# [rescue] Real amp draw of E3000

Patrick Finnegan pat at computer-refuge.org
Thu Sep 21 13:28:58 CDT 2006

```On Thursday 21 September 2006 12:59, der Mouse wrote:
> > P.S. That A=W/V equation looks familiar.
>
> 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.

> 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.)

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).  A
purely inductive load (with a +90 deg phase angle) or purely capacitive
load (with a -90 deg phase angle) thus uses 0 watts, as cos(+-90deg) ==
0.  This number is also called the "power factor" of the device, and
optimally should be very close to 1.

However, Volts x Amps always is equal to the VA (volt-amp) rating of a
device, and is useful for sizing breakers, or UPSs.  Unlike Watts,
though, you can't generally directly add VA ratings, because they
aren't normalized to a particular phase angle.  You have to take the
phase angle/power factor into account, which means yet more math....

Will, any chance you got any of that? :)

Pat
--
Purdue University Research Computing ---  http://www.rcac.purdue.edu/
The Computer Refuge                  ---  http://computer-refuge.org

```