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In order for the ballast resistor to drop 6 volts (V = I x R) there must be current flowing through it ya know. That means there needs to be a good and wired correct n working set of points and that they be CLOSED and that the coils primary is also good and continuous and conductive. That entire ignition circuit (ballast plus coil plus closed points) from the switch needs to be wired right and working in order that the ballast drops 6 volts.... If the points are open,,,,, ,,,,or not wired correct,,,,, ,,or the distributor isnt grounded there never would be current flow through the ballast resistor in which case there would be 12 volts in and out of the ballast. NO VOLTAGE DROP ACROSS BALLAST until its conducting coil current i.e. points must be wired correct n closed Sure all is wired correct?? The series ignition circuit would be out the ignition switches IGN terminal (hot 12 volts when ON) ,,,,, to and through the ballast resistor,,,,, ,,,,to and through the coil,,,,, ,,,,,to the distributor,,,,, ,, then via its closed points to ground. On the high input (NOT to distributor) coil terminal or else the low output (NOT from switch) ballast terminal theres another wire up from the starter switch thats hot ONLY while shes cranking ya know. IFFFFF F shes cranking then you get full unballasted battery voltage on the coil and NOT any ballast voltage drop cuz its effectively by passed while the starter is engaged. A volt meter on the distributors high input terminal should read around 6 volts when switch is on assuming a good working and closed set of points and a good n continuous ballast and good coil but it would read higher while the starter is engaged cuz the ballast is then by passed. Better check for proper wiring and the points are good n working n wired correct n closed and the ballast and coil are both okay John T
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