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So I'm trying to measure the power consumption of my home PC. I used to have a kill-a-watt meter which I can't find, so I wired up my multimeter to measure the current.

Now I'm not sure what to do next.

Is it just .6Ax110v = 66W? Or do I have to do one of those sqrt(2) adjustments?

Seems like a silly question to ask.

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I suggest you read this answer to a previous question.
For a resistive load

\$P = V \cdot I \$

so the 66W would be correct. No need for the \$\sqrt{2}\$; you want the RMS (Root-Mean-Square) value, not the peak value. However, your PC's power supply isn't a pure resistive load and then

\$P = V \cdot I \cdot cos(\phi) \$

where \$\phi\$ is the phase difference between current and voltage. You need to measure both simultaneously to determine the phase, something your kill-a-watt does. Since a cos is maximum 1 your power may be lower than the 66W.

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    \$\begingroup\$ What if the power supply does power factor correction? If the power factor is very close to 1.0, then wouldn't the load appear resistive? \$\endgroup\$ – W5VO Nov 4 '11 at 13:21
  • \$\begingroup\$ @W5VO - Yes, but that's only worthwhile for heavy loads with very poor cos(\$\phi\$). I wouldn't apply this for a mere PC's power supply. \$\endgroup\$ – stevenvh Nov 4 '11 at 13:28
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    \$\begingroup\$ A lot of computer power supplies have active PFC. I think in some countries they charge residential customers for the VA consumed instead of just Watts. I would think this would give the current (A) reading a better chance of being in phase with the voltage. \$\endgroup\$ – W5VO Nov 4 '11 at 14:43
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    \$\begingroup\$ I remember reading somewhere that all modern (last few years) PC supplies must have a PF of >0.9. Assuming this one does the calculation should be pretty accurate. \$\endgroup\$ – Oli Glaser Nov 4 '11 at 18:55
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More specifically, for DC, the power is constant, and the formula is just

  • \$P = V \cdot I\$

For AC, the power is actually a changing waveform, but you want the average power over one cycle. To calculate this for a resistive load, you use the RMS value for voltage and current:

  • \$P = V_{RMS} \cdot I_{RMS}\$

The VRMS of your wall outlet is assumed to be 110 VRMS, and your multimeter in AC mode most likely tells you an approximation of RMS current, so yes, 0.6 ARMS ⋅ 110 VRMS = 66 W.

If you had measured the peak voltage of your power cord, or the peak current, instead of the RMS value, and you knew that the thing you were measuring was sinusoidal, then you could use a √2 formula to calculate the RMS value, and then plug that into the above equation.

The actual value of the power cord is probably not 110 (it varies from place to place), and the actual RMS value of current is probably not exactly 0.6 (because it's not a pure sine wave and the multimeter doesn't measure true RMS values), and there may be a power factor discrepancy, so 66 W is just an approximation.

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The procedure you performed to calculate the power was true and

Power = Voltage(V) Current(i) or Current(I)2 x Resistance(R).

The power consumption for an PC will be ranging from 60-500WATTS.
and it won't be perfect answer because the circuit with pure resistive only will applicable with the above formula(p=VI).

where as your as calculating the wattage for your PC and it is not an pure resistive in nature so
you have to consider another formula

\$Power(P) = Voltage(V) \cdot Current(I )\cdot cos(\phi) \$

where Phi = Phase difference.

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  • \$\begingroup\$ This would only be correct if the PC's PSU is a resistive load, quod non. See stevenvh's answer. \$\endgroup\$ – Federico Russo Nov 4 '11 at 12:23

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