Originally Posted by
tunerjetta29
Looks to me like your running 9V through each LED only rated for 3.6V. They don't share voltage in a parallel circuit. Also your Ohm's law formula is the calculation for total resistance in a series circuit when you have clearly made a parallel circuit. Try again.
We have a voltage drop across the LED of 3.6V so there is 5.4 volts left to acount for. The voltage drop across the resistor is caculated by V=IR
Voltage = Current X Resistance). With the 330ohm resiter there will be a voltage drop of 6.6V, Idealy a 270 ohm resistor will give you the 5.4 volt drop needed but with the % of tolerance 270 ohm resistor may allow too mutch current to pass so the next higest value is 330 ohm.
By the way I know ohms law, may be you should study it a bit more.
http://www.the12volt.com/ohm/ohmslaw.asp