# Earthing resistance [closed]

I don't understand a thing about ground resistance.

Let's say that the power plant servicing my house is placed at a distance of 100 meters from it. If we hypothesize a resistivity of 100 Ω*m, the overall ground resistance seen from my house should be:

Rg = 100 * 100 + (Cable resistance + earth stake resistance)

If the above formula is right, we get for Rg a value of (10000+20)Ω. The problem is that with a ground resistance this high the grounding isn't much helpful. With a resistance of 10200 Ω a human touching a grounded chassis will take all the current.

My question is: how can an installation with such a high ground resistance be able to protect the user?

• You forgot to ask a question. It appears that English is not your first language because some of the words are not quite correct. e.g. "central of current". Please add your location / language to your user profile and we'll try to improve the question. Oh, and ask a question. – Transistor Apr 8 '18 at 16:00
• And the question is...? Why your actual earth resistance is less than that you computed? I guess you have earth stakes around you house, no? If you are in a country with the TN sistem, the transformer will have the neutral earthed near it, and you will have ground earthed near your house. – Sredni Vashtar Apr 8 '18 at 16:00
• @Sredni Vashtar sorry, corrected. – A. Rossi Apr 8 '18 at 17:45
• There is still a problem with your question: you indicate a value of 100 ohm*m for soil resistivity (even though you wrote resistance before my edit, I have to say) and then use that value as if it was resistance per unit length (ohm/m or ohm per meter). What did you mean there? – Sredni Vashtar Apr 8 '18 at 23:00

## false assumptions

• wire resistance per meter is too high by 1k times practical. (look for mΩ/m)
• ground stake 20Ω may be too high if you are in a frequent lightning strike exposed area.

Your Goal is to meet IEEE advice of 5 Ohm earth bond resistance and your regional requirements. There are 4 test methods.

If 1 mm wire = AWG18 = 21 mΩ/m * 100m gives 2.1 Ω This requries ground rod to reach moist soil at some depth for all seasons.

The resistivity of the soil may vary from 1 to 1000 ohm-m depends on the nature of the soil. Fertilizer and organic salts affect moisture resistance greatly. Also wet clay if frozen rises in R.

For AWG, like half power is -3dB, AWG#-3 reduces resistance by 50% e.g. AWG18 to AWG15 or AWG33 to AWG30 and AWG18 = 1mm diameter (1.024mm)

Remember this. Someday you won't have internet avail. to look it up.

Also remember that typical wire is 1uH/m so this can also add voltage drop from transient current in 1us.

• Tony, I believe the OP meant 100 ohm * m to be the average soil resistivity (a value that is in the ballpark of the graph you show). He mixed up that with resistance per unit length. So, the question should be, how can I have low earth resistance between the stake at the power plant and the stake in my back yard? It seems to me you are assuming he meant copper wire resistance per meter - but in a TN system only live and neutral are carried by copper. Ground is left to stakes at plant and house. – Sredni Vashtar Apr 8 '18 at 22:56
• Why are you explaining what I already know electronics.stackexchange.com/questions/367295/… – Tony Stewart Sunnyskyguy EE75 Apr 9 '18 at 1:23
• The question was unclear before my editing, and it may still be unclear now, since there are inconsistencies in the unit and the terminology originally used by the OP. Your answer assumes ground at home is connected to ground at the plant via copper wire. I believe that not to be the case. That's the sole purpose of my comment. – Sredni Vashtar Apr 9 '18 at 1:41
• We differ, In Canada, breaker panel is connected to earthed underground water supply . inspectapedia.com/electric/0505ss.jpg – Tony Stewart Sunnyskyguy EE75 Apr 9 '18 at 2:21