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DC circuit ground safety.

Unimatrix_001
Unimatrix_001 used Ask the Experts™
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Hello,

Please see the attached image (from the allaboutcircuits ebook series). In that circuit, it shows that the person receives a shock, which I find a little confusing, but I also have these questions:

1) As the ground is connected seemingly directly to the negative terminal then this seems to ruin the circuit, as the electrons wouldn't flow from the negative terminal to the positive, but would surely flow directly from the negative terminal straight to the ground (blue arrow)?

2) It states the person would get a shock and assuming they did, would this be because of electrons flowing from the person-ground to the positive terminal (red arrows)?

3) It seems that either current can flow through the load or the person, but is this diagram assuming that the load has a larger resistance than the person?

Thanks,
Uni
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Business Consultant (Owner)
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Commented:
The load resistance is sufficiently high that the voltage on the top rail is still high. At that point there is enough current ability in the source to provide high voltage to the person and the person becomes another parallel path to ground. Two household lightbulbs in parallel on the same circuit if you will. Both light up. Think about it like this and it is not confusing.

If the resistance above were nearly a short circuit, then there would be no high voltage.

... Thinkpads_User

Author

Commented:
Ok, I think I see what you're saying, although I'm still a little confused. Would you mind if we split things up a little. First part:

There must be a potential difference between the negative terminal and the ground (the negative terminal having excess electrons and the ground acting as an infinite reservoir for charge), so in the diagram would electrons just flow from the negative terminal straight to ground?

Thanks,
Uni
JohnBusiness Consultant (Owner)
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Commented:
There does not have to be such a potential difference between ground and negative. If the lower green line were extended out for the person to stand on, they would still get shocked.

In reality, there probably is a potential difference, but likely small, and does not impact the problem.

Go back to my example (which uses house current). One side of the house wiring is connected to the water pipe. Somewhere else, there is a good connection to ground (another water pipe), the household current is still very dangerous.

I think the diagram above merely assumes good grounding everywhere. ... Thinkpads_User
ozo
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Commented:
the negative terminal is at the same potential as the ground,
as it is at the same potential as the bottom half of the wire.
the positive terminal is at a different potential so the electrons try to take the least resistance path between the negative and positive terminals

Author

Commented:
Hi both,

the negative terminal is at the same potential as the ground, as it is at the same potential as the bottom half of the wire.
That makes sense, but does this statement assume that the battery negative terminal discharges instantly to the ground?

the positive terminal is at a different potential so the electrons try  to take the least resistance path between the negative and positive  terminals
This sort of makes sense, although as I understand things, electrons will want to travel to the positive terminal, so they could take any of the following paths:

1) From the negative terminal through the load to the positive terminal (this assumes that the PD between the negative terminal and ground isn't instantly 0).

2) From the ground through the load to the positive terminal (this assumes that the PD between the negative terminal and ground is instantly 0).

3) From the ground through the person to the positive terminal.

In any of the cases, is the idea of the person receiving a shock based on the principle of the person having a much lower resistance than the load?

Thanks,
Uni
JohnBusiness Consultant (Owner)
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Commented:
1. For two places at the *same* electrical potential, there is no discharge of current (electrons).

2. So now the electrons will flow through a path of least resistance to back to ground (the resistor in your diagram), but not ALL the electrons, or it would be a short circuit.

3. NO, it does not assume the person is lower resistance than the load. It mearly means the load is not taking all the current.

... Thnkpads_User

Author

Commented:
For two places at the *same* electrical potential, there is no discharge  of current (electrons).
I'm struggling to see this :( The negative terminal has an excess of electrons and the ground is considered perfectly neutral. How can there not be a PD, as electrons will from from the negative terminal to the ground until there is a neutral charge?
JohnBusiness Consultant (Owner)
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Commented:
BTW, a common misconception I hear (and you are somewhat assuming it above) is that you can *push" current. You cannot. A load will only draw what it needs. In the case above, the implicit assumption is that the load is not a short ciruit. Therefore there are source electrons left over (not a good choice of words) to servive the next load (the person). ... Thinkpads_User
JohnBusiness Consultant (Owner)
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Commented:
The negative terminal has no where except back into the source to discharge its electrons because the negative terminal is connected to ground.

Thiink about this for a minute. The negative terminal of a car battery is connected to the car body. There is no great rush of electrons flying nowhere (otherwise our batteries migh all be dead). The electrons flow out through the positve terminal in response to a load and then back to source through the ground. But the ground is a no relative potential difference to ground. .... Thinkpads_User
ozo
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Commented:
If a few electrons flow from the negative terminal to the earth when they are connected,
the loss of charge would drop the potential of the terminal to the same potential as the earth.
The potential difference can be restored if the electrons can be replenished by flowing through the circuit.

Author

Commented:
BTW, a common misconception I hear (and you are somewhat assuming it above) is that you can *push" current.
Nah, that's something I'm quite okay with...

The electrons flow out through the positive terminal
Ah, I hadn't realised you use conventional current even for something like this... New question coming up...

Author

Commented:
I'm very sorry thinkpads_user, ozo I'm really struggling with this...

Okay, using conventional current and simple terms:
The negative terminal wants electrons for a neutral charge and the positive terminal wants to get rid of its electrons to achieve a neutral charge.

Consider the negative terminal first. It can get electrons from either the positive terminal or the ground, but as the connection to the ground offers next to zero resistance compared to going through the load, then it will gain the required electrons from the ground. Is this correct?

Consider the positive terminal. It can lose electrons either through the load into the ground or through the person into the ground. Okay so far? Now as for the person getting a shock, if the load has a resistance of 0.01 milliOhm, but the person has a resistance of 1 Ohm, then it is clear that the current flowing through the person will be next to nothing compared to the current flowing through the load. Is this right?

Thanks,
Uni
JohnBusiness Consultant (Owner)
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Commented:
>>> it will gain the required electrons from the ground. Is this correct?  More or less.

>>> Consider the positive terminal. It can lose electrons either through the load into the ground or through the person into the ground. Okay so far?   I would not say it this way.

Current flows from positive to negative always, not the other way around. Also remember my earlier post. It will "lose" electrons through the load, but only what the load wants, so the power source (of electrons) still has lots of electrons left.

>>>  Now as for the person getting a shock, if the load has a resistance of 0.01 milliOhm, but the person has a resistance of 1 Ohm, then it is clear that the current flowing through the person will be next to nothing compared to the current flowing through the load. Is this right?

First, bad numbers. 0.01 milliOhms is basically a short circuit and using a short ciruit for a load changes the problem. Since you don't know the voltage of the source leave the value of ohms out of the picture.

The load will draw what it wants. The amount of current the person draws is reasonably independent of this. If the person is well grounded and has wet fingers, then the person will present a low resistance to the power source and get shocked. But this has nothing to do with the load. We are assuming the power source can supply all the current demanded from it.

The load is a bit of a red herring in the example above. It does not need to be there for the person to get a shock. And since it is (assumedly) not a short circuit, putting it there does not change things.

 ... Thinkpads_User

Author

Commented:
Excellent - thank you! :)
JohnBusiness Consultant (Owner)
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Commented:
You are most welcome. I was pleased to assist. ... Thinkpads_User

Author

Commented:
You've got a lot of patience, I can say that. :)
"The negative terminal wants electrons for a neutral charge and the positive terminal wants to get rid of its electrons to achieve a neutral charge." The negative terminal does not want anything.
If you really want to work with circuits (as distinct from understanding them on an atomic scale) forget about electrons and consider only  potential and current.
Potentical is relative (ie one considers potential differences). To ground an circuit means to set the zero for that circuit. You can connect any ONE point in a circuit to ground (with no current flow) and that point becomes V = 0. In your case that is the end of the green line.
Now the conventional current flows from the higher potential to the lower potential. In the usual teaching circuits batteries are assumed to have no internal resistance (unless explicitly referenced). so the person's hand is at a high potential. The persons feet are at ground (0) potential so there is a very high potential across that person (call an ambulance).
The bird is at a high potential but has NO contact with a lower potential so the potential difference between any part of the bird is 0.   I = V/R.  V = 0 so the bird is fine.
It makes no difference for this teaching circuit what is  the resistance of the load or person or bird.
---
Now the circuit shows the electron current as distinct from the conventional current but all of the above still holds.

---
"Now as for the person getting a shock, if the load has a resistance of 0.01 milliOhm, but the person has a resistance of 1 Ohm, then it is clear that the current flowing through the person will be next to nothing compared to the current flowing through the load. Is this right?
That is right BUT as the battery has no internal resistance, the person is still dead (and the resistance is probably red hot). For the teaching purpose of the diagram the resistance ratio is immaterial.

Author

Commented:
Aburr, thank you - as you can see this question is closed, but you brought up a couple of points that I was going to address in a new question at a later date, but as you've brought them up now would you mind looking at this question please?

http://www.experts-exchange.com/Q_26305560.html

Thanks,
Uni