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active| newest | oldest@fthies can you plz tell me what formula have you used...8.7/.020??

@Harisiqbalengr.

(Supply voltage - LED voltage drop) devided by desired current in amps (20 milliamperes = .020 amps)= resistance needed to limit the current through the LED. So for a 12volt system and a 2 volt drop LED the formula would be (12-2)/.020= 10v / .020= 50 ohms.

The power ratting for the resistor is calculated with this formula. Voltage (10v) multiplied by the current through it (.020). Which is (10v * .020a=.2w watts) so a 1/4 watt resistor would get pretty hot. I would error on the high side and go with a 1/2 watt resistor.

I like powers of 10. I like the brown right next to the black. I think it's a very attractive arrangement of colors.

I think you'll be happy with it too.

You already have the source voltage (12v) and the diode voltage drop (3.3v).

Just find out the current in mA and you'll get your resistor value!

The battery will "fry" that very quickly indeed, the current needs limiting, but to what?

L