Introduction to Electrical Power and Power Electronics

215Fault Current Analysis

Solution:

From the infinite main bus to the fault, with all cable impedances negligible, the only

impedance is that of the transformer, which is 5%. Therefore, the bus contribution

to the fault current I

fault.bus

= 100% ÷ 5% = 20 pu (note that this is pu, not percent).

The rated current on the transformer secondary side = 2,000,000 ÷ √3 × 480 =

2406 A/ph. This and all currents below are symmetrical rms values.

Therefore, CB4 current = generator contributions through 2-MVA transformer

secondary

= 20 × 2406 = 48,120 A.

motorrated current =

W500 746

3 460 092090

5

×

×××

=

..

665A

.

Using multiple M = 3 for a 460-V motor as per Section ...