Mathcad - 05-26

Mathcad - 05-26 - KVA kW KVAR kW ohm Problem 5-26 solution...

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KVA kW KVAR kW ohm
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I fault 1.146 10 4 × amp = I fault I Lrated Z pu = so I Lrated 801.875 amp = I Lrated S 3 V s = If a fault occurs on the secondary with rated voltage applied, then all of the voltage must be dropped across the equivalent impedance. Thus the current in per unit, will be the rated line current divided by the per unit impedance
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Mathcad - 05-26 - KVA kW KVAR kW ohm Problem 5-26 solution...

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