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Date:         Fri, 26 Jul 1996 14:06:35 -0700 (PDT)
Sender:       Vanagon Mailing List <vanagon@vanagon.com>
From:         magicbus@value.net
Subject:      Re: basic electronics question

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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