Simply look at the spec's on a Wire Gauge Table in Google to see that the voltage drop on the thin short wire is very small since the current is very low. If you don't use a current-limiting resistor then the LED will blow up.
It is easy. 32AWG wire has a resistance of 538 ohms for 1km. I got it from the first link in Google when I searched for Wire Gauge Table. Then only 2m has a resistance of 538/500= 1.1 ohm but there are two wires so the total resistance of the wires is 2.2 ohms. It can safely pass up to 91mA continuously.
If you have a 3.3V blue LED and want it to pass 20mA then the current-limiting resistor value is (12V - 3.3V)/20mA= 435 ohms. If you use 470 ohms then the current is (12V - 3.3V)/470 ohms= 18.5mA. The resistor will dissipate 18.5mA squared x 470 ohms= 0.16W so use a 1/4W resistor.
The voltage drop in the 2.2 ohms wires is 18.5mA x 2.2 ohms= 0.041V.
Wire gauge size question.
Posted - Apr 26 2009 : 6:03:34 PM
