As already stated, LED's need current-limiting resistors. You stated it well when you said the LED acted like a fuse. That is good so nothing else was damaged.

Here is information I copied from an earlier post of mine. If a person follows the information below, they should have no problem designing LED's into practically any situation. This should be made into a How-To:

You always need resistors with LED's to limit current when a constant voltage source is powering them (like when replacing regular light bulbs). Some LED's come with built in resistors, but their packaging or spec. sheet will say that if that's the case. Assume it has no resistor built in if it doesn't say.

Decide on how many milliamps you want going thru the LED at full voltage. (mcd = millicandellas is the measure of brightness at a given current).

R = (V(b) - V(LED)) ÷ C

where:

R is in ohms

V(b) is your car's power system voltage (assume 13.8 with it running)

V(LED) is the voltage drop across the LED (assume 1.2 unless the spec. sheet tells you)

C is the current you want in amps. 20 mA = 0.02 amps

So for example: For V(b) = 13.8 volts, V(LED) = 1.2 volts, and desired current is 20 mA (0.020 amps), the resistor would be 630 ohms - find the nearest value to that - 620 is a standard value I think.

Lower resistance will give you more current, more brightness, shorter LED life.

Some of these bulb locations may have lower source voltages, so for a given location, measure the voltage with the brightness turned all the way up with the original bulb installed, and use that number for your V(b).

For multiple LED's in series, use their total voltage for V(LED) in the formula. For example, if one LED voltage drop is 1.2 volts, and you are putting 2 LED's in series, use 2.4 volts for V(LED) in the formula. For 3 LED's, put 3.6 volts in the formula. Do not stack so many LED's that their total voltage gets more than 4 or 5 volts from the source voltage. Because the source voltage varies some, you need to have a resitor in there so that the current (and the LED brightness) is not going to be super-sensitive to those variations. If you use a voltage regulator, you still need a resistor, but you can get a little closer to the source voltage (in that case, the output voltage of the regulator) because the regulator will give less variation in its output voltage than the vehicle's system.

To determine the wattage needed in the resistor, use the formula:

P = C^2 x R

where:

P = power in watts

C = current in amps

R = actual resistor value in ohms

So for our above example of 20 mA (milliamps) and 620 ohms, the resistor would dissipate 0.248 watts. That's right at 1/4W, but you would not use a 1/4W - you want to go up one size which would be 1/2W. Always try to use at least 50% safety margin on the wattage (for one thing, the wattage rating gets de-rated at higher ambient temperatures). You need some safety factor on that so your resistor lasts forever. If your wattage calculation came out to, say, 0.15 watts, you could safely get by with 1/4W. Any more than that, move up to 1/2W. If wattage comes out more than 0.33 or so, use more than 1/2W (or parallel two 1/2 watters of twice the resistance value each for the same net resistance value).