Electrical resistance of Earth

  • Thread starter jaykay99
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hey guys;

In the lep collider of geneva, there vagabond currents! I want to assess the effect of the vagabond currents on the beam energy! Therefore i only have one question:
The current starts at the rail and then he goes through the earth to the collider. I only want to know how big the electrical resistance of this part of earth is, when the distance between collider and rail is about 1 km. I also know the resistivity of earth is [tex]100\Omega m [/tex]

I know: [tex]R=\rho \cdot \frac{l}{A}[/tex] but i dont know A!
What is it?
 
5,598
39
Don't know the answer.....can't imagine anyone would know it without an actual measurement....seems like it is highly dependent on soil type, perhaps also temperature and moisture content.....an irregular path of low resistance might also result in a much lower resistivity than an average figure.....as might the ion characteristics of the local soil

A basic question seems over what area a large electric current will spread...

Why do you need to asses earth resistance or current or power to understand beam energy?? Surely there are power meters for that....

also, what's a "vagabond" current....I understand it's a sailing dinghy in Australia but never heard of it in connection with electricity...
 
Literally, a "vagabond" is a "drifter", a person who "wanders", so I suppose a vagabond current is some kind of "drift current", but I don't enough about the lep configuration to answer the question.
 
the problem is that is want to simulate the earth resitance with a wire.
How big could the cross-sectional area of the wire. Can i say e. g. it is a sqare: A=(100m)² .

Is that logical?

pls help me
 
5,598
39
well you did not answer my questions, :
In the lep collider of geneva, there vagabond currents! I want to assess the effect of the vagabond currents on the beam energy!
so let me put it this way:
There is likely NO effect of any drift currents in the earth on beam energy unless you mean that there is some power loss due to the resistivity in the earth....

By what logic would you think otherwise??? Seems to me it's like asking the same question relative to a power transmission line using earth as the return ground.....
 
Normally you wouldn’t consider calculating earth. Earth is a quit good conductor and is considered as a close to ideal resistance. When doing calculations on safety systems, such as HFI and PHFI you only use the transition resistance, which is locally defined by the soil, moist, type of cable and so on. Here where I an it is app 30-40ohm's.
If you were to take a cable wit ex. 200V and put it in the ground, you would see, in a distance of 10cm, that you maybe would have 180V, at 20cm 170V, at 50cm 165V, at 2m 162, at 10m 161v, at 20m 161V and you could go to the other side of the earth and get a reading of maybe 160V. This is because the electricity spreads out through all of mother earth. You can consider “A” to be inf. or see erth as a realy big cable with a diameter of equator.

Dr. "Watts"
You are right. A wagabond current is a current going its own way. It could be a bad insulatec cable on a wet grass lawn!
 
Thank you for your answers!
@ Great-dane: i can follow your considerations.
let us make it concrete:

we have an electric potential difference of 5 V. The current emerges from a point and migrates 1 km to the LEP(Aluminium ring with a radius of 4km) . 1km is the distance from the point to the LEP-ring. The resistivity of earth is [tex] \rho= 100\Omega m[/tex].

What is the overall resistance of the part of earth and what current is flowing?

Thank you for helping me.

Jaykay99

p.s. does it make sense to use the http://en.wikipedia.org/wiki/Gauss%27s_law" [Broken] for the calculations?
 
Last edited by a moderator:
I made a drawing of my problem. I made it as pdf as an attachment.
I want to know what current could flow.
Can anyone help me?
 

Attachments

Hay again
I don’t think Gauses law can be used here, I never liked him so normally I don’t use him ;). I am sorry but I don’t think I can help you. If what I said earlier is true (the earth being a big cable with an A=inf.) ohms law suggests that there wouldn't be no loss! Are you sure that the 100Ohm aren’t meant as a single resistance between earth and your electrode?
 
Thx for answers

I am sure that it are 100Ohmm !

What law could i use instead of Gauses law?
Help pls
 
noone who can help me?
 
Thx for you answer but this is not that i want to know

I simply want to know how can i calculate the resistance of a medium(rho=100 ohm meter) between to 2 points which are about 1 km away.
 

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