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Running Electricity over CAT5e - please advise

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djexct

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Hi
I want to wire out my pc with cat5e cable only to make it look better inside.

Every wire coming out from the PSU will be connected to its own core inside the cat5e.
This means 2 x 4 pin power connectors will be running through approx 30 - 50 cm of cat5e.
This puts out 2 x 12v, 2 x 5v, 4 x gnd.

Am i turning my pc into an electrical fire?
Can someone who has maybe tested cat5e please advise on power + heat limits, resistence etc.

Thanks and Regards
dJ
 
CAT 5 is 24 AWG and has a resistance of about 25 ohms per conductor per 1,000 feet (about 304 Meters on the roll I measured). Calculate the voltage drop and power dissipation in the wire for the current you will pass through it to know if it will work.

Don’t forget to double the above number to get the loop resistance. Power dissipation is P= I² R and voltage drop is E= IR. Note the critical factor is the current (I) in both equations; that is the one thing you have to know to determine if it is going to be safe and effective.

It isn’t going to do you much good if only 3 of the original 5 volts come out the far end or if you dissipate enough heat in the wire to melt the insulation.
 
Thanks Gary
I dont understand 95% of what that means but the 5v dropping to 3v already confirms something else ive read, and that will mess the system up.

Over what distance is that drop measured though?
 
Im cool with Not doing it, to my own pc at least. I did test 12v over 2 strands of 30cm cable now, the fan on the other end was spinning fast as usual, but i didnt take a reading so wouldnt know.
Thanks for your advise too bro
 
Im cool with Not doing it, to my own pc at least. I did test 12v over 2 strands of 30cm cable now, the fan on the other end was spinning fast as usual, but i didnt take a reading so wouldnt know.
Thanks for your advise too bro

The fan takes very little current, it's the other items (motherboard, HDD etc.) which take high currents and will give problems.
 
It depends upon the current. How many amps are you going to try to push down the wire? The drop is over the length of the CAT 5, however long that is. You now know that it has a loop resistance of about 0.164 ohms per running meter. Multiply that times the number of meters to get the loss in one pair. Divide by the number of pairs you are going to use for one voltage to get the total resistance.

Now, you can do several things. For instance, if you can stand to lose half a volt over the run, divide that by the resistance to get the maximum current you can pass down the cable. Or if you have to pass a specific maximum current, multiply the calculated total resistance by that current in amps to get the amount of voltage you will lose in the wire.
 
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