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Fault in the Final branch circuit

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NickParker

Electrical
Sep 1, 2017
411
Is my approach right?
Fault_current_in_final_circuits_bve24v.png


The purpose is to ensure that there is enough fault current available for the breaker to trip.
 
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The fault current path will be through the branch circuit conductor and returning by the equipment grounding conductor.
In Canada, for a 15 Amp circuit, you would use the length and impedance of a # 14 AWG conductor from the transformer to the point of the fault and an equal impedance of a #14 AWG returning equipment grounding conductor.
For a 20 Amp circuit, you would use the length and impedance of a # 12 AWG conductor from the transformer to the point of the fault and an equal impedance of a #14 AWG returning equipment grounding conductor.
Our code, the CEC, gives an impedance table from which the voltage drops may be extrapolated rather than using the DC resistance.
This assumes that the fault is to equipment grounded by an equipment grounding conductor.
I am not familiar with grounding methods at your location.
But, the ground return path impedance should be considered.
If the ground return path is through grounded structural metal, then that part of the impedance may be lower than an equipment grounding conductor, but the impedance of the hot conductor may be greater.
Magnetic encirclement separating the supply conductors from the ground return path must be avoided. Such an arrangement may raise the impedance of the ground path alarmingly.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
The fault current does not flow through Branches 1 & 3, so only the impedance of Branch 2 is taken into account (in series with the 10m from 0001 to JB).
 
Using the CEC voltage drop tables, you show 122 M from the JB to the point of the fault.
If there is a solid ground at the point of the fault, and an infinite supply at the JB, the fault current will be 100 Amps.
If the fault current returns on an equipment grounding conductor the impedance will be doubled and the fault current will be 50 Amps.
Either case will trip a 15 Amp breaker, but it is unlikely that either one will trip on instantaneous.
Assumptions for simplicity:
Infinite current at the JB.
table value of 61.4 M rounded to 61 M. 0.66% error.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
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