True power measurement

I believe you are wrong.

Phase and polarity must be taken into account.

If the frequency of one is twice the other then over a half cycle of the lower frequency the circuit both absorbs and returns power to the source and you can not just multiply the instantaneous magnitudes to get the power.

There would be no net power delivered to the circuit under these conditions.

That is where power factor comes into play. At times the power delivered is negitive or the circuit feeds power back to the source.

Phase must be taken into account and you can not just multiply instantaneous values or magitudes. . If you take into account the negitive values then of course it would work and I do recognize that a digital circuit can do these computations. . . I DO NOT FOLLOW MANY OF THESE NEWS GROUPS To answere me address mail to snipped-for-privacy@aol.com

Reply to
BUSHBADEE
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No, he's right, Bush.

...and they are, if the current and voltage is multiplied.

Sure, a negative current times a positive voltage is a negative power; the load is returning power to the system.

There is power delivered when both the current and voltage are the same sign. When they aren't, power is being delivered back into the system (reactance).

When the voltage and current waveforms have a different sign, power is fed back to the source.

Phase is taken into account. The power is negative during the portion of the cycle where the waveforms are of different sign. Increase the phase angle of the two waveforms (reactance of the load) and this portion of the cycle increases. Math works. ;-)

Forgetting the negative values is never a good idea. ;-)

Quite easily and cheaply these days.

Reply to
Keith R. Williams

only in a Euclidian way. . . I DO NOT FOLLOW MANY OF THESE NEWS GROUPS To answere me address mail to snipped-for-privacy@aol.com

Reply to
BUSHBADEE

instantaneous

I think you and I have different ideas when we say sampling instantaneous voltage. I'm thinking about sampling at several times the frequency of that found in the voltage/current. In this way, if the current is a harmonic, then the calculation from each sample will reflect the alternating positive/negative power flow and still yeild the correct results.

Perhaps you're thinking about just measuring RMS for a sample that is several cycles in duration and using that in the traditional calculation of E*I*cos(theta). If that is the case, that would explain our disagreement. I agree, that just sampling RMS and trying to calculate power won't work well if the current is an even harmonic of voltage and/or non-sinusoidal.

But many modern watt-meters sample many times per cycle and calculate the power for just that short sample interval. Then alternating power results caused by non-sinusoid or multiple harmonic current is correctly calculated and when each sample's resulting power calculation is averaged, the correct value for average power is found.

daestrom

Reply to
daestrom

Ah, perhaps that does explain the difference in opinions/assumptions. After all, P==V*I at any given instant. Average that over any period and one gets, *average power*. ;-)

Reply to
Keith R. Williams

---------------------- Daestrom is right. If you took instantaneous samples of the voltage and current to get instantaneous power and then integrated and averaged over the cycle, the result would be the true average power. This is the basis of a digital wattmeter. A conventional electrodynamic wattmeter produces an torque depending on the instantaneous power but this torque is integrated and averaged naturally by the inertia of the movement so it does the same thing. It doesn't account for phase as it only "sees" the instantaneous product of voltage and current. It doesn't measure rms voltage and current and the phase angle to determine power but works at a more fundamental level.

----------

The process of multiplying instantaneous values then integrating and averaging will do the job. The quote above did not mention the averaging etc and that is misleading if you are referring to average power- However, instantaneous power is the instantaneous product of current and voltage and has no phase- just sign.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

Reply to
Don Kelly

I think you a nd I are in exact agreement. But I believe the origonal poster was refering to multiplying the voltage times the current and I do not believe he was thinking of a sampling device.

By the way talking about sampling devices, I have an HP Sampling Scope that is flat to at least 3 Gigahertz I believe. It uses active tunnel diode probes and comes with two of them. Also a delay line comes with it to delay the signal to the input to start the beam before the signal reaches the input. I also have watt meters and resistance bridges and Incutance boxes and ratio trans, and cap boxes made by Arnold Magnetics (Their pay for a HV supply I designed for them) Also have a 5000 volt hi pot tester and some Techtronic scopes which I do not believe are worth much. Does any one have any interest in them. . . I DO NOT FOLLOW MANY OF THESE NEWS GROUPS To answere me address mail to snipped-for-privacy@aol.com

Reply to
BUSHBADEE

An example for one common {GE} meter is: "...contains six independent, fully integrated analogto digital converters, one for each current and voltage signal. The converters are continuous time, delta-sigma analog to digital converters digitizing each signal 1.68 million times per second. Each input signal has a dedicated converter, multiplexing is not used. The converter over sampling rate of 512 provides approximately 3280 complete sets of decimated samples per second. For a full three element meter, 6 sets of 16-bit samples are processed, providing more than 54 decimated sample sets per line cycle."

--s falke

Reply to
s falke

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