Faults and Ground

pegggu

Member
Location
Los Angeles
hi, definitely a noobie question here, but if someone can help me understand how faults work.
so anytime there is a fault of any kind whether it be a short, lightning strike, animal occurances, anything for that matter.

when a fault occurs how does the current "know" to go through the ground as opposed to continuing through the circuit as per usual?

say in a distribution system, say there is a fault on the line, how does the current know to go through a ground conductor, rod, plate into earth rather than say going to a customer (ignoring protection, such as fuses and such).

thanks in advance!
 

charlie b

Moderator
Staff member
Location
Seattle, WA
For starters, current has no interest in finding its way to ground. There is a well known (and I can hope it is soon forgotten) notion that current will take the path of least resistance to ground. If you have heard that before, please forget that you heard it. It is a lie. Two lies actually. First, current does not just take one path, the one of least resistance. It takes any and all paths that it can find. Secondly, current is seeking a path that will bring it back to the source that set it into motion in the first place. That path might take it from the "hot conductor" (i.e., the ungrounded conductor) to a fault point, into a conduit or other metal surface, into the floor surface, into the dirt, along the dirt (yes, dirt is a good conductor), to the location of the ground rod, up the ground rod and the grounding electrode conductor to the main panel, into the neutral bar in the main panel, and from there to the neutral of the transformer that had started current flowing in the first place. At the same time, current will continue to flow through the tool or whatever component had the short circuit. So both paths, the normal path through the tool and the fault path through dirt, may be carrying current at the same time.
 

PetrosA

Senior Member
hi, definitely a noobie question here, but if someone can help me understand how faults work.
so anytime there is a fault of any kind whether it be a short, lightning strike, animal occurances, anything for that matter.

when a fault occurs how does the current "know" to go through the ground as opposed to continuing through the circuit as per usual?

say in a distribution system, say there is a fault on the line, how does the current know to go through a ground conductor, rod, plate into earth rather than say going to a customer (ignoring protection, such as fuses and such).

thanks in advance!
Depending on the fault, the current doesn't necessarily go through ground. A bolted short between the hot and neutral of a circuit or between legs or phases of a circuit will not travel back to the source via the ground.

Lightning is not a fault per se, but an electrical occurrence on a massive scale compared to any electrical installation. If you think of the sky (or atmosphere) as being one plate of a capacitor, and the Earth as the second plate of the capacitor, lightning is an arc between the plates, and no teeny tiny fuses or circuit breakers we install are going to mean anything to that arc.

In your distribution example, a fault to ground is only that because the system is grounded in multiple places and the dirt conducts electricity between various points of the system. If you hooked up batteries in series to get 1000V and touched the first or last lead standing in your bare feet on a wet lawn, nothing would happen because the terminal at the other end is not connected to earth. If you ran a jumper from that terminal to a ground rod or plate and did the barefoot experiment, it would be your last.
 

cowboyjwc

Moderator
Staff member
Location
Simi Valley, CA
If you've ever seen a house that's been hit with a major fault (lighting, high voltage wire contact) you would be amazed at what you do and don't find.

Remember it's electricity, it does what ever it wants. At least that's how it was explained to me.

I would also listen to Charlie, I had heard the same thing you had until a couple of gentlemen that I highly respect, set a whole bunch of us straight at an inspectors meeting one time.
 

GoldDigger

Moderator
Staff member
when a fault occurs how does the current "know" to go through the ground as opposed to continuing through the circuit as per usual?
Well, it may seem like a frivolous analogy, but if you look deeply into the electromagnetic theory and quantum and relativistic effects it actually holds up in some ways:

You know how a single ant, wandering away from the group may find something, then leave a scent (pheromone) trail behind it that all the other ants follow, and pretty soon there is stream of ants along that path?

Think of electrons doing that. The easier it is for electrons to follow a particular path, the more of them following behind will take that path.

:)
 

DWEames

Member
Location
New Orleans, LA
so I was recently tested and asked what determines fault-current, the amp rating of a breaker or the method of grounding or bonding, or the resistance to ground....?
 

dereckbc

Moderator
Staff member
Location
Plano, TX
when a fault occurs how does the current "know" to go through the ground as opposed to continuing through the circuit as per usual?
I will take a different approach. It is real simple if you know and understand Ohm's Law and simple parallel circuits. Here is a drawing of 3 circuits that could be found in any residential service. Two are non compliant, and one is a safe compliant circuit. Can you tell which are which?



The compliant circuit is the one on the right we will call number 2. Now look at the one on the left or #1 with two 5 ohm grounds. In number 2 if line faults to chassis, it is connected back to the source on the green equipment grounding conductor. It essentially is a dead short from L-G. A dead short is extremely low resistance, thus very high current will flow and operate the fuse. Take note earth or ground was in no way involved.

Now take a look at circuit 1. It has no equipment grounding conductor, just connected to a ground rod with 5 ohms of resistance. At the source or service we have the neutral bonded to a different ground rod with 5 ohms. Now we apply the same fault. So what happens? Not much. We are applying 120 volts to 10 ohms resistance. That will induce 12 amps of current 120 volts / 12 ohms). What will a 20 amp breaker do with 12 amps of current flowing? Nothing because 12 amps is normal current. However if you were to touch the chassis you get hit with 60 volts and will suffer a mild electrocution.

OK above is a low voltage below 600 volt example. Let's move on to high voltage that utilities use which changes everything. So I don't have to make another drawing let's use circuit 1 again. Except this time change the AC voltage to 13.2 Kv, the load is now a transformer that steps down to single phase 240/120 service like your house. We now change the fuse size to say a 3 amp fuse for a 50 Kva transformer fed with 13.2 Kva. Now we apply a primary line fault of 13.2 Kv to the transformer case. What happens???? Well Ohm's Law tells the story. 13,200 volts / 10 ohms = 1320 amps of fault current. What happens to a 3 amp fuse with 1320 amps flowing?????

In the last example earth or ground does play a role. In high voltage applications you can use earth as a conductor and utilities use it just for that purpose. But in NEC applications using earth as a conductor is forbidden. It is forbidden because it is low voltage and earth resistance is too high to be of any use.

Hope that helps.
 
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dereckbc

Moderator
Staff member
Location
Plano, TX
so I was recently tested and asked what determines fault-current, the amp rating of a breaker or the method of grounding or bonding, or the resistance to ground....?
None of the above. Voltage and Impedance determines the fault current. It is a horrible incorrect question. The answer they were looking for is Resistance to Ground which is false. It is the impedance to the source.
 
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pegggu

Member
Location
Los Angeles
thanks for the replies but i'm still a bit confused.
so i read an article that stated: "the instant a fault occurs, this causes the current to flow through the ground wire/conductor, causing the breaker to trip or the fuse to blow".

so what i'm getting at is, why is that during normal operation the current will flow through the line, through the load, through the neutral and back to the source.

but why is it that all of a sudden when there is a fault, it goes through the ground conductor/rod/plate, etc rather than continuing through it's normal operation and back to the source?

hopefully i'm making some sense. thanks!
 

GoldDigger

Moderator
Staff member
Let us look at the case of a resistive load, specifically a heater coil.
Normally current from the ungrounded conductor through the resistance to the grounded conductor and back to the source .
Let's further assume that the voltage is 120V and the resistance is 120 ohms. The normal current will be one amp.
Now a fault develops between the hot end of the coil and the neutral wire and that has a resistance of .1 ohm.
If the voltage holds at 120 until the breaker opens we will still have 1A flowing through the load. Nothing has happened to prevent the electrons from following that path!
But an additional 1200A will now be flowing through the fault, for a total of 1201 amps.
We often talk as if all of the current is now going through the fault because 1A is so small in comparison , but it is still there.
A fault from the middle of the heating coil to the metal case which is connected to the EGC will produce different numbers but will still involve the fault path being in parallel with part of the normal load path.

Tapatalk!
 

USMC1302

Senior Member
Location
NW Indiana
If you haven't done so already, you might want to check out the excellent power point on Grounding and Bonding that's posted on that forum. That may be very helpful. I review it once in awhile, just because.
 

PetrosA

Senior Member
thanks for the replies but i'm still a bit confused.
so i read an article that stated: "the instant a fault occurs, this causes the current to flow through the ground wire/conductor, causing the breaker to trip or the fuse to blow".

so what i'm getting at is, why is that during normal operation the current will flow through the line, through the load, through the neutral and back to the source.

but why is it that all of a sudden when there is a fault, it goes through the ground conductor/rod/plate, etc rather than continuing through it's normal operation and back to the source?

hopefully i'm making some sense. thanks!
The article is describing one specific type of fault - a ground fault. This could be a hot wire in a box that touches a grounded surface, or the hot and ground in an extension cord getting damaged and touching, or any other situation where the hot comes into contact with ground somewhere in the circuit. The article doesn't state it goes through the ground rod or plate, but through the grounding conductor - big difference since only the grounding conductor will offer a good enough path back to the panel to trip the breaker or fuse. If there is no grounding conductor or if the metal around the fault is not bonded to ground properly, it will become energized without tripping the breaker or blowing the fuse and pose a shock or electrocution hazard.
 

StarCat

Senior Member
Location
Moab, UT USA
Modern Times

Modern Times

Pegggu, as mentioned above the statement is worded very poorly. It says " a fault," but not " a ground fault."
What helped me in former days was to understand the different types of faults and how they are dealt with from a standpoint of opening the protective device. There are some very good threads on this board that deal with misconceptions especially regarding the ground conductor and ground rod to earth. As an example, an earth ground will not do you any good at all if the transformer neutral is not bonded and a ground fault happens, because there is no path.
As in the 12A example, the chassis and or conduit runs would then be energized because the fault has no way back to its source. We are unfortunately living in a time when correct passing on of these matters has degenerated and fallen into the hands of people writing things down that they do not properly understand.
I remember a mentor telling me one day when we were working on an RTU with respect to the on board wiring....." Don't assume its right."
Golden advice.
Many times in the field you will come across things that are not right, and in your search to get at the truth.....you will find that you may be the only one involved that knows for sure something is amiss.

All the best
 

dereckbc

Moderator
Staff member
Location
Plano, TX
That's good, but I'm not seeing a parallel circuit.
When the fault occurs the EGC is now in parallel with the load. Let's redraw the circuit. and take a look.



In the first circuit is normal operation with 5 amps of load current flowing through a 24 ohm load via L and N. The green EGC is an open circuit, thus no current flowing and holding the chassis at ground potential voltage.

In circuit 2 we have a L-G fault inside the equipment. This now puts the EGC in parallel with the load. The resistance of the EGC is extremely low compared to the load thus shorting out the load. The only resistance is now only the wiring resistance of L, Breaker, and EGC for a total of about 1 ohm resulting in 120 amps of fault current which operates the breaker.

Note Earth had nothing to do with clearing the fault, only the EGC did that which is only referenced to Earth. We could remove the earth reference and the circuit would still work normally just like it does in a car. All the earth reference does is hold the chassis at earth potential during normal operation.
 
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Little Bill

Moderator
Staff member
Location
Tennessee NEC:2017
Occupation
Electrician
When the fault occurs the EGC is now in parallel with the load. Let's redraw the circuit. and take a look.



In the first circuit is normal operation with 5 amps of load current flowing through a 24 ohm load via L and N. The green EGC is an open circuit, thus no current flowing and holding the chassis at ground potential voltage.

In circuit 2 we have a L-G fault inside the equipment. This now puts the EGC in parallel with the load. The resistance of the EGC is extremely low compared to the load thus shorting out the load. The only resistance is now only the wiring resistance of L, Breaker, and EGC for a total of about 1 ohm resulting in 120 amps of fault current which operates the breaker.

Note Earth had nothing to do with clearing the fault, only the EGC did that which is only referenced to Earth. We could remove the earth reference and the circuit would still work normally just like it does in a car. All the earth reference does is hold the chassis at earth potential during normal operation.

I fully understand how an EGC works in relation to a ground fault. I'm not the one (OP) that asked the question.

I was just pointing out that your circuits as drawn are not parallel circuits but series.
Yes the current on the EGC is in parallel with the load conductors, but the circuit, as drawn, under normal conditions is a series circuit.
 

kwired

Electron manager
Location
NE Nebraska
If you've ever seen a house that's been hit with a major fault (lighting, high voltage wire contact) you would be amazed at what you do and don't find.

Remember it's electricity, it does what ever it wants. At least that's how it was explained to me.

I would also listen to Charlie, I had heard the same thing you had until a couple of gentlemen that I highly respect, set a whole bunch of us straight at an inspectors meeting one time.
It would seem like it does whatever it wants sometimes, but it still follows basic electricity theory. Problem is impedances of some objects to the voltage and frequency of a lightning strike is not the same as it is for 600 volts or less AC 60 Hz, which is what we are generally accustomed to recognizing.

Well, it may seem like a frivolous analogy, but if you look deeply into the electromagnetic theory and quantum and relativistic effects it actually holds up in some ways:

You know how a single ant, wandering away from the group may find something, then leave a scent (pheromone) trail behind it that all the other ants follow, and pretty soon there is stream of ants along that path?

Think of electrons doing that. The easier it is for electrons to follow a particular path, the more of them following behind will take that path.

:)
That is an analogy that has some merit to begin with. Those electrons decide to follow a particular path based on pressure behind them(voltage) and which path is easiest to be pushed by that pressure (resistance/impedance). Sure there may be paths of less resistance, but they are already congested by many electrons already flowing in those paths, so some of them will find secondary paths as there is still enough push (voltage) trying to make them go somewhere.

Not necessarily entirely correct but an analogy that does help explain to some extent.

so I was recently tested and asked what determines fault-current, the amp rating of a breaker or the method of grounding or bonding, or the resistance to ground....?
What determines fault current is same thing as what determines normal operation current, the applied voltage and the total impedance of the circuit - including the impedance of the source. In a short circuit situation theoretically there is unlimited amount of current that will flow, but the source does have an impedance that will determine a maximum possible amount of current that will flow, then the conductors involved will have resistance as well giving us additional current limitation.

thanks for the replies but i'm still a bit confused.
so i read an article that stated: "the instant a fault occurs, this causes the current to flow through the ground wire/conductor, causing the breaker to trip or the fuse to blow".

so what i'm getting at is, why is that during normal operation the current will flow through the line, through the load, through the neutral and back to the source.

but why is it that all of a sudden when there is a fault, it goes through the ground conductor/rod/plate, etc rather than continuing through it's normal operation and back to the source?

hopefully i'm making some sense. thanks!
Try to remember that the word fault is generic, and basically means something not right or normal. There are many different fault scenarios and each will have unique properties compared to other types of faults.

A line to line fault will have the impedance of the source and the resistance of the conductors involved in determining how much current flows.

A line to a grounded conductor is much the same way - but there will be some current (but typically small enough to be disregarded in most cases) that can flow into a grounding electrode connected to the grounded conductor and will follow earth back to another grounding electrode and rejoin the grounded conductor at a different point. Impedance at these electrodes will generally be high enough that this "parallel path" through the earth is limited and the amount of current flowing in this path does exist but is rather insignificant compared to the lower impedance path through the intended conductors.

Then we have low level ground faults that have low level fault current but still give us shock potential hazards - we commonly use GFCI's to protect us from those hazards. The fault current is usually very low, but is high enough to cause harm to us and that is the reason for such low level protection in those instances.
 

dereckbc

Moderator
Staff member
Location
Plano, TX
I was just pointing out that your circuits as drawn are not parallel circuits but series.
Yes the current on the EGC is in parallel with the load conductors, but the circuit, as drawn, under normal conditions is a series circuit.
The EGC is an open circuit under normal conditions. Under fault conditions with a L-G fault it is in parallel and thus shunts current around the load is what I am driving at.
 

Little Bill

Moderator
Staff member
Location
Tennessee NEC:2017
Occupation
Electrician
The EGC is an open circuit under normal conditions. Under fault conditions with a L-G fault it is in parallel and thus shunts current around the load is what I am driving at.
I didn't mean to imply that your intended message is wrong, because it is spot on.:thumbsup:

I just thought that someone just learning might be confused with you calling the drawing a parallel circuit. If you had drawn the load somewhere before the the end of the square and just went straight down from H to N but continued the line around the square, that would be a parallel drawing.

I don't have any way of making a drawing, but imagine the square as drawn but without the load. Instead of putting the load at the end, just put it somewhere in the middle of the square from H-N.
 

J.P.

Senior Member
Location
United States
To the OP. On this site there is a basic electrical theory dvd/book set. It will Not be a waste of your time.

They explain lots of things very well. I bought it and some other courses a while back when studying for my journeymans test and I have to say I learned more than a thing or two. I use those things they covered day to day in some form or another.
 
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