I was looking at UPS backups such as this:
I noticed the VA rating and Watt rating are different.
I thought a Watt was a Volt-Ampere. Anyone know why the VA rating and Watt rating are different?
Thanks.
Real ( W ) and Apparent ( VA ) Power.
Graham
W=V*A*cos(angle of V to I). So it depends what load it is driving. (secret: a _perfect_ capacitor or inductor would consume 0 (zero) Watts. Why?)
Have fun
Stanislaw Slack user from Ulladulla.
Watts are delivered and consumed. VA might be delivering watts, or they might be delivering energy that is borrowed and returned every half cycle. VA do not deplete storage so fast, if a lot of what is causing the amperes is short term loans that end up back in the battery every half cycle. So the VA rating is often higher than the watt rating.
On 19 Aug 2006 17:51:33 -0700, in message , snipped-for-privacy@coolgroups.com scribed:
A Watt is identical to a volt-amp, if the load is purely resistive. If the load becomes reactive to any degree, then the source must deliver to both the resistive and the reactive vectors (volt-amps), but only the resistive vector consumes power (Watts). Here's a very good basic explanation:
VA = rms Voltage x rms Current.
Watts = true power output / consumption.
** A UPS is rated for use with electronic equipment power supplies that ( normally) draw current only in brief pulses at the crest of each peak in the incoming voltage waveform.This increases the *rms* value of the current draw but NOT the true power being supplied.
The UPS in question can supply 350 /120 = 2.91 amps rms of such current pulses.
The resistive load rating (ie a lamp) is only 210 /120 = 1.75 amps rms.
BTW:
It has ** NOTHING ** to do with reactive loads, cos phi etc.
....... Phil
The difference is something called "power factor"
Bye. Jasen
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