Why can current flow from power line to earth ground?

[DoobleD]

Take some high voltage power line. Place a conductor between the wire and Earth ground. Current goes to ground.

Why? I understand what is voltage, resistor, circuit etc. I do know that the wire is at a higher voltage than the ground. But isn't Earth also a kind of huge resistor? Thus preventing current to flow.

And when it flows into ground, where does it go? Does it go back to the power plant generator?

And finally: If current flows, it means the circuit must be closed. What if the power plant generator is isolated from ground? How can there any closed circuit including the ground in such a case?

This has probably a simple answer, but I can't find it. :)

 

[mfb]

The resistance is not that huge, but the voltage is of the order of 100 kV. This leads to a significant current flow.
Simple example: a resistance of 10 kOhm leads to a current of 10 A, which is a power flow of 1 MW, or 10 MW if you go to 300 kV. Enough to evaporate a lot of soil.

The Earth has a huge capacitance - even if you could perfectly isolate it from the whole power grid, you would get significant current flow with AC current.

 

[DoobleD]

Thank you for answering!

Ok I see for the resistance of earth. Indeed with such high voltages the current is high anyway! In addition I've just looked up resistance of Earth (I could have done that earlier...) and it is indeed suprisingly small. I've read 5 ohms for instance.

However I didn't get the part with the capacitance of Earth affecting the current even with isolated generator. Can you explain this to me?

Also, for the second part of my question, concerning closed circuit, I think I actually said something totally wrong : I wrote that if current flows, then the circuit must be closed. Actually, current can flow without that, it just needs voltage difference...Right? But then, a bird touching a wire doesn't get current, while if Earth "touches" the wire, it does...I'm even more confused now. :D

 

[NascentOxygen]

The Earth has a huge capacitance - even if you could perfectly isolate it from the whole power grid

Just 710 µF for an Earth-size conductive sphere in isolation. https://en.m.wikipedia.org/wiki/Capacitance

 

[DoobleD]

New enlightenment! I'll try to organize a bit this post to make things clearer since I had many questions.

1. About a non grounded power plant

http://www.allaboutcircuits.com/textbook/direct-current/chpt-3/shock-current-path/ (found here BTW) explains that, in theory, if the whole circuit (power plant included) is not grounded, then current would not flow to ground when, say, a person touches a wire of the circuit. But in practice, "ground faults" can appear on the circuit (for instance, a tree touching a wire), making unpredictable if the circuit is "safe" (you could touch it without closing the circuit, thus without being shocked) or not, and where it is safe and where it is not.

The article explains that power plants are actually grounded, so we can always be sure what part of the circuit is safe, and what part is not. It also seems to suggests from the pictures that if current flows to ground, it does go back to the power plant.

Ok, that would answer my first questions. Except maybe for the current going back to the power plant, still not 100% sure about that...

2. About current flowing without closed circuit

In my last post, I wasn't sure anymore that a circuit actually needs to be closed for the current to flow. Let's examine the confusion with two cases: a discharging capacitor, and a bird landing on a power line.

Case A: a capacitor

Take a charged parallel plate capacitor. There is a potential difference between each plate. Now connect the plates: current flows.

Case B: a bird around a power line

Take a bird at 0 V (with respect to ground) flying above a power line at 100 kV (still with respect to ground). The bird now lands on the wire. The bird is NOT shocked (see the article mentioned above as well). Current does not flow?

The problem

Well, what's the difference? In both cases, there is an initial potential difference, and in both cases there is no closed loop. Why is the outcome different in those two cases?

I would suggest that in the case of the bird, current actually flows from the wire to the bird, BUT it is very brief. Not long enough for the bird to be shocked, just long enough to make the bird 100 kV.

Seems to be a weird answer, can't be right. :D What is the truth?

EDIT: If we model the bird has being a resistor in parallel on the circuit (each "foot" being one end of the resistor), then current actually flows continuously trough the bird, but only in a very small amount if the bird has a high resistance. Thus the bird is neither at 0 V or 100 kV. But there is a small potential difference between each of the bird "foot". Is this approach correct?

 

[CWatters]

Take a bird at 0 V (with respect to ground) flying above a power line at 100 kV (still with respect to ground). The bird now lands on the wire. The bird is NOT shocked (see the article mentioned above as well). Current does not flow?

When the bird lands it's voltage (wrt ground) changes from 0 to 100kV. Some current will flow because the bird has capacitance wrt ground. Don't forget most high voltage lines are AC not DC. Fortunately for the bird the capacitance is small and at 50 of 60 Hz the impedance is quite high.

Elsewhere on the internet I've seen a comment that the current can be enough to make birds feel uncomfortable so they are rarely seen on 132kV lines and never on 400KV lines. I don't know if that's true.

 

[CWatters]

EDIT: If we model the bird has being a resistor in parallel on the circuit (each "foot" being one end of the resistor), then current actually flows continuously trough the bird.

No that's wrong. No current flows because both feet would be at the same voltage¹. The bird would only be in trouble if he stepped off the line one foot at a time onto a wet tree branch or metal roof (not that you should build a metal roof that close).

Note 1: The voltage on an AC power line can be considered the same at every point. At least for distances as short as those between the feet of a bird.

 

[mfb]

Just 710 µF for an Earth-size conductive sphere in isolation. https://en.m.wikipedia.org/wiki/Capacitance

710 µF, charged with 100 kV and discharged again 100 times per second (120 in the US), gives several kA of current.
Anyway, a perfect isolation from the power grid is unrealistic.

Note 1: The voltage on an AC power line can be considered the same at every point. At least for distances as short as those between the feet of a bird.

I think that was DoobleD's point. The resistance of the wire between the bird's feet is tiny compared to the resistance of the bird, so DC-like current flow through the bird is tiny compared to the current flow in the wire.

 

[DoobleD]

Thank you guys for your answers.

I think I am missing something when you both talk about capacitance and AC here...All I know regarding AC current with capacitor in a circuit, is that a capacitive reactance is produced, delaying the current with respect to voltage. Does this has anything to do with what you discuss? What is this thing of capacitance with the Earth (and the bird?) you are referring?

It is still not clear to me what happens to the bird.

 

[William White]

birds...are rarely seen on 132kV lines and never on 400KV lines. I don't know if that's true.

absolutely true - the large voltage gradient around the conductor

 

[William White]

Take some high voltage power line. Place a conductor between the wire and Earth ground. Current goes to ground.

Why? I understand what is voltage, resistor, circuit etc. I do know that the wire is at a higher voltage than the ground. But isn't Earth also a kind of huge resistor? Thus preventing current to flow.

the resistance is often sufficiently low to allow significant currents to flow in the earth. Depends on the type of soil, moisture etc. Power systems protection schemes are used to detect this current and disconnect the circuit.

And when it flows into ground, where does it go? Does it go back to the power plant generator?

It goes back to the point where the circuit is (deliberately) earthed. There is an Earth point at every voltage level on the system.

And finally: If current flows, it means the circuit must be closed. What if the power plant generator is isolated from ground? How can there any closed circuit including the ground in such a case?

If the electrial system is completely isolated from the earth, a single earth-fault will not see any current flowing through Earth because there is no circuit. A second Earth fault will. This is one of the reasons power systems ARE earthed. You want ANY fault to be detected and the circuit disconnected. If it was not earthed, a fault would go undetected.

Some circuits are isolated from Earth for safety reasons - obvious example being SELV circuits. Other examples include circuits in explosive atmospheres (some mines).

 

[William White]

No that's wrong. No current flows because both feet would be at the same voltage¹. The bird would only be in trouble if he stepped off the line one foot at a time onto a wet tree branch or metal roof (not that you should build a metal roof that close)..

its a particular problem for birds with large wing spans, such as raptors - many get killed by power lines because their span is big enough to touch one conductor and earth. or two conductors.

 

[DoobleD]

Thanks William, this confirms the answers to my initial questions! :) Any idea regarding what is happening to the bird?

 

[William White]

Thanks William, this confirms the answers to my initial questions! :) Any idea regarding what is happening to the bird?

Simply (and one can over complicate things) current flows between two points when there is a potential difference between those two points.

If the bird's feet are at the same potential, it does not get zapped.

If the bird opens its wings, and touches the earthed-steelwork of the tower carrying the conductor, then there is a potential difference between its feet and its wings, and it gets zapped.They don't like EHV conductors because the air immediately around the conductor is charged. This gives a very steep voltage gradient around the conductor. As the bird gets closer, it starts to feel this (just like you get a tingle if you lick a 9V battery!) and flys away.

 

[DoobleD]

So the bird cannot be modeled as a resistor in parallel when it has only the two foot on the same wire? I find it hard to believe. No current at all through it, even a tiny? Crappy drawing below. :D (Haven't draw transformers etc for simplicity)

 

[Nugatory]

So the bird cannot be modeled as a resistor in parallel when it has only the two foot on the same wire?

Sure it can - it's just that the voltage difference between the wire under the left foot and the wire under the right foot is fairly small, so the voltage across the bird/resistor is also small.

 

[DoobleD]

Sure it can - it's just that the voltage difference between the wire under the left foot and the wire under the right foot is fairly small, so the voltage across the bird/resistor is also small.

So there is a tiny current? Even too small for the bird to actually notice.

 

[William White]

So the bird cannot be modeled as a resistor in parallel when it has only the two foot on the same wire? I find it hard to believe. No current at all through it, even a tiny? Crappy drawing below. :D (Haven't draw transformers etc for simplicity)

the bird is connected at two points...what is the potential difference between these two points?

just say (for arguments sake) the volt drop along the conductor is 0.5mV per metre and the distance between the birds feet is 2cm

the potential difference between the birds feet is then 10 uV.

now say the birds resistance is 100 ohms (I don't know what it is, it could be thousands)

so the current flowing is i = v/r = 100 nA. This is about as close to zero as you can get

 

[DoobleD]

All right I think I get it. :)

Also, even if the wire would be a perfect conductor, without any potential drop, there would still be a (very) small potential drop between each foot I think. Since the other parts of the bird body act as a (huge) resistor.

 

[William White]

for simplicity, along an electrical conductor, the potential difference between two nearby points is zero.There are regulations which minimise the volt drop. In the UK a 230V supply can vary from 253V to 216V : a drop of 37V along the entire conductor (for those of us that live close by to the transfomer, we blow a LOT of light bulbs!. My measured voltage is 250V!)

 

[William White]

All I think I get it. :)

without any potential drop, there would still be a (very) small potential drop

re-read that.

If there was NO volt drop, there is no volt drop!

In theory, that is impossible (except in a superconductor); but in practice its best to think that the potential between two points on a conductor that are near each other is zero.

 

[DoobleD]

Oh sh*t, you are right. I think I just completely got something wrong: the bird is not actually in parallel on the circuit. If I replace the bird by a voltmeter and if the wire is a perfect conductor, the voltmeter does read 0 V because no current is going through it.

Is this finally correct?

 

[William White]

well it IS in parallel, just like the voltmeter is in parallel, but it is in parallel over a voltage of 0V (or 10 microvolts as we quickly worked out earlier)

 

[DoobleD]

Arrr, and I thought I was starting to understand EM and circuits! Let's go back to the basics.

In the picture below, are those statements correct (assuming perfect conducting wires):

1 - R1 is in parallel with the wire below it
2 - But R1 and R2 are not in parallel with each others (they are not both linked directly to the + and - sides of the battery, only R2 is)
3 - Anyway no current flows to R1 (makes sense, why would current go to an additional resistor when it can avoid it)

 

[William White]

1) yes
2) R1 and the wire it is connected to are in SERIES with R2 (you can replace R1 and the wire it is connected to with a single resistor of LOWER resistance).
3) Yes - you have short circuited it

 

[DoobleD]

you can replace R1 and the wire it is connected to with a single resistor of LOWER resistance

Assuming the wires are perfect conductors, to replace R1 and the wire below it while still having the same situation, we would have to replace them by a single perfect conductor (or equivalently, simply remove R1)?

 

[William White]

if you wanted to replace R1 and have the same situation, you would replace the wire with a wire of a bigger x-section

Two wires of 1mm^2 each in parallel carries the same current as one wire of 2mm^2

Putting wires/conductors in parallel allows more current to flow.

 

[DoobleD]

Hm, not sure I understand that last point. Since the wires are perfect conductors, why do their surface area matters?

 

[William White]

its not the surface area (well IT IS in some instances, by forget about that) it is the X-section.

There is no such thing as a perfect conductor (except superconductors).

So a conductor is typically rated by how much current it can carry (before it starts to melt!)Lets make up some numbers (to make things simple, rather than using real data)

Say you have a wire of x-section 1mm^2 and it can carry 1 amp
If you have two wires of x-section 1mm^2 then each wire can carry 1amp, so the total carried by the two wires is 2 amps.

Two wires of 1mm^2 is the same as one wire of 2mm^2

So you could replace the two 1mm^2 wires with one 2mm^2 wire and be able to carry 2 amps

Likewise, if you had a ten x 10mm^2 cables; that can carry the same as ONE 100mm^2 cable(now this is not strictly true, because you have factors such as heat dissapation, so lots of cables with air around them allowing them to cool will be better than one big cable, but don't worry about that for the time being)

 

[DoobleD]

Got it, I thought we were still in the idealized case. Thank you very much for your help William, much appreciated. Especially since my mistakes were huge basics!