Are you sure it's 12V? If the voltage is higher than 12V, that could be the problem. Also, it would help to know what LED strips you're using. Are you sure

*they* are 12V?

And, regarding power, and power dissipation, there's another formula:

P=IEEasy to remember because it spells "PIE"

And, when you combine the two formulas -- Ohm's Law and this power formula -- along with a bit of Algebra, you get the following suite of formulas, that can be used to figure out just about everything you need for DC electronics:

E = IRP = IEP = I^{2}RP = E^{2}/RSo, for instance, if you want to come up with how much power a resistor is going to dissipate, for a particular voltage across that resistor: P = E

^{2}/R ... so, if your LED strip has, for instance, 150Ω resistors, and there are 3 LEDs in series with one of those 150Ω resistors, and each LED is running at, say, 3.2V, then

P = (12V - 3 * 3.2V)^{2}/150Ω = 38mW -- [Thus:

38mW is dissipated by each

150Ω resistor]

If you want to know why that potentiometer is burning up: assuming your strip has 20 sets of 3 LEDs, and, again, the LEDs are running at 3.2V, then each 3 LED set is drawing the following amount of current:

I = (12 - 3 * 3.2V)/150Ω = 16mAand, the total power, for the strip, is:

P = 10 * IE = 10 * 16mA * 12V = 1.9WSo, if you put that in series with a 1/4W potentiometer, worst case, you're probably going to dissipate around a watt in that poor thing, and thus all the smoke and fire!

Now, some "LED"s used in LED strips, are actually little mini-COBs ["COB" = "Chip On Board"], made up of THREE LEDs! So, our above calculation would, then, become:

I = 3 * (12 - 3 * 3.2V)/150Ω = 48mAP = 10 * IE = 10 * 48mA * 12V = 5.8W!!!So, unless you're using one of those big-ol high wattage potentiometers, it probably doesn't have a chance in hell!!

BTW: Your heating problem could also have to do with putting the LED strip in a closed box. You might need to provide some sort of ventilation -- especially if you are using a strip with those 3-LED COBs.