I'm trying to size the Neutral to Ground resistor (NGR) in the Wye of a 138 KV to 4160 V Delta-Wye Transformer. In order to find the proper value I have estimated the charging current of the system by following the instructions in this page section “Estimating”. And later reading the article "Charging Current Data for Guesswork-Free Design of High-Resistance Grounded Systems" supplied as reference in that page. After my study I realized that in case a fault to ground occurs in any of the 4160 phases (2400 V phase to neutral) a lot of current will flow through ground (soil) back to the star point of the Wye and this will be limited by the NGR. Let's say I will allow 150 A (low resistance grounding) to flow. At 2400 V that is a 16 Ohm NGR.

Now reading the standard CSA M421-16 Use of Electricity in Mines they say "In resistance-grounded systems, it is standard practice to assume zero-impedance fault paths and consider only the neutral-grounding resistors to calculate prospective ground-fault currents."

By reading this, anyone may think that earth(soil) has no resistance to the pass of current. So I started to study that. First I selected my worst case scenario, a fault on my furthest point, which is 1.2 miles (1931 m) from the transformer. And then I tried to find out earth resistivity. Turns out it varies depending on many things (mineral composition of the rocks, humidity, salinity, temperature, the mood of the guy making the measurements, etc ).

For instance a couple of the articles explaining soil resistivity can be found here and here. Let's assume for an example that the soil is composed of sand with a resistivity of at least 100 Ohm per meter. The soil equivalent resistance would be 1931*100 = 193100 ~= 200 kOhm. That is not something to ignore compared to a 16 Ohm NGR!!! Earth will stop charging current much more that the NGR.

  1. Why isn't this considered?

Maybe my assumption about earth(soil) is wrong, I kind of see that with distance earth is a series conductor but also many parallel conductors, so maybe the 2 effects will compensate and resistance value will tend to a constant value (a friend of mine named this Remote Earth model)

  1. Is there a mathematical way to model my remote earth and to know in practice how much resistance will my soil have at such a distance? If I assume my soil is sea water that is still 1 Ohm/meter. I guess that it may affect my decision of which NGR to select.

Update: It is always easier (and the standard does this) to skip such a complicated measurement (the equivalent resistance of the soil). Even more when its value changes depending on so many factors and rely only on the NGR for limiting the current. Anyway this is the worst case and you can never go wrong if you are protected against the worst case. But, am I wrong by thinking that without an NGR when a ground fault occurs, thousands of amps will flow through ground? If this is the case then the soil itself must not be of such high resistance value. Has anyone seen such thing as a study saying how many ohms a mile of dirt is equivalent to? I would die to see such thing.

Update 2: checking this document it says :

The ground electrode is surrounded by earth which conceptually is made up of concentric shells all having the same thickness. Those shells closest to the ground electrode have the smallest amount of area resulting in the greatest degree of resistance. Each subsequent shell incorporates a greater area resulting in lower resistance. This finally reaches a point where the additional shells offer little resistance to the ground surrounding the ground electrode.

They don't quote any reference for claiming this. Does anybody know where they got this from and if I can make this same assumption for very long distances ?

  • \$\begingroup\$ archive.epa.gov/esd/archive-geophysics/web/html/… \$\endgroup\$ Commented Oct 19, 2018 at 15:20
  • \$\begingroup\$ You might go back to your assumptions. When they say "assume zero-impedance fault paths", are they referring to the worst case scenario? If zero impedance faults are the worst case and any other fault is less taxing, it makes sense to run your analyses on the worst case. \$\endgroup\$
    – Selvek
    Commented Oct 19, 2018 at 19:47
  • \$\begingroup\$ @Selvek Of course you are right, if you are protected against the worst case scenario then nothing can go wrong (I updated my question to reflect this). What I'm wondering here is how unnecessary expensive it is to make that assumption. An NGR is a expensive device. If it must handle more current wouldn't it cost more? Why would I have to buy an NGR to handle a current that soil (rough estimate equivalent) will never allow to flow? \$\endgroup\$
    – VMMF
    Commented Oct 21, 2018 at 3:58
  • \$\begingroup\$ Quick note; HV feeders usually have metallic earth conductors of some kind, whether they be earthwires on overhead lines, screens of cable cores, or overall metallic armouring of cables. Some or most of your earth return current will flow on the metallic conductors. \$\endgroup\$ Commented Jun 12, 2019 at 13:49

1 Answer 1


What you have is a resistivity field. Depending on how complicated it it, and what your boundary assumptions need to be, there may be closed-form solutions, or you might need finite element analysis techniques to get a grip on your situation.

It's been many moons since I've done math like this (propagation of signals in nervous system tissue), so its a bit over my head these days. https://archive.epa.gov/esd/archive-geophysics/web/html/resistivity_methods.html might serve as a good intro, and resistivity field would be a good search term.

  • \$\begingroup\$ Is it incorrect to multiply resistivity x distance to get a rough estimate? I can't perform any direct measurements as suggested in the link \$\endgroup\$
    – VMMF
    Commented Oct 19, 2018 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.