few charges on a capacitor. But in the end, both graphs show V and Q being equal for both the battery and capacitor. Then, if voltage is expressed as joules per coulomb, what has changed that accounts for the 1/2 factor? It seems that one would entirely miss this detail.
    capacitors-17-638.jpg
    Perhaps I need to think of the battery instead. Is the battery graph not the exact opposite of the capacitor graph simply because a battery has both charge and voltage being maintained (at least temporarily) by a sustained chemical reaction? I hope I am being clear.
     
  9. May 31, 2016 #8
    The battery indeed provides energy E=QV.
    Half of this energy provided by the battery is dissipated. Half end up in the capacitor.
    Imagine that you draw a diagonal for the top green triangle. One of the triangles is the energy dissipated, the other half is the one in the charged capacitor.
    There is nothing more to it.

    But be aware that the top graph does not represent all the energy stored in the battery. It is just the small amount used during charging the capacitor. The energy of the battery decreases a little after charging the capacitor but this does not reduces the voltage of the battery. It reduces the amount of charge it can provide after that. You can imagine the top graph extended to the right, up to a value of Q given by the battery's "capacity", usually given in Amperes hour.
    Charging the capacitor takes just a small "slice" of this long,long band.
     
  10. May 31, 2016 #9

    anorlunda

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    It sounds as if you are thinking of a battery as a type of capacitor. Put Q in and V increases.

    Most batteries get their voltage from chemical reactions. They store no Q at all. An ideal battery has constant V, the mechanism of where that V comes from is unspecified.
     
  11. May 31, 2016 #10
    Thanks to all. I'll just have to keep thinking about it and maybe something will click.
     
  12. May 31, 2016 #11

    cnh1995

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    Half of the energy supplied by the battery gets "lost" in the resistance of the charging path as heat energy(i2Rt ).
    For constant V and C, energy lost as heat is same for all values of R.
    Edit: I see Sophiecentaur has already said that in detail:-p.
     
    Last edited: May 31, 2016
  13. May 31, 2016 #12

    sophiecentaur

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    Try the analogy of comparing an electric motor and a spring. The electric motor is providing its designed power all the time but the spring is delivering less and less as it runs down. In well designed systems, clockwork motors (and supercapacitors used as energy reserves) are not allowed to run down completely but operate from 'full' to, perhaps 'half full'. By operating near the 'top end' they will be delivering much more constant power than if they were allowed to run down completely. This can be compensated for with suitable technology. I have an ancient (wind-up) Wall Clock that has a Fusee movement. The spring drives a tapered drum, with a light chain that operates on a varying diameter so that the tension is the same over the range of energy that the spring is allowed (26 turns of the key, so it happens). The clock stops running long before the spring is exhausted but whilst it is going, the actual power delivered to the movement is kept nearly constant, which provides more accurate timekeeping. Damned clever, those Victorians!
     
  14. May 31, 2016 #13

    David Lewis

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    The source impedance of a large battery or regulated supply may be too small to meaningfully affect practical circuit behavior. Likewise for connecting wires. If current through and resistance of the circuit is low enough then you can safely neglect the voltage drops and concommitant energy dissipation.
     
  15. May 31, 2016 #14

    sophiecentaur

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    Whatever the value, half of the energy is lost in it. Draw a circuit with a voltage source, a switch, a series resistor and a series capacitor. Then calculate the energy dissipated in the resistor when the switch is connected. It will be half the total energy supplied and equal to the stored energy in the capacitor. It's only the RC time constant that will be affected by the value of R. It's all in the top half of the 'triangle' graph.
    My statement was correct, although there are other things at work - including the radiation resistance of the circuit which will cause power to be radiated too.
     
  16. May 31, 2016 #15

    David Lewis

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    I performed the calculations and came up with the same answer you did. Thanks for correcting my mistake.
     
  17. May 31, 2016 #16
    There will definetly be electro magnetic radiation from the circuit while the current is changing. This can be detected as radio waves
     
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