High impedance EGC.

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Miguel c

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Suppose we have a line at 120 V 200 ft long 3 AWG wire, impedance 0.05 ohms
Circuit breaker lets say is a 200 A one
Now there is a grount fault and we have a 14 AWG EGC wire, 200 ft long, (violation) the EGC's impedance is 0.6344 ohmios, therefore the graunt fault current is 175 A, hence the breaker does not trip, now the metal parts are energized. Would that provide the potential for electric shock taking into account we still have an effective ground fault current path and the neutral is bonded to ground?
 
In my opinion the EGC would no longer be considered an effective ground fault path.

Because as your math shows it would not activate the OCPD to open and leave all metal parts connected to where the ground fault occurred energized.

But why is the 14AWG the EGC for a 200amp OCPD? Or just a question for theory’s sake.
 
The metal parts of the equipment will have a voltage to earth that is equal to the voltage drop on the EGC...until the EGC burns open, then the voltage on the equipment to earth will be the circuit line to neutral voltage.

That is not an effective ground fault current path because it does not flow enough current to cause the OCPD to open the circuit.
 
Have a look at the table midway through:

American wire gauge - Wikipedia

en.wikipedia.org en.wikipedia.org


166 amps melts #14 in 10 seconds, so 175 amps would melt open the EGC in a bit faster than 10 seconds.


This is why you need to upsize the equipment grounding conductor in Table 250.122 whenever delayed clearing is present.

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But to answers you question using a resistive divider:

Upon initiation of the fault voltage between exposed metal at the fault point and earth will be 111 volts.

As the EG conductor heats up resistance will go up, at 300*C the resistance will double to about 1.2688 ohms resulting in 115 volts to earth.

One the EGC melts open 120 volts will be present to earth.
 
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