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Determine the approximate short circuit current available Jump to latest post
1.  Junior Member
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May 2016
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Reputation Determine the approximate short circuit current available

Using the given information, determine the approximate short circuit current available at the transformer secondary: 13,800-277Y/480V, 2500 kVA, 5% Impedance

60,000 A
54,500 A
42,000 A
37,500 A
25,000 A

first should be 480Y/277 not 277y/480.

basically this would be the calculation. I= (2500*1000)/(480*1.732)= 3007.08
z= (100/5)= 20
short circuit = 3007.08*20= 60189

so i would just go with the closest answer correct? and also they should fix the error on 277Y.  Reply With Quote

2.  Junior Member
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Jan 2015
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Minnesota
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Reputation Agreed

I work it out to the same answer. Seems you did it right and I would go for the closest answer. Originally Posted by rasilva Using the given information, determine the approximate short circuit current available at the transformer secondary: 13,800-277Y/480V, 2500 kVA, 5% Impedance

60,000 A
54,500 A
42,000 A
37,500 A
25,000 A

first should be 480Y/277 not 277y/480.

basically this would be the calculation. I= (2500*1000)/(480*1.732)= 3007.08
z= (100/5)= 20
short circuit = 3007.08*20= 60189

so i would just go with the closest answer correct? and also they should fix the error on 277Y.  Reply With Quote

3. The questions clearly states "determine the approximate short circuit current". Go with the closest answer.  Reply With Quote

4.  Junior Member
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May 2016
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Reputation  Originally Posted by madMAX The questions clearly states "determine the approximate short circuit current". Go with the closest answer.

yes sir but its wrong wording that what confused 277y/480 doesn't exist, 480y/277 exist  Reply With Quote

5.  Junior Member
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Apr 2017
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1
Reputation fault current Originally Posted by rasilva Using the given information, determine the approximate short circuit current available at the transformer secondary: 13,800-277Y/480V, 2500 kVA, 5% Impedance

60,000 A
54,500 A
42,000 A
37,500 A
25,000 A

first should be 480Y/277 not 277y/480.

basically this would be the calculation. I= (2500*1000)/(480*1.732)= 3007.08
z= (100/5)= 20
short circuit = 3007.08*20= 60189

so i would just go with the closest answer correct? and also they should fix the error on 277Y.
2500 x 100
1.732 x .480 x5

250,000/4.1568= 60,142  Reply With Quote

6. Explain the math

Could someone explain the math a little more in depth? Never mind I was confused in Z, but looking over it again I understand.

Last edited by Ronwilson1801; December 29, 2018 at 09:27 AM.  Reply With Quote

7.  Junior Member Pro Subscriber
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Feb 2021
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Reputation A simple question hopefully

480 *1.732= 831.36

why do you multiply 480 by 1.732?  Reply With Quote

8. Originally Posted by carl21 480 *1.732= 831.36

why do you multiply 480 by 1.732?
Original equation is for single phase transformer:
IFLA=(KVA*1000)/EL-L

To calculate for 3 phase voltage, the square root of 3 (1.732) must be multiplied by the voltage:
IFLA=(KVA*1000)/(EL-L*1.732)

EL-L for both of theses transformers is 480V, one is single phase and the other is 3 phase.  Reply With Quote 