[rescue] Real amp draw of E3000

[der Mouse]

> P.S. That A=W/V equation looks familiar.

It should; it's just a slight rearrangement of W = V W A.

However, it's good only for resistive loads or instantaneous
measurements; 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.  (For example,
a purely capacitive or purely inductive load would draw current but
consume no power - not that such a thing actually exists in practice.)

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