1
GATE EE 2003
+1
-0.3
A round rotor generator with internal voltage $${E_1} = 2.0\,\,$$ p.u.and $$\,X = 1.1\,\,$$ p.u. is connected to a round rotor synchronous motor with internal voltage $$\,\,{E_2} = 1.3\,\,$$ p.u. and $$\,X = 1.2\,\,$$ p.u. The reactance of the line connecting the generator to the motor is $$0.5$$ p.u. when the generator supplies $$0.5$$ p.u. power, the rotor angle difference between the machines will be
A
$${57.42^ \circ }$$
B
$${1^ \circ }$$
C
$${32.58^ \circ }$$
D
$${122.58^ \circ }$$
2
GATE EE 2003
+2
-0.6
A generator delivers power of 1.0 p.u. to an infinite bus through a purely reactive network. The maximum power that could be delivered by the generator is 2.0 p.u. A three-phase fault occurs at the terminals of the generator which reduces the generator output to zero. The fault is cleared after $${t_c}$$ seconds. The original network is then restored. The maximum swing of the rotor angle is found to be $${\delta _{\max }} = 110$$ electrical degree. Then the rotor angle in electrical degrees at $$t = {t_c}$$ is
A
55
B
70
C
69.14
D
72.4
3
GATE EE 2003
+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
+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
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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