Whenever you make a voltage measurement, one must consider the effect of creating a voltage divider. What is the resistance on the other side of the switch? For instance, if the switch measures 1E10 ohms and it is connected to a 100 mega ohm (1E8) resistor and 10,000 volts is applied to the other end of the switch away from the resistor, and series circuit is set up such that some of the voltage will be dropped across the switch and some will be dropped across the 100 mega ohm resistor. One has a series circuit set up with basically two resistors in series. One resistor is the switch at 1E10 ohms and the other the load resistor 1E8 ohms. When applying 10,000 volts to this circuit approximately 1 µA of current will flow to the open switch and through the load resistor. Simply using ohms law the 1 µA will generate 100 volts across across the load resistor. Now if the insulation resistance across the switch is 1E11 ohms the voltage across the resistor would only be 10 volts. However, if the insulation resistance across the reed switch is 1E9 ohms, then the voltage across the load would be up to 1000 volts.I hope this all makes sense to you. Obviously, the insulation resistance across the reed switch is very important as is load resistance. Hope this explains better what you and the customer are seeing.
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