OK, you dropped a decimal point and rounded things off a bit much, and you're reading in the wrong place.So i Have a 460ohm resistor that i put in between the 12v+ and the positive lead on the Light. The meter read 5.09DCV Which means there is a 7V drop across the resistor. So if i do OHM's Law i take 7/460=0.0152 so does that mean each Light(6 LED's) Draws .02 amps or 20uA per light?
If i hook up the suply and the ground to the curcuit pannel on the truck i should be ok.Don't forget that you will get a different reading when the engine is running at a fast idle vs engine off. This is because there will be more voltage available when the alternator is generating current output.
Yes its to run them two at a time( --oooooooo, oo--oooooo,oooo--oooo, and so on, - is on and o is off)You've only shown some of the 4017 outputs used - is that how it is really wired?
5.7.0By the way, what version of Eagle did you install?
Yes i bought them from a friend... Ill get a 1 Ohm and let you know tomorrow i think i did it right just wrong resistor!Besides, are you certain that's a 460 Ohm resistor? That is not a standard value. It is more likely to be a 430, 470, or 510 Ohm resistor.