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The RMS value of a signal is in words:
"The root, of the mean, of the square".
This means first we square it (algebraic squaring), then we take the mean (integral), then we take the square root (nothing more than taking the square root). So it is three separate steps each involving a different kind of calculation.
For a general time domain signal like i(t) this is nothing more than the square of i(t) followed by the mean of that, followed by the square root of that.
If you have i(t) defined by something else like v(t)=i(t)*R then you might first solve for i(t) like:
i(t)=v(t)/R
But in any case you always do:
square root of:
(1/T)*Integral( f(t)^2 ) dt
So lets say v(t) is known to be sin(wt). If we have v(t)=i(t)*R then we have i(t)=v(t)/R, so the integral becomes:
Integral( (v(t)/R)^2 ) dt
and since R^2 is a constant we can take that outside the integral and get:
(1/R^2)*Integral( v(t)^2) dt
so the whole thing is:
sqrt(1/(R^2*T)*Inegral( v(t)^2 ) dt )
which equals:
(1/R)*1/sqrt(2)
If we had a different function we would have gotten a different value at the end.
But in cases where we dont know what v(t) or i(t) is, then we have to first solve for that if we can. If we cant solve for that function then we cant integrate. We also have to remember to integrate over the appropriate time period(s). If we find symmetry we can take advantage of that too.
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:
MrAl said:
so the whole thing is:
sqrt(1/(R^2*T)*Inegral( v(t)^2 ) dt )
Do you mean v(t)=i(t)*R? or do you mean V=I*R say for a DC circuit. In the latter case V is Vrms, however we normally associate RMS with an AC signal, but we don't have too. e.g. 12 VDC = 12 Vrms
Another definition of RMS is the DC value that would generate the same amount of heat as the AC waveform into the same resistive load R.
Sine waves are special cases that are encountered often, so we know the p-p and RMS values of common sin waves (line voltage) and we need those numbers (e.g. p-p/RMS of secondaries) to design a linear power supply. So when v(t)=Asin(t), t=0, 2∏ you can plug into the rms formula for one period and get the special case equation. e.g. https://www.wolframalpha.com/input/?i=root+mean+square+sin(t) for A=1
Cheap meters measure the average value of rectified AC and multiply it by a fudge factor valid for a sin wave only input to get RMS. High end TRMS or True RMS meters will have options for the AC and the AC+DC waveform. Averaging is easy: you full-wave precision rectify and filter.
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.
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?
Have you tried https://www.lyx.org/ to make equations? It's pretty simple. Just create a new document and make the equation. Then do a copy of the equation and enclose between [latex] [/latex] tags. Go advanced and Preview the result before posting.
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?
It seems like there has been miscommunication. Seriously I wasn't complaining about any typos etc. As a matter of fact I didn't know if you had made any mistakes there. I was merely saying that you had failed to see the angle of my question and part of the blame could be laid upon me - perhaps I didn't state my question clearly. Right now I don't have much time to rephrase my original question because I have a lot of assignments due next week and that question is part of series of questions which I ask for my own understanding so it can be postponed for sometime. Right now I just wanted to make it clear that I wasn't making any complaints about what you wrote. Thank you for the understanding.
Yes i guess you'll have to take a moment to try to restate the question. Usually the question is just how to calculate the RMS value and that's it. You must have thought about something related and want to know how it works and that's understandable, so now you have to convey exactly what it is you want to know.
Some of the other usual questions are:
How to calculate the RMS value of a triangle wave (comes up quite a bit).
How to calculate the RMS value of a triangle riding on a DC bias.
How to calculate the RMS of a pulse.
Relating RMS to power.
The page you show makes it clear that the intention is to find an equivalent DC current that would deliver the same average power into the resistor. This is why using v=Ri is not the correct way to go. Ohms law does not give you power, so it is not good for finding the effective voltage and current.
When you use the power formula I^2R or V^2/R, you directly get the RMS current and RMS voltage formulas. Hence, the fact that the operations of square root, integration (averaging) of the square is a direct result of the definition of the expressed problem. In other words, you state the problem (find the effective value that gives the same power) and you determine the answer by applying the power formula and you discover that root-mean-square is the value that works.
Finding the RMS value via the defined calculation is a separate problem, and solutions for common waveforms (usually periodic ones) are well-known. I hope this is clear.
The page you show makes it clear that the intention is to find an equivalent DC current that would deliver the same average power into the resistor. This is why using v=Ri is not the correct way to go. Ohms law does not give you power, so it is not good for finding the effective voltage and current.
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