Although the link given explains it..but one other thing is....
If you short the secondary terminals of this transformer and apply 5.4% of the primary voltage i.e 702 volts, you will have the name plate secondary current.
The maximum short circuit current at the output of the transformer is limited by the impedance of the transformer.
The inverse of impedance is the maximum short circuit current, so in your case it will be..
1/5.4 = 18.5 times the rated secondary current.
Cheers.
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He must be very ignorant as he answers all the questions he is asked. Voltaire