I know similar question have been asked several times, but I didn't get exactly the answer I was looking for.

I actually want to ask something that perhaps can worded as a practical question, but I think it would not clearly represent the thing I wanted to ask, so I'm framing this as a hypothetical question.

Assume earth has been cut into two pieces somehow (and for some reason, gravitational force is on vacation so 2 parts are hanging around as shown and are not moving), the one half has 3 phase generator with neutral earthed and it's one line and neutral are supplying a single phase load provided with adequate earth fault protection on other side.

Now suppose, I've connected an ammeter between 'earth' in two halves, and earth fault takes place on load causing current to flow to the earth, will my ammeter show some reading (to an astronaut I've recruited)?


1 Answer 1


Yes, it will.

If we assume zero capacitance between the two half earths, then it will show the full fault current.

If we assume some capacitance (it might be amusing to calculate the capacitance of two parallel plates, half-earth size, 1m apart) then some of the AC current will flow through the capacitance, bypassing the ammeter (assuming the ammeter has a finite impedance).

Of course we don't need to suspend gravity and assume a Vogon with a large chainsaw to get isolated 'earth' regions. If you have a large mass of granite, or other impervious rock, separating the bit of soil you've driven an earthing conductor into from some other remote conductor, then you will be able to generate an appreciable voltage if you try to force a current from one to the other. Good earthing benefits from damp conditions down to the water table, dry rock is a lousy conductor.

  • \$\begingroup\$ Haha, Yeah I had thought that I should edit question mentioning that capacitance is negligible, I'm so glad I didn't. Had I done that, I'd never have calculated that capacitance for 1 m separation is around 1100F!!!! For 1km separation, I would have shunt resistance only of around 3.18 mOhm.. 😲😳😱 \$\endgroup\$
    – Deep
    Mar 24, 2019 at 7:49

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