OK,so i have made this very easy for some one well versed in calculating power factor into the below results.
You will see that i raised or lowered the voltage value by small increments of each channel,so as both wave forms are even-close to V/D line-making the job easier
Each pic below shows the scope positions,and attached is the associated scope shot with the values of each channel.
So,who is well versed enough to calculate-->
1-Total P/in
2-Dissipated power of R1+L1
3-Dissipated power of R2+L2
Brad
Brad, can your scope measure the Phase Shift between channels? Or can you post one shot with just a single full cycle of both channels horizontally so we can estimate it visually?
Anyhow... for the power calculations.... I'll take a shot at it.
I think for the moment we can disregard the dissipation in the Coil portions and just look at the resistors themselves.
In your first shot, to get the input power, we are looking at the voltage across 200 ohms so this is shown by the CH2 measurement, which is 6.20 Vrms, giving a current of 0.031 Arms.
So the total power IN is P=I
2R = 0.192 watts.
EDIT: Do we need to make any corrections for possible phase shift at this point?
For the second shot, looking at the Vdrop across R1, again we read CH2 at 1.36 Vrms, which gives a current value of 0.0136 Arms, and a power dissipation of P=I
2R = 0.0185 watts.
For the third shot, looking at the Vdrop across R2, we read CH2 again, at 2.56 Vrms, for a current of 0.0256 Arms, and a power dissipation of 0.0655 watts.
So we have Pin = 0.192 W and total dissipation in the resistors of 0.0185 W + 0.0655 W = 0.084 W dissipated in the resistors, compared to 0.192 W input.
HOWEVER... The picture changes if we take the R+coil readings on CH1 and do that math. I get a total "dissipation" value of 0.2371 W which would lead us to conclude that we have an OU COP of about 1.23.
Cheezburger Time !!!
But is this really correct? We know that the pure resistors are dissipative loads that give off power as heat, lost to the system. But what about the coils? Neglecting their DC resistance of course... is the inductive load actually dissipating power like the resistors do ... or is it _storing_ it?
I've been working on this for about an hour, and it's late and I am coffee-deprived, so PLEASE check my math and assumptions everybody!