Question about the resistor wire.

Great68 · May 12, 2005

I decided to take a look at my ignition wiring the other day. I have been running my MSD coil and Ready to Run distributor straight off of the wire that goes to my coil and It actually has seemed to run fine thus far.

However, what I can't understand is that when I have the key in the "on" position, and with a voltmeter connected to the positive side of the coil it shows a full 12 volts. But when the motor is running it only shows 7 volts.

How does it go from 12 to 7 volts like that? Is it just from the load the coil is drawing with the motor running or what?

And instead of replacing the wire all the way from the ignition switch, can I just run a supplemental wire from a switched power source in addition to the stock wire? Or will that mess something up?

I don't know much about electrical circuits, so bare with me.

Thanks!

1965pony · May 12, 2005

ignition wire

Hey great,
well your right, the voltage drop is due to coil saturation usage.
Ohms law states it takes 1 amp to push 1 volt through 1 ohm of resistance. So that length of wire takes a 5 volt drop just due to resistance in the wire.

When you check it with your volt meter there is no current flow, but as soon as you put a load on it the resistance goes up so your voltage goes down.

To get your ignition working to it potential, you can remove the stock resistor wire and replace it with a 14g wire.It just runs from your firewall bulk head to the coil. Just undo your bulkhead, pop out the old wire, get a replacment connector, crimp and solder it and run it to your coil B+. Now remember you should have 2 wires going to your coil B+. One from the ignition switch and one from your starter relay. This wire from your starter relay (R terminal) is the resistor bypass so your coil gets full B+ while cranking. You could keep this wire there or remove it and rewire your ignition switch to supply full B+ while cranking. Right now you stock set up should be 0 volts while cranking. Remember to disconnect your battery when digging in your harness. Vince

65fastback2+2 · May 12, 2005

1965pony said:
Hey great,
well your right, the voltage drop is due to coil saturation usage.
Ohms law states it takes 1 amp to push 1 volt through 1 ohm of resistance. So that length of wire takes a 5 volt drop just due to resistance in the wire.

When you check it with your volt meter there is no current flow, but as soon as you put a load on it the resistance goes up so your voltage goes down.

To get your ignition working to it potential, you can remove the stock resistor wire and replace it with a 14g wire.It just runs from your firewall bulk head to the coil. Just undo your bulkhead, pop out the old wire, get a replacment connector, crimp and solder it and run it to your coil B+. Now remember you should have 2 wires going to your coil B+. One from the ignition switch and one from your starter relay. This wire from your starter relay (R terminal) is the resistor bypass so your coil gets full B+ while cranking. You could keep this wire there or remove it and rewire your ignition switch to supply full B+ while cranking. Right now you stock set up should be 0 volts while cranking. Remember to disconnect your battery when digging in your harness. Vince

and if i remember too, it uses 12v for the initial start and drops to 7v for running

Max Power · May 12, 2005

Isn't the resistor wire under the dash rather than under the hood?

Question about the resistor wire.

Great68

1965pony

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65fastback2+2

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Max Power

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Question about the resistor wire.

Great68

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