Win Hill: Inverse Marx Generator ??
in [Design]
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From: JosephKK on 13 Jul 2010 01:53 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! Gosh, in all the semiconductor physics i have seen it is "pair generation". No net charge change involved.
From: AM on 13 Jul 2010 02:09 On Mon, 12 Jul 2010 18:27:16 -0500, "krw(a)att.bizzzzzzzzzzzz" <krw(a)att.bizzzzzzzzzzzz> wrote: >Because he won't accept the corrections and is otherwise useless. Pay >attention. You have never made any such correction. You are too goddamned stupid to make any post that is contributory... ever. You haven't done so in years, if you ever did.
From: MeowSayTongue on 13 Jul 2010 02:20 On Mon, 12 Jul 2010 21:31:39 -0700, "JosephKK"<quiettechblue(a)yahoo.com> wrote: >On Sun, 11 Jul 2010 09:28:52 -0500, John Fields ><jfields(a)austininstruments.com> wrote: > >>On Sun, 11 Jul 2010 07:20:02 -0500, "Tim Williams" >><tmoranwms(a)charter.net> wrote: >> >>>"John Fields" <jfields(a)austininstruments.com> wrote in message news:3o7j36d5jvgeg5276nkr2t1fuibdmd6fij(a)4ax.com... >>>> In your example the current will be one ampere when the resistor is >>>> first connected, but will have decayed to about 368 mA after one >>>> second has passed, so there's no way you'll get one ampere-second out >>>> of it. >>> >>>Instead of clucking around, you could actually do some math. >>> >>>Definition: >>>q_tot = integral I*dt from 0 to infty >>>Equation: >>>I(t) = (V/R) * exp(-t/RC) >>> >>>So: >>>q = V/R * integral exp(-t/RC) dt from 0 to infty >>>= [-RC * V/R * exp(-t/RC)] from 0 to infty >>>= -VC * [exp(-infty/RC) - exp(0/RC)] >>>= -VC * [0 - 1] >>>= VC >>>V = 1V and C = 1F so q = 1C. QED. >>> >>>This is only highschool calculus, how embarrassing. >> >>--- >>Indeed, since: >> >>q_tot = integral I*dt from 0 to infty >> >>should read: >> >>q_tot = integral I*dt from 0 to t, >> >>I believe. ;) >> >> >>From: >> >>http://www.thefreedictionary.com/ampere-second >> >>"ampere-second - a unit of electrical charge equal to the amount of >>charge transferred by a current of 1 ampere in 1 second." >> >>So, for one coulomb of charge to be transferred through a one ohm >>resistor in one second, the voltage would have to remain at one volt >>for one second. >> >>Such is not the case when a one farad capacitor is charged to one volt >>and connected across a 1 ohm resistor for one second, since the >>voltage will decay from 1V to 0.368V during that time and there'll be: >> >> Q = CV = 1F * 0.368V = 0.368 coulomb >> >>still left in the cap when it's disconnected. > >JF: Dude, an ampere*second (== 1 coulomb) is a product unit. It can be >one kA for 1 ms or 1 uA for 1e6 seconds. Or any other combination, >including non-constant current, whose time integral of current equals one >ampere*second. That is what *HE* said. Remember the cap that takes a LONG time to *fully* discharge through the resistor? I cannot believe some of you guys are so blind!
From: John Devereux on 13 Jul 2010 03:07 "JosephKK"<quiettechblue(a)yahoo.com> writes: > 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! > > Gosh, in all the semiconductor physics i have seen it is "pair > generation". No net charge change involved. But in normal electronics usage, we would say that the battery or capacitor was charged by the solar cell! "Charge separation" or "pair generation" - there *is no* physical net charge in reality. Yet we always talk of the "charge of a capacitor" or "charging a battery". It is what this whole thread has been about (and I believe it is this usage that Larkin had in mind). -- John Devereux
From: George Jefferson on 13 Jul 2010 05:50
"JosephKK" <quiettechblue(a)yahoo.com> wrote in message news:1gsn36lhlos6n2664ku1ka0t8keuojuok1(a)4ax.com... > On Sun, 11 Jul 2010 13:54:02 -0500, John Fields > <jfields(a)austininstruments.com> wrote: > >>>Good grief, you *don't* understand this stuff. >>> >>>You, like AlwaysWrong, are certainly smart enough to learn the basics >>>of electrical circuit math, but for some emotional reason you have >>>chosen not to. I see that a lot in techs. They compensate by attacking >>>people who can do the arithmetic, calling them eggheads or >>>"inexperienced" or argue over definitions and third-order effects to >>>obscure the fact that there *are* calculable answers. >> >>--- >>Typical Larkinese. >> >>I usually show my work, while you, on the other hand, are the one who >>always waits until close to the end of the thread to start >>"explaining" what everyone's already laid out, pretending that it was >>your answer in the first place and issuing gratuitous slurs in order >>to try to demean your detractors. > > I have to grant the truthfulness of this to you JF. Larkin has some need to prove he is intelligent. Generally people that are like this are not intelligent and hence the reason Fields is correct. How else can Larkin "prove" he is intelligent if he is not. He can't come up with the right answer so he waits until someone else does then pretends it was his so he can claim that he is intelligent. He has to do this repeatedly to keep proving he is intelligent because he does things that are not intelligent(because that is what he really is). |