Current and RMS

Thread Starter

incognita

Joined Mar 26, 2009
4
Dear forum.



I hope someone can tell why in the RMS we square the physical quantity (e.g. voltage or current) and then we take the square root? Why do we do this? What is the relevance of calculating the rms value for a certain wave/signal? What kind of information it gives to us? Does it helps us to differentiate between two very similar waves?

Why we write in current/voltage as a cos function and not as a sin function? Is there any special reason?
Examples: i(t) = 10 cos (wt -10) and not as i(t)= 10 sin (wt-10) ...


Thanks for the answers.
 

Ratch

Joined Mar 20, 2007
1,070
incognita,

I hope someone can tell why in the RMS we square the physical quantity (e.g. voltage or current) and then we take the square root? Why do we do this?
This is ground that has been plowed until it is pulverized. Any good electrical science book or any number of internet sites including this forum can tell you why it is calculated the way you discribe. All you have to do is search. It has to do with the fact that power varies as the square of voltage or current.

Why we write in current/voltage as a cos function and not as a sin function? Is there any special reason?
Examples: i(t) = 10 cos (wt -10) and not as i(t)= 10 sin (wt-10) ...
Sines and cosines are the same except for the phase difference. I have seen sinusoidal waves represented in sine for as well as cosine format.

Ratch
 

studiot

Joined Nov 9, 2007
4,998
I hope someone can tell why in the RMS we square the physical quantity (e.g. voltage or current) and then we take the square root?
The RMS is a measure of average for the wave.

Now look at a sine wave, or any other with half negative and half positive, and take the arithmetical average. You will get zero.

But both the negative and positive halves contribute to the power in the wave and if we square something negative and take the square root we get a positive number, which we can add to the positive half to get a measure of the average.
 
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