Magic capacitors!
in [Design]
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From: John Larkin on 4 Aug 2010 22:33 On Wed, 04 Aug 2010 18:46:21 -0700, CIC <cicel(a)iinet.com> wrote: > >On Wed, 04 Aug 2010 18:00:20 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>On Thu, 05 Aug 2010 08:41:59 +1000, Grant <omg(a)grrr.id.au> wrote: >> >>>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. >>>> >> >>Right. And battery chargers are an elaborate public fraud, since >>batteries don't store charge. >> >>And power mosfet data sheets are all wrong. >> >>John >> > >I don't see how this applies to my flux capacitor though... > >http://kalecoauto.com/index.php?main_page=product_info&cPath=6&products_id=28 Isopropyl alcohol can fix that. John
From: Nunya on 4 Aug 2010 22:39 On Aug 4, 6:03 pm, John Larkin <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > On Wed, 4 Aug 2010 16:24:58 -0700 (PDT), Nunya <jack_sheph...(a)cox.net> > wrote: > > > > >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. > > Maybe where you work. > > > > > 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. > > Cool. You can somehow put 600 volts across (or into?) a 10,000 uF cap > without using any power to do it? I'm impressed. > > John You really don't know why those shorting wires get put onto caps, do ya? It all depends on how fast you want it to get filled. In the long run, it does require the same power in both cases. So, yes, the internal resistance of the cap would determine the power required to fill it quickly. You missed the point.
From: John Larkin on 4 Aug 2010 22:42 On Thu, 5 Aug 2010 02:40:12 +0100, "markp" <map.nospam(a)f2s.com> wrote: > >"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >news:ci3k56d0kga1776gghosaq09q2e0i2ahhq(a)4ax.com... >> On Wed, 4 Aug 2010 16:55:16 +0100, "markp" <map.nospam(a)f2s.com> wrote: >> >>> >>>"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. >>> >> >> That's not what they taught us in college, and that's not the way we >> do engineering. We say that a capacitor stores charge, the amount >> being C*V in coulombs, and it works. My whole point, which has evoked >> such ranting, is that when you use this convention, be careful about >> designing using the concept that (this kind of) charge is always >> conserved. >> >> John >> > >With all due respect, we don't, and shouldn't, say a capacitor 'stores >charge'. What do you mean by "we"? Electronic design engineers do this all the time, with reference to capacitors and batteries. You pump X coulombs into a cap; it becomes stored, coincidentally as C*V. You can extract those coulombs later, and the accounting is correct. The math works. The gear works. The misconception comes from the use of the word charge when >talking about putting energy into a capacitor, and more explictly the >significant lack of clarification given on this when being taught. This is >compounded by a confusion of the q=C*V equation which actually relates to >the charge on the plates, but one of the plates is of the same value but >opposite in polarity, so the sum of those is zero. This is an extremely >popular misunderstanding unfortunately, and leads to conclusions that >electrical charge is not conserved. In fact, in a closed system where no >electrical charge can get in or out, within that system electrical charge >*is* conserved, it's actually a fundamental law of physics (along with >conservation of energy and momentum, again for closed systems). > >The same current flows in and out of a capacitor when it is being 'charged' >(I assume you are not going to deny that). Note I said the same current, but >they are not made of the same electrons because those can't cross the plate >barrier. The same amount of electrical charge that goes in comes right out >again. How can the capacitor possibly end up with a net charge in it? If it >can, where has the electrical charge come from? Have electrons just been >conjured up out of nowhere? It's a different convention. Words. But the units work and the numbers work, so we use it. Call our kind of charge "charge separation" or "plate charge differential" if it makes you happier. Do you design electronics? Do mosfet data sheets refer to stored gate energy, or stored gate charge? John
From: John Larkin on 4 Aug 2010 22:46 On Wed, 4 Aug 2010 19:39:55 -0700 (PDT), Nunya <jack_shephard(a)cox.net> wrote: > It all depends on how fast you want it to get filled. In the >long run, it does require the same power in both cases. >So, yes, the internal resistance of the cap would >determine the power required to fill it quickly. You >missed the point. That's wronger than Always Wrong! That's Olympic class wrongness! John
From: markp on 5 Aug 2010 08:20
"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:889k5654o0h9qfgs3cej7gfe99ahsg42am(a)4ax.com... > On Thu, 5 Aug 2010 02:40:12 +0100, "markp" <map.nospam(a)f2s.com> wrote: > >> >>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in >>message >>news:ci3k56d0kga1776gghosaq09q2e0i2ahhq(a)4ax.com... >>> On Wed, 4 Aug 2010 16:55:16 +0100, "markp" <map.nospam(a)f2s.com> wrote: >>> >>>> >>>>"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. >>>> >>> >>> That's not what they taught us in college, and that's not the way we >>> do engineering. We say that a capacitor stores charge, the amount >>> being C*V in coulombs, and it works. My whole point, which has evoked >>> such ranting, is that when you use this convention, be careful about >>> designing using the concept that (this kind of) charge is always >>> conserved. >>> >>> John >>> >> >>With all due respect, we don't, and shouldn't, say a capacitor 'stores >>charge'. > > What do you mean by "we"? Electronic design engineers do this all the > time, with reference to capacitors and batteries. You pump X coulombs > into a cap; it becomes stored, coincidentally as C*V. You can extract > those coulombs later, and the accounting is correct. The math works. > The gear works. > > The misconception comes from the use of the word charge when >>talking about putting energy into a capacitor, and more explictly the >>significant lack of clarification given on this when being taught. This is >>compounded by a confusion of the q=C*V equation which actually relates to >>the charge on the plates, but one of the plates is of the same value but >>opposite in polarity, so the sum of those is zero. This is an extremely >>popular misunderstanding unfortunately, and leads to conclusions that >>electrical charge is not conserved. In fact, in a closed system where no >>electrical charge can get in or out, within that system electrical charge >>*is* conserved, it's actually a fundamental law of physics (along with >>conservation of energy and momentum, again for closed systems). >> >>The same current flows in and out of a capacitor when it is being >>'charged' >>(I assume you are not going to deny that). Note I said the same current, >>but >>they are not made of the same electrons because those can't cross the >>plate >>barrier. The same amount of electrical charge that goes in comes right out >>again. How can the capacitor possibly end up with a net charge in it? If >>it >>can, where has the electrical charge come from? Have electrons just been >>conjured up out of nowhere? > > It's a different convention. Words. But the units work and the numbers > work, so we use it. Call our kind of charge "charge separation" or > "plate charge differential" if it makes you happier. > > Do you design electronics? Do mosfet data sheets refer to stored gate > energy, or stored gate charge? > > John Yes, I'm actually an electronics design consultant. Take a look at this: www.irf.com/technical-info/appnotes/mosfet.pdf Note the equivalent circuits, which show capacitance between the gate, source and drain. They talk about 'gate charge' as being a conveient way of relating the capacitance charging and discharging (energy, not electrical charge) with current, and hence time. Again, confusion can arrise because they use the word 'charge' in two contexts. The fact is that since current flows in and out of these capacitors in equal amounts the net stored electrical charge on each one is zero. However the Q=CV equation relates to the magnitude of charge that each of the plates of these capacitances carries, but for each capacitor there is another plate with equal and opposite charge. Here is a good derivation of the elecrostatic forces between the plates of a parallel plate capacitor. Note that the electrical charge on each plate has the same magnitude Q, but one is positive and the other negative. If you think this is not correct maybe you should contact the University of Pennsylvania and tell them :) http://dept.physics.upenn.edu/~uglabs/lab_manual/electric_forces.pdf Mark. |