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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?

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

linking website, simple hint also welcome.

thanks.

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1 Answer 1

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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.

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  • \$\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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