The cutout relay is basically a solenoid type switch which was used on the early tractors before voltage regulators. It's function is to disconnect the generator output to the battery when the engine is not running to prevent battery discharge. In the absence of a voltage regulator, the method used to control the generator output was to restrict the ground leg of the field coil by means of the field resistor. Without lights, the generator "F" terminal is either wired directly to ground (under the mounting screw of the cutout relay) which provides the high output of about 10 amps, or it is wired to the "F" terminal of the cutout relay which in turn runs through a 2.8 ohm resistor on it's way to ground. This is the low output, around 4 amps. With lights, they simply moved the field resistor to the switch where the operator can select the output without moving any wires. The resistor has the same rating whether it is located in the cutout relay or the light switch but the physical size differs between the two. I have one from the relay which won't fit the switch because it is slightly too long. Sorry for the long answer. Geoff
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Today's Featured Article - More Memories of a Field Service Engineer - by Brian Dye. Introduction: The author worked for one of the largest Ford tractor dealerships in England as a trouble-shooter on tractor engines, hydraulics and combine harvesters from 1963 to 1972. In 1972 he became a dealer manager and finally left the dealer net in 1975 to run his own electronic design and manufacturing company. [Editor] We live right in the middle of Norfolk. The dealership that I worked for sold Fordson Tractors from the 1920's and some of t
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