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Ok, so I will keep it as short as I can.
5v dc power supply
3A load
Traveling over CAT5e, distance = 275'
2pr pos, 2pr neg.

At device point, reads 5.25-5.28v on meter (device not connected) Is the "Voltage Drop" I read about, only when "Load" (Device) plugged in? since I don't see a drop on the meter?

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    \$\begingroup\$ Yes, voltage drop is only seen under load. If you can find a 5 Ohm resistor, try it as a load. From memory, cat5 cable is 0.1 Ohm/m, so a short circuit will only draw about half an amp through all that wire. Yoir access point is unlikely to work, sorry. \$\endgroup\$
    – tomnexus
    Commented Feb 3, 2015 at 21:38

1 Answer 1

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Using 'official data' to check Tom's back of envelope figurings.
Alas, Frank's mental magic was very close to actual.

Wikipedia Cat5e data here

Resistance = 0.188 Ohm per loop metre.
275 feet = 84 metres 84 x 0.188 = 15.8 Ohms
Two parallel pairs = 7.9 Ohms -> say 8 Ohms.

At short circuit I = V/R = 5.25/8 = 0.66A.

To get 3A over this feeder at say 0.5V drop needs
V = IR = 3 x 8 = 24 V So total feed voltage needs to be about 24+5 ~= 30 Volts. Of the 90 Watts input about 85% is dissipated in the cable.

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