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Question

# A 500 kVA, 2.5 kV generator with transient reactance of 8% is connected to a bus through a circuit breaker as shown in below figure. The synchronous motor each rated 250 kVA, 2.5 kV. The sub transient reactance of each of the motor being 20%. If a 3-Î¦ short circuit fault occur at point A. Then the current contribution of each motor in fault current (in p.u.) is (Assume prefault voltage is rated voltage)

A
-j7.5
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B
-j2.5
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C
-j12.5
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D
-j5
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Solution

## The correct option is B -j2.5Let the base kVA be 500 kVA and base voltage be 2.5 kV, Per unit transient reactance of generator, X′g=j8100=j0.08p.u Per unit subtransient reactance of each motor, X′′m=j0.2×500250=j0.4p.u. Per unit reactance diagram is shown below, Thevenin reactance when viewed from fault terminals, Xth=j0.43×j0.08j0.43+j0.08=j0.05 p.u. At fault location Vth= rated voltage Fault current at F,If=1j0.05=−j20p.u The generator contribution is, Ig=−j20×j0.43j0.43+j0.08 Ig=−j12.5p.u. Contribution of motors, 3Im=If−Ig=−j20−(−j12.5) 3Im=−j7.5 Im=−j2.5p.u.

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