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

[davrom]

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?

 

[busbar]

FWIW = for what it's worth

 

[IRstuff]

I think wutz is simply a version of "what's"

TTFN

 

[davrom]

Thank you for your replies. I still have reserves for "wutz" (due to German influence - it would have been very easy to be used "what's&quot, but I am happy the matter has been solved out.
Sorry for inadvertence.

 

[jbartos]

Suggestion to jghrist (Electrical) Dec 8, 2003 marked ///\\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.
///True.\\ Yet the evil Utility charges for the rms value of something that averages out to nothing!!!
///Not quite. For example, a heating element will produce amount of heat during positive area proportional to the plus area and during the negative area amount of heat proportional to the negative area. Now, I am being charged by the Utility for areas plus and minus for Energy EWhrs,rms=Vrms x Irms x time, which amounts to more than EWhrs,av=Vav x Iav x time = Vdc x Idc x time = Ewhrs,dc.
It is hard to beat this engineering and scientific fact, which has been around for long time. I would much appreciate to have the Utility supplying EWhrs,dc instead of EWhrs,rms for my heaters. I feel more comfy around the wormer heaters. What about others?\\

 

[jghrist]

jbartos wrote,

"...which amounts to more than EWhrs,av=Vav x Iav x time = Vdc x Idc x time = Ewhrs,dc."

Except that EWhrs,av does not equal Vav x Iav x time. The average power equals the average of the product V·I which equals V_rms·I_rms, not the product of the averages.

In the case of current through the heater, W = I²·R. The average of I² is I_rms² (by definition of rms). Average energy is Whrs_avg = I_rms²·R·time.

 

[electricpete]

davrom - Why wutz not what's?
Nothing to do with German, just sounded funny at the time.

I think the subject of rms just makes me giddy. There are some classic discussions in the following links.

thread237-68161

http://www.brazosport.edu/~pschimpf/forums/vitreousenamelledwirewoundresistor.mht

I am tempted to take up the discussion of rms but based on past experience maybe I better just pass. I recognize the name jghrist has been around a long time and very capable of bringing order to the chaos.

 

[stevenal]

Jbart,

Try the square root of the area under the square of the sinusoid. Wutz a wormer heater? Sounds nasty.

 

[jghrist]

 

[stevenal]

Yep, nasty. Thanks.

 

[busbar]

Wasn't Faber College’s Dean Wormerheater a character in the 'Animal House' movie?

 

[jbartos]

Suggestion to jghrist (Electrical) Dec 10, 2003 marked ///\\jbartos wrote,

"...which amounts to more than EWhrs,av=Vav x Iav x time = Vdc x Idc x time = Ewhrs,dc."
///This is extracted out of context.\\Except that EWhrs,av does not equal Vav x Iav x time.
///Not true. For dc where dc=av, it does. By any chance, have you ever designed any electronics circuits where ac and dc are frequently superimposed or combined?\\ The average power equals the average of the product V•I which equals Vrms•Irms, not the product of the averages.
///Yes, however, in this case you are addressing average rms power not average dc power. There is a big difference between those two. Please, what is your educational level?\\In the case of current through the heater, W = I²•R.
///Not clear what you mean by I. Do you mean Idc=Iav or Irms?\\ The average of I² is Irms² (by definition of rms).
///Yes, the average is Irms average of sinusoidal wave not the absolute value of sinusoid, i.e. |sinwt| often obtained from a full wave rectifier. However, this is an average of Irms not Idc=Iav. It would be better to go to the engineering paper with squares and see those differences. There is a big difference between (sinwt)^2 leading to Irms and obtaining Iav from |sinwt| (including Iav^2).
Average energy is Whrsavg = Irms²•R•time.
///Yes, however, this is rms energy, i.e. Whrs,rms=Irms^2 x R x time, not av=dc energy of |sinwt| Whrs,av=Iav^2 x R x time. Please, notice that electronic engineering books have not changed anything in this area, only electrical power engineering started downplaying the form factor and differences between Irms and Iav=Idc. I know why, do you? Please, notice that this has been around for some time and will stay in here for some time, like it or not, since this is a scientific fact, not any engineering interpretation or twisting. Please notice that Irms = sqrt [(1/T) x Integral (Imax x sinwt)^2 dt] = (Imax / Sqrt2) is different from Irms = sqrt [(1/T) x Integral {[(1/T) x Integral (Imax x |sinwt| x dt)]^2 x dt}] = (2 x Imax / pi) = 0.6366 Imax = sqrt2 x 2 x Imax / (sqrt2 x pi) = (sqrt2 x 2 /pi) x Irms = 0.9 x Irms = Iav
As it can be seen, the Irms of Imax x |sinwt| is different from Irms of Imax x sinwt.
See Reference:
L.J. Giacoletto “Electronics Designers’ Handbook,” 2nd Edition, McGraw-Hill Book Company, 1977, Table 4.3 Characteristics of Periodic Waveforms.\\

 

[Skogsgurra]

jb,

I can see that you are busy convincing the engineering community that they have been using the wrong definition of RMS for about one hundred years. Would it not be worth while to sit down and think. You might find that you and Don Quixote are in the same business.

 

[jghrist]

jb wrote:

"(quoting me)'In the case of current through the heater, W = I²•R.'
///Not clear what you mean by I. Do you mean Idc=Iav or Irms?\\\"
I mean that the instantaneous power is equal to the instantaneous current squared time the resistance. It doesn't matter if the current is dc, sinusoidal, exponential, square wave, or worm shaped. To get the energy, you integrate the instantaneous power over whatever time period you want. The result is the integral of I²·dt over that time period.

Also,
"Please notice that Irms = sqrt [(1/T) x Integral (Imax x sinwt)^2 dt] = (Imax / Sqrt2) is different from Irms = sqrt [(1/T) x Integral {[(1/T) x Integral (Imax x |sinwt| x dt)]^2 x dt}]"
Actually, no it isn't because the square of sin(&[ignore]omega[/ignore];t) is equal to the square of what you call |sin(&[ignore]omega[/ignore];t)|. Even someone with my limited educational level knows that the square of -n equals the square of +n.

When I said "Except that EWhrs,av does not equal Vav x Iav x time", I was not completely correct. They would be equal in the unique case of constant voltage and current. But this is the only case.

 

[electricpete]

Another thread going on right now:

thread237-81315

 

[allmend]

"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."

Hi Buzzp

This is not true in a circuit with reactive loads (C,L). If your current is a mix of many frequencies (harmonics), your RMS-Current depends on the harmonic mix and in this case the heating effect of different harmonic currents with the same RMS-Current-Value may be different.

If you have a pure C-Load, you will not have any DC heating in your circuit. I think that this is obvious.

 

[buzzp]

Allmend,
For the case presented and as stated time and time again, this is assuming a purely RESISTIVE LOAD. Whatever the rms level, if you get the rms level and apply the same amplitude, DC, to a purely resistive load, the steady state temperature will be the same as with the AC rms current. The discussion of reactive components has never been brought up unless I overlooked a post.
As far as harmonics, the rms current takes into account harmonic current. This is the reason it is used because when you have harmonics you do not have a pure sine wave so finding the average a scaling it as though it were a pure sine wave results in errors. Even with the ugliest waves around, if you find the real RMS value, apply this to a resistive load and apply the same value in DC, the heating effect will be the same.

 

[electricpete]

I agree with buzzp 110%.

Whether we are talking dc current, ac sinusoidal current or non-sinusoidal current, only the resistance in the circuit generates heat (not the L or C).

As long as we specify the current (we are talking about heating as a function of current), the existence of L and C elements is irrelevant.

Maybe you are thinking of a circuit with constant voltage applied. Add C's and L's in series and you change the ac and dc currents and therefore the resulting heating. That does not seem relevant to a discussion of heating as a function of current, where current is the independent/controlled variable under discussion.

If the resistance is fixed and the rms of the current is specified, those TWO quantities are all that is needed to determine the heating (Irms^2*R). It doesn't matter whether that rms is associated with dc or ac or non-sinusoidal waveforms.

For dc currents, the rms value is the same as the dc value.
For ac currents the rms value is the Peak/sqrt(2)
For general currents the rms value is
sqrt(<i(t)^2>) = sqrt{(1/T) Integral(i(t)^2)dt}

BTW, depending on the size of the cable and frequency of the current, skin effect could alter the effective resistance. It is a completely different subject from what we have discussed here but important for large cables.

 

[allmend]

„RMS of any ac signal has the same heating effect as the same value in DC, period.“

„Even with the ugliest waves around, if you find the real RMS value, apply this to a resistive load and apply the same value in DC, the heating effect will be the same.“

buzzp,

You didn’t mention in your first posting that you apply in your model a resistive load and simulate the “ugly current” in your model.

I agree that the heating of a resistive load is an equivalent of RMS-Value of the current and that the RMS-Value can be calculated or measured.

But this fact has only a theoretical value.

Current is always a function of the impedance and voltage (I=U/Z). In a real circuit there is no pure resistance or inductance, and there is a feedback between the non-linear load and the voltage source. RMS-Values of a current is just a figure with little practical value.

 

[ScottyUK]

allmend,

How can you say that [blue]RMS-Values of a current is just a figure with little practical value[/blue]? The RMS value governs the thermal losses in a circuit, whether it is the resistance of a 'pure' resistance or the inevitable resistance of a real-world inductance. The resistances of all the non-ideal components we are forced to live with in this imperfect world will dissipate energy as heat when current flows through the resistance in question. The energy dissipation will depend on the RMS value of the current, and on the resistance.

 

[buzzp]

Temperature ratings are extremely important and RMS current is needed for these calculations. I do not feel the need to defend the value of rms current. If you don't understand the value of it then some reviewing is in order.

I do understand that impedances and such are important in calculating the current. This discussion has nothing to do with calculating the current and everything to do with measuring the current. Take your cheap clamp on meter and measure the output of a drive. Use that to calculate temperature rise of cable and equipment and you will be in serious trouble. Take your rms clamp-on and do the same measurement and calculations and you will be right on (ignoring skin effects).