[Lynn Eades]

I dont know where you are at with this. I cant answer your original question, BUT!!!
One side of the cut out has two wires, the other has one.
One side has a light wire that comes from the switchable side of the ign. sw., goes through the ign. warning light and away to the cut out, and connects on the same side as the heavy wire coming out of the dynamo. The single wire on the otherside of the cutout is wired into the permanent live feed from the solenoid.
I'm pretty hazy on this bit about how it works, but here is something to think about.
The small current flowing through the ign. warning light is the power to excite the field coils to start the generator. when the voltage rises up to say 6.8 volts, the cut out will start to vibrate open and closed, and thus regulating the out put.
If you take the top off of the cut out you should be able to work out how it works.
Remember with these fine British systems, it is the opposing voltage from the other end that puts out the ign. light.
I hope this helps