The percent impedance is obtain via a short circuit test. The secondary terminals are shorted and the %Z is actually the %IZ (percent voltage) that is necessary on the primary to create rated full load current to flow on the secondary. The rated full load current is based on the base / air cooled rating so that means the % impedance is also based on the base rating.

This is used for the

*infinite bus short circuit calculation*.

If %Z (primary voltage) results in the rated full load current (FLA)

Then 100% voltage results in the maximum short circuit amperes (SCA) on the secondary.

- %Z results in FLAsecondary
- 100% results in SCAsecondary

Rearranging the two relationships produces the well known (?) equation:

- Maximum SCA = (FLAsecondary x 100) / %Z

Many refer to this as the infinite bus short circuit calculation.

The FLAsecondary is the based on the lower/base kVA rating as others have noted.

Here are two examples to illustrate what would happen if the higher kVA rating is used:

**Example 1:** Short circuit current using the 3000 / 3360 kVA AA/FA and 5.75%Z - I assumed 480 volts

AA/FA where AA is the air cooled (base) rating and FA is the fan cooled rating

Using the air cooled rating (AA):

- FLA = 3000kVA / [0.48kV x sqrt(3)] = 3608.4 Amps

- SCA = [3608.4 x 100] / 5.75% = 62,760 amps (rounded up)

**Example 2: **If the 3360 kVA FA fan cooled rating was used:

The %Z would be higher. Remember it is the % Primary Voltage that results in the FLA. If the larger kVA is used, the larger FLA is also used which results in the %Z = 6.44% (scaled by the kVA ratings)

In this case:

- FLA = 3360 kVA/[0.48kV x sqrt(3)] = 4041.5 Amps

- SCA = [4041.5 x 100] / 6.44% = 62,760 (rounded up)

If the transformer was tested with the larger kVA rating, the %Z would be greater but the end result of the short circuit calculations would be the same.

Hope this helps shed a little more light on this topic (or confusion - it's a fine line

)