Calculating VA and Watts The terms VA (voltamps) and watts are frequently used interchangeably when discussing the power consumption of an electronic device. This tendency is understandable when the total power consumption of the load is small and the value of VA and watts is nearly the same . Nevertheless, it is important to understand the distinction between VA and watts in the event system power consumptions become very large or when numerous small loads are combined on a single source of power such as a UPS. VA is an expression of “apparent power” and watts is an expression of “true power” in an AC circuit. When the load is resistive, power dissipation in VA and watts will be the same.
See the following example. Figure A is a simple AC electrical circuit. In this circuit, the power source is 120 volts, and the load is a simple light bulb with 240 ohms of resistance. The circuit current (I) can be calculated using Ohm’s Law by dividing the voltage (E) by the resistance (R).

I = E/R  I = 120/240  I = .5 amps 

In this case, a current of .5 amps will flow in the circuit. The power (P) consumed by the light bulb may be calculated using one of these formulas: P = E x I or P = I^{2} R. 
P = E x I   P = I^{2} R  P = 120 x .5  or  P = .5^{2} x 240  P = 60 watts   P = 60 watts 

In either case, the power consumed by the light bulb is equal to 60 watts. Because the load is purely resistive, the power consumption in VA will also be 60 VA.


Things change, however, when the load becomes electronic. The constantly changing amplitude and polarity of AC power gives rise to reactive components in an electronic load. There are two types of reactance – inductive and capacitive – and they are opposite in nature. Together with resistance, they represent an opposition to AC current flow called impedance. VA and watts are no longer the same because circuits with impedance exhibit a characteristic called power factor (pf).
In AC circuits, VA is referred to as APPARENT power or what power appears to be flowing in the circuit. Watts are referred to as TRUE power or an indication of the power that is truly being dissipated by the load. In addition to the power that reactive loads actually dissipate, a certain amount of power is absorbed by the reactive load and then once again released to the circuit. The power that is absorbed and then released again to the circuit is know as reactive power, and it is the difference between apparent power and true power.
In Figure B, the same AC circuit is powering a computer, which is a reactive load. The computer’s impedance is known to be 60 ohms. If we simply applied Ohm’s Law, the current flowing in the circuit would be equal to I=E/R or 2 amps. Again applying Ohm’s Law, the power consumed in the circuit would appear to be:

P = E x I  P = 120 x 2  P = 240VA. 

Since the computer is a reactive load and not a resistive one, the power factor of the computer must be considered in order to determine the watts dissipated by the computer as follows: 
P = E x I x pf  P = 120 x 2 x .65  P = 156 watts 

The difference between the 240 VA apparent power and the 156 watts of true power is the reactive power or 84 VAR or voltampsreactive.


Most UPS products are rated in VA and also have a power factor rating that is prominently published as part of the product specification. In many cases, UPS power factors are designed to approximate computer power factors. In the example above, a 350 VA UPS with a power factor of .65 would deliver 227 watts, which would satisfactorily power the computer in question with about 72 watts to spare.
At low power levels, the differences between VA and watts are often slight. However, understanding the difference between VA and watts at higher power levels is very important to make sure the power protection device is compatible with the load. 