Thanks for the responses guys.
jrichker, Re: Go ahead and write up the math if you don't mind. I've taken quite a few physics courses in college as well as an intro to electrical engineering course where we built everything from high and low pass filters to rectifiers and even a breadboard computer. I'm just rusty on all this stuff, haven't touched a circuit in a few years.
You asked for it...
I= Currrent
E= Voltage
R= Resistance
W= Watts
Two 55 watt fog lamps =110 watts. Find the current in the circuit
I= W/E
110 watts/14 volts = 7.85 amps for fog lights alone.
Since the exterior lighting circuit supplies headlights, taillights, and parking lights, etc.
56 watts 2 each GE Part # L3156 corner light 28 watts each
90 watts 2 each GE Part # 9004 headlight 45/65 watts each (low beam)
63 watts 2 each GE Part # L194 parking light 31.5 watts each
56 watts 2 each GE Part # L3157 tail light 28 watts each
265 watts Total
Total other exterior lighting current
I=W/E
265 watts/14 volts = 18.92 amps
18.92 amps other exterior lighting current
+ 7.85 amps fog light current
26.77 amps with all exterior lights and fog lights on.
The 12 gauge power feed wire to the exterior lighting switch is rated at 20 amps
26.77
- 20.00
6.77 amps excess current
7.85 amps used by fog lights
-6.77 amps excess current
1.08 amps to run the fog lights left if you stay within the 20 amp limit of the wire.
With 1.08 amps of current, the fog lights probably won’t produce any useable light.
Added resistance required to reduce fog light current to permissible 20 amp limit
R=E/I
14 volts/6.77 amps = 2.06 ohms
Resistor wattage
W=EI
14 volts x 6.77 amps = 94.78 watts
You would need a 2 ohm, 100 watt resistor.
Light bulb ratings from
http://www.roadparts.com/catalog/section30.pdf
Radio shack resistor catalog - no matches , and no combinations that could be used to make a 100 watt, 2 Ohm resistor.
See
RadioShack.com
Ohm’s law – in case you have any questions about my formulas -
Ohm's Law Calculators