Electrical schoolwork...

[Huw Powell]

> V = I*R
> (13.6V - 2.0V) = (.015A)*(R ohms)

Where is that 0.015A figure coming from?

I suspect there is an error in the data we are working with (garbage in, 
garbage out...).

150,000 ohms at 3 volts yields 0.00002 amps, according to my trusty 
desktop calculator.  Which is only, what, .02 mA?  Something seems fishy.

You and Ti have both made assumptions about the current requirement of 
the LED not specified in the question.  This problem should be solvable 
with any of the iterations of Ohm's Law - you can chuckle at "syljay"s 
results all you want, but they are absolutely correct - given the data 
we have to work with.

So, Louis-Alain, which number is wrong?  Is the stated resistance of the 
LED *really* 150k ohms?

> If I were to install some red 3V LEDs to backlight the numerous switches
> around the instrument panel of my 1983 urQuattro, what is the resistance
> value I must introduce to this circuit to lower the 12V to something more
> acceptable?
> 
> Resistance value of the LEDs: 150 000 ohms Ideal voltage at the LEDs: 
> between 2.5V and 3V.
> Electrical voltage at the switches: 13.6V, permanent, switched by the
> ignition, not the IP rheostat.

-- 
Huw Powell

http://www.humanspeakers.com/audi

http://www.humanthoughts.org/