If you measured this 40 watts using a power meter (such as a Kill-a-Watt) inserted between the wall outlet and the PC's power supply, then yes.

Electricity deals with the flow of electrons (charged particles). The amount of the flow (aka *current*) is measured in amperes. The electromotive force that causes the current to flow is called *voltage*, which is measured in volts. Voltage always needs a reference point, such as (electrical) ground. The material that the current flows through is called a conductor, and has a property/attribute called *resistance* which impedes the flow of electrons. The relationship of voltage (v), current (i) and resistance (r) is expressed by Ohm's Law as *v = ir*, voltage is equal to the product of current and resistance.

For DC (direct current where the voltage is not varying) electrical power, the voltage (in volts) multiplied by the current (in amps) equals the power (in watts). For AC electrical power, voltage multiplied by the current equals the *apparent* power in **volt-amps** (VA) (there's a selection for that on the Kill-a-Watt). The volt-amps is calculated as if the alternating current is in phase (synchronized with) the alternating voltage. This would be true for purely resistive loads like a incandescent light bulb. For a reactive (inductive and/or capacitive) load like a power supply or CFL bulb, current and voltage may not be in phase (not synchronized), so a *power factor* (which is less than unity) has to be applied to come up with the *real* power expressed as the **wattage**. That is *P = fvi*, where *f* is the power factor, and is equal for 1 for DC or resistive loads and less than 1 for reactive loads.

Wattage is the power number that really counts rather than volt-amps, since that is what the power company measures and uses as the basis for your electrical bill.