Let's see hmmmm? If you have two resistors in series(one is the LED)and 
you know the voltage drop across one which is 2.6,and the total current 
is .05 amps then. 12v-2.6v=9.4v would be the voltage drop across the 
other resistor.Then E/I=R or 9.4/.05=188. 188 ohms. Right? Somebody check 
me on this. 

On Fri, 26 Jul 1996, Dan Houg wrote:

> 
> jeez, why didn't i pay attention more back in high school?
> 
> question:
> 
> i have a super bright , jumbo LED that reqires 2.6v @ 50ma
> 
> how do i figure the resistance of the dropping resistor for a 12v 
> source?
> 
> frustratedly yours,
> -dan
>