They say 200kAIC on the load side of the transformer? That means you have either a transformer way bigger then needed, or a transformer impedance that is unlikely small. In that regard, I believe there are some questions which you can ask the utility that will help you figure this out.

1. Verify the 200kAIC is the number they want you to use as the "Available short Circuit Current" on the 208V side of the utility transformer.

2. What is the size of the transformer for your service in KVA or MVA, and the impedance of the transformer. This is going to tell you the maximum amount of fault current available for your installation. Also, get the HV and LV side rated voltages.

(Keep in mind the answer from No.1 above may not match what you obtain by calculation from No.2. If that's the case, it could be they are giving you the ultimate long-term worst case)

Add your cable into the mix. This should drop your current quite a bit, as the impedance of the cable will act like a choke.

The MVA Method of calculating fault current works quite well for this situation; following is an example:

Isc = 200KA @ 208V, 3 phase. Therefore you have:

MVAsc = 1.732*208*200KA = 72MVAsc

Now determine your cable MVA rating. Assume (6) 750kcmil per phase and the run is 200ft;

MVAcble = (208)^2/((0.048/1000)*200) = 4.5MVA / ph and 4.5MVA*6 = 27MVAcble

Series MVA combine like resistance in parallel, so;

(72MVA*27MVA)/72MVA+27MVA) = 19.64MVA @ the service

Isc=19.64MVA/(208*1.732) = 54.5kA

Calculating backwards, you would need somewhere in the vicinity of a 1400ft run of (6) 750KCMIL per phase to get you down to 10KAIC. Not practical.

All in all, I'd say the numbers don't add up. Additional clarification is needed from utility.

**"Just because you're paranoid, doesn't mean they're not out to get you"**

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