Fault to Ground and Current Path

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Sierevello

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I am trying to understand the path of current in a ground fault situation.

In a fault situation the current flows back to the panel via the ground wire. When it reaches the panel it is bonded to the neutral. The panel, neutral, and ground should be at 0v and there is the potential difference causing the current to flow back to it's source. We all know that current will flow on everything connected to the circuit and not just the "path of least resistance" even though the lowest resistance path will get the most current.

For arguments sake let us suppose that a main breaker malfunctions and fails to trip: Will everything connected to ground not then become hot including the bonded water pipes etc?


Thanks, Steve
 

G._S._Ohm

Senior Member
Location
DC area
I am trying to understand the path of current in a ground fault situation.

In a fault situation the current flows back to the panel via the ground wire. When it reaches the panel it is bonded to the neutral. The panel, neutral, and ground should be at 0v and there is the potential difference causing the current to flow back to it's source. We all know that current will flow on everything connected to the circuit and not just the "path of least resistance" even though the lowest resistance path will get the most current.

For arguments sake let us suppose that a main breaker malfunctions and fails to trip: Will everything connected to ground not then become hot including the bonded water pipes etc?


Thanks, Steve
According to my idea of the wiring diagram, if:

the hot lead and the ground lead have 0.1 ohm resistance and the hot lead shorts to the ground lead at the appliance

and the service entrance conductors each have 0.05 ohm back to the transformer

then until the few milliseconds it takes the breaker to trip the series loop [tranformer secondary high side, SE conductor, Romex, short, Romex, SE conductor, tranformer secondary low side] will carry 120/(0.1 + 0.1 + .05 + .05) = 400 A.

With respect to the pole ground, the load center ground will then be at 400 A x 0.05 ohm = 20 V.

I'm neglecting the small current in the neutral back to the load center since appliance resistance is much more than these values.

You can confirm this by putting a 10 A load from the hot to ground and measuring the load center ground voltage with respect to the pole ground. It should read 20 x 10/400 = 0.5 V.
 
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ron

Senior Member
You are correct, the current will return via the EGC, across the G-N bond and back to its source via that grounded conductor.

Current will also flow on all paths back to the source, but the grounded conductor will be the lowest impedance making the "other" paths (pipes etc) carry current back to the source, but due to the very high impedance, the current will be very low on the "other" paths almost negligible relative to the grounded conductor.
 

charlie b

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Location
Lockport, IL
Occupation
Retired Electrical Engineer
For arguments sake let us suppose that a main breaker malfunctions and fails to trip: Will everything connected to ground not then become hot including the bonded water pipes etc?
It will not. Fault current will flow from the source to the damaged equipment via the ungrounded conductor, then back to the source via the grounded (neutral) conductor (as is normal) as well as via the EGC. Once it gets back to the panel, it will have reached its source. There is no driving force to cause the current to flow from the ground bar in the panel, out of the panel via the EGCs for other circuits, and via bonding jumpers to the enclosures or cases of equipment throughout the facility. Instead, the source itself, be it the utility service or a distribution transformer somewhere in the building, is the driving force that will cause the fault current to continue to flow in the loop from source to fault point and back again. This will go on forever, or until the breaker trips, or until the connection at the fault point vaporizes and opens the circuit.


Welcome to the forum.
 

G._S._Ohm

Senior Member
Location
DC area
From Bob Ludecke's power point presentation

100' of #2 AL = 0.032 ohms from transformer
25' of $4 AL = 0.13 service drop
25' of #2 CU = .005 from meter to load center
100' of #14 CU = 0.31 branch circuit

Feel free to modify my numbers so they are more in keeping with these average values.

The transformer is right outside my house so for me the first 100' value can be done away with.

BTW, I get 0.25 ohm for 100' of a single #14 copper @ 20 C. I haven't checked the other values.
See
http://en.wikipedia.org/wiki/American_wire_gauge
A length of aluminum offers 1.7 ohms for the same length of copper that gives 1.1 ohms.
 

don_resqcapt19

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Staff member
Location
Illinois
Occupation
retired electrician
Everything connected to the electrical grounding system will have an elevated voltage under fault conditions that is equal to the voltage drop on the system bonding jumper and the service grounded conductor back to the supplying transformer. The voltage on the equipment where the fault is will have an elevated voltage equal to the voltage drop back to the utility transformer. With high current faults, this voltage can be in the hazardous range. This elevated voltage will exist until the fault is cleared.
 
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