Win Hill: Inverse Marx Generator ??
in [Design]
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From: Bitrex on 10 Jul 2010 04:08 Jim Thompson wrote: > On Fri, 09 Jul 2010 20:46:10 -0700, "Paul Hovnanian P.E." > <Paul(a)Hovnanian.com> wrote: > > [snip] >> Oooh. Its the old "where did the energy go" two cap puzzle. >> > [snip] > > Which is trivial to solve :-) > > ...Jim Thompson The excess energy goes into the "KA-POW" you hear when you try the experiment with two 75V 10,000uF caps. :)
From: amdx on 10 Jul 2010 08:03 -- MikeK "Bitrex" <bitrex(a)de.lete.earthlink.net> wrote in message news:4pOdnYatdqqEsaXRnZ2dnUVZ_jmdnZ2d(a)earthlink.com... > Jim Thompson wrote: >> On Fri, 09 Jul 2010 20:46:10 -0700, "Paul Hovnanian P.E." >> <Paul(a)Hovnanian.com> wrote: >> >> [snip] >>> Oooh. Its the old "where did the energy go" two cap puzzle. >>> >> [snip] >> >> Which is trivial to solve :-) >> >> ...Jim Thompson > > The excess energy goes into the "KA-POW" you hear when you try the > experiment with two 75V 10,000uF caps. :) I heard it on my radio two blocks away! Mike
From: George Jefferson on 10 Jul 2010 09:32 >>>>>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 :-) Larkin fails to realize that V1 = V2. V1*Q1 = V2*Q2 => Q2 = V1/V2*Q1, V1 = V2 so Q1 = Q2. You would think with all his vast knowledge that would understand basic 101 electronics. +----+ | | C1 C2 | | +----+ If you discharge C1 into C2 then the voltages across them will be equal after a given time. The total charge will still not have changed. The total charge Q = Q1 + Q2. As electrons move from the cap of higher voltage to lower voltage we end up with Q = (Q1 -+ de) + (Q2 +- de) = Q.
From: John Larkin on 10 Jul 2010 11:16 On Sat, 10 Jul 2010 08:32:31 -0500, "George Jefferson" <phreon111(a)gmail.com> wrote: >>>>>>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 :-) > >Larkin fails to realize that V1 = V2. > >V1*Q1 = V2*Q2 > >=> Q2 = V1/V2*Q1, V1 = V2 so Q1 = Q2. > > >You would think with all his vast knowledge that would understand basic 101 >electronics. > >+----+ >| | >C1 C2 >| | >+----+ > >If you discharge C1 into C2 then the voltages across them will be equal >after a given time. The total charge will still not have changed. > > >The total charge Q = Q1 + Q2. As electrons move from the cap of higher >voltage to lower voltage we end up with Q = (Q1 -+ de) + (Q2 +- de) = Q. > > > 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
From: tm on 10 Jul 2010 11:35
"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. --- news://freenews.netfront.net/ - complaints: news(a)netfront.net --- |