Magic capacitors!
in [Design]
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From: Jim Thompson on 21 Jul 2010 14:50 On 21 Jul 2010 13:35:02 -0500, The Phantom <phantom(a)aol.com> wrote: >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). >> >>If you want to make it easier assume C2 is initially discharged. >> >>Is the energy before and after the same? If not explain why and why it is >>not a violation of the conservation of energy law. > >This is an old puzzle that has been dealt with before. The "lost" energy can be >carried away from the circuit in various ways. Even if the conductors have no >resistance, electromagnetic radiation will carry away energy. > >See these papers: > >http://www.hep.princeton.edu/~mcdonald/examples/EM/boykin_ajp_70_415_02.pdf > >http://www.hep.princeton.edu/~mcdonald/examples/EM/mayer_ieeete_36_307_93.pdf > >http://iopscience.iop.org/0143-0807/30/1/007/pdf/ejp9_1_007.pdf > >http://arxiv.org/PS_cache/arxiv/pdf/0910/0910.5279v1.pdf Yep. There is no such thing as black magic, energy IS conserved AND charge IS conserved. In Message-ID: <3b893612tjjndo8o4v1evro050nonjgp41(a)4ax.com> John Larkin says, "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." So John is either a liar who likes to hear his own head roar, or he's dumb as a stump. Proof is forthcoming. John thinks I'm bluffing, but I'm prettying it up so I can use it in my book as an example of extreme asininity in engineering thought. (For John's lawyers: Exact recitation of fact IS NOT libel or slander :-) ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Spice is like a sports car... Only as good as the person behind the wheel.
From: Tim Williams on 21 Jul 2010 15:10 <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. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: John Larkin on 21 Jul 2010 15:14 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
From: Jim Thompson on 21 Jul 2010 15:30 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 :-( ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Spice is like a sports car... Only as good as the person behind the wheel.
From: The Phantom on 21 Jul 2010 15:30
On Wed, 21 Jul 2010 14:10:57 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: ><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). In the R = 0 case, L is not zero and the energy is lost by radiation. See: http://www.hep.princeton.edu/~mcdonald/examples/EM/boykin_ajp_70_415_02.pdf >3. Because the answer is not equal under different conditions, the limit is undefined. QED. The final result is the same in both cases. The "hypothetical mathematical model" George proposes still has an energy loss mechanism as explained in the Boykin, et al, paper. There is no mystery. > >Tim |