Magic capacitors!
in [Design]
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From: Don Lancaster on 3 Aug 2010 18:26 On 7/21/2010 12:30 PM, Jim Thompson wrote: > 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 If you have a resistor between the capacitors,half the energy is dissipated. The math does NOT depend on the resistance value. In the case of zero resistance, you get infinite current and a huge spark that radiates the equivalent energy. However, you can charge the second capacitor more efficiently with an inductor in a resonant circuit. -- Many thanks, Don Lancaster voice phone: (928)428-4073 Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552 rss: http://www.tinaja.com/whtnu.xml email: don(a)tinaja.com Please visit my GURU's LAIR web site at http://www.tinaja.com
From: JosephKK on 3 Aug 2010 23:49 On Tue, 03 Aug 2010 15:26:13 -0700, Don Lancaster <don(a)tinaja.com> wrote: >On 7/21/2010 12:30 PM, Jim Thompson wrote: >> 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: >>> <snip> >> >> 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 > > >If you have a resistor between the capacitors,half the energy is >dissipated. The math does NOT depend on the resistance value. Sorry Don. that does not stand up to the physics or the arithmetic. > >In the case of zero resistance, you get infinite current and a huge >spark that radiates the equivalent energy. Energy may be radiated, but not by the switch. Nor is there a requirement for a spark. > >However, you can charge the second capacitor more efficiently with an >inductor in a resonant circuit. Not part of the question.
From: Jim Thompson on 3 Aug 2010 23:56 On Tue, 03 Aug 2010 20:49:23 -0700, "JosephKK"<quiettechblue(a)yahoo.com> wrote: >On Tue, 03 Aug 2010 15:26:13 -0700, Don Lancaster <don(a)tinaja.com> >wrote: > >>On 7/21/2010 12:30 PM, Jim Thompson wrote: >>> 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: >>>> ><snip> >>> >>> 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 >> >> >>If you have a resistor between the capacitors,half the energy is >>dissipated. The math does NOT depend on the resistance value. > >Sorry Don. that does not stand up to the physics or the arithmetic. Sorry, JosephKK, But it does. Shame :-( >> >>In the case of zero resistance, you get infinite current and a huge >>spark that radiates the equivalent energy. > >Energy may be radiated, but not by the switch. Nor is there a >requirement for a spark. >> >>However, you can charge the second capacitor more efficiently with an >>inductor in a resonant circuit. > >Not part of the question. Do the math JosephKK. You are a disappointment :-( Newbies. If you think differently, go into sales ;-) ...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... Performance only as good as the person behind the wheel.
From: Nunya on 4 Aug 2010 01:11 On Aug 3, 8:49 pm, "JosephKK"<quiettechb...(a)yahoo.com> wrote: > On Tue, 03 Aug 2010 15:26:13 -0700, Don Lancaster <d...(a)tinaja.com> > wrote: > > > > > > >On 7/21/2010 12:30 PM, Jim Thompson wrote: > >> On Wed, 21 Jul 2010 12:14:04 -0700, John Larkin > >> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > > >>> On Wed, 21 Jul 2010 12:17:41 -0500, "George Jefferson" > >>> <phreon...(a)gmail.com> wrote: > > <snip> > > >> Let the hedging begin... > > >> In Message-ID:<3b893612tjjndo8o4v1evro050nonjg...(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 > > >If you have a resistor between the capacitors,half the energy is > >dissipated. The math does NOT depend on the resistance value. > > Sorry Don. that does not stand up to the physics or the arithmetic. > > > > >In the case of zero resistance, you get infinite current and a huge > >spark that radiates the equivalent energy. > > Energy may be radiated, but not by the switch. Nor is there a > requirement for a spark. You are an idiot. If the two contacts are to be closed, and the resistance is zero, there WILL be a spark BEFORE closure occurs. And in ANY case of ANY spark, there WILL be radiation. > >However, you can charge the second capacitor more efficiently with an > >inductor in a resonant circuit. > > Not part of the question. As if you had any grasp of even that.
From: markp on 4 Aug 2010 11:19
"Nunya" <jack_shephard(a)cox.net> wrote in message news:52dc8249-58f7-424b-a63e- <snip> >You are an idiot. If the two contacts are to be closed, and the > resistance is zero, there WILL be a spark BEFORE closure > occurs. And in ANY case of ANY spark, there WILL be radiation. <snip> A 'spark' needs some medium to form, in air this spark is caused by electron avalanche. Since it involves electrons moving and being accelerated it will take time. The question didn't say how fast the switch was closed, it could be much faster than than the time for the avalanche to happen. Also there's no requirement for any air at all, a vacuum is a pretty good insulator. With relatively low voltages and an absense of any free electrons (like no sharp points like dust) there wouldn't be a spark in a vacuum: http://www.newton.dep.anl.gov/newton/askasci/1993/physics/PHY102.HTM Mark. |