text: CEC - IC rating of breaker required | Electrician Talk
  IC rating of breaker required
  I've gone through my notes and past books but for the life of me I can't remember how to properly do a SYMMETRICAL RMS current calc to determine the min IC rating I need for a breaker both 3 phase and 1 phase.
  Could someone help me out with the required formulae? I would forever be grateful.
  Are you thinking of ball-park available fault current for a transformer?
  Divide the secondary full load current by the transformer % impedance, e.g.: 50kVA 208V 5% impedance = 139A 1Ø / 0.05 = 2.76kA
  Welkom trying to figure out why the company that built this building with 22ka ratings where I believe it only needs 10ka.
  The disty transformer is 150kva 600-120/208.
  Trying to figure out for branch ccts ranging from 2p30 and 2p and 3p 40
  Devices are mostly wood shop tools like planet and table saws and sanders.
  If the transformer had a low impedance and the panels are close by it could push you over 10kA pretty easily:
  416FLA / 0.03 = 13.9kA
  The 0.03 is what part f the equation? Is that the constant multiplied by Load?
  So if I'm understanding, then with this xfmr having an %z of 4.56 percent then it would be 416.667/0.0456=9137.434 symmetrical amps? Then a 10ka breaker would suffice?
  That calculation is also worst-case-scenario: It assumes infinite line-side capacity and zero load-side cable impedance. Since neither one of those are true, your true available fault current is always going to be lower than that.