Calculate available short circuit current at the transformer secondary and at the end of a conductor run. Required for selecting properly rated breakers, fuses, and equipment.
Please check your inputs.
Affiliate link – we may earn a small commission at no cost to you
The available short circuit current is the largest current that can flow in a fault, and every breaker, fuse, and panel must be rated to interrupt it safely. For a three phase transformer, the full load current divided by the transformer impedance as a decimal gives the fault current at the secondary terminals. A 75 kilovolt ampere transformer at 208 volts has a full load current of about 208 amperes, and with 2 percent impedance the fault current is about 10,400 amperes.
The current falls along a run of conductor, because the wire adds impedance, so the value at the end of a long circuit is lower than at the transformer. The calculation here assumes a very strong source on the primary side, which gives a conservative, higher answer.
Equipment ratings must be equal to or greater than the available fault current, and the results are used to select breakers with a suitable interrupting rating. This work affects safety, so the calculation should be checked by a qualified engineer who has the utility data, and it must follow the local electrical code.