Ineffective ground fault current path

[Miguel c]

Hi guys, here I have this hypothetical scenario.
We have an effective ground fault current path, an excellent 8 copper EGG, suddenly someone who didn't know much about electricity decided to change the EGC to another type of EGC material with an impedance of 9.5 ohms, but the phase conductor is an excellent 8 AWG wire and has an impedance of 0.05 (9.5 + 0.05 omhs = 10 omhs)
Now there is a ground fault, 120V ÷ 10 = 12A, but the OCPD breaker is rated 30A, therefore it does not trip, now we could say that we no longer have an effective ground fault current path because the installation does not meet the code requirments for the EGFCP definition. However, the question is: Would someone be exposed to receiving a possible electric shock? or Could we assume that the installation continues to be safe since the neutral and the EGC are linked together at one point (MBJ)? What would be the load metal enclouse voltage to earth? 0 V?

 

[ptonsparky]

There will be a raised potential but how much IDK.

12 amp of fault current will never trip a 30 amp breaker. It is just a load.

 

[Miguel c]

There will be a raised potential but how much IDK.

12 amp of fault current will never trip a 30 amp breaker. It is just a load.

So you mean the load's metal enclousure would be closed to 120 V to ground but the difference in potential would be 0 V?

 

[mbrooke]

119 volts at the fault point to earth based on a 120 volt supply.

A low impedance EGC reduces touch voltage.

 

[suemarkp]

Do you know what a voltage divider circuit is? Two resistors (wires) in series (in this case 9.5 ohm and .05 ohm) with 120V across them, the divider midpoint voltage will be what is placed on metal objects. You'll get about .63V of drop from the line wire, so 119.4 volts would be on the metal enclosure (120 - 120 * .05/9.55). If the line wire was 9.5 ohms and the equipment ground was .05 ohms, then the voltage on the metal enclosure would only be .63.