That is a good try. I agree with all that you say except for one key item. 1). No negative current flows due to the diode. 2). The instantaneous power being delivered to the resistor is I^2*R as you suggest. 3). The DC rated clamp on meter should measure the total RMS current provided it can handle distorted AC waveforms. Now, here is where you have a problem with the measurement. You say to multiply the true RMS current by the voltage and that is where the problem arises. I am confident that you realized that as soon as you said it!
All of the power that is applied to the resistor comes from the wall socket. The voltage at this location is a sinewave at the frequency supplied by the electrical service. There is no DC voltage component, so the DC power being supplied is 0. The AC component of the input frequency is the only one that can have power supplied and that can only be given to current at its fundamental frequency. So, to determine how much power the resistor absorbs you must take the fundamental current component and multiply it by the fundamental voltage supplied by the wall socket. This needs to be corrected for phase shift if any exists with the product by the cosine of the difference in phase of the two components. Therefore, the DC flowing through the rectifier does not contribute to the measurement. Dave -----Original Message----- From: Duncan Cumming <[email protected]> To: vortex-l <[email protected]> Sent: Mon, May 27, 2013 2:56 pm Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments 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) 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
