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#### l0sts0ul

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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.

#### Big John

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##### Donuts > Fried Eggs
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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

#### l0sts0ul

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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.

#### Big John

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##### Donuts > Fried Eggs
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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

#### l0sts0ul

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The 0.03 is what part f the equation? Is that the constant multiplied by Load?

#### 99cents

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Estwing magic
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The 0.03 is what part f the equation? Is that the constant multiplied by Load?
Percent impedance

#### l0sts0ul

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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?

#### Big John

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##### Donuts > Fried Eggs
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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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