Editing Root mean square (section)

====Average electrical power{{Anchor|Average power}}====
{{Further|AC power}}
Electrical engineers often need to know the [[power (physics)|power]], ''P'', dissipated by an [[electrical resistance and conductance|electrical resistance]], ''R''. It is easy to do the calculation when there is a constant [[electric current|current]], ''I'', through the resistance. For a load of ''R'' ohms, power is given by:
:<math>P = I^2 R.</math>

However, if the current is a time-varying function, ''I''(''t''), this formula must be extended to reflect the fact that the current (and thus the instantaneous power) is varying over time. If the function is periodic (such as household AC power), it is still meaningful to discuss the ''average'' power dissipated over time, which is calculated by taking the average power dissipation:

:<math>\begin{align}
P_\text{Avg}
&= \left( I(t)^2R \right)_\text{Avg}  &&\text{where } (\cdots)_\text{Avg} \text{ denotes the temporal mean of a function} \\[3pt]
&= \left( I(t)^2 \right)_\text{Avg} R &&\text{(as } R \text{ does not vary over time, it can be factored out)} \\[3pt]
&= I_\text{RMS}^2R                     &&\text{by definition of root-mean-square}
\end{align}</math>

So, the RMS value, ''I''<sub>RMS</sub>, of the function ''I''(''t'') is the constant current that yields the same power dissipation as the time-averaged power dissipation of the current ''I''(''t'').

Average power can also be found using the same method that in the case of a time-varying [[voltage]], ''V''(''t''), with RMS value ''V''<sub>RMS</sub>,
:<math>P_\text{Avg} = {V_\text{RMS}^2 \over R}.</math>

This equation can be used for any periodic [[waveform]], such as a [[sine wave|sinusoidal]] or [[sawtooth wave]]form, allowing us to calculate the mean power delivered into a specified load.

By taking the square root of both these equations and multiplying them together, the power is found to be:
:<math>P_\text{Avg} = V_\text{RMS} I_\text{RMS}.</math>

Both derivations depend on voltage and current being proportional (that is, the load, ''R'', is purely resistive). [[Electrical reactance|Reactive]] loads (that is, loads capable of not just dissipating energy but also storing it) are discussed under the topic of [[AC power]].

In the common case of [[alternating current]] when ''I''(''t'') is a [[sine wave|sinusoidal]] current, as is approximately true for mains power, the RMS value is easy to calculate from the continuous case equation above. If ''I''<sub>p</sub> is defined to be the peak current, then:
:<math>I_\text{RMS} = \sqrt{{1 \over {T_2 - T_1}} \int_{T_1}^{T_2} \left[I_\text{p} \sin(\omega t)\right]^2 dt},</math>

where ''t'' is time and ''ω'' is the [[angular frequency]] (''ω''&nbsp;=&nbsp;2{{pi}}/''T'', where ''T'' is the period of the wave).

Since ''I''<sub>p</sub> is a positive constant and was to be squared within the integral:
:<math>I_\text{RMS} = I_\text{p} \sqrt{{1 \over {T_2 - T_1}} {\int_{T_1}^{T_2} {\sin^2(\omega t)}\, dt}}.</math>

Using a [[list of trigonometric identities|trigonometric identity]] to eliminate squaring of trig function:
:<math>\begin{align}
I_\text{RMS} &= I_\text{p} \sqrt{{1 \over {T_2 - T_1}} {\int_{T_1}^{T_2} {{1 - \cos(2\omega t) \over 2}}\, dt}} \\[3pt]
		   &= I_\text{p} \sqrt{{1 \over {T_2 - T_1}} \left[ {t \over 2} - {\sin(2\omega t) \over 4\omega} \right]_{T_1}^{T_2} }
\end{align}</math>

but since the interval is a whole number of complete cycles (per definition of RMS), the sine terms will cancel out, leaving:
:<math>I_\text{RMS} = I_\text{p} \sqrt{{1 \over {T_2 - T_1}} \left[ {{t \over 2}} \right]_{T_1}^{T_2} } = I_\text{p} \sqrt{{1 \over {T_2 - T_1}} {{{T_2 - T_1} \over 2}} } = {I_\text{p} \over \sqrt{2}}.</math>

A similar analysis leads to the analogous equation for sinusoidal voltage:
:<math>V_\text{RMS} = {V_\text{p} \over \sqrt{2}},</math>

where ''I''<sub>P</sub> represents the peak current and ''V''<sub>P</sub> represents the peak voltage.

Because of their usefulness in carrying out power calculations, listed [[voltage]]s for power outlets (for example, 120{{nbsp}}V in the US, or 230{{nbsp}}V in Europe) are almost always quoted in RMS values, and not peak values. Peak values can be calculated from RMS values from the above formula, which implies ''V''{{sub|P}}&nbsp;=&nbsp;''V''<sub>RMS</sub>&nbsp;×&nbsp;{{radic|2}}, assuming the source is a pure sine wave. Thus the peak value of the mains voltage in the USA is about 120&nbsp;×&nbsp;{{radic|2}}, or about 170 volts. The peak-to-peak voltage, being double this, is about 340 volts. A similar calculation indicates that the peak mains voltage in Europe is about 325 volts, and the peak-to-peak mains voltage, about 650 volts.

RMS quantities such as electric current are usually calculated over one cycle. However, for some purposes the RMS current over a longer period is required when calculating transmission power losses. The same principle applies, and (for example) a current of 10 amps used for 12 hours each 24-hour day represents an average current of 5 amps, but an RMS current of 7.07 amps, in the long term.

The term ''RMS power'' is sometimes erroneously used (e.g., in the audio industry) as a synonym for ''mean power'' or ''average power'' (it is proportional to the square of the RMS voltage or RMS current in a resistive load). For a discussion of audio power measurements and their shortcomings, see [[Audio power]].