1
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 25 English
A
$$5.78\,\angle \, - {30^ \circ }$$
B
$$5.78\,\angle \, - {90^ \circ }$$
C
$$6.33\,\angle \, - {90^ \circ }$$
D
$$10.00\,\angle \, - {30^ \circ }$$
2
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
3
GATE EE 2003
MCQ (Single Correct Answer)
+2
-0.6
A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive sequence, negative sequence and zero sequence impedance of the alternator are j0.1, and j0.04 respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive, negative and zero sequence impedances of the transmission line are j0.1 and j0.3 respectively. All per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage of the alternator neutral with respect to ground during the fault is
A
513.8 V
B
889.9 V
C
1112.0 V
D
642.2 V
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 27 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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