Ignition resistor question

zookeeper

Founding Member
Aug 25, 2001
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Rogue River, Oregon
Over the weekend, a friend asked me to help install his Pertronix 2 ignition. I stopped by his house today and looked through the instructions. His car is a '68 Mustang (same as mine) so I checked to see if I could find a resistor wire or a ballast resistor. I found no ballast resistor so I checked and found a pink resistor wire on the back of the ignition switch. I told him we'd have to bypass the wire so the coil would see 12 volts, but when I checked it with a mulitmeter to show him it only got 9 volts or so, I tested and got 12 full volts from the wire with the key on. Anyone know why that would be? I don't have time to do the install until the weekend, but if I don't have to screw around with the wiring I'd rather not. Anyone else find this? Anyone just plug in a Pertronix 2 and Flamethrower coil without bypassing the resistor wire? Would it even start?
 
The wiring harness looks like a virgin to me. In fact the entire car is all stock and looks to have led a pretty pampered life. What do you think, should I leave well enough alone since I read 12v at the coil feed? I'd hate like heck to take a perfectly good-running car (that doesn't belong to me) and create problems by doing unneeded wiring mods.
 
Thanks, that's what I hoped you'd say. BTW, it seems like I've read that the first generation Pertronix didn't need to have the resistor wire bypassed at all, why does the Pertronix 2 specify 12 volts at the coil? Thanks again! Jim
 
I would expect it to show 12V. With the key on, there is no power flowing through the coil. There is only power flowing through the coil when the points ground out the - side of the coil and it fires the spark plug.

Resistors drop voltage across them depending on the amount of current as much as it depends on the resistance of the resistor. No current means no voltage drop. Really it comes down to ohms law.
V = IR , Voltage = Current * Resistance.

This all assumes that you are measuring voltage at the coil with the points open. If the points were closed(- side of coil is grounded) you would see current flowing through the coil, and a lower voltage at the + terminal of the coil.
 
zookeeper said:
Actually, I just popped the wire off the coil so I could measure the voltage. Shouldn't it read 9v if (assuming it has the resistor in place) regardless?

No. It's kinda hard to explain, but with the wire disconnected from the coil you have an open circuit. When measuring the voltage through a resistor in an open circuit you will always read the same voltage as the supply voltage. This is always true unless it's a very large resistance (in the millions of ohms range), and even then it's only because your meter acts like a high resistance to ground (thus it's no longer a open circuit).
 
When a resistor wire goes bad, does it take the voltage up, or down? I remember reading my shop manual and it had a procedure for replacing the resistor wire and I seem to think that voltage went up when it failed. Is that right?
 
SoCalCruising said:
When a resistor wire goes bad, does it take the voltage up, or down? I remember reading my shop manual and it had a procedure for replacing the resistor wire and I seem to think that voltage went up when it failed. Is that right?


I can only see a resistor wire failing to an open situation or a higher resistance. Which would mean you'd see lower voltage or no voltage at the coil. Basically the same thing that would happen if it was a ballast resistor that was failing.