When selecting circuit breakers and fuses for the protection of an electrical system, the amount of short-circuit current available must be known at the terminals of the transformer. This is to determine the mechanical withstand rating of the circuit breaker to withstand interruption during short circuit condition.
Percent impedance describes that percentage of the rated voltage required to produce full load current while the transformer output is shorted. (Eaton, 2015).
Thus, the formula of short circuit using percent impedance, Isc = IFL x (100 / %Z)
- 75 kVA
- 240 V secondary
- 5% impedance
Full load secondary current = 75,000/240 = 312 Amperes
Applying formula: Isc = IFL x (100/%Z).
Isc = 312 / (100 / .05) = 6,240 Amps.
The highest short circuit current in this example is 20 x the FLA ( 6240 / 312).
The figure above simply tells us that if the % impedance of the transformer decreases, the SC current will also increase proportionally.
In this example, a decrease of transformer impedance to 2.5% also makes the SC current increase to 40x.
...in this example we need to select a circuit breaker with kAIC rating of not less than 6,240 amps for 5% transformer impedance. The improper selection of circuit breaker kAIC rating will completely destroy the equipment and may cause huge disaster.
Note: This is just a brief example on how % impedance affects SC analysis because there are also several factors that need to be considered in computing SC in the system. For more detailed SC analysis you ay read How to Solve Short Circuit Calculation Using Point to Point Method.
Any comments are welcome