Re: non-dissipative circuitry

[Bernard Cleyet <anngeorg@PACBELL.NET>]

I tried this once -- of course the circuit elements were less than ideal
-- fortunately the circuit was fused!!   (The 30 A. circuit breaker)

I suspect David intended to open the circuit at max. voltage in order to
leave the cap. charged.  If I remember correctly this will be at zero
current.

Another matter:  The resonance formula is derived from diff. eq. that
has transient and steady state solutions.  Does this have any effect on
this use of the resonant circuit?

bc  who may be an example of "A little knowledge is a dangerous thing."

David Bowman wrote:

> Regarding John D.'s comment:
>
> > ... There really are low-dissipation ways of charging a capacitor.
>
> and Chuck B.'s response:
>
> > and the variable capacitance method that you related seems the
> > simplest and most straightforward of them.
>
> How about the following method/scenario?
>
> Suppose we have an ideal AC voltage source which supplies a sine wave
> waveform.  The local electric utility could be used in principle.
> Next suppose we have a nearly ideal inductance and an nearly ideal
> capacitance (possibly made with some superconducting substances).
> Choose the value of the 1/(2*pi*sqrt(LC)) resonant frequency of this
> L and C to be to the exact frequency of the AC power source.  Next
> connect the L and the C in series and connect one end of this network
> to one for the two terminals of the AC power supply.  As soon as the
> voltage on the power supply happens to be executing a zero crossing
> throw a switch which connects the other end of the series network to
> the other terminal of the AC power source thus effectively placing
> this series LC circuit across the power supply at the moment of this
> zero crossing.  Since this (nearly) infinite Q network is being
> driven just *at* resonance the amplitude of the voltage and current
> excursions in the L and the C elements will be growing linearly with
> time.  The energy in the circuit will be growing proportional to the
> square of the elapsed time.
>
> After precisely an integer number of half-cycles of the input voltage
> the switch (that initially connected the network to the power source)
> is again thrown open at another zero crossing of the input voltage.
> At this moment 100% of all the energy will be in the capacitor and no
> current will be flowing either (since at resonance there is no phase
> shift between the applied voltage and the current through the
> network).  If we wait for an exact integer n half-cycles the voltage
> on the capacitor will be (n*pi/2)*V_0 where V_0 is the voltage
> amplitude of the voltage source.  Using this method we can nearly
> dissipationlessly charge our capacitor to an *arbitrarily high*
> voltage with a finite voltage amplitude power source by just waiting
> for more half-cycles before disconnecting the circuit from the power
> source.
>
> Of course in real life we might have some problem finding an infinite
> Q network and exactly tuning the L*C product so the setup is running
> at exact resonance.  In real life after a finite number of cycles
> the average rate of energy input to the L & C parts of the circuit
> will level off and all the subsequent input energy will on average be
> dissipated in the series R (however small which would include some
> radiation resistance) of the circuit.  At this point the circuit will
> be dissipating all the input power but will keep sloshing a large
> amount of energy between the L & C, but the amplitude of the sloshes
> will cease to grow any more.  But if we can make the Q of the circuit
> extremely high we still can make the method work to dissipate very
> little energy by just disconnecting the circuit before the rate of
> increase in the amplitude of the oscillations begins to appreciably
> level off.
>
> David Bowman