Power supply current draw?

[CountZero]

I was asked this at work the other day I couldn't remember the details to ensure that I was right since I haven't taken a class dealing with this stuff in like a decade.

So if I have a power supply (like a standard PC one) that can deliver at most X watts DC power to a system and it has efficiency E. And I have an outlet that can deliver at most C amps at G volts. Now G is in RMS volts for the AC outlet and all the AC stuff in 60Hz. C isn't clear if it is RMS or peak but it is connected to a breaker that will trip at C amps.

So my thought was that the max current draw for the PS is ((X/E)/G) and as long as that is less than C we are fine but I can't help feeling like I'm missing a sqrt(2) somewhere.

So am I missing a sqrt(2) somewhere?

Thanks.

 

[Paperdoc]

No problem. you're doing it right. The whole reason for expressing volts and amps both as RMS values is so that, when they are multiplied, you get the time-averaged (mean) POWER in watts.

 

[Paperdoc]

(delete this accidental duplicate reply)

 

[Mark R]

The RMS current draw would be

Irms = X/E/G/Q

where Q is the power factor. (Q = 0.96 for a PFC power supply; Q=0.6 for a non-PFC supply, older or low-end models for the US only market. PFC is mandatory in Europe, Oceania and parts of Asia).

Peak current draw would approximated by (for PFC PSUs):
Ipk = Sqrt(2).Irms
For non-PFC PSUs it is approximated by
Ipk = C.Irms
Where C is typically between 2.5-4.

Outlets and breakers are typically rated for rms current (this is the big advantage of the rms measurement - as the Irms^2 is proportional to thermal power losses)