I have a dumb question for y'all (hopefully dumb, and not: "oops, I
discovered a P.O. fubar and didn't know it!")..
What does the ballast resistor DO?
I know this has been brought up before, but reading the wiring
diagram, checking voltage at the coil, I'm not getting it. Would
somebody mind trying again?
My roadster service manual doesn't show the resistor in the wiring
diagram! So, looking at a 510 diagram I have, I see a resistor between
the ignition switch wire "IGN" and coil. Ignition sw "R" goes straight
to coil. MY understanding was that when ignition was on START (that's
"R"?) we went straight to coil, and when we're on RUN (that's "IGN"?),
it goes thru the ballast resistor, such that the *resistor* reduces
the voltage going to the coil (somebody gave me the idea it was 9V).
OOKKKAAaaaayyyy..
So I start my car, and voltmeter check the "+" on the coil -- it
reads 12V! Huh? Well, whaddyer need a resistor for then? unless it's
honked up wiring by a P.O.? I've driven it this way for 40,000 miles,
so I'm guessing it's _right_, and I just don't understand why..
I also don't get why there's no juice (0V at coil "+") when ignition
is "ON", when car isn't running.
Just out of curiosity, why does coil need a "-"? Why doesn't it just
ground to chassis?
Thanks,
Dan Neff '69 2000
Colorado Springs, CO
p.s. if somebody gets into this stuff and yer on a roll, how 'bout
explaining to me how the coil works? I understand how a 'coil'
creates an electromagnetic field, but don't know my electronics well
enough to understand how that automagically results in 30,000V from
12V! Let alone how a primary and secondary coil make this happen.
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