On 29/07/18 12:19, R.Wieser wrote:
> The voltage is that high because that way the voltage at the end of the
> (long?) ethernet cable will be high enough to be still usable (the voltage
> drops because of the internal restance of the cable - the longer the cable,
> the further the voltage goes down).

Its more a case that a higher voltage needs less current for the same power.

I.e the drop at constant power is less absoluetley  at higher voltages,
not just less in proportion to the voltage

Power loss is I^2.R and power is V.I fraction lost is IR/V

So I = P/V so loss is P^2.R/V^2


Now the 132KV lines make sense...

Power loss is proportional to power SQUARED divided by voltage SQUARED.







But


--
I would rather have questions that cannot be answered...
...than to have answers that cannot be questioned

Richard Feynman

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