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How do you go about finding transformer secondary side fault current via infinite buss method?
 

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(100*kVA)/(1.732*kV*%Z)
 

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You need the kVA of the transformerand the % impededance. Take the full load amps and multiply that by 100 and then divide that answer by the % impedance
 

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The percent impedance is how much current has to be applied to the primary to get FLA on the secondary. So if you're apply 100% to your primary, all you gotta do is multiply your secondary FLA by 100/name-plate impedance.

E.g.: 278A(100/4%) = 6,950A SCC
 

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Never seen a 7000A fuze before !
Any one else seen one?
Yeah, you just stick a bunch of 2500A fuses together. :whistling2:

The short-circuit current rating has to do with calculating whether the equipment will withstand a bolted fault, the only thing that it's used for when selecting fuses is to make sure the AIC is greater than that value so the fuse doesn't rupture.


Somebody posted a great picture of a chart they'd made that showed the fusable points of household objects. I think "ricotta cheese" was 0.5 amps. :laughing:
 
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