Win Hill: Inverse Marx Generator ??
in [Design]
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From: oopere on 12 Jul 2010 04:07 John Larkin wrote: <snip> > > That simple riddle is ancient, possibly even older than JT. > > Obviously charge is conserved in this circuit, independent of what > impedance is used to bridge the caps or of when you observe the > system. I certainly wouldn't dispute that. > > What I *did* say is conveniently right at the top of your post. > > > 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 > > I had to dig up L.O. Chua et al. "Linear and Nonlinear Circuits" which is the reference for "unusual" circuits. start quote In elementary circuit courses, a) the capacitor is defined by i=C�dv/dt where C is a constant called the capacitance. b) the inductor is defined by v=L�di/dt where L is a constant called the inductance. These definitions ara valid only if the elements are linear and time-invariant. To generalize these definitions to the nonlinear and possibly time-varying case , it is necessary to invoke the following dual variables: a) The charge q associated with a two-terminal element at any time t is defined by q=integral(-inf,t)(i(t)dt where i denotes the branch current. The unit of charge is the coulomb. .... end quote For time-varying circuits, charge and flux are _the_ variables to be used. This may provide some light here. Pere
From: o pere o on 12 Jul 2010 06:32 John Larkin wrote: > 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 > > Hi John. Just trying to give some insight. q(t) is _defined_ (see my previous post and the reference therein) as the integral of i(t) for _any_ two-terminal element. For a capacitor it only equals C�v(t) if C is time-invariant. If C is time-varying, i(t)=C(t)�dv(t)/dt+dC(t)/dt�v(t). In this case, for instance, a discontinuity in C(t) causes an impulse in i(t) which may be overlooked. Pere
From: John Fields on 12 Jul 2010 09:10 On Mon, 12 Jul 2010 10:00:39 +1000, Grant <omg(a)grrr.id.au> wrote: >On Sun, 11 Jul 2010 11:20:41 -0500, John Fields <jfields(a)austininstruments.com> wrote: > >>On Sun, 11 Jul 2010 08:36:11 -0700, John Larkin >><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>> >>>How many ampere-seconds would you get if it was discharged by a 10 ohm >>>resistor? >> >>--- >>Depends on how long you left it on there. > >What, you have some different definition of 'discharged'? --- You have to watch that tricky Larkin and his definitions, so since his question was ludicrous on its face, I chose "disharged" not to mean "totally discharged", so my answer still stands. Of course, if the resistor were to remain on the cap forever, the correct, confusion-free answer would be: "one coulomb."
From: John Fields on 12 Jul 2010 09:42 On Sun, 11 Jul 2010 22:31:43 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Sun, 11 Jul 2010 22:24:30 -0700, >"JosephKK"<quiettechblue(a)yahoo.com> wrote: >>Tell that to the designers of the GPS system. > >How does E=mc^2 relate to GPS? --- First hit from Googling "GPS and relativity": http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html
From: John Fields on 12 Jul 2010 09:44
On Sun, 11 Jul 2010 22:54:17 -0700, "JosephKK"<quiettechblue(a)yahoo.com> wrote: >I understand your will to forgiveness. It is the same poster with some >newfound civility. Still just as tiresome in the same way. >You seem to have issues with JL. Maybe you might consider not ragging on >him. --- Maybe... JF |