# Clarification on RMS value

1. ## Clarification on RMS value

The statement in wikipedia about RMS value is
"For alternating electric current, RMS is equal to the value of the direct current that would produce the same average power dissipation in a resistive load." My understanding about the statement is, if a sine signal of peak value is 5V then its RMS value is 5/Sqrt(2) = 0.707*5 = 3.535Volts. The power it will dissipate in resistance for example 1 Ohm is (3.53) *(3.53) = 12.5Watts. Am I correct in my understanding? Please advise.

2. ## Re: Clarification on RMS value

Hi,

Yes.
The other way round:
* 3.53V DC will cause the same power in a resistor as
* 3.53V AC RMS, which - if pure sinewave - has an amplitude of 5.0V

"If pure sinewave":
RMS value is not related to sinewave only. You may use any random waveform, but then the relationship of: peak_value / RMS_value is not sqrt(2).

Klaus

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Ratch

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4. ## Re: Clarification on RMS value

Originally Posted by Ratch

Ratch
The correct statement will be over a complete period; not over half-cycles.

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5. ## Re: Clarification on RMS value

c_mitra is correct, the integral should be over a full cycle
0 to 2 pi, or -pi to pi or ...

for signals other than sin, replace 5 sin(theta) with the appropriate other function

the pi (should be 2 pi over a full cycle) in the denominator is from integral with same limits
as integral in numerator, but the argument is 1, so can take divide the squared "sum" (integral)
by the "number of things added up" (integral from 0 to 2pi d(theta)

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6. ## Re: Clarification on RMS value

Originally Posted by wwfeldman
c_mitra is correct, the integral should be over a full cycle
0 to 2 pi, or -pi to pi or ...

for signals other than sin, replace 5 sin(theta) with the appropriate other function

the pi (should be 2 pi over a full cycle) in the denominator is from integral with same limits
as integral in numerator, but the argument is 1, so can take divide the squared "sum" (integral)
by the "number of things added up" (integral from 0 to 2pi d(theta)
As long as the wave form is symmetrical (after being displaced for half its period) about the x-axis, the interval need only be taken for its half-period. The correct value will still be obtained if the interval is a integral value of the half-period. See the example below where interval is 5 times the half period. If you go for the full cycle, then you are doing 2 times the half-period interval and getting the correct answer, but doing twice as much integration as necessary for a symmetrical wave.

Ratch

7. ## Re: Clarification on RMS value

Ratch is right too. If the waveform is symmetrical, the the effect that the positive excursion has will be the same as that of the negative excursion.

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Notice that he integrated for a half cycle and he averaged over a half cycle still.

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8. ## Re: Clarification on RMS value

If Symettrical implies no DC component, then this is correct. Otherwise not.

9. ## Re: Clarification on RMS value

Mater of fact, as long as the waveform can be discerned to be symmetric with respect to its partial period, that is enough period needed to determine its RMS. For instance, a sinusoid can be determined to be symmetric from a quarter of its period. Therefore, its RMS can be calculated from a quarter period as shown below.

As shown above, the RMS can be calculated from 0 to π/2 because a similiar symmetric sinusoid can be constructed from a quarter sinusoid by repositioning and flipping the quarter sinusoid areas.

Ratch

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Originally Posted by Swrmeter
If Symettrical implies no DC component, then this is correct. Otherwise not.
Ifa DC component is present, square the DC value, square the symmetric RMS AC value, add the sums, then take the square root for the composite RMS DC-AC value.

Ratch

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