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

erm......
wouldn't it be 135ohms and the best resister to use is a 150 ohm 1/4 watt?
The leds are in parellel so you have to work on the Rtotal and then presume, as per kirchoffs laws, that the current =0
You would also only need to use one resistor:blush:
.

No, because in that aplication you are running the current of both LED's through 1 resistor and in my aplication the resistors are only seeing the current of 1 LED each.

Not to give all my secrets away but there are lots of sites out there with LED calculators. First example I came too.

http://ledcalc.com/

BTW, I chose not to do it that way because if 1 LED was to fail you then you would have too much current going through the last LED and it would fail as well.

Yes, that is where I confirmed my calculations but some are not the electronic type so I chose to draw a schematic for him to folow to help make it simple ( I hope).

ok cool, that works.

you could also build in a 555 timer to ensure that you have the leds working and if they don't then they will click over to secondaries............

are we starting to overengineer here?

You started it LOL!

lol :blush:
No i didn't i simplified it by removing a component!!!

Read the link that HB Moose posted, LED's need some kind of current limmiting.

This is a better one.

http://www.daycounter.com/Calculator...lculator.phtml

Sorry, I was refuring to your first post, not the one with thw schematic. You are correct in it is a bit simpler but for the reason I all ready stated, the addition of 1 resistor I think is a better solution.

Again, I apologize for my misunderstanding of your post.

I try my best helping others out when I can but I will make mistakes once in a while.

all good mate, ozzy sarcasm