I used lan cable for light LED up on my bread board power source is PC's power supply.

without any resistor, the lan cable melt!

I reallize I have to keep OHM's law in my mind and use proper resistor for safety.

on lan cable's jacket,

it says

"0270M LS Cable UTP 4PR 24AWG CM CAT . 5E 75C KS Verified KS C 3342 HDPE 2009 V8"

how much current do this 24AWG cable tolerate?

could somebody tell me the relationship between temperture, current, thickness, and material?

linking website, simple hint also welcome.



1 Answer 1


You must have a short somewhere.

24AWG is 0.511 mm in diameter and has a resistance of 84 mOhm/m. That means that a 1 A current will generate 84 mW in heat loss per meter.
Copper has a specific heat of 0.385 J/g, this means you have to 0.385 Watt-second to heat 1 g of copper 1 °C.
Specific mass of copper is 8.93 g/cm^3, or 1 m of 24AWG weighs 1.83 g. So you need 0.705 Watt-second to heat 1 m of 24AWG 1°C, provided it can't loose this heat. Now we're generating 84 mW, this means that having 1 A for 1 second will heat -- well, warm -- your cable up by 0.12 °C.
In 1 minute your cable will be 7 °C warmer, in practice less, because the cable can conduct/convect/radiate part of this heat.
The insulation is HDPE (High Density PolyEthylene), which has a melting point around 120 °C.

In conclusion: even 1 A will not melt your cable, and a LED uses far less than that.

  • \$\begingroup\$ From OP's previous question it appears that he was running the diode without a resistor. \$\endgroup\$
    – AndrejaKo
    Apr 10, 2011 at 10:06

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