Inverse Marx generator
in [Design]
|
Prev: New Product Idea
Next: SCHOLARLY TESTIMONIAL VIDEO : Joseph Moshe (MOSSAD Microbiologist) Swine flu vaccine 1
From: Jim Thompson on 6 Jul 2010 16:16 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 -- | 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: A reincarnation of Nixon, narcissistically posing in politically-correct black-face, but with fewer scruples.
From: Adrian Jansen on 6 Jul 2010 18:18 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 -- Regards, Adrian Jansen adrianjansen at internode dot on dot net Note reply address is invalid, convert address above to machine form.
From: krw on 6 Jul 2010 19:01 On Tue, 06 Jul 2010 10:40:37 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> wrote: >On Tue, 06 Jul 2010 10:16:23 -0700, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>On Tue, 6 Jul 2010 11:33:11 -0500, "Tim Williams" >><tmoranwms(a)charter.net> wrote: >> >>>"John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:jtl6361te3ku4ukh0jhp05tab4tvhjjqru(a)4ax.com... >>>> Connect an inductor across C1 until you've bled it down to half its >>>> charge. Now connect that inductor to C2 and charge it up to the same >>>> charge as C1 has. Now disconnect the inductor. If you keep the L >>>> shorted, you can save the residual energy for reuse later. >>> ^ ^ ^ ^ >>>Ha, so charge wasn't conserved after all. See? ;-) >>> >>>Tim >> >>All my cases involved controlling the charges on two capacitors. I >>never proposed violating conservation of energy. My point is that >>inductors can be useful in some situations. >> >>John > >Your "point" is, as usual, vagueness and subterfuge... and BS. > >For amusement, newbies are invited to MATHEMATICALLY analyze this >simple case... > >Two equal value capacitors (C), one charged to Vo, the other uncharged >(zero volts). > >Connect together with a switch, start with a finite resistance value, >analyze; then reduce the resistance, re-analyze; continue this >analysis, approaching zero in the limit. > >Then scratch your head in surprise... where did the energy go ?:-) Where it always goes.
From: m II on 6 Jul 2010 19:13 StickThatInMyAssandRotateIT! wrote: > snipped retardedly formatted text. > like an idiot > Mr. Useitasitcomesoutoftheboxtotalretard. So, Archie....how's that Exorcism coming along? Able to control your involuntary obscenities yet? Probably not. Keep trying. Jesus Loves YOU! ============================================ But I say to you that everyone who is angry with his brother will be liable to judgment; whoever insults his brother will be liable to the council; and whoever says, �You fool!� will be liable to the hell of fire. Matthew 5:22 http://www.photosforsouls.com/hells-fire-duncanlong181.jpg ============================================ father mike
From: JosephKK on 6 Jul 2010 23:57
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 By analogy, consider a billiard ball striking another (at rest) ball 1/4 diameter away from the line of centroid motion (glancing blow / cut). Does energy and momentum get conserved? Show the math. |