# Electric shock through earth

1. Jul 4, 2011

### RedX

Say you touch the hot wire and a piece of plumbing simultaneously. Then current flows through the hot wire, through you, to the ground, and I assume (though I'm not sure) to where the ground wire is tied to the neutral wire (at the breaker panel). But is the current in the ground mainly through the resistance of the earth, or the capacitance? I've attached a picture to clarify what I mean:

In the picture a short has occurred, which puts a high voltage across the ground resistor and capacitor (the load is unaffected by this short and is irrelevant - I just drew it to add context to the picture). The current through the resistor is about 120/Rground - I do not know if this current is high because I have no idea what the resistance of the ground is. The capacitor is now charging up from 0 V to 120 V. The current through the capacitor would be charging an RC circuit, 120/Rwire[1-exp(-t/RwireCground)]

So the total current would be the sum:

I=120/Rground+120/Rwire[exp(-t/RwireCground)]

So which of the two terms would be bigger? I have no idea what the capacitance of the earth is. I don't even know if I'm modelling it correctly. The idea is that the earth in general is an infinite sink for charge since it's so big. However, charge won't flow from the hot wire to the earth unless the same amount of charge can jump back from the earth into the circuit, or else the powerline would start to lose electrons. So I'm modelling this as having one plate of a capacitor at the place where the hot wire shorts to earth, and the other plate of the capacitor at the place where the ground is tied to the neutral, so whenever electron flows into one plate, it flows out of the other plate.

For the situation I outlined at the beginning, you would add the resistance of the human to all resistances to get your current.

I guess I just have a hard time believing that the resistance of the ground is small, so I want to believe it's capacitance instead, but I'm not sure.

*Actually, to be more accurate, the current should be:

I=[120/Rground](1-Rwire[exp(-t/RwireCground)])+120/Rwire[exp(-t/RwireCground)]

but I'm more interested in if it's the steady state value of the resistor current, or capacitor current that is responsible for shock.

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2. Jul 4, 2011

### gsal

You refer to it as a shock because typically the human will let go immediately or, hopefully, the fuse will open up....other than that, this could be a steady state scenario...and with home electricity being A/C ...I don't see how a capacitor would come into play...

By the way, the way you are envisioning the capacitor surprises me....when it comes to earth's capacitance, I don't think of a capacitor fully embedded inside the ground; instead, I think of capacitance BETWEEN earth and something else...like clouds or power transmission lines or something like that, in other words, the earth is one side of the capacitor and the other is the cloud...and, yes, compared to a cloud, the earth probably has infinite capacitance and when the voltage in the capacitor is high enough and the dielectric strength of the air cannot take it anymore...then you have lightning...and now I am getting waaay out of topic...or maybe not...capacitance...

3. Jul 4, 2011

### RedX

That's what sort of confuses me. Say you rub two objects together, so that one is positively charged and the other negatively charged. Then you hand one of the objects to another person, and that person flies to some place far away. Then at a later time you two simultaneously touch the objects to earth. Then no matter how high the resistance of the earth, the electron from your object will flow to their object and both objects will be neutralized, very quickly. In fact, the other person doesn't even have to touch their object to the earth for your object to be neutralized, so there doesn't seem to be a causal relationship here. And it doesn't matter how much charge is on your object - the earth will gladly take all of it since it's so big (a small conductor would only take so much as to have it and the object at equipotential).

I just learned that if you stick an ohmeter in the ground, you'll measure a large resistance, but that there is a distance after which the resistance doesn't increase. So maybe the resistance of the earth isn't that bad. After all, there are power systems that have a single wire, and use the earth as the return wire, so it can't be that bad. What's strange is that the resistance is highest near where the wires enter the ground, and falls off in the ground as you get away from the wires - how does the earth know if it's close to a wire? Rather than say resistance varies, I think it makes more sense to say Ohm's law is no longer linear, but I don't know.

4. Jul 4, 2011

### gsal

Here are a few thoughts...

Resistance is not always constant...for the typical case, it depends at least on the temperature...so, a copper wire has one resistance at 25C and a different resistance at 100C; this is well known and it is accounted for with what is called temperature coefficient of resistance. This is mostly linear.

So, if a copper wire is at 25C when it starts conducting electricity...its resistance has one value, but as time passes, it will warm up (due to I2R) as the copper absorbs some heat and its resistance changes...yielding a different solution to the circuit (different current, if the voltage is the ruling quantity).

Then, of course, there are materials that are not linear and their resistance changes in different ways.

Now, copper wires have very low resistance, but even when you have two wires joined/overlapped to make a longer one, the resistance at the interface is going to be the highest as the inter-molecular spaces are the largest and electrons will find themselves looking for places to jump across from one wire to other one.

So, let's say for the moment that the interface between the 2 wires has a resistivity 10 times larger than copper itself...then, if you measure the resistance across the interface from one point in one wire to another point in the other wire, if the points are close to the interface the measured resistance will seem rather high for copper wire, but this is because the resistance of the interface is proportionally significant; but, as you measure the resistance from two points further and further apart from the interface, and as you have more and more wire (while the interface remains constant)...then the resistance of the interface start making less and less of a difference and the measured resistance starts approaching that of copper wire alone.

...maybe this is kind of thing that is happening when you stick a wire into the ground...the interface between wire and ground will be a component of the measured resistance...that's why it makes a difference how much wire you bury into the ground..the more wire, the more places for electron to jump out ( in parallel ) from the wire to the ground...of course, after certain point it does not make a difference how much wire to bury since the wire also has a cross section that limits electron flow, too...

hope this helps some

gotta go, for now

5. Jul 4, 2011

### Naty1

Earth is basically a resistance connection.

and also note the link to the single wire earth return power supply system.

"hopefully, the fuse will open up....other than that,"

NO way any fuse is designed to do that: for electrouction protection a GFCI will trip if the hot and neutral current becomes unalanced....

6. Jul 4, 2011

### RedX

How good are GFCIs? If someone were taking a bath, and drops something electrical (say they were making toast with a toaster while in the bath) in the water, would the GFCI save them, assuming it's working properly? You always see it in the movies, that the killer tosses something electrical into the bath electrocuting the victim, and I'm always wondering if the outlet has a GFCI.

7. Jul 4, 2011

### RedX

That makes sense. Would you model the interface by it's own resistance, Rinterface, in series with the resistances of the two mediums it separates, or would you calculate it like a transmission line problem, where the interface reflects some of the current and voltage, and you have to calculate the resistance (impedance) that the voltage source sees using a nonlinear formula (that involves things like the characteristic impedance of the line)?

8. Jul 4, 2011

### gsal

Replying to whether GFCI protect: I guess it would...but I don't know what the wiring code says...my house was built in 1997 and it only has GFCIs along the counter in the kitchen for toaster, microwave oven, blenders, coffe makers, etc....nowhere else in the house do I have any more GFCIs, other than maybe the garage, if I remember correctly...not even in the bathrooms do I have GFCIs where we have shavers, clippers, blowdryers, etc...even the outlet right next to the bathtub is a regular one...go figure.

9. Jul 4, 2011

### gsal

Replying to modeling of the interface between two conductors: Actually, I have limited experience on this matters and for the cases that I have had to deal with this, the system is rather small, DC, and no, there is no such thing as reflection/refraction....everything is simply in series.

Check this http://www.copperinfo.co.uk/busbars/pub22-copper-for-busbars/sec7.htm" [Broken] out.

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10. Jul 4, 2011

### RedX

My parents' house is older than yours, and all bathrooms have GFCIs, although some of the GFCI's don't have those two buttons on them that allow you to test if the GFCI is working. I don't know if it's dangerous to stick one lead of a voltmeter into the ground plug, and the other lead into the hot plug, and check if the voltmeter reads zero (if the GFCI is working properly). I mean the wires of a voltmeter are insulated. But anyways, my guess is that the reason there are testers on GFCI's are that they break a lot, so are unreliable.

11. Jul 4, 2011

### gsal

Having a safety device in proper working condition is very important since it can save a life...so, I am not entirely sure that the reason why GFCIs have testers is because they break a lot...it is simply in order to be able to check often whether they are still working or not...same with smoke detectors...it's a safety thing.

And yes, you could stick a voltmeter lead into the hot and the other into ground and it will read 120 volts or so...I just read that a GFCI will trip if the current going out of the hot wire is 0.005 different of what is coming back in through the neutral...so, even though voltmeters are designed to not draw current, they need just a bit, so, if this current is detectable by the GFCI, then, even sticking the voltmeter terminals into hot and ground may trip the GFCI (since the tiny current is not coming back through the neutral)