# New NCEES #504



## le.boot (Apr 11, 2009)

I know this is probably really simple, but I'm having a little trouble with this problem.

"The diagram below represents the Thevenin equivalent of a single phase distribution system. A fault occurs between point X and ground. Rf represents the fault resistance. The current If is 3600A when Rf is 0 ohms. If Rf is changed to 1 ohm, the current If (amperes) is most nearly:"

I get that Xs is 2 ohms. What I don't get is how the solution goes from 1+j2 to (sqrt (1 squared + 2 squared))?

Thanks!


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## Flyer_PE (Apr 11, 2009)

Given the fault current with RF = 0, you know that the line impedance is j2 ohm.

Given a fault resistance RF of 1 ohm, the total impedance expressed in rectangular form is then R + jX = 1 + j2 ohm

The magnitude of the impedance in polar form is then sqrt(R2+X2) = sqrt(12+22).


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## cabby (Apr 14, 2009)

Attached is how I approached it.


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