Hi MrAl
You have missed the point of my question there. I wasn't asking what the RMS value is rather my query was quite different: Can we find RMS value using the relation V=IR? I would request that you give the question another read.
By the way, I don't get it where you say:
How come does it equal (1/R)*1/sqrt(2)?
Thank you for the help.
Regards
PG
Hi,
Yes i made a few typos in that last post. I'll have to try to get back to that and correct those.
This is unfortunately something you have to put up with sometimes because no author writes
everything perfect ALL the time. I run into this quite a bit while looking stuff up on the web,
and most of it is so much worse than this that you wouldnt believe how bad it can be. I see
equations that are so totally incorrect that it makes no sense to me why the well respected
author would even bother to take the time to present such a thing if there was going to be
no regard for accuracy. Without accuracy in some of this stuff we end up with nothing unless
we can communicate with the author and see what the problem really was. I even see
information left totally out of the text, which is vital to understanding what the author
intended in the first place! Why bother right?
For now let's just start over. We'll do the example of the sinusoidal wave which is usually shown with the
required peak value out in front Ep. Ep is the peak value of the sine (or cosine) wave.
Ep*cos(w*t)
Evaluated with w=2*pi*f:
Ep*cos(2*pi*f*t)
To start the RMS procedure, we first square this:
Ep^2*cos(2*pi*f*t)^2
Now we find the mean which first means integrating from 0 to T, so we get:
(Ep^2*(sin(4*pi*f*T)+4*pi*f*T))/(8*pi*f)
then dividing by T:
(Ep^2*(sin(4*pi*f*T)+4*pi*f*T))/(8*pi*f*T)
and since the period T is equal to 1/f we evaluate that with T=1/f and we get the simple result:
Ep^2/2
And finally taking the square root, we get:
Erms=Ep/sqrt(2)
So the RMS value of the wave is the peak value divided by the square root of 2.
And this is true whether we look at voltage V or current I, so we get either:
Vp/sqrt(2)
or:
Ip/sqrt(2)
Depending on either the voltage or current waves.
Now if we wanted to find the RMS voltage from the equation:
v(t)=i(t)*R
we would just substitute i(t)*R for the voltage. Since R is constant if the current
wave is sinusoidal then the voltage wave is also, so if we use i(t)=cos(w*t) again
we get for the RMS of the current wave:
Irms=Ip/sqrt(2)
but since it is multiplied by R (to create the voltage wave) we have to multiply that by R:
Vrms=R*Ip/sqrt(2)
This works because R is not a function of t and so can be taken out of the integral so the
whole integral simplifies to that of finding the RMS value of the current and then simply
multiplying by the resistance R. If we kept R inside the integral we would just end up
squaring it and then later taking the square root so we'd still end up with the final
result being the same where Ip/sqrt(2) is just multiplied by R.
If you are talking about a DC value rather than AC, then the DC value can be taken out of
the integral because it is not a function of time. So we'd end up with Vdc^2/f and then
divided by 1/f which would give us Vdc^2, and then taking the square root we'd get
just Vdc back again with no multipliers. So for a DC voltage we just end up with:
Vrms=Vdc
In this case f would just be any f such that T=1/f is any period we care to choose (because a
DC voltage is the same over all time).
If this doesnt help with your question about V=IR then you have to present some example
about what you are talking about because that would mean that your question isnt clear
enough by just writing V=IR and asking about that.
BTW im not sure about what you said 'does not come out right'. What did not come out right?
We could do other examples here too no problem.