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this question has been bothering me for a long time

Say a person touch the live wire as shown in the picture, the current will flow through him and dissipates into the ground in "rings" as shown in the picture right?(since the system is grounded at the source so the voltages are not "floating" as in an isolation transformer) the fault current through a person does not really “return” to the source as drawn in pencil correct? at least not all of it anyway (in the sense that all the current that went through the person actually traveling back to the source since the current dissipates into the ground in rings), is it possible that the grounding rod at the source can "suck" up charges from the ground to the neutral wire to make up for the current that went into the ground through the person? Just how exactly does the current return to the source? assuming that the resistance of a person is 10 ohms. Much appreciated everyone.

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  • $\begingroup$ Is this really supposed to be 10 volts and not 110 volts? $\endgroup$ – Bob D Aug 1 '19 at 9:52
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For comparison check out the engineering of single-wire Earth return energy distribution.

In addition check out high-voltage direct current.

This is just for comparison, your question is for the case of household voltage.

The transit of electrons in a conducting wire is like bucket brigade transport. For every bucket emptied on the fire there a long row of buckets in the process of being passed along.

Pushing the bucket brigade model:
A direct current flowing in a two-wire system is modeled by a two chains bucket brigage, passing full buckets in the direction of the fire, the other chain returning empty buckets.

Pushing the bucket brigade model for the case of single wire direct current:
While the Earth is not particularly good conductor, it is obviously a very wide conductor.

So: single wire direct current: That's like a bucket brigade with one line passing full buckets in the direction of the fire, and the return line consisting of a huge crowd of people, each carrying one empty bucket.

Empty buckets get passed to the return line. Since there is a crowd of people the return transport of empty buckets seems very slow, but that's just because the flow has widened. If there would not be any motion of buckets back to the bucket filling point then the circulation would grind to a halt.

Let's say the fire is put out in a manner of minutes, about the time that it took to pass a filled bucket from one end of the line to the other. Then any individual bucket travels only a little distance in the return line. Still, we can count that as a closed circuit. The crowd is able to accept buckets because at the other end that same crowd is able to pass buckets to the bucket filling station. The two are connected.

Specific to your question:
You are asking about household electricity, which is alternating current. So there is not much time between current direction reversals.

The thing to think about is not whether individual charge carriers make it all the way through the ground.

What matters is whether the voltage difference makes it from one end to the other.

When the voltage makes it across the setup acts as a closed circuit.

I'm not sure this addresses the doubt that is nagging you. As I understand it the thing that is on your mind is that the Earth-as-conductor-of-current is a very wide conductor, with a correspondingly small current in individual cross sections.

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  • $\begingroup$ Hey Cleonis, thank you very much for the detailed explanation, the bucket brigade model made things much clearer for me and yes, "As I understand it the thing that is on your mind is that the Earth-as-conductor-of-current is a very wide conductor, with a correspondingly small current in individual cross sections." was what was bothering me, but as you said as long as there is a voltage difference all the current will return right? $\endgroup$ – Jason Aug 1 '19 at 14:41
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Let me start by pointing out the minimum nominal voltage of household electrical power system is 110 vac, not 10 vac. 10 vac, 60 Hz is considered low voltage with respect to electric shock risk.

Say a person touch the live wire as shown in the picture, the current will flow through him and dissipates into the ground in "rings" as shown in the picture right?

Current will flow through the person and back to the grounded connection of the voltage source through the soil.

(since the system is grounded at the source so the voltages are not "floating" as in an isolation transformer) the fault current through a person does not really “return” to the source as drawn in pencil correct?

That is not correct. Current will actually flow through the person and return to the source through the soil, as shown, but possibly through various paths.

at least not all of it anyway (in the sense that all the current that went through the person actually traveling back to the source since the current dissipates into the ground in rings),

The term "dissipate" in your picture simply refers to the fact that there are multiple potential parallel paths in the soil for current to flow as it spreads out. However all of the current returns to the source. It should be noted, however, that power distribution systems are connected to earth ground at multiple locations. Therefore, the current flowing through the body may be split up and return to multiple ground connection points.

Just how exactly does the current return to the source? assuming that the resistance of a person is 10 ohms.

As already pointed out, the current returns to the source through the earths soil. How much current flows through the body depends on multiple factors.

One factor is the body's electrical impedance, which varies with voltage. At 10 vac60 Hz the total body impedance from hand to foot, under dry contact conditions, is over 5000 ohms with most of the impedance in the skin. The minimum internal impedance of the body (impedance minus the skin) is over 500 ohms. So for all practical purposes in your example the current through the body to ground would be less than 2 mA, assuming zero impedance in the ground path. Another factors is whether or not the person's foot wear is acting as insulation. Also the impedance of the can vary depending on local soil conditions.

The thing that really confuses me is how would current return to the source, but as you said as long as there's a voltage potential all the current WILL return to the source right? I was confused about since current spreads out pretty much uniformly after it enters the ground why would all of it return to the source since the earth is so big and the voltage potential at say 50m away from the person's feet is already 0 –

You know, of course, that there does not need to be voltage difference for current to flow in a conductor. Similarly, there does not have to be a potential difference between the persons foot and the grounding rod for current to flow.

If one were run a copper wire connected to your persons foot under the ground and connect the conductor to the grounding rod that connects the power source to ground, both the persons foot and the ground rod will be at the same potential. We call it zero simply because that is what we assign the potential to be in the earth. Yet current will readily flow through the person and return to the power source through the copper conductor. The reason, of course, is the potential difference is between the persons hand and ground, the latter of which is theoretically at the same potential as the grounding rod (but really isn't because there is always some impedance in the earth path). The soil is simply a replacement for our copper wire (though it is never as good as the copper wire).

Hope this helps.

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  • $\begingroup$ Hi Bob, ty for the answer, just used those numbers for simplicity's sake, The thing that really confuses me is how would current return to the source, but as you said as long as there's a voltage potential all the current WILL return to the source right? I was confused about since current spreads out pretty much uniformly after it enters the ground why would all of it return to the source since the earth is so big and the voltage potential at say 50m away from the person's feet is already 0 $\endgroup$ – Jason Aug 1 '19 at 15:19
  • $\begingroup$ @Jason I have updated my answer to respond to your follow up question. Hope it helps. $\endgroup$ – Bob D Aug 1 '19 at 21:25

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