Each electric circuit involves some elements in closed rings. Electrons via electric filed of electrical source(s) move in the circles. Generally, current and voltage are requested in electric circuits. One must select an earth to measure the potential of each node because potential is a relative quantity. So, it needs to compare with a reference in circuit where is called "Earth".

Potential of earth node is zero and there is not current into earth because current move through circles.

One has to distinguish between "ground" (GND) and "earth".

GND defines, as Mohammad Soroosh writes, a reference potential in electric circuits but all GND points are connected by metal (rails, wires, PCB traces etc.). The potential of GND of a device can be quite different from the potential of earth (e. g. the internal GND rails of several power supplies connected in series). Depending on the application, the ground connections can indeed carry high currents.

Earth means literally a connection to the soil. Three common applications are

- lightning rods (reaching some meters into the soil),

- protective earth (PE) connections in electrical installations (usually a connection to water pipes etc. large parts of which are buried underground), and

- single-wire earth return signaling and power grids in rural areas.

In the first case, if the rod is hit by lightning there flows indeed a high current into earth but only for a short duration. Since lightning is the balancing of charges there is at first glance no closed circuit. However, according to Maxwell, we have to take the changing E field ("displacement current") between the clouds and earth into account, and both currents together form a closed loop (in 3D space).

In the second case, if there is a neutral line (N) it is connected to earth as well inside the building ahead of the circuit breakers, and all metal cases are connected to PE. In case of a short circuit, e.g. live conductor (L) to metal case, a high current causes a circuit breaker to blow but the high current does not flow into earth. In case of a person touching a live connector and earth (e. g. by standing barefoot on wet ground) simultaneously, a relatively low current flows through the closed loop including earth (a few meter), and hopefully a residual current circuit breaker will be triggered.

The third case is the only one where large distances are covered, and, in the case of power grids, large currents are flowing. While it is more inexpensive than multiple wire systems, the power quality can be bad, and it does not work with very dry soil (deserts or stony areas). In contrast to wires, increasing the distance between two earthed rods does not necessarily increase the resistance between them because the effective cross-section of the current can be increased, too.

Since large bodies of pure metal near the surface of earth are rare, the current is carried nearly exclusively electrolytically.

Hi Joerge, Thank you very much. It was the third case that I was interested in. Electrolytically would appear to be the answer I was looking for. Am I correct is assuming that you are saying that the resistance will be quite high in these cross-country power cables, hence meaning that the current will be low, and that most of the power will be transferred capacitatively and hence losslessly?

Hi Frederick, afaik the current is usually limited to 7 A .. 15 A which yields a power roughly between 90 kW and 300 kW at voltages of 13 kV resp. 19 kV, provided cos phi is near 1. I think that the capacitance between power source and load can be neglected: The relative permittivity of soil is 3 to 10, depending on the water content. Without calculation, I feel that the earthing structure might "see" a capacitance (against the surrounding earth) of some 10 nF, which results in a reactance at 50 Hz much larger than the resistance.

Hi Joerg, Thank you. So we have a high resistance and a much higher reactance. We have perhaps 15A being driven by 19kV. Well we all know that we want the current to be low in order to minimize resistive losses. So how then is the bulk of the power being transferred? It must be through some kind of electrostatic wave in the wire and in the immediate vicinity of the wire.

Hi Frederick, the soil is a relatively poor inhomogeneous conductor but on long distances the current is usually distributed over a cross-section of more than 1 km2 (depth 1,5 km, width more than 0.5 km). Assuming a specific resistance between 102 Ohm*m (wet soil) and 103 Ohm*m (dry soil), a copper wire (1.7 * 10-8 Ohm*m) with a cross-section of 170 mm2 downto 17 mm2 had the same resistance. In this case, a current of 15 A causes a voltage drop between 1.5 V to 15 V per km which results in about 1% to 10% of the initial voltage on a length of 100 km.

More critical are the endpoints where the current density is much higher, of course. The slides pointed to by the following link give not much explanation but some realistic numbers:

http://www.stonepower.se/Images/SWER.pdf

The power is transferred as usual, power density S according to Poynting: S = E x H, i. e. by the combination of those parts of the electric and the magnetic field which are perpendicular to each other (and to the direction of the power flow). S is indeed highest in the vicinity of the wire (inside the wire, S decreases fast with increasing distance from the surface), and low but distributed over a large area in the region of the earth current.

I agree Joerg Fricke

Joerg, Thank you once again. So as well as the current through the Earth being low, the current density is very low. But if the return was a copper wire instead, the current would be much higher, so much more power would be lost to resistance. In this respect, the SWER has two benefits, one being avoiding the need for a return wire while the other is the obtaining of a natural enduring high resistance.

But as regards the rate of energy flow being div S where S = ExH my own researches have concluded that this can only apply to wireless EM radiation, while E.J applies to conduction current. I see the transmission line as involving neither of these, but rather a kind of capacitance wave in the immediate vicinity of the wire, but not in the wire and as yet without any agreed mathematical formulation. H would depend on the current, and since the current is low yet the power is high, this can't be right.

Joerg, On further thoughts, div(ExH) is simply an alternative expression for the power expression VI more familiar to schoolboys. The current I is directly proportional to the induced magnetic field H. The voltage V is directly proportional to the electric field E. The greater V is at the power source, the greater will be both the lossless power and the power lost to resistance. So in this respect, div(ExH) will indeed represent a proportionate measure of the lossless power, but that is not what it is actually measuring. It is actually measuring the power lost to resistance. The lossless power is a capacitative phenomenon which is more closely related to the expression 1/2CV2 . The expression div(ExH) is only ideally applicable to wireless EM radiation, because its derivation involves Maxwell's two curl equations, meaning that Faraday's law must be applying across two separate closed circuits. That is not the case when power is travelling along a transmission line. A transmission line is a single circuit. The E field comes from the power source and is not an induced E field as per Faraday's law.

Frederick, certainly one amazing (and very helpful) aspect of circuit theory is that one can arrive at correct solutions when treating the components as lumped elements, neglecting the length of wires, and neglecting any fields (outside the components) at all (provided the spatial dimensions of the circuit are roughly smaller than 1/20 of the shortest wavelength involved). And in this context, p(t) = v(t) * i(t) holds generally or P = V * I for DC.

Due to the relations between E(t) and v(t) resp. H(t) and i(t), one could obtain the same results by taking the fields outside the components and wires into account, but this way would be generally much more involved (thinking and calculating in 3D space instead of thinking about a circuit diagram flattened to 2D and calculating with no spatial dimensions at all).

In my opinion, div(E(t)xH(t)) always denotes a source of field energy (div(S(t)) > 0) or a drain of field energy (div(S(t)) < 0), independent of the amount of radiated or incident energy. The difference between "true power" (not restricted to resistive loads, could involve a transformation to/from mechanical or chemical power as well) and "reactive power" is that true power is based on energy flowing one way only while the direction of the flow of reactive power is changing in such a way that in a periodical process, the time integral over the power (= energy) delivered or received by a reactive component during one period is zero.

Sorry, I don't see why Maxwell's curl equations should not be applicable to a single circuit. I understand that a prerequisite of Maxwell's equations is that there is only one "kind" of E field and one "kind" of H field. (Otherwise, there would be 4 equations missing: Denoting the LHS of the known equations rot E1, rot H1, div E2, and div H2, what about div E1, div H1, rot E2, and rot H2? A vector field - lacking a homogeneous component - is completely described by its div and its curl component.) In other words, in the framework of Maxwell's equations one cannot distinguish between different E or H fields based on their sources or their accompanying phenomena. E. g. rot H is the same, regardless of whether it is accompanied by nonzero J or nonzero dE/dt or both. As a consequence, S = E x H is the same, regardless of whether E is divergent or rotational or both. Of course, everybody is free to doubt Maxwell's equations *) but then the first logical step would be to establish the need for a better theory by demonstrating a failure of Maxwell when dealing rigorously with a certain situation.

*) According to Feynman, in the distant future the Maxwell equations will be regarded as the most important accomplishment of the 19th century; but he also said - and I guess most of us agree - that doubt is necessary for scientific progress.

Joerg, I support Maxwell's equations totally and one of the curl equations is Faraday's law (time varying aspect). Faraday's law is about actions in a primary circuit affecting reactions in a secondary circuit. I suppose it could also be applied to an induced back EMF in a single circuit, but when it comes to combining Faraday's law with Ampère's Circuital law for the purpose of deriving the EM wave equation, we can't do this within the context of a single circuit because the E in each equation is different. In the Faraday curl equation, the E is the EMF induced by the changing B field while the E in Ampère's Circuital law exists in the current that causes the B field in the first place. The two E's are 180 degrees out of phase. And likewise in Poynting's theorem. I'm sure that ExH only applies to wireless power transfer between circuits. I would say that the power transfer mechanism in a transmission line is a capacitative effect which has never been correctly analyzed and that even though ExH serves to all practical intents and purposes, it denies the true deeper physical picture.

Anyway, your answer to original question about the nature of the current flow inside the Earth has confirmed my suspicions that the resistance is high, hence contributing to a low current and low resistive losses, and that the power is carried in the form of electrostatic pressure inside the wire and/or dielectric polarization outside the wire.

I'll ask another question about AC power transfer on a new thread.

Frederick, only today I discovered that your thoughts expressed in this thread are just the "tip of the iceberg", you being the author of a large collection of essays.

Just to understand your position: From your point of view, does an electrical engineer - applying the equations in his mainstream textbook - arrive always at the correct results (but with a background of wrong ideas and explanations), or do you know of a case where his results differ substantially from your results? ("Results" being numbers denoting measureable quantities like voltage, current, dissipated power.)

Joerg, I don't know of many cases in electrical engineering where the results differ substantially from those predicted by mainstream textbooks. There is however an issue about discharging a capacitor into an open ended transmission line where there is a discrete cut off. This can be discussed later. But as a theoretical physicist I think there are some big problems relating to applying the correct equations to the wrong situations. In particular the question mark hangs over the physical nature of the propagation of voltage along a transmission line.

Just to clear up my point about self induction within a single circuit, I agree with you that div S relates to sources and drains. We can apply it within a single circuit when the current is changing, such as in the moments after the power is switched on, but in reality it is adding nothing new to the input power. It is merely a backfire aspect of the input power relating to self inductance. In my mind div S would never be involved in power transmission along a transmission line.The only time it would be involved in the transmission of power would be when power is being transmitted between two circuits. The two curl equations then both apply to the secondary circuit with the magnetic field having originated in the primary circuit. And this in my mind is also the basis of wireless EM radiation.

To add my little bit:

I've never heard of power applications going without a proper return wire...

The only applications I am aware of that were regularly operated with a single conductor (return via "earth") were telephone applications - operated @ 60 V or 120 V "loop voltage", thus allowing for some voltage drop on the earth "wire".

There proper grounding was somewhat critical (as current density is max. at the grounding points - often lowered by the injection of salt water.

Conduction in the soil is mostly by ionic conduction - requiring 6moisture (water) plus some salt(s) that would represent some electrolyte.

After all, this application was limited to no more than 20 mA per communications line.

Hello Uli, I was of the same opinion, and thought that even telegraph lines with ground return were limited to developing nations or otherwise the times of "Wild West". But I learned that even today single wire power lines are in operation in Australia, New Zealand, Alaska, and elsewhere. (Just google "swer power line".)

Dr. Dreher, Joerg Fricke sent me this link about Single Wire Earth Return http://www.stonepower.se/Images/SWER.pdf

It's an accepted phenomenon, but I'm wondering if a current does actually travel through Earth for the long distances involved, bearing in mind that the power transfer mechanism is capacitative and hence an actual return current wouldn't be necessary at all. Is there any evidence of electrolytic effects, such as electroplating around the earth connectors? I'm speculating on the possibility that the situation at the return Earth connection point may in fact be identical to the situation that arises in a lightning conductor, and that the Earth itself is actually just a sink. Joerg mentioned that Maxwell's displacement current completes the circuit in the case of a bolt of lightning from a cloud to Earth. I wanted to discuss this matter further with him. Displacement current is my specialty and I can't see how displacement current is involved in lightning. With lightning, all I can see is a radial current sinking into the Earth. If there was a solenoidal return to the clouds, it would defeat the purpose of the discharge.

@ Joerg

I think we can consider these rural regions of AU, NZL etc. "developing areas" - no "reasonable" grid available. Or consider them "Wild West" - fits just too well.

I'd be really careful regarding the grounding points where dangerous voltage may occur without pre-warning.

@ Frederick

I'm sure you may watch electrocorrosion (do not expect electroplating) in the presence of suitable anions like Cl-, SO2-- etc. if the coupling resistance is high enough.

Uli and Joerg, Based on what both of you are saying, I'm now speculating that SWER leads to a higher resistance, and that this in turn leads to both a reduction of resistive losses and also a reduction of lossless power delivered. In other words, even though SWER saves on the cost of wire and reduces resistive losses, it also reduces overall efficiency by reducing the lossless power transmitted. What's your opinion on that?

And this aside, I still suspect that the Earth acts as a sink rather than as a return. I'd still like to hear Joerg's response regarding Maxwell's displacement current around a bolt of lightning.

Frederick, at 50 Hz we have a wavelength of about 6000 km while the length of the power line might be some 100 km in the worst case. Therefore I think we can treat voltage and current as "virtually DC" (of course with cos phi and all that but RF effects like wave propagation and reflection as well as nonzero currents in open circuits are nearly absent).

As to the lightning, if the current of the lightning bolt had not to be concentrated to a cross-section of some dm2 or less but were distributed over the whole area under the charged cloud, there would be a similarity to the problem of the disconnected plate capacitor which is (slowly) discharged through its weakly conducting dielectric: There should not be any magnetic field involved because dD/dt = -J, so curl H remains zero. In this case, we have no discrete closed circuit but current and "displacement current" (= dD/dt = return path) are distributed over the same volume of space.

In the case of lightning, the capacitor plates are still formed by earth and the cloud. Since the field is distributed over the whole space under and near the cloud but the current of the lightning bolt forms a thin thread, the 3D circuit seems to be donut-shaped, consisting of infinitely many, infinitesimally thin closed circuits which all pass through the lightning bolt and somewhere back through the decreasing D field.

Much has been written about the perhaps untoward term "displacement current" but may I assume that we both agree that in vacuum (or in pure aether) it denotes just the changing D field?

An increase in loop resistance certainly results in a reduction of the overall power deliverable (given a constant input voltage) as well as an increase (not decrease) of the relative resistive losses. (Absolute figures are irrelevant when it comes to losses!) As total power deliverable decreases and relative resistive losses increase, net power delivered ("lossless power) decreases as well.

The cost of the wiring is a one-time invest, the losses accumulate over time. Thus SWER might seem appropriate in the short term. On the long run, SWER is inferior.

And - as these circuits can be considered "isolated" (in some weird way), earth (aka ground) acts as the return wire, definitely not as some "sink".

Uli, OK we're agreed on the first bit. Thank you for clarifying this. On the second bit though, are you sure that the Earth isn't acting as a sink rather than as a return? Why would the Earth connector at the far end behave any differently than a lightning conductor? Why would the current need to travel 100 miles back to the power source again when it could just dissipate into the Earth? Especially if there were alot of rocks and tarmac along the return route.

Joerg, The capacitor analogy is fine, but you seem to be under the impression that a conduction current and a displacement current in the opposite direction would be present at the same time in the same locality. I think I get your donut picture which suggests that you see a closed solenoidal circuit with conduction current forming the central line of a downward path and a virtual displacement current fanning out and returning in an upward coaxial cylindrical path wrapped around the downward conduction path. Having studied Maxwell's displacement current extensively, I don't think that it ever exists additionally to conduction current in the same locality. I think that before the lightning bolt strikes, there will be a displacement current occurring as the charge builds up in the cloud. This will then be superseded by the conduction current in the form of bolt of lightning. It's an ongoing downward phenomenon with no return.

In order to read Maxwell's original thoughts on displacement current, you can read the preamble to part III of his 1861 paper "On Physical Lines of Force" http://vacuum-physics.com/Maxwell/maxwell_oplf.pdf

It's basically polarization current in a dielectric except that he alludes to tangential stress and a twist displacement rather than a linear displacement.

The modern idea that displacement current cancels with the conduction current in a charging or discharging capacitor seems to have been devised as a means of deriving Maxwell's displacement current in the absence of a dielectric medium. I think it misses the point completely. I further don't even think that the displacement current that occurs between the plates of a charging capacitor is Maxwell's twist-displacement current that is used with Ampère's Circuital Law when deriving the EM wave equation. It's a subject which is badly misunderstood.