@Mannyrock posted:But, . . . what'a'bout dc powered led lights? If I hook a 3 volt led light to a 9 volt battery, will the light burn up?
Most definitely, trying to do its best imitation of an old-fashioned flash bulb! 
No offense intended, but I think you need to seek out a basic explanation of the relationship of voltage (measured in volts), resistance (in ohms), and current (in amps). As a visual pneumonic representation of the relationship between them, I like the classic Ohm's Law pyramid:
Cover the term you're trying to calculate ("V" for voltage, for instance), and the pyramid tells you the calculation you need to do ("I" for current times "R" for resistance, in this example).
Spec's differ from LED to LED, but most 3 volt LEDs are intended to draw about 20 mA, or 20/1,000th of an amp. In your example, you need to figure out a resistance large enough to drop the 9 volt supply voltage to 3 volts across the LED (or, stated differently, what resistance will allow a current of 20 mA when exposed to 6 volts, which is the amount by which it must 'drop' the voltage and split it with the LED). Since R is the term sought, cover the R, and the other side of the equation is V/I, or 6 (volts) divided by 0.02 (amps), which yields a value of 300 ohms for the current-limiting resistor. Simple, right? Even easier, just use the DigiKey calculator to do the same thing!
