# I want to wire a single led off a motorcycle battery [duplicate]

I want to use my motorcycle battery to power a single led that has a maximum voltage of 3.6 volts. what resistance do i need to drop the voltage from 12 volt to 3?

Using your given forward voltage of 3.6 volts, I will assume it is the simple LED that uses only 20 mA of current, then:

$$V=IR$$

$$\frac{V}{I}=R$$

So voltage in this equation is voltage of the source minus the voltage lost or used by the LED = 12-3.6=8.4

SO:

$$\frac{12-3.6}{20mA} = 420 Ohms$$

• 420 ohms at 3.5 watts. A standard 1/8 watt resistor will quickly turn to smoke... Commented Feb 27, 2015 at 7:35
• @MarkU, how are you getting 3.5W for the resistor? At 20mA, I'm getting 0.02^2*420 = 0.168W. Commented Feb 27, 2015 at 8:05
• @mark 8.4 V × 0.02A is 0.168 Watts. A 1/4 Watt resistor is needed. How did you get 3.5 Watts? Commented Feb 27, 2015 at 8:06
• Whoops, must have gotten something mixed up... after re-checking I agree with 0.168 watt i.e. 1/4 watt resistor rating. Sorry for the false alarm. Commented Feb 27, 2015 at 9:45