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I have a vacuum cleaner on which is proudly written:

12.0 Amps

1300 W

Assuming a 120 V (RMS) power supply, how would the power consumption of the vacuum cleaner be calculated given the 12.0 A current draw?

It seems that \$120 \mathrm{\,V}\cdot12 \mathrm{\,A}\ne1300 \mathrm{\,W}\$ and \$120 \mathrm{V}\cdot\frac{12 \mathrm{A}}{\sqrt{2}}\ne1300 \mathrm{W}\$, so then where might the power rating "1300 W" come from?

Community
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    I guess the 12A is a _maximum_, as you need a max to determine what devices can be on one group. The power consumption is an _average_, because you need an average to calculate how many energy it's going to consume hence how much the device will cost. –  Apr 28 '13 at 12:15
  • I read somewhere that power can range from 110-130V AC and 110*12= 1320 (close but not 1300), so it's possible that they didn't use 120 for voltage. Also, we don't know if there is an internal regulator/resistor/divider/transformer/etc. that makes the motor/internal parts a lower wattage, but it still uses 12 amps. Like @CamilStaps said, it could be a max and average, but I think that would be stupid of them to put it there and not label it max/average. – Anonymous Penguin Apr 28 '13 at 13:52
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    If its proudly displayed along with the brand name, the 12.0 Amps is probably labeled "cleaning power" or some other such empty phrase. It's marketing nonsense. (Hmm, that's redundant, isn't it?) Find the nameplate, and get the **actual** current draw from that. – Pete Becker Apr 28 '13 at 16:23

3 Answers3

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There are many unknown factors. Both power and current will vary with load and there is only one way to find out how they relate: measure them.

One of the most important properties of AC power you neglected is the power factor cos(φ) with inductive loads:

\$P = U \cdot I \cdot \cos(\varphi)\$

\$\begin{align} \cos(\varphi) & = \dfrac{P}{U \cdot I} \\ & = \dfrac{1300\text{W}}{120\text{V} \cdot 12\text{A}} \\ & \approx \boxed{0.9} \end{align}\$

which sounds about what I'd expect for a vacuum cleaner.

Check this Wikipedia article on electric power for more background details.

jippie
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An "alternative" answer: -

120V x 12A = 1440W

BUT because there's a fair to reasonable chance that it is an AC motor, the power factor will reduce the 1440W to something less.

Andy aka
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Some kinds of loads will take power from the AC source during some parts of each 60Hz cycle but feed some power back to the source during other parts. The power which is fed back to the source will be added to the reported current for the device, but will be subtracted from the total power. The current and power are reported separately because circuit breakers measure the former, and utility company meters measure the latter.

supercat
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