The ground return resistance may be on the order of 5 to 10 Ohms. A good ground resistance is 5–10 ohms which can be measured using specialist earth test equipment.
The big question is, for how long a distance is this resistance valid for?
I would expect it to not vary much and for the resistance to be dominated by the neighborhood of the ground connections. For instance, in 2D films point-to-point resistance only increases as the log of the separation between the electrodes since the farther they get apart the more width the current can spread out over. In 3D there is even more space for the current to spread out over. Earth isn't a uniform conductor so I assume you actually have to take into account the geology of the land, but in normal circumstances the current density and therefore voltage drop far from either end should be extremely low.
Yes, the distance doesn't matter much. The earth has a huge cross-sectional area. It's the contact resistance between the earth and the electrode, which dominates the resistance. Once the current is flowing through the earth, it can spread out over a huge volume.
Suppose we take a plastic cube shaped box, with a volume of 1m
3, glue two 1m
2 copper electrodes to the inside of each side and fill it with the local soil and measure the resistance, which is around 1k. Now imagine the current flowing through the soil between two large electrodes, placed 100km apart, spreads out over a an area of of 10km
2, the distance is 100km but cross-sectional area is now 10km
2, so the resistance is 100k*1k/10k
2 = 1Ohm.
In reality, the above example is flawed. The current can take a much larger cross-section than 10km
2 and the electrodes won't be that big. The further the electrodes are apart, the more the current can spread out, so the resistance of the earth over most of the path, will be insignificant, compared to the earth immediately surrounding the electrodes.
I wonder what influence does a capacitance have. For example if one takes an ideal wire (ignoring it's resistance and impedance for simplicity) and runs this wire to the Moon, it should be possible to transfer AC power just because of Moon (and Earth) self capacitance (even if Moon to Earth capacitance would be zero, which is not). I mean it should be possible to send some power over one wire without DC or even AC return path. Wikipedia writes that Earth self capacitance is around 710uF. That is a lot even for 50Hz (around 4.5 Ohms impedance).
That's irrelevant for this discussion, as the earth is acting as a return current path, but you seem to be a bit confused about self-capacitance.
Normally when we talk about capacitance, we really mean mutual capacitance. A large capacitor from a power supply might have 100 000µF printed on the label, which is 0.1F, but that's the mutual capacitance measured between its plates. The self-capacitance will be several orders of magnitude lower, around 100pF. When we talk about charging a capacitor, electrons from the cathode are moved to the anode. The net charge never changes, because the number of electrons inside it, remains the same! This means that if we measure the capacitance between the terminals, it will be 0.1F, but between the earth and the capacitor it will only be 100pF. The earth has such a large self-capacitance, compared to the capacitor, it can be ignored. Now if we take two of those capacitors, place them far enough apart that the mutual capacitance is insignificant and measure the capacitance between them, it will be 50pF, equivalent to them both connected in series.
Well, a capacitor *is* an AC return path. For electrostatics the self capacitance of the two bodies just add in series if you neglect the mutual capacitance (which is only a small correction in this case). However, this isn't electrostatics as the earth-moon distance is nearly 100 wavelengths at 50 Hz. So you have to look at this as a transmission line + antenna problem and I think this falls apart pretty badly. I'll admit I haven't thought a lot about what happens here, but I would guess that even if you manage to launch some power into the impedance presented by the wire it would almost all be radiated rather than make it to the moon.
Yes, of course capacitor is an AC return path, I was not trying to say otherwise. But my point is that it should work without any return also. If there is an RF wave going througth open space (and open space I believe is also a transmittion line) it can obviously transfer power with no return. For example if there is antenna on the Moon surface (ground plane).
In case of one wire telephone signals (higher frequencies), impedance of self capacitance of Earth may be dominating, not the full ground return path impedance. But thats just my guess (I have no clear idea how to calculate).
When a circuit is connected to an antenna, electrons are moved to and fro, between the circuit and the antenna. At very high frequencies, the self-capacitance of the circuit itself, is high enough not to matter, lower frequencies require a large ground plane to increase the self-capacitance of the circuit and act as a source/sink for the electrons, much lower frequencies require a physical connection to the earth, otherwise the ground plane would have to be enormous to increase the self-capacitance to the required level.
Old telegraph systems used the earth as a return. The self-capacitance of a cable will be tiny, compared to the earth, so no, the self-capacitance of the cable will dominate that of the earth.
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