I hate to say this, but what you say is absolutely wrong. You could only do as you describe if the "voltage" being averaged is the RMS voltage. You cannot take the average voltage and multiply it by the average current to get average power. For example, suppose that the voltage was V=1_0.5sin(wt). The average of this voltage is 1. Lets say we have a constant current of 1A. By your method, the power would be 1 watt. However, the actual power is:
P = (1A) sqrt(mean(1+0.5sin(wt))^2)) = (1A) sqrt(mean(1 + sin(wt) + 0.25sin(wt)^2)) = (1A) sqrt(1+.25 (mean(.5 - .5cos(2wt)))) P = (1A) sqrt(1.125) = 1.0607 Watts On Mon, Oct 27, 2014 at 11:58 AM, David Roberson <[email protected]> wrote: > The instantaneous power being delivered by the source is equal to the > product of the current and voltage. When the current is constant, only DC > voltage loads can accept power and thus energy from the source. All of the > AC voltages that appear across the source terminals integrate to zero > during a full cycle and do not enter into the input power equation. This > understanding seems to escape most people until they review the theories > carefully. I had to prove roughly the same issue to several skeptics that > thought that DC due to load rectification of the AC power source could be > used to sneak extra power into the earlier ECAT. They thought this was > possible since the input power meter did not monitor DC directly. > >
