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 >
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