1
GATE EE 2006
MCQ (Single Correct Answer)
+2
-0.6
The $$A, B, C, D$$ constant of a $$220$$ $$kV$$ line are:
$$A = D = 0.94\,\angle \,10,\,\,\,B = 130\,\angle \,730,\,\,\,C = 0.001\,\angle \,900.\,\,$$ If the sending end voltage of the line for a given load delivered at nominal voltage is $$240$$ $$kV$$, the % voltage regulation of the line is
A
$$5$$
B
$$9$$
C
$$16$$
D
$$21$$
2
GATE EE 2006
MCQ (Single Correct Answer)
+2
-0.6
A generator feeds power to an infinite bus through a double circuit transmission line. A $$3$$ phase fault occurs at the middle point of one of the lines. The infinite bus voltage is $$1$$ pu, the transient internal voltage of the generator is $$1.1$$ pu and the equivalent transfer admittance during fault is $$0.8$$ pu. The 100 MVA generator has an inertia constant of $$5$$ MJ/MVA and it was delivering $$1.0$$ pu power prior of the fault with rotor power angle of $${30^ \circ }\,\,$$. The system frequency is 50Hz.

If the initial accelerating power is $$X$$ pu, the initial acceleration in elect deg/sec2, and the inertia constant in MJ-sec/elect deg respectively will be

A
$$31.4$$$$X,$$ $$18$$
B
$$1800$$ $$X,$$ $$0.056$$
C
$$X/1800,$$ $$0.056$$
D
$$X/31.4,$$ $$18$$
3
GATE EE 2006
MCQ (Single Correct Answer)
+2
-0.6
A generator feeds power to an infinite bus through a double circuit transmission line. A 3 phase fault occurs at the middle point of one of the lines. The infinite bus voltage is 1 pu, the transient internal voltage of the generator is 1.1 pu and the equivalent transfer admittance during fault is 0.8 pu. The 100 MVA generator has an inertia constant of $$5$$ MJ/MVA and it was delivering 1.0 pu power prior of the fault with rotor power angle of $${30^ \circ }\,\,$$. The system frequency is 50Hz.

The initial accelerating power (in pu) will be

A
1.0
B
0.6
C
0.56
D
0.4
4
GATE EE 2006
MCQ (Single Correct Answer)
+2
-0.67
Three identical star connected resistors of $$1.0$$ $$p.u$$ are connected to an unbalanced $$3$$ phase supply. The load neutral is isolated. The symmetrical components of the line voltages in $$p.u.$$ calculations are with the respective base values, the phase to neutral sequence voltages are
A
$${V_{an1}} = X\angle \left( {{\theta _1} + {{30}^0}} \right),\,\,{V_{an2}} = Y\angle \left( {{\theta _2} - {{30}^0}} \right)$$
B
$${V_{an1}} = X\angle \left( {{\theta _1} - {{30}^0}} \right),\,\,{V_{an2}} = Y\angle \left( {{\theta _2} + {{30}^0}} \right)$$
C
$${V_{an1}} = {1 \over {\sqrt 3 }}X\angle \left( {{\theta _1} - {{30}^0}} \right),\,\,{V_{an2}} = {1 \over {\sqrt 3 }}Y\angle \left( {{\theta _2} - {{30}^0}} \right)$$
D
$${V_{an1}} = {1 \over {\sqrt 3 }}X\angle \left( {{\theta _1} - {{60}^0}} \right),\,\,{V_{an2}} = {1 \over {\sqrt 3 }}Y\angle \left( {{\theta _2} - {{60}^0}} \right)$$
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