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Re: 50%
I've been reading this thread about the 50% energy loss and it reminded
me of a experiment I'd done about 15-20 years ago. I'd just been given
a neat 5,000pF 10,000V Jennings variable vacuum capacitor. I wish I
still had it now! I'd made a calibrated turns counting dial on it so I
could adjust it to whatever capacitance I wanted.
I set it to 5,000pF and charged it to 5,000 volts. Then I cranked it
out to 2,500pF and used one of those neat 1000:1 40KV Tektronix scope
probes to measure the voltage. It was about 10,000 volts. The
measurement was a bit hard to make since the time constant was only a
couple seconds when I connected the probe to measure the voltage but a
storage scope helped. I also did the reverse of this, charge the
2,500pF capacitor to 10,000 volts and then adjust it in to 5,000pF. It
then measured about 5,000 volts. My readings weren't the most accurate
due to things like leakage on the glass and likely some corona discharge
in the air but I got in the ballpark. Now do the math on the two:
(5000V)**2 x (5000pF/2) = 0.0626j
(10000V)**2 x (2500pF/2) = 0.125j
Same charge, but the capacitance of the capacitor was changed. Now,
if the mechanics were frictionless I could have extracted mechanical
energy when I let the plates of the capacitor move closer together
(5000pF) since the plates have opposite charge and are attracted to each
other, and it would have taken mechanical energy to move them farther
apart (2500pF) for the same reason. Is this mechanical energy equal to
the energy change from 0.0625j to 0.125j? I've often wondered about
this but don't know how to do the conversions.
Now that I think about it, due to the actual mechanical construction
of that capacitor (many concentric cylinders with alternate ones
connected together) I don't think the actual attraction between the
plates is all that well translated to the actuator.
August
--
August Johnson KG7BZ AMI 733 http://www.whitemtns-dot-com/~kg7bz
P.O. Box 795
Pinetop, AZ 85935