Transformers and Impedance Percentages

Out put of transformer short circuited and input connected to some varying voltage until its output current equals to the rated current of transformer.

(some voltage/rated voltage)x100= %z impedance.

Also see the previous posting http://cr4.globalspec.com/thread/41265

- MS

By definition; The %impedence voltage is the three phase voltage expressed in percentage of the rated voltage, which when applied at one winding, with the other winding all three phases shorted (bolted), produces rated current in the winding to which voltage is applied.

If you are calculating it, it is nothing but impedance of primary plus impedance of secondary as measured at primary. The copper loss/coreloss parallel path also is to be considered, but effect is negligibel when calculating short circuit impedance as much lesser impdedance is in parallel.

Now the use of % impedence: The short circuit current through the trasformer is required for relay settings calculation, busbar/breaker adequacy calculation etc.

Say %Z of rated voltage drives rated current in primary and secondary winding. So if we apply full rated voltage it will be (1/%Z)x100 times the rated current.

For eg. If 20% is the impedance voltage of a transformer, then (1/20)x100=5 times rated secondary current would be the current from the transformer to a 3 phase fault at a bus directly connected to it.

Single phase to ground and three phase are also calculated some what in simillar way, but is more complex.

Hope clarifies.