[joeypoconos]

Bill - The cut-out insulators are in place as you describe.

I hate to be a pest and continue to ask the same type of questions over and over again, but I still need to understand how the cut-out operates internally. I’m sure the terminology I’m using to describe my plight leaves much to be desired.

In any event, here I go again;
The cover on my cut-out contains the FORD script and is embossed “BATT” above the post where the wires that lead to the lights, junction box, etc. are attached.

The opposite post is embossed “ARM”.

I removed the cover from the cut-out and inspected the guts and found points very similar to the points in the distributor. The BATT terminal is a simple metal device that serves as the STATIONARY side of the points. This post gets its electricity from the battery via one of the wires connected to its external post. Unless you have a fuse mounted on your starter, or somewhere else in the system, with the fuse installed this post will be HOT. If you have an un-fused system, the post is always HOT.


The ARM side is also a metal plate. This plate is soldered to a fairly thick coiled copper wire. The opposite end of the coiled copper wire is soldered to the SPRUNG side of the points.

Based on the way this thing is put together it appears when the points are OPEN there is no electricity flowing from the BATT side to the ARM side. That means the battery is the source of power to the electrical system. I believe the battery sends out power in one direction (battery outward) which causes it to discharge. When the voltage from the battery drops below the output voltage produced by the generator it needs to be charged. This condition causes the cut-out to “CLOSE” the points allowing current to flow from the BATT terminal to the ARM terminal and into the generator where through some type of magic the generator takes over as the source of power to the electrical system. As the power generated by the generator is bi-directional it serves to charge the battery and provide the juice to power everything. When the battery is sufficiently charged, or the car is turned off, the points open breaking the circuit and allowing the battery to take over once again.

Now, if what I am assuming (and we all know what that means) is correct, with the car running and the points closed there should be voltage at both the BATT and ARM side of the cut-out.