i was under the impression that the resistor was on the negative wire to cut the power before it goes to the distributor but after the power has gone through the coil so the coil gets the full 12V but the distributor does not, but i have seen some that say you cut the power with the resistor by putting it on the positive side from the battery before it goes to the coil...which is right? or will either accomplish the same thing? it seems to me youd want the full battery power to the coil but use the resistance between the coil adn distributor to cut it down.
The resistor goes between the 12 volt ignition source and the + terminal of the coil. The resistor cuts the voltage seen at the coil (and the points) down to about 8 volts. The function of the ballast resistor is to extend the life of the points. However, when starting the car, you want the full 12 volts to go to the coil. That's why the starter solenoid has the "R" terminal on it. A wire runs from the "R" terminal to the positive directly to the + terminal on the coil so that the coil gets a full 12 volts while cranking, bypassing the ballast resistor.
Your 71 does not have a separate ballast resistor - the wire going to the "+" side of the coil is a resistor wire, and has the correct resistance to drop the voltage to the coil. The reason for the dropped voltage is 2-fold:
1. To extend the life of the points. The points will not last long with a full 12 volts supplied continuously
2. To prevent the coil from burning up if you leave the key in the "on" position when the points are closed. In this configuration, you have a direct current path to ground from the ignitiion switch and through the coil. If there were no additional resistance in the circuit, you'd fry the coil in minutes.
As long as nobody has cut the "+" coil wire out of your wire harness, you do not need to add a ballast resistor.
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