AC peak current measurement

I am using a Valhalla Scientific digital power analyzer to measure power current and voltage. Unfortunately, the instruction manual for this unit has been lost, so I need some help understanding what this thing is measuring.

The unit displays true power, amps in true rms and volts in true rms.

1.) What is meant by "true power"? How are the calculating this? 2.) I know that the current is not sinusoidal. Is there a way to calculate the peak current using this meter?

Thanks, David

Reply to
David N.
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True power means the average value of the instantanious power. At each instant, you multiply the voltage by the current and then you average the results. Very often some of the values you are averaging are minus because energy is actually being returned to the mains at that time.

If you merely multiply the RMS current by the RMS voltage, you get a number that is often higher than the true power.

Reply to
MooseFET

Some references on true power, etc.:

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I think Valhallas do typically measure peak inrush current, for instance. I would get the model number and use Google to see if you can find the manual. You can also check the Valhalla website to see what you can find or you could just call the company. I remember talking to some people there years and years ago and they were helpful.

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Reply to
mgkelson

In general, is there a way to find the peak current, given Vrms, Irms and P (true power) when the V and I waveforms look as follows?

. . . . . . . . V . . . . . . . . . . . . . . . . . I ... ... .... . . . . . . . . .

If the meter reads true rms for Irms, then I can't simply use Im = Irms/0.707. Is it possible to get Im from these known values?

Thanks, David

David N. wrote:

Reply to
David N.

This is the averaged value of the instantaneous product of line volts and load current. If both are sines, that would be equivalent to

Erms * Irms * cos(phase angle)

Not unless you know the exact waveform.

John

Reply to
John Larkin

Is this a measurement you will have to make often, or is this a one time need? How do you know the current waveform is as you have shown? Normally, one would need an oscilloscope to know the waveform and if you have a scope it can make the measurement.

If you will need to do this often, it might make sense to buy an instrument for the purpose. I see a number of Fluke 189 meters for sale on ebay; $250 buy it now or starting bid of $100 and take your chances. This meter can make the peak measurement you need.

Reply to
The Phantom

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