1
GATE EE 2012
+1
-0.3
The sequence components of the fault current are as follows:
$${{\rm I}_{positive}} = j1.5\,pu,\,\,{{\rm I}_{negative}} = - j0.5\,\,pu,$$
$${{\rm I}_{zero}} = - j1\,\,pu.$$ The typeof fault in the system is
A
$$LG$$
B
$$LL$$
C
$$LLG$$
D
$$LLLG$$
2
GATE EE 2008
+1
-0.3
A 3-phase transmission line is shown in figure: Voltage drop across the transmission line is given by the following equation: $$\left[ {\matrix{ {\Delta {V_a}} \cr {\Delta {V_b}} \cr {\Delta {V_c}} \cr } } \right] = \left[ {\matrix{ {{Z_s}} & {{Z_m}} & {{Z_m}} \cr {{Z_m}} & {{Z_s}} & {{Z_m}} \cr {{Z_m}} & {{Z_m}} & {{Z_s}} \cr } } \right]\left[ {\matrix{ {{i_a}} \cr {{i_b}} \cr {{i_c}} \cr } } \right]$$\$
Shunt capacitance of the line can be neglect. If the line has positive sequence impedance of $$15\,\,\Omega$$ and zero sequence in impedance of $$48\,\,\Omega ,$$ then the values of $${{Z_s}}$$ and $${{Z_m}}$$ will be

A
$${Z_s} = 31.5\,\Omega ;\,\,{Z_m} = 16.5\,\Omega$$
B
$${Z_s} = 26\,\Omega ;\,\,{Z_m} = 11\,\Omega$$
C
$${Z_s} = 16.5\,\Omega ;\,\,{Z_m} = 31.5\,\Omega$$
D
$${Z_s} = 11\,\Omega ;\,\,{Z_m} = 26\,\Omega$$
3
GATE EE 1997
+1
-0.3
For a fault at the terminals of a synchronous generator, the fault current is maximum for a
A
3-phase fault
B
3-phase to ground fault
C
line-to ground fault
D
line-to-line fault
4
GATE EE 1996
+1
-0.3
For an unbalanced fault, with paths for zero sequence currents, at the point of fault
A
The negative and zero sequence voltages are minimum.
B
The negative and zero sequence voltages are maximum.
C
The negative sequence voltage is minimum and zero sequence voltage is maximum.
D
The negative sequence voltage is maximum and zero sequence voltage is minimum.
GATE EE Subjects
Electric Circuits
Electromagnetic Fields
Signals and Systems
Electrical Machines
Engineering Mathematics
General Aptitude
Power System Analysis
Electrical and Electronics Measurement
Analog Electronics
Control Systems
Power Electronics
Digital Electronics
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