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?
0
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
4
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:
$$\frac{12-3.6}{20mA} = 420 Ohms$$
-
\$\begingroup\$ 420 ohms at 3.5 watts. A standard 1/8 watt resistor will quickly turn to smoke... \$\endgroup\$– MarkUFeb 27, 2015 at 7:35
-
\$\begingroup\$ @MarkU, how are you getting 3.5W for the resistor? At 20mA, I'm getting 0.02^2*420 = 0.168W. \$\endgroup\$– Dan LaksFeb 27, 2015 at 8:05
-
\$\begingroup\$ @mark 8.4 V × 0.02A is 0.168 Watts. A 1/4 Watt resistor is needed. How did you get 3.5 Watts? \$\endgroup\$– PasserbyFeb 27, 2015 at 8:06
-
\$\begingroup\$ 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. \$\endgroup\$– MarkUFeb 27, 2015 at 9:45