paul said:

Hello,

Today I plugged in a power meter I bought a while ago, and decided to

check out how much power my old Pentium 233MMX was using, here are the

results (UK Power):

Normal Operation

244 Volts, 0.21 Amps, 31 Watts, Power Factor = 0.60

But if I disconnect the hard drive, I get:

250 Volts, 0.17 Amps, 41 Watts, Power Factor = 1.00

How could it be using more power when the hard drive is disconnected?

41 Watts is how much power it is using, right? Which one (if any) is

accurate?

Thanks!

paul

Based on my limited knowledge (having just finished an AC circuit theory

course at my local community college), I believe there are 3 ways to measure

the power drawn by a load from an alternating voltage source:

1. Apparent power. This is RMS AC Volts x RMS AC Amps. It is reported not

in Watts, but in "VA" = Volt-Amperes.

2. True power. This is the resistive component of Apparent Power, and is

reported in Watts. If the load has no capacitive or inductive elements,

then all the load is resistive. True power is the power that is dissipated

in heat through your load.

3. Reactive power. This is the inductive and capacitive (= reactive)

component of Apparent Power, and is reported in "VAR" = Volt-Amperes

Reactive. Reactive power isn't dissipated in heat, but cycles back and

forth between you and the power company as alternating current flow.

Apparent power is the vector sum of True Power and Reactive Power: the

square root of (True Power**2 + Reactive Power**2).

The ratio of True power to Apparent power is the Power Factor.

Motors (hard drives, fans, etc.) are typically inductive loads.

Now let's look at the figures your meter is reporting:

Normal Operation

244 Volts, 0.21 Amps, 31 Watts, Power Factor = 0.60

244 V x .21 A = 51.2 VA apparent power. Of that, the 0.60 Power Factor x

51.2 VA = the 31 Watts true power being consumed by the resistive loads in

your computer and hard drive. At the same time, there is a reactive

(inductive and capacitive) power draw of sin(arccos 0.60) x 51.2 VA = 41

VAR. The square root of (true power**2 + reactive power**2) is indeed 51.4

VA, which matches your figure of 51.4 VA.

But if I disconnect the hard drive, I get:

250 Volts, 0.17 Amps, 41 Watts, Power Factor = 1.00

250 V x .17 A = 42.5 VA apparent power. Since the Power Factor is 1.00, all

of that apparent power is due to the resistive loads in your computer and

hard drive; the true power is 42.5 Watts, or approximately the 41 Watts you

reported. There is little or no inductive and capacitive load. That makes

sense, because the hard drive motor was a major part of the inductive load

when it was connected.

So, you can see that the apparent power draw does decrease, from 51.2 VA to

42.5 VA, when you disconnect the hard drive.

Can somebody else explain why the true power (resistive load) -increases-

from 31 Watts to 41 Watts when the hard drive is disconnected?

Wayne