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Discussion Starter · #1 ·
Hey all.

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.

Thanks in advance
 

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Discussion Starter · #3 ·
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.
 

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Discussion Starter · #7 ·
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?
 

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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?
Looks good to me.

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.
 
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