LL - NY larrycleung at gmail.com
Tue Nov 4 05:37:12 PST 2008

```Since V = IR

R = V/I

and P = VI

I = P/V

substitute in and we get R = V^2/P  or R = 12V^2/55W = 144V^2/55W = 2.6 Ohms
per lamp.

I'd venture if you could find any generic 2 - 3 ohm minimum 24V rated
resistor (it's going to need to be able to handle 55W's though) it'll not
only work, but work for a long time. Any lesser Power rated resistor will
eventually fail. (I don't think I've ever seen retail resistors that are
rated for lesser voltages, so I'd not worry about breakover voltage, and any
computer circuit board rated 3 or 5 V resistors wouldn't be capable of
handling +55W.

LL - NY a.k.a. the Physics Teacher.

On 11/4/08, Grant Lenahan <glenahan at vfemail.net> wrote:
>
> bulb watts = volts x amps
> so amps = watts / volts
>
> now you have amps
>
> set resistor such that volts / ohms = amps  --or-- ohms = volts / amps
>
> Grant
>
> On Nov 3, 2008, at 10:14 PM, Tom Love wrote:
>
> > I built a relayed headlight harness and now it seems I need to fool
> > the auto check system into thinking there are still bulbs
> > connected. Using a resistor seems to be the way to go but  how do I
> > figure the size to fool the system into thinking the bulbs are
> > still connected? TIA
> > _______________________________________________
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> > quattro at audifans.com
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>
>
> Grant Lenahan
> glenahan at vfemail.net
> (201) 602-4702 mobile
>
>
>
>
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>
```