Brad, can you confirm that your FG "black" output lead is not grounded back through the chassis and to the mains ground pin?
Also, my estimate of 72 degrees phase shift is just that: an estimate based on the scope's graticule markings, the trace zero crossings and peaks. Also, your coil has relatively few turns of pretty heavy wire I think, so your coil's DC resistance is probably very small. Can you get a reading of that somehow? A more careful estimate of the phase shift, and including the coil's DC resistance, may make some difference in the calculations.
. Maybe Verpies can help out here. Really, what we are calling Input power should actually be considered the total power dissipation of the load circuit, and I think my calculations breaking down the power dissipations of the various parts of the circuit show that this is correct.
Brad, can you confirm that your FG "black" output lead is not grounded back through the chassis and to the mains ground pin?
TK
I have always stated that my FGs black lead is grounded,and so is the scopes ground lead's.
This is why i have to use my inverter to run my scope,so as i get isolation between the two common ground's of the scope and FG.
So we are really back to square one, since we don't actually know the impedance of the power source (the FGs we are using). I'd like to believe that my ElCheepo DDS is 50 ohms impedance but I don't think it actually is.
With my FG,you can either choose 0 or 50 ohms impedance.
I'm not sure how the Max Power Transfer Theorem affects these calculations or conclusions, except that the true Input Power is likely to be larger than what we are calling Input Power
Unless the power factor is 1,then i think the real power value will be less than the apparent power we are measuring.
I carried out a load test today,using a simple load.
The load is a FWBR,with a cap across the output of the FWBR,and an LED as a load(also across the cap)
With the FG placed on the AC input of the FWBR,i can get 2.82 volts across the cap/LED-BPC circuit not in play.
This is the maximum power the FG can deliver to the load(LED)
If i then hook the FG leads to the BPC circuit,minus the input resistor R1(R2 still in play-10 ohms),and then place the AC inputs of the FWBR at a certain turn of L2,and the other at the input of L1,i can get 2.91 volts across the cap/LED,and 700mV across R2
Not only is the dissipated power greater in the LED,when the BPC is in play,we are also dissipating 49mW of power from R2. But this is only when one side of the FWBR is on a certain turn of L2--i think it was the 3rd or 4th turn in from the outer turn.
If i go a turn either side of the correct one,then the power being delivered to the load drop's to a value below that of what the FG it self could deliver to the load.
Thought that was interesting.
Will shoot a video tomorrow after work,and post here.
Brad
Never let your schooling get in the way of your education.