No Subject

• To: tesla-at-pupman-dot-com
• From: daniel_hess-at-VNET.IBM.COM
• Date: Thu, 31 Oct 96 18:44:23 CST

     If the circuit settles down to a steady state (no oscillation),
50% of the energy is indeed lost, but charge is indeed conserved.
Elaborating: E=0.5CV^2  You can see that a cap at half its original
voltage has only 1/4 of its original energy. Two identical caps at
half the original voltage of one has still only half the original
energy.
     BUT, since Q=CV, you can see that if you double C and halve V,
Q (charge) remains conserved.

Malcolm

I find it easier to understand this argument if I relate it to the
inverse-square law, i.e.; If I have a capacitor, any capacitor,
charged to a certain voltage, then that capacitor represents a
certain quantity of energy. If I wish to double that quantity of
energy then I must either quadruple the capacity, (charged to a like
value,) OR quadruple the voltage of the capacitor. Right?

Daniel.

• Prev: What efficiency?!
• Next: Re: 50%
• Index(es):
  • Main
  • Thread