# Help needed for NCEES #530



## AD_power (Sep 19, 2012)

If somebody knows from what magic hat they pulled additional impedance of 0.025??? Transformer connected to the bus, it should be pretty simple and straightforward solution:

Isc= I on 480v side (1203A) / Z of transformer (0.04). I'm confused.


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## DK PE (Sep 19, 2012)

Your solution is only valid if for the "infinite bus" on the utility side. In this case they give you the capability of that side.


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## Flyer_PE (Sep 19, 2012)

There are a couple of threads that cover this: Thread 1, Thread 2


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## AD_power (Sep 20, 2012)

thanks a lot. Now I get it.


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## danderson (Sep 20, 2012)

As stated by other members here, the MVA is SO much easier on these type problems.

See...

http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf


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## JB66money (Apr 9, 2014)

When using the per-unit method there is an impedance associated with the fault duty. This obtained first by getting the per-unit apparent power which is found by using the 1000kVA rating of the transformer the system's base apparent power and converting it into MVA, which is 1MVA. Next the fault duty's per-unit value is S=40MVA / 1MVA = 40p.u. . Next the per-unit impedancnce associated with the fault duty is z = 12 / 40 = .025pu. Next we find the per-unit fault current is

if = 1 / (.04+.025) = 15.38pu.Next the base current on the secondary side of the transformer where the fault occurs is Ib = (1000*103) / (sqrt(3) * 480 ) = 1202.81Amps

Finally we determine the acctual fault current by multiplying the per-unit fault current by the base current on the secondary side of the transformer.

IF = if * Ib = (15.38)*(1202.81) = 18499 Amps.


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