I’ve had this idea that I recently realized is an assumption. It seemed to me that the average of the instantaneous power (I*V) is always the real power dissipated by the load. However, if you have an AC voltage centered around ground with a resistive load, you’ll get a current waveform with the same shape and phase as the voltage. If you average the product of the current and voltage in this situation, you’d get zero. So here’s my question:
If you have an AC voltage waveform centered around gnd over some complex load that causes distortion/deformation in the current signal so that the current is not purely sinusoidal, how would you measure the power?
What if the AC voltage waveform was slightly offset?
Best Answer
Think again...
I think you need to look at the top left diagram (resistive load).
Picture from this answer. See also these answers: -
What's the most economical way to digitally measure 240V mains voltage, current and power factor?
Average Power Versus Real Power
Measuring AC power usage
Quite simply instantaneous v multiplied by instantaneous i then averaged will always give you real power consumed (i.e. what you would be billed on by your utility company).
Exactly the same answer: Average(v x i) is power.