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  1. #1
    Member Since
    Sep 2002
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    99 Silver 515
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    LED resistor

    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

  2. #2
    Member Since
    Feb 2010
    Location
    2000, Kaiser, VX, 0125
    Posts
    134
    Thanked: 0
    Thank you for explaining. Found a handy LED bulb calculator ta boot for those interested and afraid of maths:

    http://led.linear1.org/1led.wiz



    Quote Originally Posted by hughesdt View Post
    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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