Part 2:
Without the engine running, your ignition circuit consists of the battery,
the wiring to the ignition switch, the switch itself, the wiring between the
switch and the coil/Pertronix, the coil and Pertronix units themselves, the
engine block, the block ground strap, the chassis, and the battery ground
strap. Any of the wires, switches, and wire junctions can easily act as
resistances, and that is most likely the problem. A resistance causes a
voltage drop proportional to the current and the resistance: V = I x R,
where I is the current, and R the resistance. If you have just the 0.6 ohm
coil connected across the battery, then from 12 volts you would get 20 amps
of DC current: 12 = 20 x 0.6.
When you turn on the ignition, the Pertronix quickly figures out that the
engine isn't turning over, so it doesn't turn on its coil charging switch
(transistor), and no coil current flows. Because there is no current flow,
there is no actual "circuit", and you won't measure any voltage drop. You
can measure the voltage at either coil terminal, and you'll just see the
battery voltage: 12 volts.
Now when you ground the (-) terminal on the coil, you complete the circuit,
and current flows from the battery, through all the wire and ignition switch
(remember, these could be like resistances if the contacts aren't clean
anywhere), to the coil (+), through the coil, then from the coil (-) to the
chassis or engine block, and back to the battery through its ground strap.
Steve, you're getting 7 volts when you do this, so that tells me that if we
assume 0.6 ohms coil resistance, we have about 11 amps coil current, and
somewhere in the remainder of the circuit we're getting 5 volts drop. 5
volts divided by 11 amps is 0.45 ohms - the resistance in the remainder of
your coil circuit (i.e. all the wiring, the chassis, the ignition switch,
and the battery).
Still more coming... Stay tuned.
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