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What is the r.m.s. value of a current? 10

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davrom

Electrical
Nov 30, 2003
15
I responded to a question of this forum (cable selection based on duty cycle) where I afirmed that the r.m.s. value is similar to the thermal equivalent current. I realized this may be a stupidity (in fact, I am pretty sure).

I analyzed the problem and now I believe that the r.m.s. value of a current is the so called "effective" value of the current (for ac currents).

I.e. for an ac current i(t) = I*sin(omega*t - fi), the effective value is sqrt(2)*I, where I is the amplitude.

Can anyone explain me what the r.m.s. value is?
 
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The rms value is the peak value divided by the square root of 2. It's basically a way of stating the statistial average of a sine wave.

Am I missing something in your question?
 
No, you don't! Thank you. It is indeed the "effective" value of a current.
 
Suggestion: IEEE Std 100-1992 "Dictionary"
Yrms=sqrt[(1/T)(integral from a to (a+T) of y(t)^2 x dt)]
(there is a typo in the dictionary, namely t should be T)
Also,
y=sqrt[A1^2 + A2^2 + ..... + An^2]
where A1, A2, ... An are the rms values of the fundamental component, second harmonic, .... , nth harmonic, respectively.
Refresh the Fourier Series theory.
 
Note that ScottF's explanation is limited to pure sinusoids.

I tried to put the algebra on here, but it looked awful. The RMS value includes all frequencies contained in a waveform, not just the fundamental. A college-level textbook in your library should give you the algebraic form of the calculation.


 
davrom,

You are being much too hard on yourself when you say "...the r.m.s. value is similar to the thermal equivalent current. I realized this may be a stupidity (in fact, I am pretty sure)."

You are in fact correct. The RMS value is the constant (or dc) value of a time varying current that would produce the same heat. "Effective" may be a better term because it applies to real energy other than heat as well. RMS, or root mean square, is the square root of integral of the square of the time varying current over a time period divided by the time period. It is the thermal equivalent because the heat produced by a current through a resistance is proportional to the square of the current.

For a sinusoidal current over an integral number of cycles, the integration works out to make the equivalent constant value equal the peak value divided by sqrt(2).
 

FWIW, {no offense intended} IEEE 100-1992 tags the term “effective value” as deprecated.
 
This all begins with the fact that resistive loss in
any conductor is proportional to the square of either
measured value current through or voltage across.
P=I*I*R or P=(V*V)/R
RMS allows to account for the expected heating to be a
non-linear function of the current or voltage.
So which wire will run hotter.
12A at a 50% duty cycle.
6A at a 100% duty cycle.
The average current is the same but the 12A wire is hotter.
Rodar
 
In the DC world, when we calculate power loss, we use

Loss = R * I^2

In the AC world, we would like use the same formula. Therefore the concept of r.m.s was invented, such that we can do

Loss = R * I(r.m.s)^2

In certain sence, r.m.s is an equivalent current for loss calculation. Since heating is propotional to loss, sometimes people think r.m.s is the thermal equivalent current.
 
Suggestion: RMS of dc current or voltage is Irms=Idc or Vrms=Vdc
However, if the waveform is a pure sinusoid, then Irms is different from Idc. Irms is a mathematical approach to waveforms to obtain a constant value that is convenient to use, e.g. Irms of Vmax x sin(wt) = Vmax/sqrt2.
Irms = constant is often called Irms dc and
Vrms = constant is often called Vrms dc. This is by a coincidence that a constant value of some function, e.g. Vmax sinwt happens to be equal to a constant=Vmax/sqrt2.
 
RMS of any ac signal has the same heating effect as the same value in DC, period. There is nothing extraordinary about this. Its been the same way since the beginning of time.
 
Actually, buzzp, God didn't create ac until the fourth day. [angel]
 
Suggestion: Take a page of engineering paper with nice squares on it. Plot an ideal sinusoid of voltage or current. Then, convert the areas under the sinusoid to area that would represent the sinusoid area in form of rectangle. Measure the area. This area would be a dc rectangle area, e.g. battery dc, equivalent. Then, plot area that would correspond to rms value. This area should be bigger than the dc area rectangle. What this means is that rms dc rectangle is bigger. The power of dc power rectangle (Idc x Vdc) would be .81 x area of rms rectangle (Irms x Vrms). Therefore, the Utility nicely collect 19% more for rms power. That is why the Utility is so rich. Who would like to have 19% extra in ones pocket? I am just a little consumer.
 
jbartos,

The area under the sinusoid would be positive for 1/2 cycle then negative for the other 1/2 cycle. Total area over one cycle would be zero. Yet the evil Utility charges for the rms value of something that averages out to nothing!!! [soapbox]

 
busbar, what "FWIW" does mean?

electricpete, what "Wutz" does mean? Is your irony strictly limited to my English or is extended over my person too?

To the rest of the members: thank you all replying to my question. I have got the answer I have been looking for.
 
Jghrist

Indeed, because successive half cycles are positive and negative, the area enclosed by the sinusoidal waveform will equate to zero for an integer number of full cycles.

That is why root-mean-square was invented - because when you square a negative value you get a positive one. You then work out the mean of the "positively forced" waveform and square root it !!
 
tomato,

If the utility (mean, vile and nasty people as they seem to be) refused to pay you the return current (the negative half-cycles) and charged you for the RMS value of the positive half-cycles - would that help?

davrom,

I have a hard time reading the native's tounge myself (being Swedish). I think that FWIW means "For What it is Worth" and that "Wutz" is a (German influenced?) way of saying "What does". But then again, I'm more of a technician than a linguist. So I might be wrong there. But I am sure that no-one meant to hurt or ridicule you.

 
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