"....but no specific description of what those problems are, or thier cures."
You must not have checked the archives. We get at least two or three of these questions a week.
As Zane said, get the resistors correct or you"re going to melt the points & the coil.
Technology & materials being what they were in the 30"s, that square coil would melt if it ran on much more than 4 amps for any length of time. (see tip # 38 for an example). In order to get a hot spark at the same time the starter was drawing max current from the battery, a ballast resistor was added in the ignition circuit. What that did was add about .3 ohms of resistance in the circuit, added to the 1.5 ohms of the coil. That got you 3.5 amps or so at start up. As the voltage increased when the engine was running to about 7.5 volts, the resistor heated up, adding more resistance in the circuit. 1.0 ohms hot, plus 1.5 ohms of the coil got you down to 3 amps or so to keep from melting the coil. The same rule (actually, Ohm"s Law) applies to a 12v circuit. I= E/R. Current equals voltage divided by resistance.
Until you find out the resistance of the coil, you will be playing a very expensive guessing game regarding the value of the resistors you will need in the circuit to keep from melting the coil.
A digital multi-meter has two probes & a switch. Set the switch on resistance. Put one probe on the top of the coil & the other on the pigtail at the bottom. It will give you a reading in ohms. Lets just say it reads 3.0 ohms. Your OEM ballast resistor (which you must use) is about 1 ohm hot. A coil a 3.0 ohms, plus the ballast resistor at 1.0 ohm (hot) gives you 4.0 ohms resistance in the circuit. Your 12 volt alternator puts out 14.5 volts. You need to determine current (amps). 14.5 v divided by 4.0 ohms gets you 3.6 amps; close & probably ok. But, and this is the problem......what if the coil is only 2 ohms? Do the math. 14.5 volts divided by 3.0 ohms gets you 4.8 amps! Not good! And, if the coil is less than 2 ohms (and some are) it will fry quickly. So, to get it to 3.5 amps, you need another resistor in the circuit. Either that, or keep spare $30 coil around.
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