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Why doesn't wattage equal amperage times voltage?
June 10, 2008 8:49 PM   Subscribe

Another air conditioning question for the summer! 4.8 A *110V=515W?

I was reading the label on my air conditioner and noticed that it claims to use 115 V, 515 W, 4.8 A. Obviously the math I was taught in high school physics doesn't quite work out there. Why not? I'm guessing that there's some kind of safety margin built in here somewhere, but if so, where, and is it realistic?
posted by phoenixy to Technology (12 answers total)
 
Er, for clarity's sake, the typo is in the headline, not in the text of the question. It really does say 115 V.
posted by phoenixy at 9:04 PM on June 10, 2008


First thought: just because the unit requires 115V AC doesn't mean that it actually assumes that 115V AC will reach the motor (voltage drop).
posted by misterbrandt at 9:08 PM on June 10, 2008


My guess is that 515W is the average (sustained) power consumption, and 4.8A is the peak current that the AC power supply has to be able to withstand. It might draw a bit of extra current while the compressor is starting up, for instance.
posted by teraflop at 9:15 PM on June 10, 2008


Amps X volts = watts works with DC voltage only.

For AC: Power (single phase): P = Vp×Ip×cos φ as detailed on this page.
posted by Daddy-O at 9:27 PM on June 10, 2008


Complex power? I haven't really looked at many air conditioner labels or any standards behind that to see how it's listed, but in general and importantly in the case of industrial equpment the power ratings very importantly include the "real" power in watts that is transferred as well as the "imaginary" ("real" and "imaginary" are mathematical terms, do not take this to imply that "imaginary" power does not exist) power that sort of floats around in the system. Importantly, though the "real" power by and large measures the transfer of energy, a power source will have to be able to supply sufficient "volt-amps" (unit of magnitude of both
real" and "imaginary" power considered together) to account for both the "real" and "imaginary" power needs. (Calculating 115V * 4.8 A gives me 552 VA, and 515 W being less than 552 is consistent with this.)

Teraflop's answer could also be right or partially right - I don't know by what standards they're listing these labels. I know that consumer power consumption is largely considered and charged in terms of "real" power while industrial power consumption has both considered, and your air conditioner is really as close as your house ever gets to industrial power consumption.
posted by TheOnlyCoolTim at 9:41 PM on June 10, 2008 [1 favorite]


Air conditioners typically have induction motors that require more reactive power than real power. Real power is the amount that is measured by your power meter and for which you are charged by the power company. Because the motor is an inductive load, it actually requires more current than would be measured by the power meter. 515W/552W = 93% power factor which is pretty good and suggests that the air conditioner has some power correction circuitry to improve its power factor.

The watts number on the label is important because it indicates the rate of real power use for which you will be billed and indicates how much it costs you to run. The actual amperage is higher than what the real power indicates and is important for knowing how much load will be on the house circuit that it is plugged into.

Reactive power is the difference between the real power (515W) and the total power (552W). You can think of reactive power as borrowed from the power company on one half cycle due to the induction of the motor and sent back to the power company on the next half cycle. You aren't charge for this "borrowed" power but it is an important number to know because it determines how big the wires and circuit breaker need to be because of the increased current.
posted by JackFlash at 10:53 PM on June 10, 2008 [5 favorites]


That first sentence should have said that air conditioners typically have induction motors that require more total power than real power.
posted by JackFlash at 10:58 PM on June 10, 2008


JackFlash's answer adds a great deal to mine. I'll just add for everyone's greater understanding that what he named "reactive power" in his post is the same as what I named "imaginary power."
posted by TheOnlyCoolTim at 11:25 PM on June 10, 2008


TOCT and JackFlash beat me to it. It's a case of power factor, and it is the first question electrical engineers ask when sizing electrical cables supplying mechanical equipment (generally, anything with a motor): the total power can be much greater than the real power, though in this case it isn't.

Daddy-O, P=IV holds for AC single phase supplies. The formula you mention simple reminds us to remember power factor, which in the case of many circuits can be disregarded any way.
posted by nthdegx at 2:56 AM on June 11, 2008


"My guess is that 515W is the average (sustained) power consumption, and 4.8A is the peak current that the AC power supply has to be able to withstand. It might draw a bit of extra current while the compressor is starting up, for instance."

As a side note, the starting current can be much much bigger than the full load current, or running current, but normally the start-up time is short and within the trip-time of the protective device on the circuit: that is to say, the great current at start-up doesn't matter all that much.
posted by nthdegx at 2:58 AM on June 11, 2008


by the way, your outlet probably carries something north of 120v.
posted by caddis at 5:40 AM on June 11, 2008


Yes, you can't take those figures as precisely exact. The voltage is nominal voltage, meaning in name only. A standardized figure used to get calculations "close enough" and to have a margin of safety.

And AC power is different. It behaves like DC and like an RF signal at the same time.
posted by gjc at 6:21 AM on June 11, 2008


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