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