mbe@dde.uucp (Martin Berg) (06/01/89)
vaso@mips.COM (Vaso Bovan) writes: >A Paradox of Capacitor Energy Storage >I've heard several competing answers to this paradox. None is entirely >satisfactory: >Consider an ideal 2uF (for computational ease) capacitor charged by a 10 volt >source. Eventually, the energy stored is (1/2)*CV^2=100 joules. ^^^^^^^^^^^^^^^^^^^^^ You forgot that the 'u' in 2 uF stands for 'micro', that is 2*10^-6 F. So the energy stored is 100 uJ (100*10^-6 J). But this does not affect your 'Paradox of Capacitor Energy Storage'. >Consider the capacitor to be isolated from the voltage source, and then >directly shorted across an identical (ideal) capacitor. Eventually, the >voltage across each capacitor will be 5V. Now, there are two equally >charged capacitors, each storing (1/2)*CV^2= 25 joules, for a total of >of 50 joules. What happened to the other 50 joules ? The current in a capacitor is defined as: I = - dV/dt * C (that is: change in voltage pr. sec. times the capacity, forget the sign). This means that you cannot change the voltage across a capacitor instantaneously : this would inply infinite current. It also means that when you short your two capacitors across each other, then the current will rise to an level where internal resistance and wire resistance is not neglectable. When this happens then you loose energy in form of heat. Also: when you start shorting the capacitors, you will get a spark - this is also a waste of enery. If you use an inductor when connecting the capacitors then you would not loose energy (ok, ok a little if the components is not ideal). But then the resulting voltage would not be 5 V - it would be 7.07 V (for energy preservation). By the way: I'm not so sure that the resulting voltage would be 5 V in your example. -- mbe@dde.dk | "The answer is 42" or | D. Adams ..uunet!mcvax!enea!dkuug!dde!mbe |
myers@hpfcdj.HP.COM (Bob Myers) (06/02/89)
>A Paradox of Capacitor Energy Storage > >I've heard several competing answers to this paradox. None is entirely >satisfactory: > >Consider an ideal 2uF (for computational ease) capacitor charged by a 10 volt >source. Eventually, the energy stored is (1/2)*CV^2=100 joules. > >Consider the capacitor to be isolated from the voltage source, and then >directly shorted across an identical (ideal) capacitor. Eventually, the >voltage across each capacitor will be 5V. Now, there are two equally >charged capacitors, each storing (1/2)*CV^2= 25 joules, for a total of >of 50 joules. What happened to the other 50 joules ? This is one of the "classic" puzzles in electronics, and more than anything else is a prime example of how you can do things with "theoretically ideal" parts which make absolutely no sense at all in the real world. (Here's another example - an ideal 100 uF capacitor is instantaneously connected across an *ideal* 10V voltage source. How long does it take for the voltage across the capacitor to reach 10V?) Anyway, some possible answers to the above: 1. Given the "ideal" components used in this problem, we are forced to conclude that there is zero resistance in the conductors used to connect the two capacitors. However, there *cannot* be zero inductance in these conductors, if they have any physical length at all. Therefore, what we actually have is a resonant L-C circuit, which is excited into oscillation at its resonant frequency when the second capacitor is connected. With no resistance present, the oscillation is never damped out (no loss of energy via a resistive element), and so it continues forever. The original assumption, that the circuit would wind up at 5VDC across the two caps, is incorrect. (For that matter, consider the word "eventually" in the problem statement; where does the "eventually" come from in the absence of resistance?) 2. Let's suppose, though, that we further "idealize" the situation by considering that the conductors between the capacitors have no inductance. (This is possible only if they have no physical length.) For example, suppose that I simply (through the typical Amazing Blackboard Magic) double the area of the original capacitor's plates, hence doubling the capacitance. Now what happens? 2b. What if there is just a little inductance? Enough so that the theoretical resonant frequency (given by 1/(2pi*sqrt(LC))) is EXTREMELY high - say, that of UV light? (See what trouble you can get into with these "ideal" and "purely theoretical" calculations?) 3. Another go at it: this time, instead of doubling the plate area, I will instantaneously insert a sheet of dielectric with a relative permittivity of 2 (the original capacitor had air or vacuum between the plates). What happens in this case? 4. Of course, all of the above discussion ignores the fact that there will ALWAYS be a loss of energy from this circuit; if we step back from the realm of pure circuit theory, and look at the situation from a fields point of view, we have all we need (time-varying electric and magnetic fields) for this thing to lose energy via EM radiation. Exactly how this happens, of course, would depend on the physical structure of the apparatus. Now the first answrer is incorrect, and we WOULD expect a loss of energy from the system, and so a damping of the oscillations. We cannot, though, with the information given so far, describe this in detail. I certainly hope that this has further confused the issue. :-) Bob Myers KC0EW HP Graphics Tech. Div.| Opinions expressed here are not Ft. Collins, Colorado | those of my employer or any other myers%hpfcla@hplabs.hp.com | sentient life-form on this planet.
siegman@sierra.Stanford.EDU (Anthony E. Siegman) (06/02/89)
This problem is sometimes known as the "Case of the Nefarious Joule Thief". Put a small resistor, or a small inductance, or both, in series with the switch; solve the circuit equations for the transient behavior starting when the switch closes; and examine for physical significance, considering particularly the case when the series R or L tend to zero
fransvo@maestro.htsa.aha.nl (Frans van Otten) (06/02/89)
Martin Berg writes: >Vaso Bovan writes: > >>A Paradox of Capacitor Energy Storage > >By the way: I'm not so sure that the resulting voltage would be 5 V >in your example. Of course it would be 5 volt. The following formula applies to capacitors: q = C*u (the charge is the capacitance times the voltage) Math for the problem (both capacitors have capacitance C): Before: q1 = C * 10 V = 10C coulomb q2 = C * 0 V = 0C coulomb Total charge 10 coulomb. after: q1 = C * 5 V = 5C coulomb q2 = C * 5 V = 5C coulomb Total charge 10 coulomb. -- Frans van Otten | fransvo@maestro.htsa.aha.nl or Algemene Hogeschool Amsterdam | fransvo@htsa.uucp or Technische en Maritieme Faculteit | [[...!]backbone!]htsa!fransvo
tomb@hplsla.HP.COM (Tom Bruhns) (06/02/89)
mbe@dde.uucp (Martin Berg) writes: > ... > >If you use an inductor when connecting the capacitors then you >would not loose energy (ok, ok a little if the components is not ideal). > >But then the resulting voltage would not be 5 V - it would be 7.07 V >(for energy preservation). Bull. At least it would be only as it passed through. In a non- radiating, non-dissipative ("ideal" :-) system, there would be an oscillatory current that would yield (assuming the junction between the caps to be the reference terminal and the inductance all lumped between the other ends of the caps) two sine waves, 180 degrees out of phase, each going between 0 volts and 10 volts, as measured at each capacitor. Repeating a question I posted earlier: can you think of a _simple_ circuit that will let you get 7.07 volts on each cap, with no external energy source? (7.07 volts final value, after an "energy transfer" period.) Or here's an easier one: move all the charge from one of the capacitors to the other one, so the first cap (orig. 10 V) is now zero and the second (orig. zero) is 10 V. Again, no external energy sources. You might have to use "ideal" components to do it; also tell me what sort of loss you would expect in the real world. > >By the way: I'm not so sure that the resulting voltage would be 5 V >in your example. >-- >mbe@dde.dk | "The answer is 42" >or | D. Adams >..uunet!mcvax!enea!dkuug!dde!mbe | >----------
abali@parts.eng.ohio-state.edu (Bulent Abali) (06/02/89)
In article <147@sierra.stanford.edu> siegman@sierra.UUCP (Anthony E. Siegman) writes: >Put a small resistor, or a small inductance, or both, in series with >the switch; solve the circuit equations for the transient behavior >starting when the switch closes; and examine for physical >significance, considering particularly the case when the series R >or L tend to zero Exactly. A resistor R can be assumed in series with the switch, and the energy dissipated on R can be derived (easy). The resulting equation will show that the energy dissipated is R independent. One half of the original energy of the system will be lost on R, regardless of its value. If L is assumed instead of R, we can arrive at the same conclusion except that the 1/2 energy will not be lost, but will be stored in the inductor regardless of its value. -=- Bulent Abali Ohio State Univ., Dept.of Electrical Eng. 2015 Neil Av. Columbus, Ohio 43210 abali@baloo.eng.ohio-state.edu
siegman@sierra.Stanford.EDU (Anthony E. Siegman) (06/03/89)
There's a mechanical analog to the switched-capacitor "where did half the energy go" problem. A slow conveyor belt rolls past a loading chute which dumps dirt or coal or something on the belt at a fixed rate dm/dt (m=mass of dirt dumped). The dirt is instantaneously accelerated to the belt's steady velocity v as it hits the belt's surface. Force required to keep pulling the belt (no friction) = f = dp/dt where dp = dm v = momentum added to the mass dm. Work done in time dt = f dx = dm v^^2 (using dx = v dt). But kinetic energy added to (or carried by) the dirt is only (1/2) dm v^^2. Where's the other half?
estell@m.cs.uiuc.edu (06/04/89)
When doing the "one charged, one uncharged capacitor shorting" experiment, get yourself an AM radio, tune it to a frequency unused in your area, and place it close to the capacitors. It's the best way to prove to doubting students that the energy lost is mainly in the form of EMR. (For additional fun, use an AM/FM radio and do the demo twice....!)