Ok does anyone know how to do this Im sure I need some sort of resistor or something any help would be appreciated.

To run and LED of 12 Volts you will need a resistor in series with the LED.

(Input voltage - LED Voltage)/LED current in amps = Resistor value ohms

So for an LED running at 2Volts and a current of 10mA (0.01A) the resistor value will be (12-2)/0.01 = 1000ohm (1K ohm)

Also the LED must be connected with the correct polarity, the long lead is positive. If the leads have already been cut the lead closest to the flat section near the base of the LED is negative.

Rod.

Ok here is a question for ya since the voltage in a car can fluxuate is there anyway to make up for this? And what would be the rules if two or three LEDs were connected in series how would you deal with the mAs and volts?

The formula only changes slightly.

(Input voltage - LED1 Voltage - LED2 Voltage - LED3 Voltage)/LED current in amps = Resistor value ohms

So for three LEDs in series running at 2Volts each and a current of 10mA (0.01A) the resistor value will be (12-2-2-2)/0.01 = 600ohm. The nearest value that is easy to get is 620ohn and would be fine.

Make sure the input voltage is greater than the total of all the LED voltages.

Don't worry about voltage fluctuations. Most LEDs run happily from 5 - 20mA, if you calculate using 10mA any voltages changes in a car will not cause the current to go outside the 5 - 20ma range.

Rod.

You are correct the voltage in a car will fluctuate, so the value of the resister needs to be a little bit over 1000 Ohms to allow for this fluctuation [say 1100 (or 1.1 K ) ohms.
The LED may change brightness a little as the car voltage fluctuates, but the eye doesn't notice changes in intensity much less than 50%, so you may not notice the fluctuation in brightness or if you dod it will be small.

For LEDs in series, each LED will "consume" (cause a voltage drop of) about 1.7 volts, so for 3 LEDs in series, the voltage drop over the LEDs will be 1.7*3 = 5.1 volts.
This Voltage drop across the LED is called the "forward voltage" and is often quoted as a useful charateristic of the LED as some LEDS have a much higher forward voltage [like 3.4 volts].

The forward voltage (plus a little bit) is also the lowest voltage that the LED will work (light) with, so a normal LED won't light below about 1.7 volts, so thats why they don't light up when connected to a single battery cell, which is about 1.5 volts or so.

So you now with 3 LEDS in series have to rework your calculations for the original resistor and since the voltage the first LED will "see" is now 5 volts, you take the difference between the car voltage and the "voltage drop" (i.e. 7 volts) and plug it into the equation again to get the revised resistor values.

If you use the existing 1000 Ohm resistor for a 1 LED circuit and then put 3 LEDs in series, then either no LEDs will light, or they all will, but dimmer than with 1 LED in the circuit.

Either way, you won't blow anything up, so suck it and see.

Don't ask me what happens if you put them in parallel - probably they all suck the same amperage, so the resistor needs to be made smaller to allow for the higher current.

Here are the stats for the LEDs in question.
"36mA (max). Typical Voltage is 1.9, with a maximum voltage of 2.2V. Comes as package of 1."

So given this
R=(12-1.9-1.9-1.9)*0.036=175ohms

And if the first LED sees 6.3 volts will it not blow?

Yes, you still need the resistor when LEDS are in series.
The resistor value will be smaller than the 1000 ohms you would have with just one LED.

Regarding your comment re: the LED "sees" 6 volts and will blow, the same comment could be made that a single LED circuit "sees" 12 volts.

The resistor in both cases reduces the "unsafe" 12 volts voltage to a safer one.

It's best not to run the LED at their maximum current. Best to leave some headroom for voltage fluctuations.

I would suggest 20mA so (12-1.9-1.9-1.9)/0.02 = 315ohms (300ohms is the closest easy to get value).

If you calculate the resistor value at a current of 20mA the maximum LED current of 36mA will not be exceeded until the cars voltage is over 15 volts.

The top LED will not "see" 6.3 volts. Its positive lead will be at 6.3 volts and it's negative lead will be at 4.4 volts so it will only "see" 1.9 volts.

The current you chose will depend on how bright you want the LED to be, the lower the current the lower the brightness.

Rod.

And don't stack too many LEDs in series so that your sum of forward voltages get close to 12V - the closer you get (really, the less voltage over the resistor) the more affected you get by voltage fluctuations.

/Michael

Oh I only plan to have three. There are going to be for a VTEC indicator light on my friends Gen2 acura integra GS-R. When the engine switches to the higher cam profile (VTEC kicks in) the lights will come on. i plan on using a relay and tapping into the VTEC solenoid wire I'll let you guys know how it turns out.

Might be counterproductive to use a relay. A standard automotive relay (Bosch style) is
likely to draw more current than a single string of LEDs. If you want it for electrical isolation, look at "optical relays". You might be able to use a reed relay if the solenoids magnetic field leaks out of the casing...

/Michael