101130-2014 EST
Continuation to Zef:
Consider an AC circuit. From the main panel 3 wires travel to the end of the circuit. One is the hot, another is neutral, and the third is the EGC. There is no load connected anywhere on this circuit. The neutral wire is disconnected at the main panel from the neutral bus. Now the neutral is floating. There is nearly infinite resistance from the neutral to anything else. However, there is capacitance from the neutral wire to the other wire and maybe a conduit.
With the hot wire energized there will be some sort of capacitive divider network that puts a voltage on the floating neutral. Unloaded this voltage will be some constant less than 1 times the voltage on the hot wire. If you use a high impedance voltmeter to measure the voltage from the floating neutral relative to the neutral bus in the main panel, then you will read some voltage less than the hot voltage. May be very close or 10%. If you put a 25 W incandescent bulb between the floating neutral wire and the neutral bus this voltage will likely drop to virtually 0 V.
If we used the effective capacitance from the hot wire to the floating neutral as 20 pfd per foot, then to get 5 MA of current at 120 V the capacitive reactance would need to be 24,000 ohms or 0.1 mfd at 60 Hz. This would be a 5000 ft branch circuit. In the past there was a very useful paper slide rule made by Shure Brothers for inductive and capacitive reactance, and resonance calculations. I do not know if it is still available.
Most residential branch circuits won't have more than maybe 200 ft of Romex. This would be in the ballpark of 4000 pfd and that has a reactance of 0.65 megohms at 60 Hz. Max current at 120 V is then about 0.2 MA.
.
