Faults and Ground

[hurk27]

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.

To determine when a circuit is in series or parallel you have to keep the current source and the load as separate parts of a circuit, any time a single load is connected to a source it can be said it is in series or parallel as it exhibits both, but if you have two or more loads that are in parallel with each other then you have a parallel circuit in parallel/series with the supply, so generally we never identify the current source as part of the series/parallel circuit when making a determination if a circuit is series/parallel because it will always fit both terms, even if you have two series supplies once you get to the load side of the circuit it reverts to that this series supply can be thought as both parallel or series with the load since it meets both definitions.

 

[mopowr steve]

It always goes back to the source in any and all paths that will get it there. There may be something depicted in that particular article you read to state that. Most likely if the fault occurred to earth ground it would not trip a breaker except for high voltage. Earth ground resistance is often to high for 120 volts to trip a common 15 or 20 amp breaker. Typical ground resistance with an 8' cu rod in soil conditions here is right around 25 ohms. Use ohm's law and you'll see how the current of 120 volts doesn't come close to trip rating.

 

[mosseater]

To determine when a circuit is in series or parallel you have to keep the current source and the load as separate parts of a circuit, any time a single load is connected to a source it can be said it is in series or parallel as it exhibits both, but if you have two or more loads that are in parallel with each other then you have a parallel circuit in parallel/series with the supply, so generally we never identify the current source as part of the series/parallel circuit when making a determination if a circuit is series/parallel because it will always fit both terms, even if you have two series supplies once you get to the load side of the circuit it reverts to that this series supply can be thought as both parallel or series with the load since it meets both definitions.

The issue I see is that I believe you are technically correct, but as drawn, the fault to the case of the load doesn't supplant the original circuit, unless the original connection has been arced off. Without further information to go on, the drawing clearly shows two paths of current flow back to the source from the fault point onward. Even though the bulk of current flow will be following the low resistance path back to neutral, and the voltage drop at the point will be virtually 100% and likely not be enough to drive the load and generate an accessible current through it, it IS still a complete curcuit and my contention is, some miniscule current will flow in a path parallel with the fault back to neutral. Doesn't the rule of all currents in a complete circuit entering a point must leave the point still apply? Or is that off the table now too? Been a long time since theory class for me. Glad the electrons know what they're doing!

 

[GoldDigger]

Even though the bulk of current flow will be following the low resistance path back to neutral, and the voltage drop at the point will be virtually 100% and likely not be enough to drive the load and generate an accessible current through it, it IS still a complete curcuit and my contention is, some miniscule current will flow in a path parallel with the fault back to neutral. Doesn't the rule of all currents in a complete circuit entering a point must leave the point still apply? Or is that off the table now too? Been a long time since theory class for me. Glad the electrons know what they're doing!

Hopefully nobody is arguing that no current flows through the original load path, just that it is small enough to ignore for all practical purposes.

And the fallacy that current will follow only the path of least resistance has been debunked here many times.

 

[kwired]

The issue I see is that I believe you are technically correct, but as drawn, the fault to the case of the load doesn't supplant the original circuit, unless the original connection has been arced off. Without further information to go on, the drawing clearly shows two paths of current flow back to the source from the fault point onward. Even though the bulk of current flow will be following the low resistance path back to neutral, and the voltage drop at the point will be virtually 100% and likely not be enough to drive the load and generate an accessible current through it, it IS still a complete curcuit and my contention is, some miniscule current will flow in a path parallel with the fault back to neutral. Doesn't the rule of all currents in a complete circuit entering a point must leave the point still apply? Or is that off the table now too? Been a long time since theory class for me. Glad the electrons know what they're doing!

Hopefully nobody is arguing that no current flows through the original load path, just that it is small enough to ignore for all practical purposes.

And the fallacy that current will follow only the path of least resistance has been debunked here many times.

Correct, current follows all possible paths, but in proportion to the resistance of those paths. Make one path a very low resistance and the majority of current flows in that path, but there is still some current in all other available paths.