Fault Current Path

Benihana

Member
Location
Suffolk County, NY
Occupation
Electrical Engineer
Can somebody explain how fault current's flow?

My understanding is that when a ground fault occurs, it will eventually flow back to the main service entrance. Then, since it is connected to either ground, building steel or main water pipe, the small resistance creates a large current which trips the breaker. Is that true?

And does the same apply for a short condition?
 
If I understand your question correctly: You understand how a line-to-line or line-to-neutral short causes high current and tripped OCPDs, right?

Connecting the grounding system to a service conductor (usually a neutral) means a fault to a grounded surface causes a similar current to flow.
 
Yes, I'm asking how either a short or a ground fault creates high currents which trip OCPDs.

I have an idea but I want to confirm with others so I know I'm not just making stuff up in my head.
 
Yes, I'm asking how either a short or a ground fault creates high currents which trip OCPDs.
Then it's a simple matter of Ohm's Law: one volt is capable of forcing one amp through one ohm.

The current increases inversely with the impedance (resistance plus reactance) of the circuit, up to the capacity of the source.

We introduce intentional weak links in the circuit to minimize damage.


Ask for more detail where you want it.
 
In a technical sense, V=IR.

If you have a voltage and the R is SUPER DUPER small then I is a big number.

So what makes that resistance/impedance number small? Good conductors.

So then why does the OCPD trip? because energy from the source of the fault will pour into the the low impedance until something disconnects it.

That can be a thermal element melting in the OCPD or the magnetic spring opening the circuit.

Then it's a simple matter of Ohm's Law.

The current increases inversely with the impedance (resistance plus reactance) of the circuit, up to the capacity of the source.

We introduce intentional weak links in the circuit to minimize damage.

You beat me to it lol.
 
Can somebody explain how fault current's flow?

My understanding is that when a ground fault occurs, it will eventually flow back to the main service entrance. Then, since it is connected to either ground, building steel or main water pipe, the small resistance creates a large current which trips the breaker. Is that true?

And does the same apply for a short condition?
In general that has nothing to do with fault clearing. The fault clearing path is from the ungrounded conductor, to the EGC, to the main bonding jumper, to the grounded service conductor, back to the transformer where the current originated from.
 
Yes, I'm asking how either a short or a ground fault creates high currents which trip OCPDs.

I have an idea but I want to confirm with others so I know I'm not just making stuff up in my head.
When discussing "fault currents" you need to understand there is little difference between a short circuit and a ground fault.
Basically a short circuit is a fault between two normal current carrying conductors, while a ground fault involves a non-normal current carrying path. Earth/dirt does not come into play for voltages less than 2.4kV.
 
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