E = IR
I= E/R
R= E/I
You need to drop 8 volts at .0225 amps for one LED.

8/.0225 = 355.558 ohm resister.

Power (watts) = EI  .0225 X 8 = .18 watts

So you need a 355 ohm, 1/2 watt resister.

-----Original Message-----
From: Vanagon Mailing List [mailto:vanagon@gerry.vanagon.com] On Behalf
Of Edward Maglott
Sent: Saturday, December 03, 2005 9:51 AM
To: vanagon@GERRY.VANAGON.COM
Subject: Bright LED Cheap source & usage

Always looking for a good hack, I noticed recently an LED flashlight on
sale for $10 at the Eckerd Drugstore.  This has 21 bright white LEDs in
it
and uses 3 aaa batteries in series.  All the bulbs are in parallel, so
each
one is getting straight 4.5V DC.  It makes a nice little flashlight, and
I'll probably use it as one before hacking it up for the bulbs.  This
works
out to $.21/LED, which is a great price, especially if you consider most
electronics places you would have to order them from, and pay shipping,
meet a minimum order, etc.  I measured .4 amps draw or 1.8W.  .4 amps
might
run down those 3 aaa batteries pretty fast.  I think it said 10 hours on
the package.  If you buy one, be sure to use the "try it" button and
make
sure all 21 LEDs are working, one was bad on the one I bought...

Here's a link to the same item on a
website:              http://snipurl.com/kgds

Now my question.  What size resistor do I need to run one of these LED
bulbs on the 12V system of the vanagon?  I want to put one in the back
light of an aftermarket tach I have, which is very dim at night...

Thanks, and enjoy.

Edward