Win Hill: Inverse Marx Generator ??
in [Design]
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From: Jim Thompson on 11 Jul 2010 11:46 On Sat, 10 Jul 2010 09:07:22 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Sat, 10 Jul 2010 11:35:35 -0400, "tm" <noone(a)msc.com> wrote: > >> >>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >>news:o61h36lt8fvhsc00mrc9824ju0jd4hml8s(a)4ax.com... >>> >>> To celebrate the 21st century, I have composed a new riddle: >>> >>> Start with a 4 farad cap charged to 0.5 volts. Q = 2 coulombs. >>> >>> Carefully saw it in half, without discharging it, such as to have two >>> caps, each 2 farads, each charged to 0.5 volts. The total charge of >>> the two caps remains 2 coulombs, whether you connect them in parallel >>> or consider them separately. >>> >>> Now stack them in series. The result is a 1F cap charged to 1 volt. >>> That has a charge of 1 coulomb. Where did the other coulomb go? >>> >>> I think this is a better riddle. >>> >>> John >>> >>> >> >>One should not confuse charge with energy. >> >> > >Exactly the point I've been making. Some EEs seem to think that charge >is always conserved. Some physicists seem to think that energy is >always conserved. They can't both be right. > >I'll side with the physicists on this one. > >John "Side" with whomever you like. But both "laws" apply simultaneously. ...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: John Larkin on 11 Jul 2010 11:46 On Sat, 10 Jul 2010 22:51:34 -0700, "Artemus" <bogus(a)invalid.org> wrote: > >"George Jefferson" <phreon111(a)gmail.com> wrote in message >news:i1bka6$h8v$1(a)news.eternal-september.org... >> > ><snip> >> >> Suppose you have a black box. You measure the temperature in it. The >> temperature changes. We know temperature is related to heat which is related >> to energy. Hence we can see that some energy has changed in the black box. >> Where did this energy go? IT HAD TO GO OUTSIDE THE BOX!!! Why? Because >> energy is conserved and the only way it could change without going outside >> the box is if it were created or destroyed. Hence the box is not an isolated >> system. >> ><snip> > >This is great. You should be in Congress. >Art > Hilarious! I guess if you put a lighted flashlight in an insulated box, the temperature inside can't go up. John
From: Jim Thompson on 11 Jul 2010 11:52 On Sun, 11 Jul 2010 09:05:34 +0100, John Devereux <john(a)devereux.me.uk> wrote: >Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: > >> In the next few days, when I have time, I will issue a mathematical >> proof that Larkin is totally wrong. Watch for it ;-) >> >> Why haven't Win Hill and Phil Hobbs come to Larkin's defense? >> >> Bwahahahaha! > >I'm no Phil Hobbs, but isn't all this argument because we are conflating >two different usages of "charge"? > >The "charge" on a capacitor, as somone pointed out already, is really >charge *separation* (dilectric polarization). The Q=CV refers to a >*separation* of charge, not an absolute quantity. The "absolute" charge >- the total number of electrons minus the number of protons - is >normally low or zero. Unless your whole circuit picks up an >electrostatic charge from somewhere else. It is this "absolute" charge >which is conserved, the "Q=CV" "charge" of normal electronics is >not. Take a solar cell charging a battery for one obvious example. As >Larkin would say, where did the charge come from? Photons don't carry >charge! The photons entered the game from "outside the box" as someone opined. As for your "explanation" above... :-( If I'm so wrong and Larkin is so right, WHY don't Hill and Hobbs jump to his defense? ...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: John Larkin on 11 Jul 2010 11:52 On Sun, 11 Jul 2010 06:28:04 -0500, John Fields <jfields(a)austininstruments.com> wrote: >On Sat, 10 Jul 2010 21:24:16 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > > >>As an EE, I think that a 1F cap charged to 1 volt stores 1 coulomb of >>charge, namely because I can observe 1 ampere-second of integrated >>current if I connect its plates through a resistive conductor. >--- >Nope; an ampere-second is one ampere for one second. Am I to suppose that half an amp for two seconds is not a coulomb? Or that an exponentially decaying current can never integrate to one coulomb? > >Plus, (just as an aside) to completely discharge the cap would take >forever and, at the end of that time, what you'd get out of the cap >would be: > > QV 1C * 1V > W = ---- = --------- = 0.5 joule > 2 2 > >which is exactly what you'd get out of your 4 farads charged to 0.5V. >example. Yup. Energy is conserved in the restacking-the-caps case, but charge isn't. That was my simple point all along. Thanks for the confirmation. John
From: John Larkin on 11 Jul 2010 12:06
On Sun, 11 Jul 2010 09:05:34 +0100, John Devereux <john(a)devereux.me.uk> wrote: >Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: > >> In the next few days, when I have time, I will issue a mathematical >> proof that Larkin is totally wrong. Watch for it ;-) >> >> Why haven't Win Hill and Phil Hobbs come to Larkin's defense? >> >> Bwahahahaha! > >I'm no Phil Hobbs, but isn't all this argument because we are conflating >two different usages of "charge"? > >The "charge" on a capacitor, as somone pointed out already, is really >charge *separation* (dilectric polarization). The Q=CV refers to a >*separation* of charge, not an absolute quantity. The "absolute" charge >- the total number of electrons minus the number of protons - is >normally low or zero. Unless your whole circuit picks up an >electrostatic charge from somewhere else. It is this "absolute" charge >which is conserved, the "Q=CV" "charge" of normal electronics is >not. Take a solar cell charging a battery for one obvious example. As >Larkin would say, where did the charge come from? Photons don't carry >charge! Right. This being an electronics design group, and specifically a discussion about capacitors on schematics and not electrostatics, I assume that when we talk about the charge on a capacitor, it's zero if there's no voltage across its terminals, and any subsequent charge stored by the cap is equal to the integral of applied current. Q = CV = integral(I.dt) where V is the voltage across the cap C in farads, V volts, t seconds or in the rectangular case, Q = CV = IT starting from some initial point of Q=0, V=0. Q is measured in coulombs, which are dimensionally ampere-seconds. Given this definition of charge on/in a capacitor, one could further define the "total charge" in a circuit somehow. Carefully. Does anybody want to argue about that? John |