| The RMS value of a function is often used in physics and electronics. For example, we may wish to calculate the power P dissipated by an electrical conductor of resistance R. It is easy to do the calculation when a constant current I flows through the conductor. It is simply:
P = I^2 R..,..!
But what if the current is a varying function I(t)? This is where the rms value comes in. It may be trivially shown that the rms value of I(t) can be substituted for the constant current I in the above equation to give the average power dissipation:
P_..mathrm{avg}..,..! = ..langle I^2R ..rangle ..,..! (where ..langle ..cdot ..rangle denotes the arithmetic mean)
= R..langle I^2 ..rangle..,..! (R is constant so we can take it outside the average)
= I_..mathrm{rms}^2R..,..! (by definition of RMS)
We can also show by the same method
P_..mathrm{avg} = {V_..mathrm{rms}^2..over R}..,..!
By taking the square root of both these equations and multiplying them together, we get the equation
P_..mathrm{avg} = V_..mathrm{rms}I_..mathrm{rms}..,..!
However, it is important to stress that this is based on the assumption that voltage and current are proportional (that is the load is resistive) and is not true in the general case (see AC power for more information).
In the common case of alternating current, when I(t) is a sinusoidal current, as is approximately true for mains power. The rms value is easy to calculate from the continuous case equation above. If we define Ip to be the peak amplitude:
I_{..mathrm{rms}} = ..sqrt {{1 ..over {T_2-T_1}} {..int_{T_1}^{T_2} {(I_..mathrm{p}..sin(..omega t)}.., })^2 dt}..,..!
Since Ip is a positive real number:
The RMS value of a function is often used in physics and electronics. For example, we may wish to calculate the power P dissipated by an electrical conductor of resistance R. It is easy to do the calculation when a constant current I flows through the conductor. It is simply:
But what if the current is a varying function I(t)? This is where the rms value comes in. It may be trivially shown that the rms value of I(t) can be substituted for the constant current I in the above equation to give the average power dissipation:
-
..>
 |
 (where  denotes the arithmetic mean) |
|
 (R is constant so we can take it outside the average) |
|
 (by definition of RMS) |
..>
We can also show by the same method
By taking the square root of both these equations and multiplying them together, we get the equation
However, it is important to stress that this is based on the assumption that voltage and current are proportional (that is the load is resistive) and is not true in the general case (see AC power for more information).
In the common case of alternating current, when I(t) is a sinusoidal current, as is approximately true for mains power. The rms value is easy to calculate from the continuous case equation above. If we define Ip to be the peak amplitude:
Since Ip is a positive real number:
Using a trigonomentric identity to eliminate squaring of trig function:
![I_{..mathrm{rms}} = I_..mathrm{p}..sqrt {{1 ..over {T_2-T_1}} ..left [ {{t ..over 2} -{ ..sin(2..omega t) ..over 4..omega}} ..right ]_{T_1}^{T_2} }](http://upload.wikimedia.org/math/5/c/7/5c7a2a2727f8c0b96bc045f8a07add97.png)
but since the interval is a whole number of complete cycles (per definition of rms for a periodic function) the sin terms will cancel
![I_{..mathrm{rms}} = I_..mathrm{p}..sqrt {{1 ..over {T_2-T_1}} ..left [ {{t ..over 2}} ..right ]_{T_1}^{T_2} } = I_..mathrm{p}..sqrt {{1 ..over {T_2-T_1}} {{{T_2-T_1} ..over 2}} } = {I_..mathrm{p} ..over {..sqrt 2}}](http://upload.wikimedia.org/math/d/1/1/d1117f145a4716ac98572bedad2b0edf.png)
The peak amplitude is half of the peak-to-peak amplitude. When the peak-to-peak amplitude is known, the same formula is applied by using half of the p-p value.
The RMS value can be calculated using equation (2) for any waveform, for example an audio or radio signal. This allows us to calculate the mean power delivered into a specified load. For this reason, listed voltages for power outlets (e.g. 110 V or 240 V) are almost always quoted in RMS values, and not peak values.
From the formula given above, we can calculate also the peak-to-peak value of the mains voltage which is about 310 volts (USA) and 677 volts (Europe) respectively.
It is possible to calculate the RMS power of a signal. By analogy with RMS voltage and RMS current, RMS power is the square root of the mean of the square of the power over some specified time period. This quantity, which would be expressed in units of watts (RMS), has no physical significance and no practical use. However, the term "RMS power" is sometimes used in the consumer audio industry as an incorrect synonym for "mean power" or "average power". For a discussion of audio power measurements and their shortcomings, see Audio power.
In chemistry, the root mean square velocity is defined as the square root of the average velocity-squared of the molecules in a gas. The RMS velocity of an ideal gas is calculated using the following equation:
where R represents the ideal gas constant (in this case, 8.314 J/(mol·K)), T is the temperature of the gas in kelvins, and M is the molar mass of the compound in kilograms per mole.
[edit] Relationship to the arithmetic mean and the standard deviation
If  is the arithmetic mean and sx is the standard deviation of a population then
Here we can see that RMS is always greater than or equal to the average, in that the RMS includes the "error" / square deviation too.
[edit] See also
[edit] External links
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