Hello Mr RB,
Thank you for the great advice.
I have a few questions please.
1. You're actually saying that I can treat the current that comes from the capacitor as DC current, and the voltage on that capacitor as DC voltage; and then you get P = I * V .
Did i get you right?
2. I just dont quite understand why do you suggest to connect a large capacitor in parallel with the existing one?
Is it only to decrease the ripple current and voltage, and therefore being able to measure the true power as the product of two DC components (I and V) with better proximity?
3. Moreover, adding this large capacitor will only increase the apparent power that the SMPS consumes from the power plant, but wont change the true power the SMPS consumes from the power plant, right?
4. Why did you say the connect the resistor between the two capacitors, and not after them? we dont care about the current thats going to the capacitors, but only the current thats going to the system which the PSU drives.
If I understood your method right, then its a really good one, it relays on the fact that Pin = Pout (where Pin is the power that the power plant provides, and Pout is the power that the capacitor provides).
The capacitor itself also requires power from the power plant, but its reactive power therefore we dont need to measure it.
So I wonder why i was suggested to measure areas of curves which is such a hard work.