[rescue] Real amp draw of E3000

[der Mouse]

>> It should; it's just a slight rearrangement of W = V W A.
> I'm gonna guess that's a typo, and you meant W = V x A.

Actually not a typo in the usual sense of the term.  The W on the
right-hand side was, when I sent it, an ISO 8859-1 multiplication sign,
0xd7.  Something somewhere between me and list output stripped the high
bit, turning it into 0x57, W.

>> if your load is not purely resistive, your power consumed is
>> generally not your RMS volts times your RMS amps - it's the integral
>> of your instanteous volts times your instantaneous amps.
> A more common simplification is to say that the power used (in watts)
> is the voltage times the amperage times the cosine of the phase angle
> between the two (assuming your power is AC with a nice sine wave).

This works only if your current is also a nice sine wave.  With a lot
of modern loads (in particular, anything with a switching power
supply), this is far from true.

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