Question for all you EEs out there.  I'm putting a new alternator on my
Velocity.  This one will be internally regulated (and have the appropriate
"crowbar" OV protection.  The unit I am installing has two terminals
arranged in a "T".  One is "field" and the other is "Indicator" (idiot
light).  I am blessed that the builder didn't have a 20- or 22- ga wire to
string from the "Alt"  (split Alt-Master) switch, so he used both conductors
of 2-wire shielded 24 ga for the 12 ft run from the switch to the
regulator.  Anyway, I am betting that one will be enough to excite the
regulator and I can use the other for an idiot light.

Thing is, I have to determine which is which and how the "I" lead works (is
it hot or grounded when there is no output).  I aim to wire things up with
jumpers and spin the alternator by hand, so to get the charging voltage at
B+ I will need a resistor between B+ and the line back to the battery.  That
done, I can connect stuff up and see if, when all ready to go, the "I" lead
is hot or grounded, and if it goes cold or breaks ground connection when I
spin up the unit.

My questions are:  Will this work (will I be able to easily twirl the pulley
and get charging voltage if I put the proper resistor in the circuit (so it
doesn't have to generate much current in order to build up charging
voltage)?  What would be an appropriate resistance (and size) of the
resistor I am placing in the circuit?

I know I could have all this done at an alternator rebuild shop, but there's
none convenient to me

Hope this will work.  Makes sense to me at this point .... Jim S.