What you do is subtract the battery voltage from the LED voltage (13.8-3.6=10.2V) then divide my the led current (20ma), 10.2V/20ma=510 ohms. Ohms law, E=IR. If you put 2 LEDs in series 330ohms. Most LEDs are 20ma, but their voltages vary.
Dan
VXer
What you do is subtract the battery voltage from the LED voltage (13.8-3.6=10.2V) then divide my the led current (20ma), 10.2V/20ma=510 ohms. Ohms law, E=IR. If you put 2 LEDs in series 330ohms. Most LEDs are 20ma, but their voltages vary.
Dan
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