jds@m2c.m2c.org (jds) (05/26/89)
I just picked up a couple of 14ufd 25KVDC caps and am looking for something to do with them... I'm looking for good demonstations, experiments, etc. I figure these guys could be used for a nice spark-gap demo or perhaps vaporize wire. If you've got any ideas, let me know! -- john
tomb@hplsla.HP.COM (Tom Bruhns) (05/26/89)
jds@m2c.m2c.org (jds) writes: > I just picked up a couple of 14ufd 25KVDC caps and am looking for something >to do with them... I'm looking for good demonstations, experiments, etc. >I figure these guys could be used for a nice spark-gap demo or perhaps >vaporize wire. > > If you've got any ideas, let me know! > >-- john >---------- Lesseehere... That's about 8750 joules when both are fully charged: nearly nine kilowatt seconds. That's enough to do significant damage, especially if it's released quickly. You didn't mention if the caps are rated for rapid-discharge service, however. Those that aren't can be damaged by too-rapid discharge, and usually also cannot be discharged in microseconds because of their inductance. That out of the way, one neat thing to do is to flash a large flashlamp through them. You may be able to find an appropriate one as a "pull" from a laser or an airport approach system. These generally will be rated for lower voltage than your caps, but 28 uF at 5 or 6 kV is still a pretty good jolt -- or a lotta light. _OBVIOUSLY_BE_VERY_ _CAREFUL_WITH_THIS_ENERGY, WHATEVER FORM YOU CONVERT IT TO!!! The amount of light involved could make you (or a friend, neighbor, pet, etc.) blind very quickly, like in microseconds. In fact, BE CAREFUL WITH THE CAPS EVEN WHEN YOU THINK THEY ARE DISCHARGED!!! Put a shorting strap across them any time they aren't actually in use. Handle them only through a long plastic rod that you know is a good insulator, any time you think they may have any charge on them; build a discharging resistor that you can put across them to bring them down in a controlled manner -- have an indicator in the discharging resistor that will tell you when they are discharged to a safe level -- be sure that the resistor can handle the voltage and power (probably about 25 2-watt metal film resistors in series would do well -- they are noted for their ability to not open under an overload, but DONT exceed thier voltage rating). SAFETY FIRST!! Have a good day -- and a safe one!
larry@kitty.UUCP (Larry Lippman) (05/28/89)
In article <4924@m2c.M2C.ORG>, jds@m2c.m2c.org (jds) writes: > I just picked up a couple of 14ufd 25KVDC caps and am looking for something > to do with them... I'm looking for good demonstations, experiments, etc. > I figure these guys could be used for a nice spark-gap demo or perhaps > vaporize wire. > > If you've got any ideas, let me know! A brief mental calculation tells me that just one of those capacitors can store at least 4,000 joules of energy. This is more than enough energy to be lethal; I would recommend caution before attempting any "experiments" with this device. When I was in high school, I had a large World War II surplus airport runway beacon which used a large coiled flashlamp with a power supply that had an enormous oil-filled capacitor about 18 inches high. One day, out of (with some hindsight) shear stupidity I discharged the capacitor with a large screwdriver. The most notable effect was the sound of the discharge - my ears _literally_ rang for a week. As you might also guess, the screwdriver almost vaporized into two pieces. <> Larry Lippman @ Recognition Research Corp. - Uniquex Corp. - Viatran Corp. <> UUCP {allegra|boulder|decvax|rutgers|watmath}!sunybcs!kitty!larry <> TEL 716/688-1231 | 716/773-1700 {att|hplabs|utzoo}!/ \uniquex!larry <> FAX 716/741-9635 | 716/773-2488 "Have you hugged your cat today?"
jbs@fenchurch.mit.edu (Jeffrey Siegal) (05/29/89)
In article <3186@kitty.UUCP> larry@kitty.UUCP (Larry Lippman) writes:
) A brief mental calculation tells me that just one of those capacitors
)can store at least 4,000 joules of energy.
) This is more than enough energy to be lethal;
... but less than 1/200 of a Jelly Doughnut!!!
Jeffrey Siegalrdsnyder@mit-amt (Ross D. Snyder) (05/29/89)
A great way to have fun with HV caps is to build a project that is used as a demo in 6.013, the Electromagnetic Fields and Energy class at MIT. The device was invented by Prof. Harold "Doc" Edgerton, inventor of the strobe lamp and a professor here at MIT. The device is named the "Boomer." It consists of a HV supply which charges 48uF's (I think) worth of caps to 2KV (I think). A serious HV relay is used to dump the charge through a pancake coil about 10" in diameter made of something like 4-gauge copper wire. I remember the resonant frequency of this LC circuit was about 2KHz, which resulted in a damped sinusoid upon discharge. The pancake coil converts the electrical energy into magnetic energy. Needless to say, the device produces a significant short-duration magnetic field. A sheet metal disk is placed on top of the pancake coil. The magnetic field from the coil produces a current in the disk, which results in a repulsive force between the coil and the disk, thus launching the disk. The demo was done in a lecture hall with a ~30ft ceiling, and an aluminum disk could be made to hit the ceiling. A larger version of this device has been used by Doc Edgerton with a metal plate about the size of a manhole cover to create underwater sonar impulses. The professor let students come up after the lecture and put their hands over the pancake coil. I did it and could feel the current induced in my hand because of its conductivity and the presence of the magnetic pulse. A videotape of the demo was shown with Doc Edgerton doing the demo himself in a room with a ~10ft ceiling. Another professor suggested in the tape that doing the demo in that room would likely punch a hole in the ceiling tiles, but suggested that MIT might not care too much since it was Doc's company, EG&G, that had donated the entire building to MIT. :-) A word of caution: The HV supply and caps were in a steel cabinet with safety interlocks on the access panels. The relay was in another steel box within the cabinet and had an interlock on its door. Don't just throw together a kludge. If you're going to build one of these, do a good job and insulate everything and use HV wire, lots on silicone sealant, and bleeder resistors across your caps. -Ross
dant@mrloog.LA.TEK.COM (Dan Tilque;1545;92-101;OPUS_SW;) (05/29/89)
Jeffrey Siegal writes: >Larry Lippman writes: >) A brief mental calculation tells me that just one of those capacitors >)can store at least 4,000 joules of energy. >) This is more than enough energy to be lethal; > >... but less than 1/200 of a Jelly Doughnut!!! Hey, if jelly doughnuts are so powerful, why isn't everyone trying to fuse them instead of heavy water. Jelly doughnuts are much more common than heavy water. After all, JFK was one. --- Dan Tilque -- dant@twaddl.LA.TEK.COM "Lord, heal this software!" -- fundamentalist software engineering
brian@ucsd.EDU (Brian Kantor) (05/29/89)
It occurs to me that one of the most useful things you could do with these capacitors would be to develop some way to charge them up in your car so you could stack the outputs and generate enough directed EMP to fry the radio and the electronic ignition in that police car just behind you. Don't forget to smear mud all over your license plate. - Brian
willner@cfa.HARVARD.EDU (Steve Willner) (05/30/89)
From article <3806@mit-amt>, by rdsnyder@mit-amt (Ross D. Snyder): > A word of caution: The HV supply and caps were in a steel cabinet --no doubt carefully grounded-- > with safety interlocks on the access panels. The relay was in > another steel box within the cabinet and had an interlock on its > door. Don't just throw together a kludge. If you're going to > build one of these, do a good job and insulate everything and use > HV wire, lots on silicone sealant, and bleeder resistors across > your caps. The above is absolutely right; pay attention! And one more: Make sure you have a fiberglass stick, a couple of feet long, with a well-grounded contact on one end. As soon as you open a cabinet, grasp the stick (by the end _opposite_ the ground!) and touch the ground to every terminal of every capacitor. Normally this does nothing, but it produces an interesting :-) effect if any of your bleeder resistors has gone bad or if you left the power on and your interlock has failed.
frankb@hpsad.HP.COM (Frank Ball) (05/30/89)
*Subject: HV Cap Fun! * * I just picked up a couple of 14ufd 25KVDC caps and am looking for something *to do with them... I'm looking for good demonstations, experiments, etc. *I figure these guys could be used for a nice spark-gap demo or perhaps *vaporize wire. * * If you've got any ideas, let me know! Here at work they keep electronic parts in "anit-static" plastic bags. There are a couple of varieties-one of which is a plastic coated with a vary thin (somewhat transparent) layer of nickel. While 25KV will normally arc about 1/2 inch-it can create an arc about 10 inches long over the surface of one of these anti-static bags. After you have done this hold the plastic up to the light and you can see the tracks from the arcs: they vaporize the metal off of the plastic. Frank Ball 2LR-O frankb@hpsad.HP.COM Hewlett Packard (707) 794-4168 1212 Valley House Drive fax: (707) 794-4452 Rohnert Park CA 94928-4999 I'm the NRA.
dhesi@bsu-cs.bsu.edu (Rahul Dhesi) (05/30/89)
[about energy storage by capacitors]
I've always been very, very careful around capacitors ever since I
discovered to my amazement that not only can they hold a charge for a
long time, but also that: You can discharge some electrolytic
capatitors, remove the short circuit, and after a few minutes the
capacitor regains a smaller charge.
--
Rahul Dhesi <dhesi@bsu-cs.bsu.edu>
UUCP: ...!{iuvax,pur-ee}!bsu-cs!dhesi
Career change search is on -- ask me for my resumetomb@hplsla.HP.COM (Tom Bruhns) (05/30/89)
dhesi@bsu-cs.bsu.edu (Rahul Dhesi) writes: >[about energy storage by capacitors] > >I've always been very, very careful around capacitors ever since I >discovered to my amazement that not only can they hold a charge for a >long time, but also that: You can discharge some electrolytic >capatitors, remove the short circuit, and after a few minutes the >capacitor regains a smaller charge. >-- Electrolytics are notorious for this (some folk call it "soakage", or "dielectric absorption"). But be aware that you aren't safe from this effect just because you capacitor is not an electrolytic. Other caps will do it, too. Larry's tale reminded me of the FCC inspector that used to go around this area doing broadcast station inspections. He would open the doors on the back of the transmitter to "make sure" it went off the air like it's supposed to (interlocks working). He quit doing it, so the story goes, when he came across a transmitter where the interlocks not only really _were_ working but where they dropped a direct short across all the big caps. It was apparently a bit too much of a surprise for him (he survived OK, but quit checking in quite so dramatic a way...).
kencr@haddock.ima.isc.com (Kenny Crudup) (05/30/89)
In article <7487@bsu-cs.bsu.edu> dhesi@bsu-cs.bsu.edu (Rahul Dhesi) says: >I've always been very, very careful around capacitors ever since I >discovered to my amazement that not only can they hold a charge for a >long time, but also that: You can discharge some electrolytic >capatitors, remove the short circuit, and after a few minutes the >capacitor regains a smaller charge. >Rahul Dhesi <dhesi@bsu-cs.bsu.edu> Yup. They sure do. Its 'cause nature abhors a vacuum, and when you let all the energy out, the caps pull electrons out of the air. -- Kenneth R. Crudup, Contractor, Interactive Systems, Cambridge MA ...you ain't guarding the door, so what you need a gun for? Phone (617) 661 7474 x238 {encore, harvard, spdcc, think}!ima!haddock!kencr kencr@haddock.ima.isc.com
logajan@ns.network.com (John Logajan) (05/31/89)
In article <7487@bsu-cs.bsu.edu>, dhesi@bsu-cs.bsu.edu (Rahul Dhesi) writes: > You can discharge some electrolytic > capatitors, remove the short circuit, and after a few minutes the > capacitor regains a smaller charge. This is not uncommon especially in high-voltage capacitors, including vacuum tube CRTs. Since the plates of the capacitor are insulated from each other by a high resistance material (or vacuum) there can come to exist areas of induced charge within the dielectric itself (or in the CRT, on remote surfaces.) The initial discharge will not purge these because the resistance is so high that it takes a long time for the charges to leak back out to the plates. -- - John M. Logajan @ Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428 - - logajan@ns.network.com / ...rutgers!umn-cs!ns!logajan / john@logajan.mn.org -
vaso@mips.COM (Vaso Bovan) (05/31/89)
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 ?
ornitz@kodak.UUCP (Barry Ornitz) (05/31/89)
In article <3186@kitty.UUCP> larry@kitty.UUCP (Larry Lippman) writes about his high-school experience of discharging a large HV oil-filled capacitor. I too made a mistake in discharging a low inductance cap with a screwdriver. I had a 16 uF, 4KV discharge capacitor (0.1 uH series inductance) charged to 600 volts when I used a screwdriver to short it. After all, I had discharged several hundred microfarads of electrolytics before with only a small spark. Like Larry found out, the bang made my ears ring for an hour and the flash blinded me for about 10 minutes. It burned off 1/4 inch of the screwdriver. I now have another discharge capacitor: 18 uF, 10 KV, 0.04 uH series. A quick calculation says at full charge, this will hold about 300 times the energy as the 600 volt situation described. Believe me, this thing is now kept with a shorting bar permanently affixed across its terminals. Dielectric absorption, the bane of audio phreaks, can partially recharge these capacitors even after they have been momentarily shorted. While this effect is small in these oil- filled capacitors, it can still be dangerous. I don't know exactly what is lethal in terms of stored energy for these capacitors. I have recently been working with a HV power supply made by Glassman High Voltage (75 KV, 3 mA). The Glassman engineers told me that they consider 4 Joules to be fatal (although, of course, many people have survived far more - but why take chances?). At the higher voltages, skin resistance offers virtually no protection at all. Another thing to worry about with oil-filled HV capacitors is that until only a few years ago, the filling oil was a polychlorinated biphenyl (PCB). The disposal of these capacitors can be a significant problem. David Anthony pointed this out to the net once about a problem he had with the capacitors in an older broadcast transmitter. The laws governing the transport and disposal of PCB containing materials are quite strict. My overall recommendation: be careful! Barry ----------------- | ___ ________ | | | / / | | Dr. Barry L. Ornitz UUCP:..rutgers!rochester!kodak!ornitz | | / / | | Eastman Kodak Company | |< < K O D A K| | Eastman Chemicals Division Research Laboratories | | \ \ | | P. O. Box 1972 | |__\ \________| | Kingsport, TN 37662 615/229-4904 | | -----------------
naiman-jeffrey@CS.YALE.EDU (Jeffrey Naiman) (06/01/89)
In article <20772@quacky.mips.COM> 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. > >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 ? OK, here's a silly question. Why is the voltage across the two identical capacitors 5V? Volts are joules/coulomb, or more simply, energy per electron. Voltage drops when going through a resistor (sticking to DC current). Why should the energy an electron has drop when it goes to another capacitor? The formulas: U=1/2 C V^2=100 U=1/2 Q V=100 , Q=20 (Q is halved for two capcitor case) U=2 x .5 x 10 x 10=100 for two capacitors U=1 x .5 x 20 x 10=100 for one capacitor I'm obviously missing something.... - Jeff Naiman (naiman@yale.edu)
arnief@tekgvs.LABS.TEK.COM (Arnie Frisch) (06/01/89)
In article <1418@ns.network.com>, logajan@ns.network.com (John Logajan) writes: > In article <7487@bsu-cs.bsu.edu>, dhesi@bsu-cs.bsu.edu (Rahul Dhesi) writes: > > You can discharge some electrolytic > > capatitors, remove the short circuit, and after a few minutes the > > capacitor regains a smaller charge. > > This is not uncommon especially in high-voltage capacitors, including Another problem is that some capacitors can assume a charge after they have been shorted as a result of changing temperature. This is especially true of ceramic capacitors. You discharge it at 20 degrees C and then you just heat it up to - say 50 degrees C - and you find it's charged up to a significant voltage again, maybe enough to give you a nasty jolt. (It happened to me.) Arnold Frisch Tektronix Laboratories
steve@arc.UUCP (Steve Savitzky) (06/01/89)
In article <20772@quacky.mips.COM> 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. call it C1 >Consider the capacitor to be isolated from the voltage source, and then >directly shorted across an identical (ideal) capacitor. call it C2 > 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 ? ... followed by several unsatisfactory responses from several people who obviously didn't think about either the physics or the electronics of the situation. PHYSICS: Consider the case where there is only a single electron in C1. It has potential energy. When it moves over to the other capacitor, that potential energy is converted to kinetic energy. When it gets to C2, it will have have just the same amount of potential energy again. Now it sloshes back to C1. The system oscillates. Think of it as a marble between two hills. If the wires between C1 and C2 have resistance, some of the kinetic energy is converted to heat, and the oscillations eventually damp down. Think of friction on the marble. ELECTRONICS: We have the following circuit: ---------[ Z1 ]--------- | | ----- C1 ----- C2 ----- ----- | \ S1 | ----------- \----------- Look familiar? Those "wires" have inductance and (unless they're made of superconductors) resistance, represented by Z1. (Ever try to draw an inductor in ascii? :-) When you close the switch, current flows from C1 to C2, and energy is stored in the magnetic field of the inductors. When the two capacitors reach the same charge (voltage), the current keeps flowing, powered by the collapsing magnetic field of the inductors. Eventually, all the charge ends up in C2, and the current stops. Then the process starts again in the opposite direction. The system oscillates. If the wires are resistive, some of the stored energy is converted to heat, and the oscillations damp down. THE TWO EXPLANATIONS ABOVE ARE EQUIVALENT. ("kinetic energy" and "energy stored in a magnetic field" are equivalent ways of looking at the energy of a moving charged particle.) Satisfactory?
ankleand@mit-caf.MIT.EDU (Andrew Karanicolas) (06/01/89)
In article <20772@quacky.mips.COM> 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. > >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 ? Interestingly enough, this problem came around on sci.electronics about this time last year; this capacitor problem is a classic EE problem. Basically, the difficulty in determining what 'happened' to the other 50 Joules arises from an implication in the problem statement. The implication is as follows: suddenly connecting the capacitors together is tantamount to saying that the capacitor voltage in the circuit changes in a step-function manner, u(t) since the capacitor current is related to the capacitor voltage by i(t)=Cdv(t)/dt, a step change in the capacitor voltage results in the capacitor current changing in a delta-function manner, delta(t) using energy conservation, one will then implicitly attempt the evaluation of the following, undefined, integral: / +inf | | u(t) * delta(t) dt | / -inf the fact that this integral is not defined is why there is a 'paradox'. The point is that the *details* of how the capacitors are connected together determines what the final energy of the system will be. Modelling a series resistance shows that half of the original energy is dissipated in the resistor. More elaborate models of electromagnetic radiation can also be considered. A switch that has time varying resistance can also be worked into the problem. Clearly, intricate models for the switching mechanism can be devised. However, there is a fundamental explanation for the 'missing' energy in the mathematics in the case of the 'paradoxical' problem where things occur suddenly. One source to see a more complete treatment of this problem: "Circuits, Signals and Systems" Siebert, William McC McGraw Hill 1986 Andy Karanicolas MIT Microsystems Laboratory ankleand@caf.mit.edu ankleand@charon.mit.edu
timk@egvideo.UUCP (Tim Kuehn) (06/01/89)
>In article <20772@quacky.mips.COM> 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. >> I trust that C is in uF, otherwise your result would be 100E-6 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. Is it really? This would seem to violate the laws of conservation of energy, since, as is further stated: >> 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 ? [some stuff deleted] >Why should the energy an electron has drop when it goes to another >capacitor? > >The formulas: U=1/2 C V^2=100 > U=1/2 Q V=100 , Q=20 (Q is halved for two capcitor case) > U=2 x .5 x 10 x 10=100 for two capacitors > U=1 x .5 x 20 x 10=100 for one capacitor > >I'm obviously missing something.... > >- Jeff Naiman (naiman@yale.edu) Consider that in this ideal scenario, you have one capacitor charged to contain 100 J of energy. If you connect another capacitor in parallel with it (I presume that's the configuration) then part of that charge will drain off the charged capacitor into the uncharged capacitor until both are equally charged. Therefore, your system MUST end up with 100 J at the end, which does not necessarily mean that your final voltage will be 5V on the line. To wit: In the first scenario with the 2uF cap charged to 10V we have: 1/2 C * V ^ 2 = 1/2 * 2 * 10 * 10 ^ 2 = 100 J adding another capacitor on parallel doubles the C term, the energy stored must remain constant (see the law on conservation of energy!), therefore the voltage the caps will have in the final equilibrium is the only unknown. Solving the above equation for V with C = 4uF: (1/2) (4) * V ^ 2 = 100 V = (100 / (1/2 * 4)) ^ (0.5) V = 7.07 Volts which, as you can see, is not the same as the 5V you stated previously. No paradox at all, when you look at it. +-----------------------------------------------------------------------------+ |Timothy D. Kuehn timk@egvideo | |TDK Consulting Services !watmath!egvideo!timk | |871 Victoria St. North, Suite 217A | |Kitchener, Ontario, Canada N2B 3S4 (519)-741-3623 | +-----------------------------------------------------------------------------+
timk@egvideo.UUCP (Tim Kuehn) (06/01/89)
In article <349@arc.UUCP> apple.com!arc!steve (Steve Savitzky) writes: >In article <20772@quacky.mips.COM> vaso@mips.COM (Vaso Bovan) writes: >>A Paradox of Capacitor Energy Storage >> [first part of the original question deleted] > >> 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 ? > >... followed by several unsatisfactory responses from several people >who obviously didn't think about either the physics or the electronics >of the situation. And from looking at your answer I come to the same conclusion about what you posted too! :-) > >PHYSICS: > Analogy deleted.... > >ELECTRONICS: > >We have the following circuit: > > ---------[ Z1 ]--------- > | | > ----- C1 ----- C2 > ----- ----- > | \ S1 | > ----------- \----------- > >Look familiar? Those "wires" have inductance and (unless they're made >of superconductors) resistance, represented by Z1. That is true, however we're talking about IDEAL circuits here - no stray capacitance, capacitor leakage, resistance or inductance in the line, etc., so that the effects you're speaking of don't come into play. Also the original question didn't mention anything about there being an inductor in the circuit, and even if you did assume the wires had inductance, the effect wouldn't be noticible for what we're dealing with here. (after all, the inductance of a short piece of wire is somewhere in the nH or pH range which makes the effect of L di/dt *quite* small.) Again, if the wires had resistance, unless they were a high-resistance wire or a long piece (couple hundred feet) of normal copper wire the effect would be negligable. (Given a normal wire of a short length, we're probably talking about 1E-3 or 1E-4 ohms, again *way* too small to worry about here.) [explination about oscillating systems with inductors, capacitors, and resistive components deleted] > >If the wires are resistive, some of the stored energy is converted to >heat, and the oscillations damp down. > >THE TWO EXPLANATIONS ABOVE ARE EQUIVALENT. > [more stuff deleted] >Satisfactory? Nope. You are to be commended on giving an informative discussion on the nature of oscillating systems, however you should be slapped on the wrists for making the error of not making sure of the question and the parameters first, and then answer a question that wasn't asked! (insert many :-)'s here...) +-----------------------------------------------------------------------------+ |Timothy D. Kuehn timk@egvideo | |TDK Consulting Services !watmath!egvideo!timk | |871 Victoria St. North, Suite 217A | |Kitchener, Ontario, Canada N2B 3S4 (519)-741-3623 | +-----------------------------------------------------------------------------+
arnief@tekgvs.LABS.TEK.COM (Arnie Frisch) (06/01/89)
In article <20772@quacky.mips.COM>, vaso@mips.COM (Vaso Bovan) writes: > A Paradox of Capacitor Energy Storage > 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 ? In real circuits, infinite currents are not allowed - either the inductance or resistance of the circuit prevents the current from reaching infinite value. In this circuit, no resistance is specified and no inductance is specified, so I guess the current - on paper - could be infinite for zero time (i.e. an impulse). Just like I guess the energy - on paper - gets radiated into free space.
strong@tc.fluke.COM (Norm Strong) (06/01/89)
In article <20772@quacky.mips.COM> 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. Well. . . 100 microjoules. But it's the principle that counts. Right? }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 extra energy is dissipated in the form of heat when the current flows though the series resistance of the capacitor. If the series resistance of the capacitor is negligible, it will be dissipated in the form of a bright flash accompanied by a loud bang (but you won't notice it until the values get appreciably higher than 10E-4 joules.) At about 1000 joules, the flash will eat off the end of the wire you use to connect the capacitors together, and the bang will scare the hell out of you. :-) -- Norm (strong@tc.fluke.com)
strong@tc.fluke.COM (Norm Strong) (06/01/89)
In article <8739@fluke.COM> strong@tc.fluke.COM (Norm Strong) writes: }In article <20772@quacky.mips.COM> 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. } }Well. . . 100 microjoules. But it's the principle that counts. Right? } }}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 extra energy is dissipated in the form of heat when the current flows }though the series resistance of the capacitor. If the series resistance }of the capacitor is negligible, it will be dissipated in the form of a }bright flash accompanied by a loud bang (but you won't notice it until the }values get appreciably higher than 10E-4 joules.) }At about 1000 joules, the flash will eat off the end of the wire you use }to connect the capacitors together, and the bang will scare the hell out }of you. :-) [Change that to 1E-4} It also might be of interest to note that electrical discharge is used for some types of machining. It's possible to machine a corkscrew shaped hole in a solid block of metal using this method. Just try it any other way! -- Norm (strong@tc.fluke.com)
tomb@hplsla.HP.COM (Tom Bruhns) (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. > >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 ? >---------- Not a paradox at all. If you put a resistor (pick _any_ value) between the 10-volt-charged and the uncharged caps, it's easy to figure out that you loose half the energy in the resistor. That assumes "ideal" components. If you put an ideal inductor between the two (instead of a resistor), you end up with a resonant circuit that will "ring" forever. In practice, you will have both inductance and resistance, and if you didn't go out of your way to put in inductance, the circuit will most likely be overdamped and get rid of its energy in a non-oscillatory way. That's a fairly simplistic view. In a practical sense, some of the energy will be radiated as electromagnetic waves. If you do the experiment, you would probably be able to hear a "click" on a nearby properly setup AM radio. There's a practical class of circuit that's closely related to this problem. Consider a circuit wherein a capacitor is discharged quickly into a load (like in a flashlamp/strobe, or a pulsed magnetron). You want to efficiently recharge the capacitor to get ready for the next pulse. You know if you charge it through a resistance that power will be lost in that resistance -- total energy equal to the energy change in the capacitor, unless you do something special. So -- you put an inductor in series with the charging supply, and a diode so the current can't come back through the inductor. And - viola - you get a much reduced energy loss and a voltage step-up to boot. (This circuit is deceptively simple; it makes a good interview problem, because it tests the interviewee's knowledge of some very basic components in a way he likely hasn't thought of them.) This is also closely related to many switching power supply circuits.
tomb@hplsla.HP.COM (Tom Bruhns) (06/01/89)
naiman-jeffrey@CS.YALE.EDU (Jeffrey Naiman) writes: > >OK, here's a silly question. Why is the voltage across the two identical >capacitors 5V? Volts are joules/coulomb, or more simply, energy per >electron. Voltage drops when going through a resistor (sticking to DC current). >Why should the energy an electron has drop when it goes to another >capacitor? > >The formulas: U=1/2 C V^2=100 > U=1/2 Q V=100 , Q=20 (Q is halved for two capcitor case) > U=2 x .5 x 10 x 10=100 for two capacitors > U=1 x .5 x 20 x 10=100 for one capacitor > >I'm obviously missing something.... > >- Jeff Naiman (naiman@yale.edu) >---------- Charge is conserved; energy is converted from potential energy to kinetic (heat and electromagnetic) when you connect the two caps like that. The electrons have potential energy when on the cap. If they fall through a potential, they convert that potential energy to kinetic energy. That energy won't come back as an electrical potential (voltage) if you let it get out as heat or radiation. Interesting question: Can you design a simple circuit to connect the two caps and end up with 7.07 volts on each? (no other energy source) (I'll take a ckt that even manages to get 7 volts :-) Tom Bruhns tomb%hplsla@hplabs.hp.com
oconnor@nuke (Dennis M. O'Connor) (06/02/89)
With a high resistance connection and zero inductance, the losses would be in to resistive heating. In a real system ( non-zero inductance and resistance ) the system will oscilate for a some time, with the oscillations getting smaller until they are beneath the thermal noise of the resistance. Resistive losses usually dominate, but some electromagnetic radiation occurs ( as it does whenever a charge is accelerated/decellerated ) In an ideal system ( non-arcing connection with 0 resistance and the 0 inductance, capable of handling infinite current ) the system loses energy entirely by electromagnetic radiation. That's how the big EMP simulators work, I think. Try shorting a big capacitor while someone in an adjacent, sonically-isolated area listens to an AM radio low in the band : you should be able to here the EM energy ! Of course, I could be wrong. I've TOTALLY negleted any and all quantum and relativistic effects ! :-) -- Dennis O'Connor oconnor%sungod@steinmetz.UUCP ARPA: OCONNORDM@CRD.GE.COM
whh@PacBell.COM (Wilson Heydt) (06/02/89)
In article <20772@quacky.mips.COM>, 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. Please--if you want to store 100 joules, use a 2 F capacitor. > 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 ? Why do assume the result will be 5 v.? (Have you actually done the experiment?) Aren't you more likely to have 2 caps at 10 * (2 ** -2) v.? Each storing 50 joules? (Or 50 microjoules for your 2 uF units.) --Hal ========================================================================= Hal Heydt | In the old days, we had wooden Analyst, Pacific*Bell | ships sailed by iron men. Now 415-645-7708 | we have steel ships and block- whh@pbhya.PacBell.COM | heads running them. --Capt. D. Seymour
tan@ihlpb.ATT.COM (Bill Tanenbaum) (06/02/89)
< >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 ? < < < [Much correct stuff deleted for brevity] < < The point is that the *details* of how the capacitors are < connected together determines what the final energy of the < system will be. Modelling a series resistance shows that < half of the original energy is dissipated in the resistor. ---------------------------------- The final energy of the system is independent of the details of the system, assuming only that electric charge is conserved. Half of the original energy must be lost as the system decays to the final state. The time dependence of the voltage and current do of course depend critically on the details of the system. It is not necessary to assume any ohmic resistance is present. Even if all material present is superconducting, radiative losses will cause energy dissipation, although the time scale may be long. -- Bill Tanenbaum - AT&T Bell Labs - Naperville IL att!ihlpb!tan
tan@ihlpb.ATT.COM (Bill Tanenbaum) (06/02/89)
In article <62169@yale-celray.yale.UUCP<, naiman-jeffrey@CS.YALE.EDU (Jeffrey Naiman) writes:
<
< OK, here's a silly question. Why is the voltage across the two identical
< capacitors 5V? Volts are joules/coulomb, or more simply, energy per
< electron. Voltage drops when going through a resistor (sticking to DC current).
< Why should the energy an electron has drop when it goes to another
< capacitor?
<
< The formulas: U=1/2 C V^2=100
< U=1/2 Q V=100 , Q=20 (Q is halved for two capcitor case)
< U=2 x .5 x 10 x 10=100 for two capacitors
< U=1 x .5 x 20 x 10=100 for one capacitor
<
< I'm obviously missing something....
<
< - Jeff Naiman (naiman@yale.edu)
-----------------
Volts are joules/coulomb, but they are also coulombs/farad. The amount
of charge (coulombs) is conserved. The capacitance (farads) has doubled.
So the voltage (volts) must be cut in half. V=Q/C. This is what you missed.
The energy (joules), on the other hand, is not conserved, since the system
is not closed. Since U = QV, and V gets cut in half while Q is constant,
U must be cut in half. Where does it go? Mostly to resistive losses
if ohmic resistance is present, otherwise to radiative losses.
--
Bill Tanenbaum - AT&T Bell Labs - Naperville IL att!ihlpb!tanjtwarden@pawl.rpi.edu (Joseph T. Warden) (06/02/89)
In article <575@crdgw1.crd.ge.com> oconnor@nuke (Dennis M. O'Connor) writes: >With a high resistance connection and zero inductance, the losses >would be in to resistive heating. > >Try shorting a big capacitor while someone in an adjacent, >sonically-isolated area listens to an AM radio low in >the band : you should be able to here the EM energy ! > >-- > Dennis O'Connor oconnor%sungod@steinmetz.UUCP ARPA: OCONNORDM@CRD.GE.COM For anyone with experience with rf generating equipment (ie. Q-switched Nd-YAG lasers), your suggestion is one of the tricks for determining paths of rf transmission - we use a portable FM radio with a directional antenna. Joseph Warden Department of Chemistry Rensselaer Polytechnic Institute Troy, NY jtwarden@pawl.rpi.edu
raoul@eplunix.UUCP (Otero) (06/02/89)
In article <8742@fluke.COM>, strong@tc.fluke.COM (Norm Strong) writes: > In article <8739@fluke.COM> strong@tc.fluke.COM (Norm Strong) writes: > }In article <20772@quacky.mips.COM> vaso@mips.COM (Vaso Bovan) writes: > }}A Paradox of Capacitor Energy Storage > }} > }}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 ? > } > }The extra energy is dissipated in the form of heat when the current flows > }though the series resistance of the capacitor. What about superconducting materials? No resistive heat dissipation at all. No, the gentleman who posted about the inductive effects of the wires connecting the capacitors was much closer. Most real systems will have heat loss as current travels through them, but even perfect conductors have electromagnetic radiative losses: current oscillating through space produces electromagnetic radiation, which can be described as an impedance of the system. If the net impedance of the system is high enough, more it damps the system and heat and radiation take up the energy before anything oscillates. If the impedance is to low, the system oscillates, with current and charge oscillating back and forth between the capacitors. This paradox is almost identical to that of shorting across a charged capacitor. Where does the energy go? This sort of problem is also partly why good capacitors are so much easier to make than good inductors. Capacitors 'only' hold charge, so they don't have to deal with the losses and problems inductors do. Capacitors also pretty much contain their fields between their plates, where inductors' fields project beyond their cores and cause no end of grief.... -- Nico Garcia Engineer, CIRL Mass. Eye and Ear Infirmary eplunix!cirl!raoul@eddie.mit.edu
fransvo@maestro.htsa.aha.nl (Frans van Otten) (06/02/89)
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. >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 ? If the circuit was really ideal, you're doing something "illegal". The voltage over the capacitor depends on the electrons in it. They can't move in zero time so the voltage over the capacitor can't be discontinu. When you try to do that anyway, you get the same "illegal" situation as when you short-circuit a voltage-source or disconnect a current-source. So what happens: 1. There is a non-zero voltage over a zero-Ohm connection; 2. This results in a infinitely high current; 3. Infinity current times zero Ohm can mathematically be anything; 4. After zero time the capacitors are equally charged; 5. Some energy can be dissipated. Popular math: i = infinitely many Amperes R = 0 Ohm t = 0 seconds energy = power * time = voltage * current * time = R * i * i * t = 0 * infinity * infinity * 0 = ( 0 * infinity ) * ( infinity * 0 ) = ? * ? = ? (not neccisarily zero) This for the impossible theoretical question. In practice, this situation is impossible because there will always be a non-zero resistance and other non-ideal values. A calculation then shows that the amount of dissipated energy does not depend on the resistor value. -- Frans van Otten | fransvo@maestro.htsa.aha.nl or Algemene Hogeschool Amsterdam | fransvo@htsa.uucp or Technische en Maritieme Faculteit | [[...!]backbone!]htsa!fransvo
jeffw@midas.STS.TEK.COM (Jeff Winslow) (06/02/89)
We can sure tell who, responding to the "Missing Joules" puzzle, has played with circuits much and who hasn't. Trust me, the final voltage across the caps will *not* be 7.07v. 5v is much closer - very very close if the caps don't leak. OK, don't trust me. Try it. Jeff Winslow
mph@behemoth.phx.mcd.mot.com (Mark Huth) (06/03/89)
In article <20772@quacky.mips.COM> vaso@mips.COM (Vaso Bovan) writes: >A Paradox of Capacitor Energy Storage > >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 ^ No. There is no paradox. The voltage will not be 5 volts. It will be 50**.5 volts. A tad over 7. Mark Huth
mmm@cup.portal.com (Mark Robert Thorson) (06/03/89)
A friend of mine once worked at Lawrence Livermore National Laboratory where he maintained a shed full of BIG capacitors used in energy experiments. These capacitors were about the size of a small oil drum. They would put little pill bottles on the capacitors. When they would hear one rattling, they'd dive for cover, as that was an indication a capacitor was about to short out (which would be followed by the capacitor bank dumping its energy into the failed capacitor).
jbs@fenchurch.mit.edu (Jeffrey Siegal) (06/03/89)
In article <747@eplunix.UUCP> raoul@eplunix.UUCP (Otero) writes: >Capacitors also pretty much contain their fields between their plates, >where inductors' fields project beyond their cores and cause no end >of grief.... Not if the inductor is torus (doughnut) shaped. In this case the field is completely internal. Anyone know how close to "perfect" toroidal superconducting inductors are? Jeffrey Siegal
jeffw@midas.STS.TEK.COM (Jeff Winslow) (06/03/89)
I have a question for the non-5v advocates: Where did all the extra charge
come from? Q = C*V. If the final voltage is 7.07, you just made 8.28 coulombs
for free.
Maybe this will reassure some people. Look at the case where there
is a resistor R connecting the two capacitors, each with capacitance C. The
current through this resistor starts out at 10/R and decays toward zero at
infinite time by an exponential function with time constant RC/2. (See any
circuit theory book if you don't believe me.) In other "words"...
I = (10/R) * exp (-2*t/(R*C))
P = R*I^2 = (100/R) * exp (-4*t/(R*C))
To get the energy dissipated in the resistor, integrate power over time from
0 to infinity...
/oo oo
E = (100/R) * | exp(-4*t/(R*C)) dt = (100/R) * (-R*C/4)exp(-4*t/(R*C)) ]
/ 0 0
= (100/R) * ( 0 - (-R*C/4)) which is, (voila!)
= 25*C
2 Farads? There's your missing 50 joules - in the resistor.
(Note that the factor of 25 is, more generally, 1/4 of V^2, where V is the
initial voltage difference between the capacitors.)
It's not at all hard for me to believe that in the resistanceless case, the
energy is carred off by radiation - anybody want to volunteer to do that
analysis? My Maxwell's eqs are a bit rusty.
Jeff Winslowhenry@utzoo.uucp (Henry Spencer) (06/04/89)
In article <747@eplunix.UUCP> raoul@eplunix.UUCP (Otero) writes: >What about superconducting materials? No resistive heat dissipation at all. Nope, sorry, wrong: superconductors have zero resistance only for DC. Running AC through a superconductor does produce heat. -- You *can* understand sendmail, | Henry Spencer at U of Toronto Zoology but it's not worth it. -Collyer| uunet!attcan!utzoo!henry henry@zoo.toronto.edu
leonard@bucket.UUCP (Leonard Erickson) (06/05/89)
In article <20772@quacky.mips.COM> 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.
<
<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 ?
I remember seeing this in an old physics text. The answer is that it radiates
the energy as EM radiation. After all, the energy is going to surge back and
forth for a while in any *real* system.
--
Leonard Erickson ...!tektronix!reed!percival!bucket!leonard
CIS: [70465,203]
"I'm all in favor of keeping dangerous weapons out of the hands of fools.
Let's start with typewriters." -- Solomon Shortarnief@tekgvs.LABS.TEK.COM (Arnie Frisch) (06/05/89)
In article <1989Jun3.211635.2752@utzoo.uucp>, henry@utzoo.uucp (Henry Spencer) writes: > In article <747@eplunix.UUCP> raoul@eplunix.UUCP (Otero) writes: > >What about superconducting materials? No resistive heat dissipation at all. > > Nope, sorry, wrong: superconductors have zero resistance only for DC. > Running AC through a superconductor does produce heat. > -- > You *can* understand sendmail, | Henry Spencer at U of Toronto Zoology Also, superconductors can not conduct infinite currents, the maximum current is limited by the magnetic field it produces - which ultimately causes the conductor to go "normal". 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!
hall@vice.ICO.TEK.COM (Hal Lillywhite) (06/05/89)
In article <11888@eddie.MIT.EDU> jbs@fenchurch.UUCP (Jeffrey Siegal) writes: >Anyone know how close to "perfect" toroidal superconducting inductors >are? If they are carrying infinite (or even very large) current, there ain't no such animal! The critical current will be exceeded and the superconductivity destroyed. This will effectively prevent a zero resistance short between 2 capacitors. Nice to think about but according to currently understood physics impossible to achieve.
dbell@cup.portal.com (David J Bell) (06/07/89)
In article <20772@quacky.mips.COM> 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. } }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 ? [etc., etc.... much discussion omitted...] Hmmm... U = 1/2 C V^2 (V = volts) Ke = 1/2 M V^2 (V = velocity) Lessee: Two billiard balls, equal mass, one traveling... etc... In other words, discharging one cap into the other through a *real* world circuit is equivalent to elastic collision and dissipated energy. Whataya know; duality everywhere! Dave dbell@cup.portal.com
knudsen@ihlpl.ATT.COM (Knudsen) (06/08/89)
A simple mechanical analogy: Imagine a silly-putty meteor weighing 1 kg going 100 m/s. It has kinetic energy 0.5 m v**2 or 5000 joules. It has momentum mv = 100 kgm/s. It collides head-on with an identical stationary meteor, and the two stick together (perectly in-elastic collision). To conserve momentum, the conglomerated 2 kg meteor now moves at half the original speed, or 50 m/s. The kinetic energy of the pair has dropped to 2500 J. Half the original kinetic energy was dissipated inside the bodies of the meteors, as frictional vibration and ultimately heat (funny how everything ends up as heat, but you knew that). Rubber meteors would be more fun. Likewise, try connecting the two caps not thru a shorting wire but an inductor. Like slipping a stiff coil spring between the meteors before they collide, and they're covered with flypaper glue... -- Mike Knudsen Bell Labs(AT&T) att!ihlpl!knudsen knudsen@ihlpl.att.com Round and round the while() loop goes; "Whether it stops," Turing says, "no one knows!"
bold@richsun.UUCP (Jason Bold) (06/09/89)
>In article <20772@quacky.mips.COM> vaso@mips.COM (Vaso Bovan) writes: >>A Paradox of Capacitor Energy Storage >>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. Let me see if I can take a crack at this one... To start: Energy = 100 joules, c = 2uF. To finish: Energy = 100 joules, c = 4uF. Assume t = oo to forget about all of these infinite oscillation theories. So, E1 = (1/2)CV^2 = 100 joules = E2, by conservation of energy. E2 = (1/2)(4uF)(V2^2) = 100 joules V2^2 = 100/[(1/2)(4uF)] = 50 V^2 :: V2 = 5*sqrt(2) = 7.07V. Well, assuming a lossless circuit, noone out there can deny this law of physics. Let's see if conservation of charge still holds. It should. Q=CV: Q1 = C1V1 = (2uF)(10V) = Q2, based on conservation of charge. Well, if you work it out, V2 does come out to be 5V. Gee, what went wrong? We should have gotten the same answer both times. It must be in the definitions of Q1,Q2,V1,V2,E1, and E2. Q1 = the charge on C1 before the connection Q2 = the charge on C1 and C2 after the connection V1 = the voltage on C1 before the connection V2 = the voltage on C1 and C2 after the connection E1 = total energy in the system before the connection E2 = total energy in the system after the connection the system = capacitors C1 and C2 I think the problem lies in the fact that Q1,Q2,V1, and V2 are vectors and E1 and E2 are scalars. For example, what would happen if you connected another capacitor charged to 10V. in parallel with the first one (C1), but in opposite polarity? Well, I think everyone would agree that the charges would cancel. This would mean no energy left (E2). Where did it go. Well, E=(1/2)mv^2. I think it is the energy required to move the electrons from one plate to the other. If any electron moves at all, energy must be expended. That's the only explanation I can come up with. Anyone tried this experimentally (unfortunately intuitively, I think the voltage V2 is 7.07V)., I would like to know. -- ============================================================================== = Jason Bold "A trend monger is a person who dreams up a trend... and = = bold@richsun.UUCP spreads it throughout the land using all the frightening = = little skills that science has made available." - FZ =
bold@richsun.UUCP (Jason Bold) (06/16/89)
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 holds true is q=CV, q is a constant. I believe that NO electrons will go flying off into space. Look at the problem from the reverse angle. In other words, what happens if you start with two capacitors connected together at 2uF apiece, charged up to 5.0V? Does everyone agree that it will take a finite amount of work/energy to move the charge from one capacitor to the other such that the first capacitor is completely discharged? Then just remove it from the circuit by disconnecting the two. Viola, you are back at your original situation! I would guess that the amount of energy required to charge the second capacitor to 10V. and leave the first one discharged is the amount of energy that is "lost?" when the two are brought together. The REAL question is "Where does the energy go?". Energy and work are the same thing, so it is the work required to move half of the electrons from one capacitor to the other. It is somewhat akin to the following scenario: Assume you have a 10lb. ball suspended 1ft. above the ground by a string. There is potential energy stored there. If you cut the string (we are assuming that 'cutting' the string takes no energy), and the ball falls to the ground, then where did the energy go? There is no potential energy or kinetic energy left! But WAIT, you say!!! If this is an ideal system, the ball would bounce forever, or else the energy is dissipated when it goes smack against the ground. That is correct. It's the same thing everyone has been saying about the system oscillating forever. It would with no damping factor, ie. a resistor, or damping pot, in this case, soft ground. So, my conclusion is that if there is no damping factor, a resistor, the potential and kinetic energies constantly are changing forever. the "lost" energy is the energy that is being used to keep the circuit oscillating. It will require exactly this amount of energy to completely stop the oscillations. I believe someone else has already posted a proof to this. So, the case of the missing joules has been solved. The potential energy stored in the capacitor has been (50J of it anyway) converted into kinetic energy (the ocillations). Therefore, both q=CV and E=1/2(CV^2) still hold. It's just that now, with an inductor (no loss), E= 1/2(CV^2) + 1/2(LI^2). Any comments? -- ============================================================================== = Jason Bold "A trend monger is a person who dreams up a trend... and = = bold@richsun.UUCP spreads it throughout the land using all the frightening = = little skills that science has made available." - FZ =
mmm@cup.portal.com (Mark Robert Thorson) (07/22/89)
When I was a kid, I was given a rechargable flashlight that didn't work anymore. It was a crude affair on the inside, just a bridge rectifier, a cap, and a current-limiting resistor to trickle charge the batteries (no step- down transformer, just the resistor). I wired these components into a cardboard box with a set of Fahrenstock clips bolted to the top. Now I have a crude^squared battery charger. I also had two batteries, which I wedged in the clips so they could charge while I was at school. When I came home, one of the batteries was in pieces on the floor, which seemed odd. I assumed it fell off the table, but I didn't yet realize why the cap and some of the innards had come off the case. The other battery didn't seem to have any charge on it, so I took it to the garage to open it up and see how it was put together. By the way, the case of this other battery was bulging, but for some reason that didn't register in my mind. I put the battery on one of the concrete steps of the garage and hit it with a hammer. I have never heard such a loud bang as the one that battery made. My ears were still ringing the next day.