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Re: Calculating AIC ratings at service.

Here’s one simple way - not as precise as a computer program can do, and with no certifications behind it.

Use Ohm’s Law in the form: R = E/I. If a 240V source has 13,012 amps available at the secondary, then the resistance of the transformer can be calculated as 0.0184 ohms. From Table 8, the DC resistance of 40 feet of #2 aluminum is (0.319 ohms/1000 ft) x (40 ft) = 0.0128 ohms. Adding 0.0184 to 0.0128 gives 0.0312 ohms.

Now using the I = E/R version of Ohm’s Law: I = 240/0.0312 = 7,700 amps. A panelboard rated above 10K AIC would suffice. Note that if you went through the trouble to find an AC resistance (not just the DC resistance), the AIC result would be smaller. So using DC resistance is conservative.

However, I agree with Bennie: This is not the kind of process that most electricians would (or even should) be called upon to undertake for a residential service. It is the kind of thing I would do for a complex distribution system, using a fancy computer program, and issuing the results under my seal and signature. {Please note that you may not use the above information as though it were issued under my seal - it is not!} But I can’t understand why an Inspector would want such a calculation, and why an Inspector would be willing to accept it without a PE seal.

The “Other Charlie.”

Charles E. Beck, P.E., Seattle

Comments based on 2017 NEC unless otherwise noted.

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