Partnered Output Coils - Free Energy

[Red_Sunset]

If you believe that the current is unlimited at T0 then you are mistaken.  .................................
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MarkE,
  I made the assumption from a general viewpoint only  (that the time constant of the inductors is shorter than your input signal).  Looking at your reply, they obviously aren't.   
I think we are also looking at different timing diagrams (3), I see the latest that starts with the Mosfet switched on.  I started with a previous one with the mosfet in the OFF condition

1..  At power-up, the current and mag field in both coils would be max.(TP1=0v)
No they will both be zero at T0.
*Red  Granted, I was more thinking after pwr settling down  and before the signal generator is switched on

3..  1H will produce a EMF that will be dissipated through the mosfet protection diode (TP1=diode junction voltage)
When the MOSFET first turns on there is zero net flow through the MOSFET.
* Red    Please expand your answer. 
* Red   I would gather that the current drain provided by the H1 and now replaced by the mosfet wouldn't change the current condition for 2H.  We can ignore the EMF current that will drain through the mosfet assembly as misc
 
4..   At this point, the current limit of the power supply set to 1A will save the mosfet.(TP1=~0v, mosfet conduct voltage)
The voltage source was specified as ideal.  It can supply any current the circuit will draw from 1V.
* Red   The current in H2 during the "ON" period remains unchanged since no current change occurs,  therefore we have a near DC condition with a increasing current gradient . On a longer coil time constant this would naturally be regulated by the impedance value. But in all conditions it would be an increasing value !

Regards, Red_Sunset

[MileHigh]

Just a reminder:  The whole experiment really starts at two seconds.  You don't really have to worry about the "set up" period between zero and two seconds.

[MarkE]

MarkE,
  I made the assumption from a general viewpoint only  (that the time constant of the inductors is shorter than your input signal).  Looking at your reply, they obviously aren't.   
I think we are also looking at different timing diagrams (3), I see the latest that starts with the Mosfet switched on.  I started with a previous one with the mosfet in the OFF condition

1..  At power-up, the current and mag field in both coils would be max.(TP1=0v)
No they will both be zero at T0.
*Red  Granted, I was more thinking after pwr settling down  and before the signal generator is switched on

3..  1H will produce a EMF that will be dissipated through the mosfet protection diode (TP1=diode junction voltage)
When the MOSFET first turns on there is zero net flow through the MOSFET.
* Red    Please expand your answer. 
* Red   I would gather that the current drain provided by the H1 and now replaced by the mosfet wouldn't change the current condition for 2H.  We can ignore the EMF current that will drain through the mosfet assembly as misc
 
4..   At this point, the current limit of the power supply set to 1A will save the mosfet.(TP1=~0v, mosfet conduct voltage)
The voltage source was specified as ideal.  It can supply any current the circuit will draw from 1V.
* Red   The current in H2 during the "ON" period remains unchanged since no current change occurs,  therefore we have a near DC condition with a increasing current gradient . On a longer coil time constant this would naturally be regulated by the impedance value. But in all conditions it would be an increasing value !

Regards, Red_Sunset

Hopefully the diagram above clarifies any uncertainty.

The intention is to show the MOSFET as conducting at T₀.  But as in my prior explanation if we model with the MOSFET off at T₀, then it will be fully on within 90ns.  1V across 3H ramps current at 0.33A/s and in less than 90ns ramps less than 30nA.  As explained before:  The MOSFET turn-on changes the voltage across both coils simultaneously, but does not change the current instantaneously.  The magnetic field of the inductors prevents that.  Current flow begins increasing in the clockwise loop from the voltage source through the 2H inductor and the MOSFET.  The current through the 1H  inductor decreases very slowly because the voltage in its loop is nearly zero.  The current that it carries is tiny and immaterial to the circuit behavior.  The current does reach zero long before the MOSFET turns off at T₀ + 2 seconds.

The time constant of an inductor is:  L/R.  We have stipulated that R = 0.  the L/R time constant is therefore undefined and the inductors perform as ideal integrators of Vdt.

[MileHigh]

You are getting super low-level there Mark.  KISS comes to mind.

[EMJunkie]

Ok, due to demand, another: