Join Date: Mar 2002
Location: Austin, TX, USA
The bulb resistance may be 20 ohms cold, but lit it will increase significantly. But bulbs disipate their power as heat and light. Resistors on the other hand disipate power only as heat. So if you put a 20 ohm resistor in circuit, you'd be disipating 14^2 / 20 watts or 9.8 watts. That's a dang hot resistor. The bulb is probably more like a 1 watt bulb or less. So that means the bulb filement is more like 392 ohms. So the answer to your question is no, the resistor won't burn up, so long as you don't make it too small in value and too low in wattage. But you have to be practical. You could install a 2 watt resistor in there but the thing would get so hot you'd have to worry about it melting things.
You won't burn anything up by disconnecting the L line for a test. If it is in fact an exciter then the alternator won't put out and it will be evident by the battery voltage measurement. But I believe the light output is really a fault circuit created by an open collector transistor that the regulator turns on during a fault condition. And no, the resistor I recommended will not burn up the resistor or the regulator.
I am not sure what the other two terminals are for, including the F terminal. I haven't found any technical documentation specifying what it is, but I know it works without it just fine.
68 Coupe, 350 w/ Edelbrock Performer RPM heads, cam, intake, 700R4, Dave's small body HEI