Electrical Engineering · Question Bank

Faults

106 questions · page 4 of 6. Answer in your head first, then check the solution.

#61📜 EE Pre-Board March 2014📘 PowerhouseMedium
The positive, negative and zero-sequence reactances of a 20 MVA, 13.2 kV synchronous generator are 0.3 pu, 0.2 pu and 0.1 pu, respectively. The generator is solidly grounded and is not loaded. A line to ground fault occurs on phase a. Neglecting all the resistances, determine the fault current.
A2,916 A
B4,374
C2,525
D5,050
#62📜 EE Pre-Board March 2014📘 PowerhouseEasy
The reactance of a generator designated “X” is given as 0.25 per unit based on the generator’s nameplate rating of 18 kV, 500 MVA. The base for calculation is 20 kV, 100 MVA. Find “X” on the new base.
A0.0405 pu
B0.0504 pu
C0.0450 pu
D0.0540 pu
#63📜 EE Pre-Board March 2014📘 PowerhouseHard
The sequence components of the line A current are as follows: \(I_{a0} = 14.13\angle 17.34^\circ\) A, \(I_{a1} = 708.26\angle -31^\circ\) A and \(I_{a2} = 2.98\angle 10.06^\circ\) A. Find the line current \(I_b\) in amperes.
A\(695\angle 205^\circ\) A
B\(695\angle -155^\circ\) A
C\(695\angle 151^\circ\) A
D\(695\angle 209^\circ\) A
#64📜 EE Pre-Board March 2014📘 PowerhouseMedium
In a short circuit analysis, the positive, negative, and zero sequence impedances are 0.20 pu, 0.20 pu, and 0.25 pu, respectively, using a base MVA of 50. Estimate the fault current on the faulted lines if a line to line fault at the 138 kV level occurs.
A1,204 A
B2,718 A
C1,569 A
D906 A
#65📜 EE Pre-Board August 2014📘 PowerhouseEasy
What is the equivalent of \(a^2 + a + j\), where \(a = 1\angle 120^\circ\)?
A1 + j
B1 − j
C−1 + j
D−1 − j
#66📜 EE Pre-Board August 2014📘 PowerhouseMedium
In a 13.8 kV, 10 MVA base three phase system, a phase to phase fault occurs. If the Thevenin’s equivalent impedance of the system is 2.63%, what is the magnitude of the fault current?
A13,800 A
B15,900 A
C27,600 A
D7,970 A
#67📜 EE Pre-Board August 2014📘 PowerhouseMedium
The three unbalanced currents are \(I_a = 10\ \text{cis}\ (-30^\circ)\), \(I_b = 0\) and \(I_c = 10\ \text{cis}\ 150^\circ\). Find the phase C positive sequence current.
A5.77 A
B\(5.77\ \text{cis}\ 120^\circ\)
C\(8.66\ \text{cis}\ 120^\circ\)
D\(5.77\ \text{cis}\ 240^\circ\)
#68📜 EE Pre-Board August 2014📘 PowerhouseEasy
When a line to ground fault occurs, the current in a faulted phase is 100A. The zero sequence current in the case will be
Azero
B66.6 A
C33.3 A
D100A
#69📜 EE Pre-Board August 2014📘 PowerhouseEasy
In which type of faults given below all the 3 components \(I_{a0}\) , \(I_{a1}\) and \(I_{a2}\) are equal?
ASingle line to ground fault
BLine to line fault
CDouble line to ground fault
DThree phase fault
#70📜 EE Pre-Board August 2014📘 PowerhouseHard
A synchronous generator having a subtransient reactance of 0.15 pu and operating at 5 percent above its rated voltage supplies a synchronous motor having a 0.2 pu subtransient reactance. The motor is connected to the generator by a transmission line and a transformer of total reactance 0.305 pu. A sudden three-phase short circuit occurs at the generator terminals. Determine the per unit subtransient fault current.
A−j6.443 pu
B−j7.443 pu
C−j9.079 pu
D−j10.079 pu
#71📜 EE Pre-Board August 2014📘 PowerhouseEasy
For a single line to ground fault the zero sequence current is given by j3.0pu. The current carried by the neutral during the fault is
Aj1.0pu
Bj3.0pu
Cj9.0pu
Dj6.0pu
#72📜 EE Pre-Board March 2015📘 PowerhouseEasy
For a single line to ground fault the zero sequence current is given by j3.0 pu. The current carried by the neutral during the fault is
Aj 1 pu
Bj 3 pu
Cj 9 pu
Dj 6 pu
#73📜 EE Pre-Board March 2015📘 PowerhouseEasy
Fault level means
AVoltage at the point fault
BFault current
CFault power factor
DFault MVA
#74📜 EE Pre-Board March 2015📘 PowerhouseEasy
A 150-MVA, 13.8-KV, Y-connected, 3ɸ, 60-HZ alternator is operating at rated voltage and at no-load when a three-phase faults occurs at its terminals. If the substransient reactance of the alternator is j0.2 pu, calculate the current the instant fault occurs.
A10,870 A
B6,276 A
C32,611 A
D31,380 A
#75📜 EE Pre-Board March 2015📘 PowerhouseEasy
The per unit value of a 2-ohm resistor at 100 MVA base and 10 KV base voltage is
A4 pu
B0.5 pu
C0.2 pu
D2 pu
#76📜 EE Pre-Board March 2015📘 PowerhouseEasy
The reactance of a generator designated X” is given as 0.25 per unit based on the generator’s nameplate rating of 18 kV, 500 MVA. The base for calculations is 20 kV, 100 MVA. Find X” on the new base.
A0.0617 pu
B0.0405 pu
C0.0081 pu
D0.0450 pu
#77📜 EE Pre-Board March 2015📘 PowerhouseEasy
The line currents in amperes in phases a, b and c respectively are 500 + j150,100 – j600, and -300 + j600. Determine the zero sequence component of phase a current
A-100 – j50 amps
B-100 + j50 amps
C100 – j50 amps
D100 + j50 amps
#78📜 EE Pre-Board March 2015📘 PowerhouseEasy
There is no zero sequence component of the fault current for
ALine to line fault
BSingle line to ground fault
CSymmetrical fault
DDouble line to ground fault
#79📜 EE Pre-Board March 2015📘 PowerhouseMedium
At a certain point of the system network the positive, negative and zero sequence impedances are 0.25 pu, 0.25 pu, 0.3 pu, respectively. The base MVA is 100. The level at that point is 34.5 KV. Determine the zero sequence current for a one line to ground fault.
A6,275 A
B7,516 A
C8,132 A
D2,091A
#80📜 EE Pre-Board August 2015📘 PowerhouseEasy
The per unit impedance of a circuit element is 0.30. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element will be
A0.3
B0.003
C0.6
D0.006

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