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Is there something inherently more efficient about 220V power which makes it inherently more efficient and therefore deserving of more stringent standards? (Of course, the impedance load induced by wiring might be higher for 115V voltages; though I don't think that ends up being too large at typical power supply amperage and line impedance. I could be wrong there though...)

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closed as not constructive by Ƭᴇcʜιᴇ007, bwDraco, Tom Wijsman, Nifle, studiohack Nov 18 '11 at 4:23

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

Closevoters: How is this off topic? The rules say no specific product recommendations, or specific products. They don't say anything against this type of general question though. – Billy ONeal Nov 15 '11 at 2:57
up vote 3 down vote accepted

I^2*R losses (in the wiring) scale with the square of the voltage. Half the voltage means four times the I^2*R losses. Typical home wiring resistance might be .5 ohms.

So, let's take a 450W power supply that's 90% efficient. It draws 500W. At 115V, that's 4.3478A while at 220V, that's 2.2727A. So the I^2*R loss is 2.58W at 220V but 9.45W at 110V. The difference, 7W, represents an extra 1.5% of effective power supply efficiency.

Since Titanium is targeting 96% efficiency, you can't just ignore the 1.5% loss. That's 1/3 extra loss.

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Well, +1, but it's still a lot of loss at "platinum" level too... – Billy ONeal Nov 15 '11 at 3:20

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