That is not true. My analysis applies to any waveform at the wall socket. The
diode just causes DC and harmonics to flow. They do not influence the final
outcome. Take time to do the calculation yourself and you will realize that
what I am saying is true. Why the endless repetitions?
Dave
-----Original Message-----
From: Duncan Cumming <spacedr...@cumming.info>
To: vortex-l <vortex-l@eskimo.com>
Sent: Mon, May 27, 2013 3:18 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
This is only true for sinusoidal waveforms. As soon as you introduce
non-sinusoidal waveforms, such as by using a diode, then different
calculations must be used.
http://en.wikipedia.org/wiki/Power_factor
On 5/27/2013 7:49 AM, David Roberson wrote:
All of the power being delivered into the resistor from the wall
socket can be determined by taking the AC voltage which is a sine wave
and multiplying it by the fundamental frequency of the AC current(also
a sine wave). This must be adjusted by multiplication by the cosine
of the phase angle between the supply voltage and fundamental current.
