Resistors will get hot, it'* part of "resisting". If it gets hot enough, it will burn out. This is ususally caused by using too small a resistor, either in physical size or resistance (too much for the application, that is).

Here'* a link to translate the colors http://ourworld.compuserve.com/homep...n/resistor.htm

Ah ha! 240 Ohms, I was thinking right... I just figured I'd wait til some who knew what they were talking about dropped in, your linking abilities are grand MOS

What'* an angel eye?
I can help you calculate the correct resistor with some more info.

Angel eye is a model of fog light made by Hella, it has a ring around the perimeter of the foglight that lights up, the look is borrowed from BMW'* projector headlight

Ok. Well if you hook up a new (let'* say 240 ohm) resistor, and then measure the voltage across the resistor itself with a decent meter, we can figure out how much power the resistor is dissipating and then select one that will last. If it is really an LED, you could also dim the LED a bit by using a slightly higher resistance value. However, don't make it brighter with a lower resistance, because that will burn out the LED.

That'* it man. The wattage of the stock resistor must not be big enough for the load this light is using. I'm thinking a ceramic block resistor would be nice, but don't know if they make ceramics in that big of a value.
What is the formula for determining how much a resistor is dissipating?

Sure, they make power resistors in all values. Try Newark.com or Mouser.com.

The amount of power dissipated by a resistor is equal to (V^2)/R. So let'* say you have a 240 ohm resistor and you measure 10 volts across the resistor. The power dissipation is 10^2/240 = 100/240 =.416 watts. A good choice for this resistor would be a one-watt resistor. In general it'* good to have a part rated at twice the actual dissipation.

HEY!! Nobody said there'd be math!!