Re: non-dissipative circuitry

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

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
charge on the capacitor and the voltage across the capacitor will be
a local maximum.  At this same moment the voltage across the
inductance will also be at a local maximum (but of opposite polarity
thus cancelling the voltage on the cap when these voltages are added
in series) because even though the current is zero its time
derivative is quite large during such a zero crossing of the current.
thus the voltage across the inductance peaks when it has depleted
all of its magnetic energy and all that energy is in the capacitor.

> 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?

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.

Suppose we apply the voltage V_0*sin(wt) across the circuit starting
at time t = 0 and assume that initially the capacitor is uncharged
and that there is no current flowing initially.  Then while the
circuit is still ringing (i.e. before the circuit is disconnected)
the current I(t) has the form:

I(t) = (V_0/(2*L))*t*sin(w*t) where we take w = 1/sqrt(L*C).

The charge Q(t) on the capacitor is:

Q(t) = (C*V_0/2)*(sin(w*t) - w*t*cos(w*t))

and the voltage across the capacitor is just this charge above
divided by C, i.e.

V_C = (V_0/2)*(sin(w*t) - w*t*cos(w*t)) .

The voltage across the inductance is the difference between the
applied voltage of V_0*sin(w*t) and the voltage across the capacitor.
It is also the the derivative of the current above times L.  The
expression for this voltage is:

V_L = (V_0/2)*(sin(w*t) + w*t*cos(w*t)) .

The total energy E(t) in the circuit at time t is:

E(t) = (C*V_0^2/8)*((w*t)^2 + (sin(w*t))^2 - w*t*sin(2*w*t))

     = (C*V_0^2/8)*(1/2 + (w*t)^2 - sqrt(1/4 + (w*t)^2)*cos(2*w*t-s))

where the phase s of the oscillatory cosine term is given by
s = Arctan(2*w*t). After subtracting out the oscillatory term
from the growing term in the energy, we find that the trend of the
energy grows like ~ t^2.  As it grows it is alternately tossed
back and forth between the inductor and the capacitor.

David Bowman