One more RMS question (theory) 1

[QUAVIET]

If I have a unknown AC voltage impressed on a purely resistive circuit and wish to know the RMS value of this voltage but do not have a meter. In theory, if I measure the temperature of the resistive element and then remove the unknown voltage source and hook up a variable DC power supply. I then adjust the variable DC power supply untill I record the same temperature on the resistive element. Then is not the DC value of the power supply output the same as the RMS value of the unknown AC voltage source?

 

[Skogsgurra]

Yes, that is actually the definition of RMS.

 

[QUAVIET]

What I really meant to say was, what causes an X-RAY? I always thought is was one electron hopping on to another. Actually it is the deceleration. According to a Science Teacher.

 

[IRstuff]

What does that have to do with rms?

As for your original question, that's the way certain radiometers are designed. Two identical resistors are employed, one is exposed to the radiant source while the other is driven by a DC source until its temperature matches the other. Thus, by inference, you've measured the rms power of radiant source.

TTFN

 

[jbartos]

Suggestion: Visit

http://cybertron.vlsi.uwindsor.ca/85-124/Docs/85-124-Lab-2.pdf

for a typical explanation of DC versus AC RMS relationship in the measurement or meter areas. E.g.
"Average Responding Type RMS Meter
An average responding type RMS analog meter is designed for sinewave use only and rectifies (full-wave rectification) the input sinewave to form a DC waveform with an average value of 0.636 VMAX. The meter movement responds to the average DC
value. An internal multiplying scale factor of 1.11 (=0.707/0.636) is used to obtain the correct RMS value."

The above are the practical aspects of DC versus AC of sinusoidal waves.

For example, it is obvious that 10A DC supplied to a resistor produces more heat or Watts than 10A RMS = 1.11 x Idc = 1.11 x 9 ADC
where Idc = 9ADC only. Therefore, for 10A DC = 0.9 x Irms = 0.9 x 11.1 A AC , Irms=11.1 A is needed. If Irms=11.1A is not supplied, only Irms = 10A, then the Idc = 10A produces more Watts or heat than Irms=10A.

 

[Skogsgurra]

jb,

Are you actually saying that "Idc = 10A produces more Watts or heat than Irms=10A"?

Do you realise what you are saying? Why (in your opinion) would that be so?

 

[panelman]

jb

it looks wrong to me...the rms value of an ac wave is equal to the dc current that would produce the same work (or at least it was when i was in school)

 

[IRstuff]

I think jb is confusing the generally accepted as erroneous measurement process with the definition. The definition is correct. The meter gives erroneous results.

TTFN

 

[QUAVIET]

Thanks you all for the informative responces to my question. The reason I asked the X-RAY question was because when I worked as a X-RAY repairman 20 years ago I asked a PHD who calibrated our equipment what causes an X-RAY. He said what do you think causes it. I said that when the electron collides with the rotor of the tube it jumps on to a electron shell but can not remain there because it is too energetic and has to jump to a closer shell thus it releases a X-RAY. He said that is what they teach in High School. He said that it was actually the deceleration of the electron. You guys seem on top of it so I thought I would ask you.

 

[Skogsgurra]

QUAVIET,

Yes, it is the decelaration, where the energy is converted to radiation that causes X-rays. As you probably know, it was called "Bremsstrahlung" (brake radiation), which all is about deceleration. I still do not understand why you started with a question about RMS.

 

[panelman]

glad i'm not the only one struggling with the point of the starter question

 

[electricpete]

I agree with IRstuff. The statement:

"It is obvious that 10A DC supplied to a resistor produces more heat or Watts than 10A RMS..."

is incorrect. The context of the discussion before and after makes it clear that that poster is confused between measured and actual quantities.