1
GATE EE 2004
MCQ (Single Correct Answer)
+2
-0.6
A $$3$$-phase generator rated at $$110$$ MVA, $$11$$ kV is connected through circuit breakers to a transformer. The generator is having direct axis sub-transient reactance $$X'{'_d} = 19\% ,\,\,$$ transient reactance $$X{'_d} = 26\% \,\,$$ and synchronous reactance $$=130$$%. The generator is operating at no load and rated voltage when a three phase short circuit fault occurs between the breakers and the transformer. The magnitude of initial symmetrical rims current in the breakers will be
A
$$4.44$$ kA
B
$$22.20$$ kA
C
$$30.39$$ kA
D
$$38.45$$ kA
2
GATE EE 2004
MCQ (Single Correct Answer)
+2
-0.6
A 3-phase transmission line supplies $$\Delta - $$ connected load Z. The conductor $$'c'$$ of the line develops an open circuit fault as shown in figure. The currents in the lines are as shown on the diagram. The +ve sequence current component in line 'a' will be GATE EE 2004 Power System Analysis - Symmetrical Components and Symmetrical and Unsymmetrical Faults Question 26 English
A
$$5.78\,\angle \, - {30^ \circ }$$
B
$$5.78\,\angle \, - {90^ \circ }$$
C
$$6.33\,\angle \, - {90^ \circ }$$
D
$$10.00\,\angle \, - {30^ \circ }$$
3
GATE EE 2004
MCQ (Single Correct Answer)
+2
-0.6
A 500 MVA, 50 Hz, 3-phase turbo-generator produces power at 22 kV. Generator is Y-connected and its neutral is solidly grounded. Its sequence reactances are X1 = X2 = 0.15 and X0 = 0.05 pu. it is operating at rated voltage and disconnected from the rest of the system (no load). The magnitude of the sub-transient line current for single line ground fault at the generator terminal in pu will be
A
2.851
B
3.333
C
6.667
D
8.553
4
GATE EE 2003
MCQ (Single Correct Answer)
+2
-0.6
A three-phase alternator generating unbalanced voltages is connected to an unbalanced load through a 3-phase transmission line as shown in figure. The neutral of the alternator and the star point of the load are solidly grounded. The phase voltages of the alternator are
$${E_a} = 10\angle {0^ \circ }V,\,\,\,{E_b} = 10\angle - {90^ \circ }V,\,\,{E_c} = 10\angle {120^ \circ }\,\,V.\,\,\,\,$$ The positive sequence component of the load current is GATE EE 2003 Power System Analysis - Symmetrical Components and Symmetrical and Unsymmetrical Faults Question 28 English
A
$$1.310\angle - {107^ \circ }A\,$$
B
$$0.332\angle - {120^ \circ }A\,$$
C
$$0.996\angle - {120^ \circ }A\,$$
D
$$3.510\angle - {81^ \circ }A\,$$
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