# What is the significance of UPS output power factor?

A UPS's output rating is shown as 2000VA/1800W. I understand that the power factor here is 0.9, but the device or load I will connect to the UPS might have pf<0.9 or pf=1 (for light bulb.) From my understanding, the real and reactive power will vary based on my load's pf. Then why does UPS specific a fixed pf? Isn't it enough just to mention the VA rating?

• The power factor gives the losses. This power factor is different depending on the usage. In a factory, the power factor is 72%(0.72), for your device its 90%(0.9), but you will still pay for the whole 100%, unless you are a company. The fixed power factor gives the losses from the device, the losses after the device have a power factor of their own. Commented Oct 5, 2020 at 6:41
• With active PFC, the load won’t matter Commented Jun 27, 2021 at 12:03
• @CFCBazarcom That's not the case. Domestic customers pay for the real power only, and only industrial customers (currently) get charged for reactive power. Power factor is only tangentially related to losses, and a power factor of 0.7 is not the same as an efficiency of 0.7. Commented Dec 19, 2021 at 16:01

With the loads presented to a domestic UPS being resistive (incandescent lamps) / inductive (ceiling fans) / capacitive (CFL / LED lamps), the probability of the reactances cancelling out does exist.

The UPS manufacturer may hence assume a power factor of 0.9 (midway between 0.8 & 1!) and give the rating in kW in addition to kVA.

The kW rating would be of help in selecting the UPS since the loads would be rated only in watts.

There is no need to specify wattage for the UPS as VA rating is enough and complete. However, since a layman generally understands wattage best, the manufacturer has also specified its capacity as 1800 watts assuming a power factor of 0.9 for the load. If the load is theoretically assumed as completely resistive ( power factor 1), the same 2000VA UPS will deliver 2000 watts.

First reason is redundancy/simplicity. Not everybody can calculate 2000 VA/1800 W = 0.9 PF. But more importantly, it is mentioned due to marketing reasons. Most early UPS systems were designed to handle output with 0.6 PF. Marketing people realized that people associate high PF values with better products so they use high PF numbers to justify higher prices.

UPS manufacturers first tried to lure buyers using higher VA ratings. In other words, higher VA rating = much more expensive unit. But, it is cheaper to add extra components to increase VA ratings, compared to increasing the W rating which is the actual power used by the load. Designing system which can give out more actual power also costs much more \$. Using VA values was a way to inflate prices.