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From: Jim Thompson on 9 Jul 2010 20:00 On Fri, 09 Jul 2010 16:10:27 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: >On Thu, 08 Jul 2010 19:51:42 -0700, Jim Thompson ><To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: > >>On Wed, 07 Jul 2010 08:50:50 -0700, John Larkin >><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >> >>>On Wed, 07 Jul 2010 08:18:12 +1000, Adrian Jansen <adrian(a)qq.vv.net> >>>wrote: >>> >>>> >>>>Jim Thompson wrote: >>>>> On Tue, 06 Jul 2010 12:59:35 -0700, John Larkin >>>>> <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>> >>>>>> On Tue, 6 Jul 2010 12:46:20 -0700 (PDT), whit3rd <whit3rd(a)gmail.com> >>>>>> wrote: >>>>>> >>>>>>> On Jul 6, 6:53 am, John Larkin >>>>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>>>> On Mon, 5 Jul 2010 22:28:44 -0700 (PDT), whit3rd <whit...(a)gmail.com> >>>>>>>> wrote: >>>>>>>> >>>>>>>>> On Jul 5, 9:41 pm, John Larkin >>>>>>>>> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>>>>>>> You can have two caps, C1 charged and C2 not, and transfer all the >>>>>>>>>> charge from C1 to C2, without loss. In fact, you can slosh the charge >>>>>>>>>> between them, back and forth, forever. Just don't use resistors. >>>>>>>>> It has to be identical size capacitors, otherwise 'all the charge' >>>>>>>>> can't be transferred without adding/losing energy... >>>>>>>> Not so. >>>>>>> Put a microcoulomb of charge on a 1 uF capacitor. Transfer it all to >>>>>>> a 2 uF capacitor. The first state of the system holds twice the >>>>>>> energy of the second. >>>>>> >>>>>> Well, depends on words now. I can transfer "all the charge that's in >>>>>> C1 to C2" (ie, wind up with C1 at zero volts, and no energy lost) but >>>>>> the numerical amount of coulombs must change if the cap values are >>>>>> different, to conserve energy. I can move the charge back into C1, and >>>>>> return the system to its original state. >>>>>> >>>>>> My point was that you can move charge between caps, without losing >>>>>> energy, but not by using resistors. >>>>>> >>>>>> John >>>>>> >>>>> >>>>> Depends on the definition of "depends" :-) >>>>> >>>>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >>>>> >>>>> C1*V1 == C2*V2 >>>>> >>>>> ...Jim Thompson >>>>If you conserve energy, then you must have >>>> >>>>C1*V1^2 = C2*V2^2 >>> >>>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. >>> >>>John >>> >>> >> >>Would you care to prove that for us John? Mathematically, that is. No >>hand-waving. After all you do claim trivial EE101 :-) >> >> ...Jim Thompson > >Newbies will take note that Larkin has NOT responded to this request. > >Would someone out there like to mathematically prove that charge is >NOT conserved in Larkin's folly (and yet energy is ?:-) > How about you, Win Hill? What do you think? ...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 | Obama isn't going to raise your taxes...it's Bush' fault: Not re- newing the Bush tax cuts will increase the bottom tier rate by 50%
From: krw on 9 Jul 2010 22:41 On Fri, 09 Jul 2010 16:45:42 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: >On Fri, 09 Jul 2010 17:26:33 -0500, "krw(a)att.bizzzzzzzzzzzz" ><krw(a)att.bizzzzzzzzzzzz> wrote: > >>On Fri, 09 Jul 2010 07:56:02 -0700, Jim Thompson >><To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: >> >>>On Fri, 09 Jul 2010 04:16:27 -0700, >>>"JosephKK"<quiettechblue(a)yahoo.com> wrote: >>> >>>>On Thu, 08 Jul 2010 08:32:12 -0700, John Larkin >>>><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>> >>>>> >>>>>What path? Understanding bog simple circuits? Stuff like this should >>>>>be second nature to any electronics designer. It sure shouldn't need >>>>>to involve cranking up Spice. You use Spice when you *don't* >>>>>understand how a circuit works. >>>>> >>>>>John >>>>> >>>>That sounds like a sure fire recipe for getting screwed by SPICE. I have >>>>watched it happen so very many times. >>> >>>The only people who get "screwed" by Spice are those amateurs who >>>don't understand what .OPTIONS settings, particularly "time-step" >>>values can do to you. >> >>The *only* people? How about those who think their model is real, even at the >>fringes and beyond? How about those who don't believe in real components? How >>many FTL circuits have we seen here? > >I DID say "amateurs". I have also recited here my admonishing a >former employee about applying common sense to Spice results. But you limited your comments to the subset of amateurs who don't understand what the ".OPTIONS" settings do. My point is that the universe of "amateurs" is *much* larger than that. The real scary ones are the ones who believe the FTL simulations because "the computer said so". >"FTL"?? We have lots of those ;-) Yes, and it's scary. >>>Then there are those who avoid showing a Spice result for their >>>abortion... to avoid embarrassment ;-) >> >>Seems thin. > >Thin? What is thin, other than Larkin's BS ?:-) It's thin complaining about lack of Spice results to avoid embarrassment. If there were simulations there would be no embarrassment. No?
From: kevin93 on 9 Jul 2010 23:08 On Jul 7, 10:39 am, John Larkin <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > On 7 Jul 2010 09:38:56 -0700, Winfield Hill > > > > > > <Winfield_mem...(a)newsguy.com> wrote: > >Jim Thompson wrote... > >> John Larkin wrote: > >>> Adrian Jansen wrote: > >>>> Jim Thompson wrote: > >>[snip] > > >>>>> Depends on the definition of "depends" :-) > >>>>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> > > >>>> If you conserve energy, then you must have > >>>> C1*V1^2 = C2*V2^2 > > >>> 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. > > >> John says, "...charge is not conserved." > >> Newbies are invited to Google on "conservation of charge". > >> (AND run the math problem I previously posted.) > >> John is so full of it I'd bet his eyes are brown ;-) > > >> Unfortunately, Adrian Jansen mis-states the results as well :-( > > > I haven't been following this thread, but I have a comment. > > > The operative phrase must be, "and do it efficiently." > > > This is easy to do, with a dc-dc converter for example, or a > > mosfet switch and an inductor. In these cases it's easy to > > manipulate E1 and E2, C1*V1^2 = C2*V2^2. Forget about charge. > > Exactly. To say "Charge is always conserved" is absurd. It is > conserved in some situations, not in others. The context must be > stated exactly. > > Charge two identical caps to the same voltage, then connect them in > parallel, but with polarities flipped. ALL the charge vanishes. > > On the other hand, energy is always conserved. > > John Charge is conserved even in this situation - by reversing the polarity of one you have also reversed the polarity of charge so of course when you connect them the charges add up to zero. The energy within the two components however is not conserved - after the connection the electrical energy is zero it having been released as heat, radiation and sound etc. kevin
From: m II on 9 Jul 2010 23:21 Jim Thompson wrote: >> You can have two caps, C1 charged and C2 not, and transfer all the >> charge from C1 to C2, without loss. In fact, you can slosh the charge >> between them, back and forth, forever. Just don't use resistors. > John Larkin, Please explain how you do that? Magic switch? Or magic > perfect inductor ?:-) I'm wondering why C1 would want to donate more than half it's charge to c2. A vision of two connected propane bottles equalizing comes to mind. confused, in Seattle... mike
From: JosephKK on 10 Jul 2010 06:59
On Fri, 09 Jul 2010 07:59:23 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Fri, 09 Jul 2010 05:02:27 -0700, >"JosephKK"<quiettechblue(a)yahoo.com> wrote: > >>On Wed, 07 Jul 2010 19:44:14 -0700, John Larkin >><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >> >>>On Wed, 07 Jul 2010 19:26:10 -0700, >>>"JosephKK"<quiettechblue(a)yahoo.com> wrote: >>> >>>>On Wed, 07 Jul 2010 10:39:10 -0700, John Larkin >>>><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>> >>>>>On 7 Jul 2010 09:38:56 -0700, Winfield Hill >>>>><Winfield_member(a)newsguy.com> wrote: >>>>> >>>>>>Jim Thompson wrote... >>>>>>> John Larkin wrote: >>>>>>>> Adrian Jansen wrote: >>>>>>>>> Jim Thompson wrote: >>>>>>>[snip] >>>>>>>>>> >>>>>>>>>> Depends on the definition of "depends" :-) >>>>>>>>>> "Charge" IS conserved. So if you transfer Q from C1 to C2 >>> >>>>>> >>>>>>>>> If you conserve energy, then you must have >>>>>>>>> C1*V1^2 = C2*V2^2 >>>>>> >>>>>>>> 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. >>>>>>> >>>>>>> John says, "...charge is not conserved." >>>>>>> Newbies are invited to Google on "conservation of charge". >>>>>>> (AND run the math problem I previously posted.) >>>>>>> John is so full of it I'd bet his eyes are brown ;-) >>>>>>> >>>>>>> Unfortunately, Adrian Jansen mis-states the results as well :-( >>>>>> >>>>>> I haven't been following this thread, but I have a comment. >>>>>> >>>>>> The operative phrase must be, "and do it efficiently." >>>>>> >>>>>> This is easy to do, with a dc-dc converter for example, or a >>>>>> mosfet switch and an inductor. In these cases it's easy to >>>>>> manipulate E1 and E2, C1*V1^2 = C2*V2^2. Forget about charge. >>>>> >>>>>Exactly. To say "Charge is always conserved" is absurd. It is >>>>>conserved in some situations, not in others. The context must be >>>>>stated exactly. >>>>> >>>>>Charge two identical caps to the same voltage, then connect them in >>>>>parallel, but with polarities flipped. ALL the charge vanishes. >>>>> >>>>>On the other hand, energy is always conserved. >>>>> >>>>>John >>>> >>>>Well let's consider this test case you just described. There was energy >>>>stored in each capacitor before closing the switch. There is none >>>>afterwards. Where did it go? How did it get there? ^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^ >>> >>>Heat, light, e/m radiation, sound, maybe some chemical changes in the >>>switch material. >>> >>>The capacitors also lost a little bit of mass. Actually, that's where >>>the energy came from. >> >>But i asked where it went to, and HOW it got there. >>> >>>John >>> >> >>Trained speculation and NO information on the _how_ let alone the _why_. >>Or colloquially, "hand waving". > >Your question was unclear. Are you asking where the energy came from >to initially charge the caps, or where the energy went at the instant >of discharge? I answered the latter. > >If your question was the former, there's no need to answer. Charged >caps was an assumption as an initial condition. > >Please state your question more clearly. > >John After closing the switch [beginning at the closure of the switch] to discharge the caps was indeed very clear. Just to help clarify, you may assume trivial switching losses (or not), then continue with a clear explanation showing reasonable causation. Or with (some) math if you prefer. The questions still are: Where did the energy go? You mumbled something. How did it get [go to] there? No answer at all yet. ^^^^^^^ clarification added. |