Magic capacitors!
in [Design]
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From: John Larkin on 21 Jul 2010 15:57 On Wed, 21 Jul 2010 12:59:58 -0500, "George Jefferson" <phreon111(a)gmail.com> wrote: > > >"Tim Williams" <tmoranwms(a)charter.net> wrote in message >news:azG1o.20990$lS1.2654(a)newsfe12.iad... >> "Michael F�rtsch" <michael.foertsch(a)chello.at> wrote in message >> news:57c17$4c46f981$5472c223$14009(a)news.chello.at... >>> The total electrical energy after the charge has been transfered is 50% >>> of the electrical energy that was stored in C1. The residual 50% had >>> been handed to the government as a charge transfer tax. >> >> Wrong -- the inductance and resistance of the circuit is unspecified, so >> 1., there is no correct answer, and 2. it's a nonphysical circuit, so not >> only is it unspecified, it's meaningless. > >WRONG. That has nothing to do with it. We can "specify" that the inductance >and resistance is 0. Just because this is physically not possible does not >mean we cannot create such a hypothetical mathematical model. > > Sorry, I don't do singularities. They are against my religion. John
From: Tim Williams on 21 Jul 2010 16:56 "The Phantom" <phantom(a)aol.com> wrote in message news:2cie4650u0spseb6pb1rhs9oaed2eh6hk2(a)4ax.com... > In the R = 0 case, L is not zero and the energy is lost by radiation. See: The presumption is radiation is able to leave. Expressing radiation as a lumped constant equivalent series resistance, assume zero ESR. In other words, inside a superconducting box, so there is no way to measure the resonance (it's a theoretical problem, that's okay) and it resonates forever. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: George Jefferson on 21 Jul 2010 17:16 <keithw86(a)gmail.com> wrote in message news:ed6b7de0-5287-47e4-9d4b-217880b3c2e5(a)5g2000yqz.googlegroups.com... > On Jul 21, 12:59 pm, "George Jefferson" <phreon...(a)gmail.com> wrote: >> "Tim Williams" <tmoran...(a)charter.net> wrote in message >> >> news:azG1o.20990$lS1.2654(a)newsfe12.iad... >> >> > "Michael F�rtsch" <michael.foert...(a)chello.at> wrote in message >> >news:57c17$4c46f981$5472c223$14009(a)news.chello.at... >> >> The total electrical energy after the charge has been transfered is >> >> 50% >> >> of the electrical energy that was stored in C1. The residual 50% had >> >> been handed to the government as a charge transfer tax. >> >> > Wrong -- the inductance and resistance of the circuit is unspecified, >> > so >> > 1., there is no correct answer, and 2. it's a nonphysical circuit, so >> > not >> > only is it unspecified, it's meaningless. >> >> WRONG. That has nothing to do with it. We can "specify" that the >> inductance >> and resistance is 0. Just because this is physically not possible does >> not >> mean we cannot create such a hypothetical mathematical model. > > Ok, give us a mathematical model of "divide by zero". 1/0? I don't see what this has to do with the problem though... I'm sure you do... of course this assumes you are right.
From: George Jefferson on 21 Jul 2010 17:23 "Tim Williams" <tmoranwms(a)charter.net> wrote in message news:2DH1o.44763$Ls1.44697(a)newsfe11.iad... > <keithw86(a)gmail.com> wrote in message > news:ed6b7de0-5287-47e4-9d4b-217880b3c2e5(a)5g2000yqz.googlegroups.com... >> Ok, give us a mathematical model of "divide by zero". > > Since we're taking the limit of two variables (R --> 0 and L --> 0), this > is a multivariate calculus problem, and therefore it matters which > variables we take to zero first, or in what proportion. > > It is only necessary to find two different answers to prove the limit does > not exist, so we shall take simple directions (R and L axes, > respectively). > > 1. IFF the limits are equal, then the limit exists. > 2. In the L = 0 case, half the energy disappears (E = 1/2 Eo); in the R = > 0 case, the energy remains (E = Eo). > 3. Because the answer is not equal under different conditions, the limit > is undefined. QED. > You have no clue. The "lost" energy must go somewhere. It goes into moving the charge from one plate to another. If there is no resistive force then no energy is lost in the resistance. If there were a resistance then it would reduce the overall voltage and that loss would be converted into heat in the resistance. When the switch is closed the negative charge must move from one plate to another. Hence a force exists and because of the finite length between the plates work is done. Regardless of how the caps are connected the there will be work done in moving the charge. The link provided by phantom shows that in the ideal case the energy is radiated away. This should make absolute sense as the medium surrounding the wire must absorb the energy(hence it is radiated). Your "proof" is just pure nonsense. The idea is that you must put in half the energy to move half the charge(which is 1/4 the original energy) and regardless of how the energy is dissipated it must be done in some way. For an ideal wire it obviously can't be done through heat.
From: George Jefferson on 21 Jul 2010 17:30
"Jim Thompson" <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote in message news:rcie465itdu34iaajm1itdqslepu2i87r6(a)4ax.com... > On Wed, 21 Jul 2010 12:14:04 -0700, John Larkin > <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>On Wed, 21 Jul 2010 12:17:41 -0500, "George Jefferson" >><phreon111(a)gmail.com> wrote: >> >>> >>> >>>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in >>>message >>>news:dj7e465sga7fe3nq7hfl3f0uk601pvrem8(a)4ax.com... >>>> On Wed, 21 Jul 2010 11:19:31 -0500, "George Jefferson" >>>> <phreon111(a)gmail.com> wrote: >>>> >>>>> >>>>> >>>>>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in >>>>>message >>>>>news:s43e46la1p1vt11527eg3ptl9ulm44dfrj(a)4ax.com... >>>>>> On Wed, 21 Jul 2010 07:54:03 -0500, "George Jefferson" >>>>>> <phreon111(a)gmail.com> wrote: >>>>>> >>>>>>>Suppose you have two capacitors connected as >>>>>>> >>>>>>>--*-- >>>>>>>| | >>>>>>>C1 C2 >>>>>>>| | >>>>>>>----- >>>>>>> >>>>>>>where * is a switch. >>>>>>> >>>>>>>What is the total energy before and after the switch is closed(in >>>>>>>general). >>>>>> >>>>>> Energy is conserved, so it's the same, if you account for all the >>>>>> manifestations of energy. >>>>>> >>>>> >>>>>You didn't answer the question. I assume this because you don't know. >>>>> >>>> >>>> State the question unambiguously and I will. >>>> >>>> As I said, the puzzle is both ancient and trivial, so probably JT >>>> invented it. There are web sites and even academic papers devoted to >>>> it. Given all that, how could I not understand it? >>>> >>> >>>Um you don't get it. Your ignorance in basic electronics amazes me. >> >>That's funny. But people can choose to be amazed in all sorts of ways. >> >> >> Michael >>>got it(although he didn't explain where the energy went but I think gets >>>it). >>> >>>Assume the second cap is initially "uncharged" and has the same >>>capacitance >>>as the first. >>> >>>Then the initial energy is >>> >>>Wi = 1/2*C*V^2 >>>Wf = 2*1/2*C*(V/2)^2 = 1/4*C*V^2 = 1/2*Wi >>> >>>Hence the final energy of the system 1/2 what we started with. >> >>Miraculous calculation. Yours and about 300 web sites that admire this >>puzzle. >> >>You didn't wxplain where the energy went - see those 300 web sites - >>but you are assuming losses. Another solution is that no energy is >>lost, and it rings forever, in which case the final state that you >>cite never happens. The exact waveforms are actually interesting. >> >>> >>>I'd really like to hear your explanation but I know thats impossible(as >>>you'll steal someone elses). After all your the one that believes charge >>>isn't conserved... heres your change to *prove* it. >>> >>> >>> >> >>Check my previous posts. I noted the exact waveform across a resistive >>switch, for any values of C1 and C2, and an independent way to compute >>the energy lost in that switch. >> >>Given an inductor, one can move all the energy from one charged cap to >>another, uncharged one. If the C values are unequal, the C*V (charge) >>on the first cap obviously becomes a different C*V on the second one. >>I noted that here some weeks ago, too. >> >>This is all EE101 stuff. >> >>John > > Let the hedging begin... > > In Message-ID: <3b893612tjjndo8o4v1evro050nonjgp41(a)4ax.com> > > You said: > > "Right. If you dump all the energy from one charged cap into another, > discharged, cap of a different value, and do it efficiently, charge is > not conserved." > > Note the NOT CONSERVED. > > Now you say, "...the C*V (charge) on the first cap obviously becomes a > different C*V on the second one". > > Where did the charge come from/go to? > > John "The Bloviator" Larkin is totally incapable of admitting error. > > I truly suspect you're too ignorant to understand :-( > I'm glad my post got what it was suppose to get out. I kinda feel like Breitbart. |