Hi, first post on this forum. I've attempted to do my homework elsewhere with no luck. I just replaced my distributor & coil with a Pertronix II dizzy & Flame Thrower coil. It's all working, yay, and the car is running better than it ever has. I'm just really confused about a few things, and I'm hoping somebody here can explain why I saw, and what I am still seeing.
I initially attempted to splice into the pink wire near the ignition, and again near the firewall with a 12ga. wire to effectively bypass the resistance wire. Ohms Law suggests there would be virtually no current in the pink wire, and 100% in my new yellow wire. Perhaps I gave up too easily, but I decided there was no way I was splicing into the pink wire at the firewall, so I ran my new wire through the firewall and to the (+) side of the NEW coil. I left the existing lead connected as well, which shouldn't make a difference. Per the Pertronix instructions, I turned the key to ON and measured the voltage at the coil to be 10V. I was expecting 12V. When I went to turn the key off (it had only been 30-45 seconds), there was that terrible smell of electrical fire, and a hint of smoke. The 3" of insulated pink wire between the ignition and my splice were freaking hot. My little infrared temp sensor is out of batteries, so I don't know how hot, nor do I know how hot it gets normally. But I know I've never smelled that before. Since I only saw 10V at the coil, the other 2V must have been in the wire I thought I was splicing around. P=IV, and P was clearly significant, which means that there was current in that pink wire still. How is that possible??
So my next move was to completely eliminate pinkie, which I did. Disconnected from the coil, cut out at the ignition (I left the wire connected to the firewall plug, so rolling back will forever be an option). Turning to ON & measuring, I get the full 12V at the coil. However, once the new dizzy is on, and all the wires are hooked up to the dizzy & coil, I only see 10V again. So where the crap are the other 2V? Aside from the 3" of black/red wire just off the ignition, all the wires are brand new, and there are no splices other than the one at the ignition. There should be virtually no voltage drop in this portion of the circuit.
Please, help me understand.
I initially attempted to splice into the pink wire near the ignition, and again near the firewall with a 12ga. wire to effectively bypass the resistance wire. Ohms Law suggests there would be virtually no current in the pink wire, and 100% in my new yellow wire. Perhaps I gave up too easily, but I decided there was no way I was splicing into the pink wire at the firewall, so I ran my new wire through the firewall and to the (+) side of the NEW coil. I left the existing lead connected as well, which shouldn't make a difference. Per the Pertronix instructions, I turned the key to ON and measured the voltage at the coil to be 10V. I was expecting 12V. When I went to turn the key off (it had only been 30-45 seconds), there was that terrible smell of electrical fire, and a hint of smoke. The 3" of insulated pink wire between the ignition and my splice were freaking hot. My little infrared temp sensor is out of batteries, so I don't know how hot, nor do I know how hot it gets normally. But I know I've never smelled that before. Since I only saw 10V at the coil, the other 2V must have been in the wire I thought I was splicing around. P=IV, and P was clearly significant, which means that there was current in that pink wire still. How is that possible??
So my next move was to completely eliminate pinkie, which I did. Disconnected from the coil, cut out at the ignition (I left the wire connected to the firewall plug, so rolling back will forever be an option). Turning to ON & measuring, I get the full 12V at the coil. However, once the new dizzy is on, and all the wires are hooked up to the dizzy & coil, I only see 10V again. So where the crap are the other 2V? Aside from the 3" of black/red wire just off the ignition, all the wires are brand new, and there are no splices other than the one at the ignition. There should be virtually no voltage drop in this portion of the circuit.
Please, help me understand.
