collinge@uvicctr.UVic.ca.UUCP (Doug Collinge) (06/05/89)
We have pretty well established by now that the missing energy either heats the non-ideal resistance of the wire or radiates if the system oscillates due to inductance. Two questions: 1) Is is possible to make such a circuit with zero effective inductance assuming that the geometry of the circuit can be ideal? What is this gemometry? 2) Build such a "tank circuit" in some way that prevents it from radiating, such a coaxially, maybe. Dunk it in liquid He so that it superconducts. Does it now oscillate forever? Remember, this is a real device, not ideal. -- Doug Collinge School of Music, University of Victoria, PO Box 1700, Victoria, B.C., Canada, V8W 2Y2 collinge@uvunix.BITNET decvax!uw-beaver!uvicctr!collinge ubc-vision!uvicctr!collinge __... ...__ _.. . ..._ . __... __. _. .._ ..._._
myers@hpfcdj.HP.COM (Bob Myers) (06/07/89)
>1) Is is possible to make such a circuit with zero effective inductance >assuming that the geometry of the circuit can be ideal? What is this >gemometry? Off the top of my head, I don't believe that this can be done. Inductance results from the magnetic field induced around any current-carrying conductor; therefore, if charge is to be moved any distance at all, a field will be produced and so inductance will be present. Coaxial and twisted-pair conductors negate this effect by arranging the "forward" and "reverse" (or "source" and "return") currents so that the induced fields from each are in opposite directions. Given equal magnitude currents, then, the fields exactly cancel and there is no (realistically, very little) net inductance. However, I do not believe that one can arrange this circuit such that all fields cancel, in light of the fact that we're still talking about putting two capacitors in there somehow, and the current *must* physically separate by some small distance to make this connection. Thus, there must always be some net inductance. >2) Build such a "tank circuit" in some way that prevents it from radiating, >such a coaxially, maybe. Dunk it in liquid He so that it superconducts. >Does it now oscillate forever? Remember, this is a real device, not ideal. No can do, for similar reasons. Of course, this response comes after a *very* quick consideration of the matter, and there may indeed be some novel configuration which at least approaches this ideal. Anybody found it yet? 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.
cdl@mplvax.EDU (Carl Lowenstein) (06/09/89)
In article <11170010@hpfcdj.HP.COM> myers@hpfcdj.HP.COM (Bob Myers) writes: >>1) Is is possible to make such a circuit with zero effective inductance >>assuming that the geometry of the circuit can be ideal? >Off the top of my head, I don't believe that this can be done. Inductance >results from the magnetic field induced around any current-carrying conductor; As I remember, a few years ago somebody at Sandia Laboratories devised a configuration which was essentially a current path on the 'two sides' of a Moebius strip. Interesting application of a somewhat obscure mathematical contraption. -- carl lowenstein marine physical lab u.c. san diego {decvax|ucbvax} !ucsd!mplvax!cdl cdl@mplvax.ucsd.edu
ankleand@mit-caf.MIT.EDU (Andrew Karanicolas) (06/10/89)
>This puzzle has been beaten to death. There is no puzzle now, nor was >there one 30 years ago when I first heard it. There are no perfect >conductors, or capacitors, or inductors. The capacitors wind up at 5 >volts, and the extra energy goes into heat and em radiation. Let's >kill it! Beaten to death? Hardly. Most of the responses I have seen have missed the point entirely. You can model the loss any way you want, and a lot of people have done so correctly, but I have yet to see anyone else explain why the ideal situation breaks down. The original question has not been answered, instead, its assumptions were modified so that a tractable problem is possible. At this risk of being arrogant, I find this approach to solving problems annoying. The real solution to the ideal problem depends on one understanding, or at least appreciating, some of the theory of Dirac delta distributions. The 'capacitor problem' has been solved for a long time, there are a few good books out there on electric circuit theory that treat problems like this correctly. If anyone is interested, I would gladly send e-mail to them describing the solution or direct them to books where the solution is explained. Andy Karanicolas MIT Microsystems Laboratory ankleand@caf.mit.edu ankleand@charon.mit.edu
myers@hpfcdj.HP.COM (Bob Myers) (06/17/89)
>If there were an inductor inserted in the circuit, no energy would be carried >off in EM radiation, because that required perpendicular electric and magnetic >fields to create a propogating wave. Anyway, an ideal inductor won't lose any >energy either, it is stored in a magnetic field. I believe the equation that >Any comments? Yep. As long as we remain solely in the realm of *circuit* theory, and ignore *field* theory, you are correct in the assumption that the circuit would oscillate "forever", given ideal components, and not lose energy via EM radiation. However, your statment above which supposedly eliminates EM radiation as a factor is flawed. Consider the fact that any time-varying magnetic field will generate a time-varying electric field, and vice-versa. If there is a time-varying current in a conductor, there will be a time-varying magnetic field surrounding that conductor, regardless of whether or not the conductor is "ideal". *Some* energy will be lost to radiation; how much depends on the size, shape, and orientation of the components in the circuit, as well as the frequency of the signal involved. The classic circuit-theory assumption that an inductor (or any "purely reactive" component) loses (or dissipates) no energy ignores this fact, and rightly so. With typically-sized components and frequencies, the effect is probably several orders of magnitude below the unavoidable resistive losses in real components. However, it *does* exist, and this is, after all, a "ideal components and take the answer to the Nth degree" kind of problem. The oscillation will not continue at the same amplitude "forever". (If you still don't believe that energy can be lost through EM radiation, then please explain where the ubiquitous 60Hz hum comes from.) 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.