Quantum Energy Generator (QEG) Open Sourced (by HopeGirl)

[MarkE]

Mark,

I have a question for you. If you first feed a Ferrite cored tank circuit to store energy in the tank circuit,
and then switch off the input and simultaneous switch in more Ferrite core material, then will the stored
energy in the tank circuit increase because you did change the inductance to a higher value?
(Let us assume that you can switch in more Ferrite material very fast without any losses in this thought experiment.)

GL.

The energy stored in a magnetic field is the integral of B*H.  In your scenario, H is fixed, but B could potentially go up.  In order to make B go up you will have to perform work.  A complication is that the energy in the tank is alternating between the electric and magnetic fields.  When you change B relative to H, you change the time constant of the network immediately detuning it.

[Groundloop]

The energy stored in a magnetic field is the integral of B*H.  In your scenario, H is fixed, but B could potentially go up.  In order to make B go up you will have to perform work.  A complication is that the energy in the tank is alternating between the electric and magnetic fields.  When you change B relative to H, you change the time constant of the network immediately detuning it.

Mark,

De-tuning the L/C tank circuit is not a problem since the input is turned off before we add core material to the coil.
We only need to know the resonant frequency when the input is running. But I feel you have not answered my
question, does the stored energy in the tank circuit increase or decrease when we "magically" add more core
material to the coil? Let us assume we add that core material very fast and at no energy usage to do so.

GL.

[MarkE]

Mark,

De-tuning the L/C tank circuit is not a problem since the input is turned off before we add core material to the coil.
We only need to know the resonant frequency when the input is running. But I feel you have not answered my
question, does the stored energy in the tank circuit increase or decrease when we "magically" add more core
material to the coil? Let us assume we add that core material very fast and at no energy usage to do so.

GL.

Let me rephrase your question then:  "Given two tank circuits, one using larger toriods than the other, which circuit stores more energy?"  The answer is:  "There is not enough information supplied to answer the question."  A bigger form can store more energy before it saturates than a smaller toroid.  As you change the dimensions of the toroid the inductance per turn changes as does the resistance per turn.  The ratio of the inductive reactance to the resistance at the resonant frequency determines hat fraction of stored energy is lost each cycle, IE it determines the Q factor.  The inductance and the peak current determine the peak stored energy.

[dvy1214]

Mark,

De-tuning the L/C tank circuit is not a problem since the input is turned off before we add core material to the coil.
We only need to know the resonant frequency when the input is running. But I feel you have not answered my
question, does the stored energy in the tank circuit increase or decrease when we "magically" add more core
material to the coil? Let us assume we add that core material very fast and at no energy usage to do so.

GL.

Might be wrong but studying magneto restrictive materials along with basic understanding of why magnetism manifests IE organization of domains(which costs energy) plus the realization that what you are asking is actually impossible(" Let us assume we add that core material very fast and at no energy usage to do so") might help you come to a conclusion.

[MileHigh]

Groundloop:

I usually find simple examples like that fun.  There is a limitation or perhaps we can call it a flaw in the logic of your example.  However, I would prefer to discuss that later.

Permit me to give you a very simple and related example.  Just like in your example, we will assume everything is based on idealized components (zero wire resistance, etc.)  I think my example will give you some insight into your example.

You have two coils in series, Coil A and Coil B.  The "left side" of Coil A connects to the "right side" of Coil B to form a  circuit.  In other words, the current flows through Coil A, then through Coil B, and then back to Coil A.  There is no magnetic coupling at all between Coil A and Coil B.

When you start the experiment there is a zero-ohm jumper across Coil B, effectively taking it out of the circuit.  There is no current flowing through Coil B at the start of the experiment.

To start the experiment, by "magic" you have one ampere of current flowing through the circuit.   So that means there is one amp flowing through Coil A, and Coil B is being bypassed because of the zero-ohm jumper.

Now since this is an idealized example, if you do not disturb the circuit, current will flow through Coil A forever.

So the question is, what happens when you remove the shorting jumper across Coil B so that it becomes part of the circuit?

(MarkE and other regulars who know please don't answer.)

MileHigh