Fault current path and breaker tripping parallel paths

[dm9289]

I was studying and a graphic came up that confused me slightly. The graphic is attached below. the point of the graphic was if you connect the neutral and gnd together after the first means of disconnect you will cause the breaker to trip later during a fault.

My "belief" is return current path is based on impedance.
For example I=V/R so if return path is 2 ohm thru neutral 120v/1=120amp
2 return paths ground and neutral linked in sub panel or past first disconnect with alternate path 3 ohm. So my feeling would be the total impedance would be product over sum (3x1)/(3+1)= .75 ohms
I=V/R 120/.75=160amps so my belief is breaker would trip sooner with 2 paths

I realize I'm likely wrong and missing a basic concept and your help is appreciated. I do also understand this creates objectionable current which is not desired and not code compliant.

 

[don_resqcapt19]

You are correct.

There is no overcurrent device in the neutral conductor. The only overcurrent devices that see the fault current and that can act to clear the fault are in the ungrounded conductors. A lower impedance for the fault return path will result in higher current on the ungrounded conductors and result in a faster clearing time.

 

[dm9289]

You are correct.

There is no overcurrent device in the neutral conductor. The only overcurrent devices that see the fault current and that can act to clear the fault are in the ungrounded conductors. A lower impedance for the fault return path will result in higher current on the ungrounded conductors and result in a faster clearing time.

Thank you. I was confused by the teaching. Maybe I heard it incorrectly in the video

 

[dm9289]

I was studying and a graphic came up that confused me slightly. The graphic is attached below. the point of the graphic was if you connect the neutral and gnd together after the first means of disconnect you will cause the breaker to trip later during a fault.

My "belief" is return current path is based on impedance.
For example I=V/R so if return path is 1 ohm thru neutral 120v/1=120amp
2 return paths ground and neutral linked in sub panel or past first disconnect with alternate path 3 ohm. So my feeling would be the total impedance would be product over sum (3x1)/(3+1)= .75 ohms
I=V/R 120/.75=160amps so my belief is breaker would trip sooner with 2 paths

I realize I'm likely wrong and missing a basic concept and your help is appreciated. I do also understand this creates objectionable current which is not desired and not code compliant.

I corrected the normal path to 1 ohm on above post

 

[LarryFine]

I see where the problem is: it's not a Mike Holt graphic.

 

[kwired]

I corrected the normal path to 1 ohm on above post

You are correct that the parallel path should lessen impedance and that would allow more current to flow, which should result in faster trip time, may need sensitive equipment to measure how much faster in some cases though as it may not be obvious to naked eye.

That said it still allows normal operation current to flow on non desirable paths which increases potential shock hazards.

 

[dm9289]

You are correct that the parallel path should lessen impedance and that would allow more current to flow, which should result in faster trip time, may need sensitive equipment to measure how much faster in some cases though as it may not be obvious to naked eye.

That said it still allows normal operation current to flow on non desirable paths which increases potential shock hazards.

Agreed hazards related to alternate paths

 

[mikeames]

That's a pretty bad graphic for a few reasons.
1, The kerned text is way off, huge spaces.
2, More importantly its, inaccurate.