Why do you think it is 2Volts? Did you measure it as such? Or did mentally subtract 6 from 8 to get 2V? If it is a "factory resistor" it should be the frontmount distributor (and not a sidemount engine swapped in from an 8N, as I have), and the resistor should be "an infamous ballast resistor" mounted to the phenolic block on the back of the dash. If you are trying to add some sort of dropping resistor to "compensate" for the 8V battery and charging system, you need to carefully select an appropriate value from a resistor factory. There is not one available to accomplish this task from the Ford factory. If it is a front mount, and we are talking about the "infamous ballast resistor", and you are accurately measuring this voltage "soon" after turning on the key, and not after half an hour of fighting with the tractor, and not while cranking the engine, then perhaps the resistor is bad or its connections are dirty. The voltage at the coil should be something like 3 volts, maybe 4 volts with the 8V battery. But 6 volts is too much and will fry the frontmount coil. Two volts sounds a little weak, but the tractor should still start if the everything else is in order. It probably will drop from 4 volts to 3 volts or less as you crank the engine, due to voltage drop across the output resistance of the battery. I'm no better at reading minds than Dell, and maybe I shouldn't be making these guesses...
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