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Re: 50%

>Message-ID: <199611010525.WAA02083-at-poodle.pupman-dot-com>
>Date: Thu, 31 Oct 1996 22:25:03 -0700
>From: Tesla List <tesla-at-poodle.pupman-dot-com>
>To: Tesla-list-subscribers-at-poodle.pupman-dot-com
>Subject: Re: 50%

   [ snip ]

   Well, indeed, thanks to Richard and Malcolm and others commenting
   on this thread, I believe I have come to a new realization about
   the *resistive* or *capacitive* discharge of capacitors.  Funny I
   can't remember this from college.  We must have covered it....

   However, the point is made:

>Actually, No.  Richard is correct, insofar as 50% of the capacitively 
>stored energy always disappears in circuit losses (resistive and magnetic) 
>when _direct_wired_connections_ are used to connect the second capacitor.
>The cap charged to 5kV in the above example holds only _25%_ of the initial
>energy, not 50% (CV squared and and all that).
>However, the 50% efficency value is by no means an upper limit, if one is
>allowed to use components other than direct wires.  A switch in series with 
>an inductor between the two caps will do the job quite nicely:
>When the switch is closed the current starts at zero, peaks at V/ SQRT(L/C),
>then returns to zero in a nice smooth haversine.  When the current crosses
>zero the second time open the switch, and Voila!  Perfect energy transfer!
>
>I guess that the message here is "Never say never."
>
>-GL

   Indeed.  I realized that perfect capacitors and inductors do not
   USE energy.  They store it and give it back.  Also I realized that
   the energy in a tank circuit ALL leaves the capacitor and goes into
   the inductor (at the point when the inductor current is maximum and
   the capacitor voltage is precisely zero).  So the thought occurred
   to me last night (as to the above author) that therefore if we
   start with a charged perfect capacitor, and hook it up to a perfect
   inductor over perfect conductors, an oscillation will start that
   will never decay. (No such devices exist AFAIK).  However, it *IS*
   theoretically possible this way to get ALL the energy out of the
   capacitor with a (perfect) switch that switches the current
   somewhere else exactly at the point when the capacitor voltage
   crosses zero (at which point all the energy would be in the 
   inductor).  Now, do we have the same problem getting all the
   energy out of the inductor ?

   What a good thread this has been!  My respects to you all.
 
 Fred W. Bach ,    Operations Group        | Internet: music-at-triumf.ca
 TRIUMF (TRI-University Meson Facility)    | Voice:  604-222-1047 loc 6327/7333
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