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Which is a statistician's way of calculating the area under the curve.

Its the actual area under the curve that represents the flow of energy

Now a nice sign wave approximates to an RMS of the peak as you say 0.707.

Different curves obviously have different values but the value is calculated from the statistical method terms RMS.

What to know more or are you asleep now?

Times the peak voltage by 0.7071 and you get 229.8volts "RMS"

If you view 230volts rms on a ossiciloscope (sp) you will see 325 volts.

The rms voltage in ac is the equivilent heating effect of DC voltage. or is briefly shown in brian scaddon electrical instillation.

Hope this helps, by memory of college.

Then if you want you can then take them over a time period if the waveform is a regular continuous function (which is is in the case of mains electricity) and apply formulae 2

using the time frame T1 and T2 and t.

http://www.calibration-checkbox.com/img/rmsform.jpg

You still want to go on???????

Basically you are taking multiple points on the curve of the voltage sign wave.

You are squaring the value.

Adding all the values together.

Then finding the average of these squares by dividing the total by number of points you have taken.

Then and finally take the square root of this average.

And lo it should be around 0.707 of the value of the peak of the curve.

Try drawing it on a graph paper and doing it.

Note the more points you take the more accurate the answer.

Lastly and obviously the points on a mains waveform all conform to a sinusoidal waveform. Therefore if you really want to know.

A sinusoidal waveform can be represented by a formulae and this formulae can be resolved for RMS values and thus the true calculation for a true sinusoidal waveform simplifies out to be the constant you quoted of 0.707.

OK?

http://www.free-ed.net/sweethaven/modelec/acee/lessonMain.asp?iNum=0102

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You may find this helps to get your head around whats happening...if the peek voltage is 325v.what is the rms value ?.i have herad of 0.707 how does this work ?. :|

with a DC voltage connected across a load..

at ANY point in time the current should be constant. (assuming ideal conditions!)

If an AC voltage was connected across the same load

At any point in time the voltage will be either part way through rising or falling to or from a peak.

If you could take a snap-shot of the voltage you could thus measure the current at that instant in time...

for example lets say you had an AC voltage that is going to peak at 350v

connected to a 10ohm load.

If we could read values at 50v samples as it rises from 0v to 350v

we would get..

{I=V/R}

{P=IV}

0v @ 10ohm =0a =0w

50v @ 10ohm =5a =250w

100v @ 10ohm =10a =1000w

150v @ 10ohm =15a =2250w

200v @ 10ohm =20a =4000w

250v @ 10ohm =25a =6250w

300v @ 10ohm =30a =9000w

350v @ 10ohm =35a =12250w

The current and power are increasing as the voltage across the load increases.

If we worked out the average of these eight samples of current and power

we get (0+5+10+15+20+25+30+35)/8= 17.5a

and the power will give us 4375watts

If we calculate the voltage back from the average power & current

(power = I x V) so voltage =power/I = 4375/17.5 = 250v

250v is 0.714 of the peak 350v.

From this very rough calculation you can sort of see how the 0.7xx figure starts to gel into place.

REMEMBER this is a very very wide sample.. only looking at the first quarter of a full cycle...,

And this is a VERY VERY simplified concept to show the sort of principal involved with calculating the true RMS of an AC supply.

When loads more samples are taken over the FULL cycle...

and you apply all the squaring, rooting and meaning...

as Calltronics say's a Very complex sample & calculation method!

then the 0.707 figure comes into place!!

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nice link! Applaud SmileyHere is a good explanation.....http://www.free-ed.net/sweethaven/modelec/acee/lessonMain.asp?iNum=0102

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the 0.707 bit is just one of those fixed ratio figures you need to remember..

Peak x 0.707 = RMS value.

bit like ratio of 'Pi' & circles..

e.g. Diameter x 'Pi' = Circumference. where 'Pi' = 3.142!

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