Join Date: Mar 2002
Location: Austin, TX, USA
Sounds like this one is figured out. The signal on the positive size of the coil will be roughly a square wave that alternates between about 8 volts and 12 volts. It depends on the system voltage (which usually is 13 volts or more under charge). But anyway, we can assume about a 4 volt span.
The square wave signal has a duty cycle equal to the dwell reading or about 30%. So the value measured by a DC volt meter will be about 70% of this 4 volt span below the system voltage. 70% of 4 is about 2.8 volts. So if the charging system is holding the system voltage at 13.8 volts, the coil will show 11 volts on the coil + terminal as measured by a DC volt meter. In reality though, 13.8 is conservative for system voltage which can go as high as 14.5 volts or higher.
The main thing the resistor does is limit current flow. The voltage reading is an artifact of the current limiting effects. Looking at it another way, the resistor gets hot and essentially is disipating heat that would otherwise be disipated by the ignition coil itself. The coil gets pretty warm on it's own, but without the resistor it would get even hotter.
If you ever encounter someone complaining of intermittant ignition problems and see that the top insulator of the coil is cracked or the bottom is pooched out, you're probably looking at a coil that has over heated. More than likely this is because the ballast wire / resistor has been bypassed in a breaker point ignition system or electronic system which requires a current limiting resistance.
68 Coupe, 350 w/ Edelbrock Performer RPM heads, cam, intake, 700R4, Dave's small body HEI