If the resistance is a constant, then would either (V^2)/(R) or (I^2)(R) equal average power?
Harry Michel Jullian wrote: > As discussed here before, if any or both of the current and voltage is > constant then power can indeed be measured as simply as you said, otherwise it > can't. It is a common mistake (or fraud) in overunity claims to pretend that > average I times average V is equal to average I*V (power) when the signals are > both time-varying. It is also common to believe (or pretend) that they are > constant by not looking at them with the right time resolution. > > Michel
