Average Power of A.C.: Assume Phase Difference Zero?

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The average power of AC is calculated using the formula IVcosθ, where I and V are RMS values and θ is the phase difference. In common applications like lamps and computers, it may be reasonable to assume a phase difference of zero, leading to a power factor (PF) of 1 for resistive loads. However, devices like fans and motors typically have lower PFs, which can vary significantly based on the equipment used. Higher power equipment generally has a PF above 0.8, while smaller devices may have a PF around 0.9, but this can fluctuate. Overall, while some assumptions can be made for specific types of equipment, caution is advised due to the variability in PF across different devices.
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We know that the average power of A.C. is IVcosθ, where I and V are RMS and θ is the phase difference. I would like to ask in common application, such as power of a lamp, computer, fan etc. Can we simply assume the phase difference is zero and get a fairly accurate answer? Thank you.
 
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I'm sure others with more knowledge in this area will post a better response. Power Factor for switching power supplies that qualify for UL or other certs control the PF so essentially (0 - EDIT, this should be 1) for that one. Incandescent light bulbs are for the most part (0 - EDIT this should be 1) as well. Fans or other motor appliances are probably not. There are/were devices that are sold to view the power factor (watt wizzard I think was one) and other devices that are suppose to compensate to yield lower usage for those devices.

So to answer your question, no I don't think you can assume a power factor of (0 - EDIT this should be 1) because one location may have many fans, or other devices that might cause a 5 - 10% difference. However, if your location is just electronics and lights you might be able to assume PF of (0 - EDIT this should be 1).

Others will chime in and provide a better answer.
 
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The power factor of a resistive or a perfectly corrected load is 1.00
The PF of a lossless inductor or capacitor is zero.

You can assume that all higher power equipment will have a PF better than 0.8
Very small items may have a worse PF but they will only be using small amounts of AC current.

So you can assume PF is 0.9 +/– 0.1 for any approved electrical product.
You can only assume for filament lamps and resistive element heaters that PF = 1.00
 
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Baluncore, thank you. Many years since I used PF remember the concepts, but the number slipped my mind. I should have looked up. Thank you for the correction.
 
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