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RMS is defined as the AC equivalent voltage that produces the same amount of heat or power in a resistor if the same is passed in the form of a DC voltage to the resistor. But shouldn't the power in AC change continuously due to change in voltage and current and hence producing varying power in the resistor as opposed to the DC circuit where a constant power is generated. I am confused so please help me.

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    \$\begingroup\$ The instantaneous power in AC is changing all the time, but when you think about the average power, RMS is equivalent with DC. \$\endgroup\$ – Oskar Skog Sep 30 '17 at 11:04
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    \$\begingroup\$ @user42172: See also my answer to electronics.stackexchange.com/questions/328185/…. \$\endgroup\$ – Transistor Sep 30 '17 at 19:38
  • \$\begingroup\$ A quick note: The "M" in RMS stands for "Mean", which is another word for average. \$\endgroup\$ – Spehro Pefhany Oct 1 '17 at 3:52
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The powers are equal if you consider the AVERAGE power. Many of the other answers have kind of taken shortcuts without explaining all the conditions that must apply for the shortcuts to be legitimate. And you yourself have some subtle mistaken assumptions built-in to your question. If you are an EE student you should read the rest of this answer.

RMS is defined mathematically as the root of the mean of the square of a function. If the function is periodic (repeats itself) then generally, the mean calculation should be over an exact number of cycles. The function could be anything, and doesn't need to be periodic. This is the definition of RMS. It has nothing to do with DC or voltage or current at all. In fact, it is often used in statistics.

Instantaneous power in a load is simply instantaneous current multiplied by instantaneous voltage. P = V * I.

Average power is calculated by averaging the instantaneous power. For repetitive waveforms, the average can be performed over exactly one cycle (or any integer number of cycles). For non-repetetive waveforms, the average must be performed over the entire waveform, or "for a long time." Everything I have written so far is true in a fairly general way. It does not depend on any details about how the voltage or current waveforms look. You can calculate the average power of ANY waveform if you average instantaneous power over a cycle. You can calculate the instantaneous power of any waveform if you know voltage and current.

For DC circuits, it so happens that average power is just V * I.

In the special case of the sinusoidal voltage applied to a resistive load, Pav = Vrms * Irms, where Pav is average power. You can prove this, if you want, by doing the rms calculation over one cycle of a sinusoid.

But, if the load is not resistive, then that equation is not true. If the load is resistive but the voltage is not sinusoidal, then the equation is true, but the RMS voltage will not be equal to Vpeak / sqrt(2), as it is with a sinusoid.

There is one more thing worth mentioning. If the voltage is sinusoidal, and the load is reactive (inductive or capcitive), you can still calculate power if you know something called the "power factor."

For this special case, Pav = Irms * Vrms * PF (where PF is power factor, and Pav is average power).

As far as average power goes, it is often the case that average power is more important than instantaneous power. In general, this is true when the thermal time constant is much longer than the electrical period of the AC waveform. If you look at a high-speed video of an incandescent lightbulb powered by AC, you will see that its brightness does vary a bit as the AC waveform changes, but, because the filament takes some time to heat up and cool down, the perceived brightness of the bulb is based strictly on Vrms * Irms. The mass of the lightbulb itself averages the power somewhat. And your eye averages out whatever ripple remains.

If the filament were very, very tiny, it might not have enough mass to average out the power, and its brightness would vary all the way from near zero to full brightness.

I hope this clears up most of your confusion.

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Average power is what gives rise to a sustained heating effect: -

enter image description here

Power is the instantaneous multiplication of v and i.

If we translate i into v/R then power is \$\dfrac{v^2}{R}\$

And, average power is the mean of \$\dfrac{v^2}{R}\$

If we then say that R = 1 ohm (just for convenience) we can say: -

Average power = mean(\${v^2}\$)

Then it follows that if we take the square root we get RMS voltage

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  • \$\begingroup\$ After re-reading, I agree with you, so I am deleting my comment. Please accept my apology. \$\endgroup\$ – mkeith Sep 30 '17 at 21:07
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    \$\begingroup\$ @mkeith no probs dude. \$\endgroup\$ – Andy aka Sep 30 '17 at 21:17
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But shouldn't the power in AC change continuously due to change in voltage and current and hence producing varying power in the resistor

Yes, the instantaneous power in a non-constant voltage/current is not constant.

But in your definition an important adjective is missing. Average. You must consider the average electrical power:

  • in the period, for periodic waveform
  • in the signal duration, for arbitrary waveforms.
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    \$\begingroup\$ So the rms voltage will produces same average electrical power in the resistor of a dc? \$\endgroup\$ – Hydrous Caperilla Sep 30 '17 at 10:54
  • \$\begingroup\$ yes it will, by definiton. \$\endgroup\$ – next-hack Sep 30 '17 at 10:56
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Integrated power is 'easy' to measure as a consequence of the heating effect. One of the most accurate ways of measuring energy is by measuring the resultant temperature rise.

An AC signal does vary continuously, but the instantaneous information is typically hard to make sense of - it doesn't relate to anything. In all the contexts I can think of which are not quantum/semiconductor effects, what is interesting is the 'average over some time period'. (The peak voltage can be important in other contexts, as noted in comments.)

For an AC signal, you would normally want to average for at least one cycle (otherwise you get a different result).

RMS of a voltage translates directly to being equivalent to the DC voltage if you're considering power dissipation across a resistor. Since this is frequently useful, its what we conventionally use to measure AC - but isn't the only factor which will be important in any specific scenario.

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  • \$\begingroup\$ *So rms is the averaged value that we get by integrating the instantaneous power over one full cycle which will produce the same electrical power in a dc as it will produce in an ac circuit in one complete cycle......am i correct? \$\endgroup\$ – Hydrous Caperilla Sep 30 '17 at 11:30
  • \$\begingroup\$ Integrate power, or integrate v-squared. It's the same thing. Average, and translate back to a voltage. That is the meaning of RMS, and my understanding is that the term is defined to have this property, rather than constructed along this set of rules (although generally there is no difference) \$\endgroup\$ – Sean Houlihane Sep 30 '17 at 11:35
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    \$\begingroup\$ The peak voltage matters for whether a current can break down an atmospheric resistance. The peak current matters for whether a current-sensitive meter (like the nerves and muscles in a human heart) will have the signal it is reading overridden by the peak current. Edison demonstrated that by using AC instead of DC, a lower RMS voltage is sufficient to cause electrocution. \$\endgroup\$ – Jasper Sep 30 '17 at 16:03
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The RMS value is obtained as follows:

(1) The square of the waveform function (usually a sine wave) is to be determined.

(2) The function resulting from step (1) is averaged over time. This is the point where your confusion commes from

(3) The square root of the function resulting from step (2) is found.

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  • \$\begingroup\$ Is the function averaged over just one full cycle? \$\endgroup\$ – Hydrous Caperilla Sep 30 '17 at 11:15
  • \$\begingroup\$ Half a cycle will do ... but typically over many cycles. \$\endgroup\$ – Brian Drummond Sep 30 '17 at 11:45
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    \$\begingroup\$ Half a cycle will do if and only if the instantaneous power waveform is symmetrical about the half-way point in time. \$\endgroup\$ – mkeith Sep 30 '17 at 17:50
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The RMS value of a signal v(t) is,

$$v_{rms} = \sqrt{\frac{1}{T}\int^{T/2}_{-T/2} v(t)^2dt} $$ ,where T = time period of the signal v(t).

This is the mean squared value of the signal and its square root is defined as root mean squared value of the signal (RMS).

But if this signal is passed through a resistor R, we get the power dissipated in one period is:

$$Power = \frac{1}{T}\int^{T/2}_{-T/2}i(t)v(t)dt = \frac{1}{RT}\int^{T/2}_{-T/2}v(t)v(t)dt$$

Thus, power dissipated is equal to: $$Power = v_{rms}^2/R$$

Thus, if we have a DC signal of value $v_{rms}$, it will dissipate the same power as the signal v(t) when passed through any resistor.

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  • \$\begingroup\$ I'm suspicious of the units in your first 2 equations... \$\endgroup\$ – Sean Houlihane Sep 30 '17 at 11:37
  • \$\begingroup\$ Unit is V^2.... this is how signal power is defined. \$\endgroup\$ – sarthak Sep 30 '17 at 11:40
  • \$\begingroup\$ kg.m^2.s^-3 is not v^2, unless I missed something. \$\endgroup\$ – Sean Houlihane Sep 30 '17 at 11:43
  • \$\begingroup\$ Well.... v(t) is some voltage signal, so dimensions become V^2*T/T = V^2. You can confirm here: en.wikipedia.org/wiki/Spectral_density \$\endgroup\$ – sarthak Sep 30 '17 at 11:46
  • \$\begingroup\$ Volt is Joule per coulomb \$\endgroup\$ – Sean Houlihane Sep 30 '17 at 11:47

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