Jpflex
Electrician big leagues
- Location
- Victorville
- Occupation
- Electrician commercial and residential
Is the formula for available fault current for a transformer KVA divided by secondary voltage for single phase or voltage x 1.732 for 3 phase
Then the resultant secondary I amperes is divided by the transformers impedance percentage?
But if this is correct how do you compute primary utility conductors resistance/ figures and therefore small reduction to maximum available fault current for NEC required labeling?
Then the resultant secondary I amperes is divided by the transformers impedance percentage?
But if this is correct how do you compute primary utility conductors resistance/ figures and therefore small reduction to maximum available fault current for NEC required labeling?