Magic capacitors!
in [Design]
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From: markp on 4 Aug 2010 11:55 "John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:2vge46h4sragrk4jdn6sasde6hg2r52nos(a)4ax.com... > 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 Yes, Q=CV equation is somewhat misleading in this context. A capacitor doesn't store electrical charge, it stores energy. This is a very common misconception, when we say 'charge a capacitor' we don't mean put electrical charge into it, we mean put energy into it. The plates are equal and opposite in electrical charge due to an abundance of electrons on one plate and an equal and opposite charge on the other. The total stored electrical charge in a capacitor is zero, and the Q=CV equation relates to how much charge flowed *in and out* of the capacitor (in fact since electrons can't cross the barrier between the plates, it actually describes the *modulus* of the abundance of charge on each plate, one abundance is positive and the other is negative). Mark.
From: markp on 4 Aug 2010 13:13 "John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:6ese46h8ne8luliofpjilue4af6d514js1(a)4ax.com... > On Wed, 21 Jul 2010 16:30:20 -0500, "George Jefferson" > <phreon111(a)gmail.com> wrote: > >> >> >>"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. >> > > Initial condition: > > C1 2F, 1 volt, 1 joule, 2 coulombs > > C2 1F, 0 volts, 0 joules, 0 coulombs > > Now remove all the energy from C1 and deliver it to C2. An inductor > will move the energy nicely. > > Now > > C1 has 0 volts, 0 joules, 0 coulombs > > C2 has 1.414 volts, 1 joule, 1.414 coulombs. > > John Sorry but I'm afraid you've fallen into the 'capacitor stores charge' trap John. It doesn't, as I said in another post, capacitors don't store charge, the plates are equal and opposite in the excess or depletion of electrons (they can't store electrical charge, since we all know the current going in and the current coming out of a capacitor is equal, and current is Coulombs per second). The total stored electrical charge in a capacitor is zero. That negates the whole premise of this analysis. Mark.
From: Grant on 4 Aug 2010 18:41 On Wed, 4 Aug 2010 18:13:49 +0100, "markp" <map.nospam(a)f2s.com> wrote: > >"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >news:6ese46h8ne8luliofpjilue4af6d514js1(a)4ax.com... >> On Wed, 21 Jul 2010 16:30:20 -0500, "George Jefferson" >> <phreon111(a)gmail.com> wrote: >> >>> >>> >>>"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. >>> >> >> Initial condition: >> >> C1 2F, 1 volt, 1 joule, 2 coulombs >> >> C2 1F, 0 volts, 0 joules, 0 coulombs >> >> Now remove all the energy from C1 and deliver it to C2. An inductor >> will move the energy nicely. >> >> Now >> >> C1 has 0 volts, 0 joules, 0 coulombs >> >> C2 has 1.414 volts, 1 joule, 1.414 coulombs. >> >> John > >Sorry but I'm afraid you've fallen into the 'capacitor stores charge' trap >John. It doesn't, as I said in another post, capacitors don't store charge, >the plates are equal and opposite in the excess or depletion of electrons >(they can't store electrical charge, since we all know the current going in >and the current coming out of a capacitor is equal, and current is Coulombs >per second). The total stored electrical charge in a capacitor is zero. That >negates the whole premise of this analysis. > >Mark. Charge 1uF to 50V and put your tongue across the terminals, now tell me there's nothing there ;) Grant. >
From: Nunya on 4 Aug 2010 19:24 On Aug 4, 3:41 pm, Grant <o...(a)grrr.id.au> wrote: > On Wed, 4 Aug 2010 18:13:49 +0100, "markp" <map.nos...(a)f2s.com> wrote: > > >"John Larkin" <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote in message > >news:6ese46h8ne8luliofpjilue4af6d514js1(a)4ax.com... > >> On Wed, 21 Jul 2010 16:30:20 -0500, "George Jefferson" > >> <phreon...(a)gmail.com> wrote: > > >>>"Jim Thompson" <To-Email-Use-The-Envelope-I...(a)On-My-Web-Site.com> wrote > >>>in > >>>messagenews:rcie465itdu34iaajm1itdqslepu2i87r6(a)4ax.com... > >>>> 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: > > >>>>>>"John Larkin" <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote in > >>>>>>message > >>>>>>news:dj7e465sga7fe3nq7hfl3f0uk601pvrem8(a)4ax.com... > >>>>>>> On Wed, 21 Jul 2010 11:19:31 -0500, "George Jefferson" > >>>>>>> <phreon...(a)gmail.com> wrote: > > >>>>>>>>"John Larkin" <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote in > >>>>>>>>message > >>>>>>>>news:s43e46la1p1vt11527eg3ptl9ulm44dfrj(a)4ax.com... > >>>>>>>>> On Wed, 21 Jul 2010 07:54:03 -0500, "George Jefferson" > >>>>>>>>> <phreon...(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: <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 :-( > > >>>I'm glad my post got what it was suppose to get out. I kinda feel like > >>>Breitbart. > > >> Initial condition: > > >> C1 2F, 1 volt, 1 joule, 2 coulombs > > >> C2 1F, 0 volts, 0 joules, 0 coulombs > > >> Now remove all the energy from C1 and deliver it to C2. An inductor > >> will move the energy nicely. > > >> Now > > >> C1 has 0 volts, 0 joules, 0 coulombs > > >> C2 has 1.414 volts, 1 joule, 1.414 coulombs. > > >> John > > >Sorry but I'm afraid you've fallen into the 'capacitor stores charge' trap > >John. It doesn't, as I said in another post, capacitors don't store charge, > >the plates are equal and opposite in the excess or depletion of electrons > >(they can't store electrical charge, since we all know the current going in > >and the current coming out of a capacitor is equal, and current is Coulombs > >per second). The total stored electrical charge in a capacitor is zero. That > >negates the whole premise of this analysis. > > >Mark. > > Charge 1uF to 50V and put your tongue across the terminals, now tell me > there's nothing there ;) > > Grant. > > Idiot! His point was/is that it stores ENERGY, NOT charge. His detail of the process was concise enough for a monkey to understand it. The fact that you did not is quite a tell. The correct term would be "fill" a 1µF cap to 50V. Another proof is that it takes POWER to "charge" a battery. It does NOT take any power to "fill" a cap. It is merely a temporary transference point for an energy "packet". If there is a load on the cap, THEN it will take power to TRANSFER energy THROUGH the cap.
From: markp on 4 Aug 2010 19:08
"Grant" <omg(a)grrr.id.au> wrote in message news:s3rj565jd7in972eu9uufgk3cufi9ni444(a)4ax.com... > On Wed, 4 Aug 2010 18:13:49 +0100, "markp" <map.nospam(a)f2s.com> wrote: > >> >>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in >>message >>news:6ese46h8ne8luliofpjilue4af6d514js1(a)4ax.com... >>> On Wed, 21 Jul 2010 16:30:20 -0500, "George Jefferson" >>> <phreon111(a)gmail.com> wrote: >>> >>>> >>>> >>>>"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. >>>> >>> >>> Initial condition: >>> >>> C1 2F, 1 volt, 1 joule, 2 coulombs >>> >>> C2 1F, 0 volts, 0 joules, 0 coulombs >>> >>> Now remove all the energy from C1 and deliver it to C2. An inductor >>> will move the energy nicely. >>> >>> Now >>> >>> C1 has 0 volts, 0 joules, 0 coulombs >>> >>> C2 has 1.414 volts, 1 joule, 1.414 coulombs. >>> >>> John >> >>Sorry but I'm afraid you've fallen into the 'capacitor stores charge' trap >>John. It doesn't, as I said in another post, capacitors don't store >>charge, >>the plates are equal and opposite in the excess or depletion of electrons >>(they can't store electrical charge, since we all know the current going >>in >>and the current coming out of a capacitor is equal, and current is >>Coulombs >>per second). The total stored electrical charge in a capacitor is zero. >>That >>negates the whole premise of this analysis. >> >>Mark. > > Charge 1uF to 50V and put your tongue across the terminals, now tell me > there's nothing there ;) > > Grant. I didn't say a capacitor can't store energy, it certainly can. When you say 'charge 1uF to 50V' the 'charge' bit refers to putting energy into the capacitor, not electrical charge. When you put your tongue across the terminals you form an electrical circuit, the plate with excess electrons pushes electrons through the tongue and the plate with a depletion of electrons sucks exactly the same number of electrons back in. The capacitor at the end of that 'discharge' cycle has the same number of electrons in it as it had when it was 'charged'. Here's a good explanation of capacitors: http://amasci.com/emotor/cap1.html Mark. |