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21st December 2019, 03:44 #1
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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.

21st December 2019, 03:55 #2
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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).
KlausPlease don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.
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24th December 2019, 09:40 #3
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24th December 2019, 14:47 #4
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24th December 2019, 16:06 #5
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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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24th December 2019, 17:12 #6
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Re: Clarification on RMS value
As long as the wave form is symmetrical (after being displaced for half its period) about the xaxis, the interval need only be taken for its halfperiod. The correct value will still be obtained if the interval is a integral value of the halfperiod. 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 halfperiod interval and getting the correct answer, but doing twice as much integration as necessary for a symmetrical wave.
RatchLast edited by Ratch; 24th December 2019 at 17:26.
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24th December 2019, 19:17 #7
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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.
   Updated   
Notice that he integrated for a half cycle and he averaged over a half cycle still.
Akanimo.

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24th December 2019, 19:20 #8
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Re: Clarification on RMS value
If Symettrical implies no DC component, then this is correct. Otherwise not.

24th December 2019, 23:05 #9
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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
   Updated   
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 DCAC value.
RatchHopelessly Pedantic
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