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"It is also the percentage of the normal terminal voltage at on side required to circulate full-load current under short circuit conditions on other side."
The impedance of a transformer has a major effect on system fault levels. It determines the maximum value of current that will flow under fault conditions.
It is easy to calculate the maximum current that a transformer can deliver under symmetrical fault conditions. By way of example, consider a 2 MVA transformer with an impedance of 5%. The maximum fault level available on the secondary side is:
2 MVA x 100/5 = 40 MVA
and from this figure the equivalent primary and secondary fault currents can be calculated.
A transformer with a lower impedance will lead to a higher fault level (and vice versa).
The figure calculated above is a maximum. In practice, the actual fault level will be reduced by the source impedance, the impedance of cables and overhead lines between the transformer and the fault, and the fault impedance itself.
A transformer’s impedance (also referred to as impedance-voltage) is equal to the voltage, in % of its rated-voltage, that when applied to the primary-winding of a transformer will cause rated-current to flow in its shorted secondary-winding. For example, consider a transformer rated 500kVA, 6kV/400V, and 4.0% impedance. It means that when 240V is applied to the primary-winding, then the rated current, 720A, will flow in the shorted secondary.
To elaborate, consider a two-winding transformer having an equal number of turns in its primary and secondary windings. The effective transformer resistance is the sum of the primary and secondary resistances. Similarly, the effective transformer reactance is the sum of the primary and secondary leakage reactances. Finally, the effective transformer impedance is the vector addition of its effective resistance and reactance.
If additional information is required, let me know!
Regards, Phil Corso (cepsicon@aol.com)
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