# Does AC power rating stand for peak or average power?

For instance, take a light bulb. It has voltage and power rating, like "230V, 60W". However, for alternating current, which they usually consume, power is alternating too. Then, the most reasonable power ratings to use are the average power, which can be used to calculate heat, and the peak power, which can be used to calculate electrical stuff. So, which one is rated?

• These are rms values (root-mean-square). The average voltage and current would be both zero for a sinusoidal AC. For a pure resistive load the average power is zero as well.
– nasu
Jul 6, 2017 at 16:24
• @nasu TBH, I have trouble understanding your comment. I realize that the average voltage and current would be zero, but the wattage is proportional to the current squared which is always above or equal to zero. And, exactly, are both the rated voltage and wattage RMS? That would mean that the peak voltage is higher than the rated one, is that true? The average wattage could be measured using arithmetic mean and I don't think there's a reason to use RMS in this case. Or are they equal when we deal with positive-only values? I highly doubt that. Jul 6, 2017 at 16:33
• @nasu "For a pure resistive load the average power is zero as well." Nope. Because the power is the product of current of potential the power is always non-negative across resistive loads—which is as it should be because otherwise lightbulbs would violate the conservation of energy, no? Jul 6, 2017 at 16:39
• Yeah, you are right about the power. Unfortunately I cannot edit my comment. I could delete it but then the comments after the first will look weird.
– nasu
Jul 6, 2017 at 17:33

Almost all AC devices have their voltage and current capabilities stated in RMS terms (root mean square), which make it easy to calculate power with equations like $P=VI$. For a sinusoidal, this is $V_{rms}=\frac{1}{\sqrt 2}V_{max}$.

You could state the power terms in either notation, but history has shown that the RMS numbers are considered to be the most convenient for the majority of calculations, so we use it. Electrical engineers learn to put the $\sqrt2$ in wherever necessary, such as calculating peak voltages.

One example of where it is necessary is in terms of arc protection. 110V power lines are 110V RMS. That RMS voltage implies a peak-to-peak voltage of 155.54V. When designing systems to deal with the potential of an arc forming, one has to consider the peak voltage.

• And what's for the wattage? It makes sense to use arithmetic mean here, but @nasu said in a comment (I think so) that the wattage is also RMS. Jul 6, 2017 at 16:44
• @ВасилийСвинко One of the reasons RMS is so convenient is that power(wattage) is simply $P=VI$ or if you use ohms law to substitute for V, $P=I^2R$. If you integrate the sinusoid, that's simply the average power that you get. Jul 6, 2017 at 16:47
• do I understand correctly that if we multiply the rated wattage by the time the device has been working for, we'll get neither peak nor the actual (energy per time) value, but the smaller value and the device actually consumes more than one could suppose based on its wattage? That's not something I expected. Jul 6, 2017 at 16:55
• Oh. Wait. I'm starting to understand that. Thanks! Jul 6, 2017 at 17:01
• First sentence is incorrect. Voltages or currents are typically given as RMS values. Power is generally just averaged over the cycle. (making the rms voltage consistent with the power given) Jul 6, 2017 at 18:31

Then, the most reasonable wattage ratings to use are the average wattage, which can be used to calculate heat, and the peak wattage, which can be used to calculate electrical stuff. So, which one is rated?

The average is typically specified.

The instantaneous power is not as useful as you might think for calculating "electrical stuff". For example, if you want to know whether the filament of an incandescent bulb will burn out, just knowing peak power is not enough. You also need to know the thermal mass of the filament and the duration the peak power is applied. At 50 or 60 Hz, the thermal mass is enough to smooth out the cycling of temperature of the filament, so that average power is more relevant to this calculation than peak power.