Re: non-dissipative circuitry

[Bernard Cleyet <anngeorg@PACBELL.NET>]

Thanks to JD in addition to DB.

I remember a series LCR experiment in Sophomore Lab.  (My independent try
was in HS.) The R a 100?W lamp with the L having a removable (variable L)
core.  I also remember the Voltages (PDs) across the L and C at resonance
(lamp fully lighted)  were very high, 600? V.

"Since the Q is supposed to be infinite the transient decay time is
also infinite.  The circuit is completely operated in the regime
where the initial conditions still strongly matter.  If we wait until
a steady state response develops that is *too* long since by then the
dissipation will have kicked in with a vengeance and this would have
defeated the purpose of having a nearly dissipationless process."



The posts induced me to create a program that gives the current as a
function of time.  Counter intuitively, but agreeing with the above, I find
the time to reach "steady state" is inverse to the R.   e.g.  for 6 ohms ~
4.5 cycles to reach 63%,   (steady state is ~ 20 A), and ~ 15 cycles for 2
ohms.  (60 A;  period ~ 17 msec)


"No. I meant what I said.  The current in the circuit will be in-phase
with the *applied* voltage across the circuit.  The applied voltage
will be zero when the current is zero, but at such a moment the
charge on the capacitor and the voltage across the capacitor will be
a local maximum."

Duhh.  bc hits head.  An analytic truth.

bc

David Bowman wrote:

> Regrading bc's comment:
>
> > I tried this once -- of course the circuit elements were less than ideal
> > -- fortunately the circuit was fused!!   (The 30 A. circuit breaker)
>
> In order to get the R.M.S. current up to 30 A it appears that you
> may have had quite a high Q--unless that current came about because
> the growing voltage swings breached the capacitor's dielectric and
> the circuit shorted out and cooked the cap (& maybe the inductor as
> well).
>
> > 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.
>
> No. I meant what I said.  The current in the circuit will be in-phase
> with the *applied* voltage across the circuit.  The applied voltage
> will be zero when the current is zero, but at such a moment the

cut