I will give it my best shot.
Consider a diode in series with a resistor, and connected to an AC
outlet. For the first half of the cycle the diode conducts, and a
positive current flows. For the second half, the diode does not conduct
and NO NEGATIVE CURRENT FLOWS, even though a negative voltage is
present. This is the function of a diode. So what you have is an
intermittent flow of positive current, which delivers power to the load
resistors. The magnitude of this power is given by I^2*R.
If you were to measure the current using a clamp on ammeter which is DC
rated, then the current would be determined accurately and the RMS value
determined by the digital voltmeter (which must be a true RMS type, of
course). Multiplying this by the voltage gives the power dissipated in
the resistor.
If you were to measure the current using a clamp on ammeter which is
NOT DC rated, then only the fluctuations in current would be measured,
which fluctuations would be a lot less than the true value of current.
So a misleadingly low value of current would be measured, leading to a
substantial under estimate of the power dissipated in the resistor.
This is electrical engineering 101, but it shows some of the problems
involved in measuring AC power with a non-sinusoidal waveform. There are
those who assume that a commercially available "power meter" measures
just that, but in fact this is a difficult task that should be
undertaken by a qualified engineer with knowledge of the waveform that
he is measuring. The specs of such an instrument clearly indicate the
limitations to which it is subject, and one must be careful not to
exceed these limitations. An absence of DC sensing capability is one
such limitation.
In the diode example above, the diode itself does not provide any power
whatever. It merely confuses some types of power meters.
On 5/26/2013 9:51 PM, David Roberson wrote:
It does not make any difference whether or not the instrument measures
DC current through the input power cables. That issue is dead unless
someone wants to insist that Rossi or one of his partners hid a DC
supply inside the wall, or in some other place which allows the DC to
appear at the power input terminals. This would have been obvious to
anyone looking at the voltage.
Andrew or Duncan please explain how the DC current through the input
power cable is able to deliver a large power to the load resistors?
It can not be done with any type of diode hidden within the blue box.
Are you ready to concede the point?
Dave
-----Original Message-----
From: Alan Goldwater <[email protected]>
To: vortex-l <[email protected]>
Sent: Mon, May 27, 2013 12:19 am
Subject: Re: [Vo]:RE: [Vo]:Re: [Vo]:Torbjörn Hartman describes power
measurments
That is a different instrument. The one used in the tests (PCE-830
<http://www.industrial-needs.com/technical-data/power-anlayser-PCE-830.htm>)
does not measure DC.
On 5/26/2013 7:53 PM, Jones Beene wrote:
Did you actually check the PCE site?
It looks to me like all the current clamps on the PCE power analyzer
site measure both AC and DC
http://www.industrial-needs.com/technical-data/current-detector-PCE-DC-3.htm
