Re: superconducting LC tank circuit?

On 1 mar, 16:05, j...@xxxxxxxxxxxxxxxxxxxx wrote:
s...@xxxxxxxxxxxx wrote:
On 1 mar, 12:35, j...@xxxxxxxxxxxxxxxxxxxx wrote:
s...@xxxxxxxxxxxx wrote:
On 1 mar, 10:35, j...@xxxxxxxxxxxxxxxxxxxx wrote:

<snip old crap>

An LC circuit is a LC circuit no matter the physical size or arrangement
of the inductor and capacitor.
Disagreement. An isolated LC system is stand alone, not in the same
situation as one inducing energy into any surrounding other circuits.
An LC circuit is a LC circuit no matter what the construction material.
Again, an isolated LC system is not in the same circumstance as
one being tapped by an associated circuit.

Nothing has been said about any "associated circuit" or being "tapped"
by anything.

Take an inductor and a capacitor and place them in parallel and you
have a LC tank circuit.

It doesn't matter what the parts are made of, their physical size, or
arrangement, it is still a LC tank circuit.

Put this circuit in your kitchen or in intersteller space a 100 light
years from the nearest object and it is still a LC tank circuit.

Hit the LC circuit with an impulse and it will ring at a frequency
of f = 1 / ( 2 x pi x sqrt ( L x C) ).

Remove the apparatus used to provide the impulse so the LC tank circuit
is now totally isolated.
So you have begun to read on the matter. Good.

Yeah, about 35 years ago in Circuits 101.

Then you need a serious refresh.

The initial amplitude of the ringing will depend on how much energy
you dumped in with the impulse.

As time passes, the amplitude of the ringing will decrease as energy
is lost.
Through heating of the wire. Yes.
If the LC circuit is constructed with normal materials, most of the
loss will be due to resistive loss in the materials and I^R heating.
All of it in an isolated LC system.

That was said above.

Are you having comprehension problems?

You seem to have this comprehension problem. What you said before
is that there was some loss other than wire heating in an isolated
LC system. I said that all of it is through wire heating.

Can you not understand what "totally isolated" as stated above means?

I return you the question.

If the LC circuit is constructed with true supercondutors and a
lossless dielectric for the capacitor, there will be no resistive
loss of energy because there is no resistance.
And there will consequently be no loss if there is no loss to
heating, as seems to have come to the attention of the OP.

Nonsense.

You deny reality all you want. That won't change it.

Any conductor with AC current generates an electromagnetic field
at the frequency of the current.

The tank circuit has conductors which are the inductor itself,
the connections from the inductor to the capacitor, and the
capacitor plates.

In most LC circuits the conductor which forms the inductor is
physically the largest, but that is irrelevant to the arguement.

Since the LC circuit itself contains inductors carrying an AC
current, the LC tank circuit generates an electromagnetic field.
You are wrong here. Read again. In an isolated LC tank, there
is no AC current. You have displacement current, which is
entirely different and the reason why free EM energy can propagate
with LC characteristics in vacuum.

Babbling nonsense.

If a LC tank circuit is ringing, there is real AC current flowing
flowing in the LC circuit.

If there were no real current flowing, there would be no I^R loss
for real conductors.

You are wrong and ignorant.

The amplitude of the generated field is directly related to the
physical size of the conductor measured in wavelengths of the
current frequency.

Electromagnetic fields propagate away into space.

Electromagnetic fields contain energy.

Therefor, energy is lost from the LC tank circuit through the
electromagnetic field.
Your reasoning is wrong. There is no "therefore energy is lost"
for the only reason that energy oscillates in an isolated LC system,
at whatever frequency.

Babbling nonsense and almost incomprehensible English.

Perfectly clear English to people knowledgeable in electromagnetism.

Depending on the initial amplitude of the ringing, the amount of
energy being propagated, and the resolution of any measuring
equipment, it may take some time to be able to measure the loss.

That time may be seconds, minutes, or years.
I am in no hurry. Let me know when you actually get a reading.
In the meantime, I suggest you beef up on displacement current.

I did 35 years ago.

I think you did not.

OK, I finally get it, you are a babbling kook.

Who is the kook exactly? You imposed yourself on me, not the
other way around.

And I stand by the opinion I wrote in my first contribution to the
OP.

André Michaud

.

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