If I have an atom pc drawing 40 watts at peak usage, does that really mean it is consuming less power than a 60 watt lightbulb?

I'm no electrician, and I have no knowledge of how electricity is gauged so this could be a very dumb assumption, but I'm curious!

  • Mind you, they now have energy efficient light bulbs that can produce as much light as a 60W bulb for only 11W. – Hand-E-Food Oct 19 '11 at 1:58

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.

  • 2
    Actually, there's a mistake here. VA does not assume that voltage and current are in phase. It's only what people who think that watt is equal to VA do. That's the reason why VAs are used in AC systems to get the actual load on the power distribution system. No matter the phase difference, apparent power in VA will remain the same while the active power (expressed in watts) and reactive power (expressed in volt-amperes reactive, var) will change depending on the phase difference between the voltage and current. – AndrejaKo Oct 19 '11 at 15:48
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    @AndrejaKo - thanks for catching that badly worded sentence. I've replaced the "assumes that" with "is calculated as if". – sawdust Oct 19 '11 at 20:39

yes "watts is watts" Volts X Amps = Watts.

If you have a 110V bulb using about 1/2 of an Amp =55W


a 1.2V processor using ~45.8 Amps =55w

you have a relative same ammount of "Power" use.

The process to get power to your processor :-) can have losses of power in many places. etc etc.

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