300+ REAL TIME Power Systems Objective Questions & Answers 2023

250+ TOP MCQs on Comparison of Faults in Three Phase System – 1 and Answers

Power Systems Multiple Choice Questions on “Comparison of Faults in Three Phase System – 1”.

1. A three phase transformer has a name plate details of 30 MVA and voltage rating of 230Y kV/69∆ kV with a leakage reactance of 10% and the transformer connection via wye-delta. Taking a base of 230 kV on the high voltage side, turns ratio of the windings is ______
A. 2
B. 1.5
C. 6
D. 4

Answer: A
Clarification: n= V_HV/V_LV = (230/√3)/69=1.9245.

2. A three phase transformer has a name plate details of 30 MVA and voltage rating of 230Y kV/69∆ kV with a leakage reactance of 10% and the transformer connection via wye-delta. Taking a base of 230 kV on the high voltage side, the transformer reactance on the LV side is
___________
A. 176.33 Ω
B. 1763.3 Ω
C. 47.6 Ω
D. 15.87 Ω

Answer: A
Clarification: Zb(HV) = kV_b(HV))²/MVA_b
Zb(HV) = 230²/30=1763.33 Ω
XΩ =0.1*1763.33 = 176.33 Ω.

3. A three phase transformer has a name plate details of 30 MVA and voltage rating of 230Y kV/69∆ kV with a leakage reactance of 10% and the transformer connection via wye-delta. Taking a base of 230 kV on the high voltage side, the transformer reactance referred to the low voltage side in ohms is __________
A. 47.61 Ω
B. 15.87 Ω
C. 176.33 Ω
D. 157.8 Ω

Answer: A
Clarification: Zb(HV) = (kV_b(HV)²/MVA^b = 230²/30=1763.33 Ω
XΩ(HV) =0.1*1763.33 = 176.33 Ω
XΩ(LV) =176.33/(1.92452) = 47.61 Ω.

4. A three phase transformer has a name plate details of 30 MVA and voltage rating of 230Y kV/69∆ kV with a leakage reactance of 10% and the transformer connection via wye-delta. Taking a base of 230 kV on the high voltage side, the transformer reactance referred to the low voltage side in ohms is
A. 0.1
B. 0.2
C. 0.198
D. 0.4

Answer: A
Clarification: XΩ(LV) = XΩ(HV) /n²
= 176.33/(230/69)² =15.87 Ω
Xpu= 15.87/158.7 = 0.1.

5. A 200 bus power system has 160 PQ bus. For achieving a load flow solution by N-R in polar coordinates, the minimum number of simultaneous equation to be solved is ___________
A. 359
B. 329
C. 360
D. 320

Answer: A
Clarification: Total buses = 200
PQ buses = 160
PV buses = 200-160 = 40
Slack bus = 1
Total number of equation = (40-1)*1 + (160*2) = 359.

6. Two alternators A and B having 5% speed regulation are working in parallel at a station. Alternator A is rated at 15 MW while B is at 20 MW. When the total load to be shared is 12 MW, then how much of the load will be shared by the alternator B?
A. 6.85 MW
B. 5.14 MW
C. 6 MW
D. 7 MW

Answer: A

7. Two alternators A and B having 5% speed regulation are working in parallel at a station. Alternator A is rated at 15 MW while B is at 20 MW. When the total load to be shared is 12 MW, then how much of the load will be shared by the alternator B?
A. 6.85 MW
B. 5.14 MW
C. 6 MW
D. 7 MW

Answer: A

8. A 400 V, 50 Hz three phase balanced source ripples to a star connected load whose rating is S(=300+j400) kVA. The rating of the delta connected capacitor bank needed to bring p.f. to 0.9 lagging is _______ KVAR.
A. 254.7
B. 25.4
C. 84.9
D. 284.5

Answer: A
Clarification: Qc = P[tanθ1 – tanθ2]
=300[(4/3) – cos^-10.9] = 254.72 kVAR.

9. A 400 V, 50 Hz three phase balanced source ripples to a star connected load whose rating is S(=300+j400) kVA. A delta connected capacitor bank needed to bring p.f. to 0.9 lagging. The operating power factor of the system is
A. 0.8
B. 4/3
C. 3/4
D. 0.6

Answer: A
Clarification: cosФ = 3/5 = 0.6.

10. A given system to be analysed was found with the below phasor representation of the system voltages. Which of the symmetrical components will be present in the mentioned system?
A. Positive sequence components
B. Negative sequence components
C. Zero sequence components
D. All of the mentioned

Answer: A
Clarification: In an unbalanced system, all the symmetrical components will be present.

11. The phasor operator which is used to depict the unbalanced phase voltages into three phase quantities, provides a rotation of
A. 120o counter clockwise
B. 120o clockwise
C. 90o counter clockwise
D. 90o clockwise

Answer: A
Clarification: ‘α’ operator used for the conversion has the counter clockwise rotation of the quantity and it has unity magnititude.

250+ TOP MCQs on Switchgear and Protection – 4 and Answers

Tough Power Systems Questions on “Switchgear and Protection – 4”.

1. An isolated generator connected to a turbine with its continuous maximum power of 20 MW, 50 Hz. Generator connected with two loads of 8 MW, each operate at 50 Hz.. Generator has 5% droop characteristic. If an additional load of 6 MW is added then the change in the frequency will be ____________
A. 0.25
B. 0.75
C. 0.5
D. 1

Answer: A

2. For the given system the Ybus matrix is show below.
D. None of the mentioned

Answer: A
Clarification: Y11=j2+(1/5j) = 1.8j
Y12=-(1/5j) = 0.2j; Y13=0
Hence option ‘a’ is most applicable.

3. If short TL is delivering a lagging power factor load, sending end p.f. would be ________
A. (V_rcosФ_r+IR)/Vs
B. (V_rcosФ_r+IRsinФ_r)/Vs
C. (V_rsinФ_r+IR)/V_s
D. (V_ssinФr+IRcosФr)/V_s

Answer: A

4. If all the sequence voltages at the fault point in a power system are equal, then fault is __________
A. LLG fault
B. LG fault
C. Three phase to ground fault
D. Line to Line fault

Answer: A
Clarification: It is LLG fault.

5. The below mentioned graph is a recorded data of the voltage fluctuation of the network, then it can be concluded that these faults are __________
A. L-G
B. A three phase motor short
C. Feeder fault
D. Any of the mentioned

Answer: A
Clarification: The sudden dip is the LG fault in the graph as the dips are not very large and the system is maintaining its stability.

6. The inertia constant of two groups of machines which do not swing together are 4 and 20 MJ/MVA, the equivalent inertia constant of the system is _________
A. 2.85
B. 14
C. 6
D. 12

Answer: A
Clarification: For the incoherent system, the equivalent inertia = H₁H₂/(H₁+H₂).

7. A fault current of 1500A passing through the primary side of a 400/5 CT. On the secondary side an inverse time over current relay is connected whose current setting is 75%, the plug setting is __________
A. 5
B. 3.75
C. 6
D. 4

Answer: A

8. The three phase line is working at 50 Hz with conductors rearranged as shown. The Cph of TL at 132 kV is ___________
A. 8.75*10^-9
B. 8.75*10^-12
C. 26.2*10^-12
D. 26.2*10^-9

Answer: A

9. For a 100 MVA, 11 kV three phase Alternator observes a three phase fault at terminals of it. If the fault current is 2000 A, the pu value of positive sequence reactance ______
A. 2.62
B. 0.46
C. 1.31
D. 0.92

Answer: A
Clarification: PU fault current = 2000/5248.63 = 0.381 pu
0.381 = 1/X
X= 1/0.381 =2.62 pu.

10. A lossless TL having SIL of 3000 MW is provided with uniform distributed series capacitance compensation of 25% and shunt compensation of 30% then the SIL of compensated TL will be _______________
A. 3949.9
B. 3289
C. 2784.9
D. 2877.6

Answer: A
Clarification: SIL=(V² )/Z
SIL α 1/Z
Zcomp = Zs√((1-kse)/((1+ksh)(1-Klsh)))
Kse = 0.25; Klsh = 0; Kcsh = 0.3,
Zcomp =0.7595
SILcomp = 0.7595*3000 = 3949.96 MVA.

11. A lossless TL having SIL of 3000 MW is provided with uniform distributed series capacitance compensation of 25% and shunt compensation of 30% then the compensated impedance of TL will be(in pu) ____________
A. 0.7595
B. 2.88
C. 0.647
D. 3.27

Answer: A
Clarification: SIL=(V²)/Z
SIL α 1/Z
Zcomp = Zs√((1-kse)/((1+ksh)(1-Klsh)))
Kse = 0.25; Klsh = 0; Kcsh = 0.3,
Zcomp =0.7595.

250+ TOP MCQs on Fuels and Answers

Power Systems Multiple Choice Questions on “Fuels”.

1. Which of the following is correct arrangement of different types of coals according to their increasing carbon content?
A. Peat, bituminous, lignite, anthracite
B. Peat, lignite, bituminous, anthracite
C. Bituminous, peat, lignite, anthracite
D. Peat, anthracite, bituminous, lignite
Answer: B
Clarification: Coal is the most commonly used fuel in thermal power plant. Coal contain moisture, carbon, hydrogen, sulphur, Nitrogen, Oxygen and ash. Coals are classified according to their calorific value and carbon content.

2. Which of the following is most suitable fuel for thermal power plant?
A. Peat coal
B. Lignite coal
C. Bituminous coal
D. Anthracite coal
Answer: c
Clarification: Peat and lignite coals have low calorific value and high ash content. Anthracite coals are found in small quantities and are therefore costlier for power generation. Bituminous coals have enough calorific value, low moisture and ash content so most suitable for power generation.

3. Read the below statements.
I. Sulphur is one of the combustible elements
II. The Sulphur content of coal should be less than 1%
A. Only I is correct
B. Only II is correct
C. I and II are correct
D. I and II are incorrect
Answer: C
Clarification: Sulphur content is one of the factors taken into account while selecting coal for steam power plant. Sulphur is combustible element and can produces energy. But its primary combustion product, Sulphur dioxide, is a health hazard due to which sulphur content in coal should be low.

4. Which of the following coal has highest moisture content?
A. Sub-bituminous coal
B. Bituminous coal
C. Anthracite coal
D. Peat coal
Answer: D
Clarification: Peat is the first stage in the process of transformation of buried vegetation into coal. It has about 60% to 90% moisture content and low carbon content (about 5% to 20%). Peat coal is not suitable for use in power plants.

5. Which of the following coal has highest enthalpy of combustion?
A. Peat coal
B. Sami-bituminous coal
C. Semi-anthracite coal
D. Lignite coal
Answer: B
Clarification: Peat and lignite coals have enthalpy of combustion about 3000 KJ/kg and 17600 KJ/kg. Semi-anthracite coals are harder than bituminous coals and have enthalpy of combustion up to 34750 KJ/kg while sami bituminous coals have that up to 35000 KJ/kg.

6. What is the main drawback of liquid fuel in thermal power plant?
A. High handling cost
B. Hard and Adhesive oil slag
C. Difficult to store
D. High cost of liquid fuel
Answer: D
Clarification: Handling and storing cost of oil fuels are less, load fluctuations can be made easily and few operations are needed. There is no disposal of ash. Superheaters are subjected to hard and adhesive oil slag which is often difficult to remove, but other heating surfaces are almost trouble free.

7. A minimum of 4 to 6 percent of ash is required in coal.
A. True
B. False
Answer: A
Clarification: This 3 to 6% of ash is required on hand fire grates and some stokers for protection of grades against overheating.

8. Grindability of coal should be low.
A. True
B. False
Answer: B
Clarification: Grindability index is inversely proportional to the power required to grind the coal to a certain fineness. Higher the grind ability factor index better the coal is.

9. The selection of coal for power plant does not depends on _________
A. Calorific value
B. Nitrogen content
C. Sulphur content
D. Grindability index
Answer: B
Clarification: The selection of coal for power plant depends on the number of factors such as calorific value, weather ability, sulphur content, ash content, particle size, grindability index, and caking characteristics. Nitrogen content is not needed for the selection of coal for power plants.

10. Read the following statements about coals.
I. Lignite coals can be used in thermal power plants
II. Super-anthracite coal are best for thermal power plants
A. Only I is correct
B. Only II is correct
C. I and II are correct
D. I and II are incorrect
Answer: A
Clarification: Lignite have 30 to 50% moisture and 20 to 40% carbon content and can be used in power generation. Super-anthracite coals are very hard with shiny black surface. It is non cooking and is very difficult to ignite and so, it has little importance in power generation.

250+ TOP MCQs on Inductance of a Single-Phase Two-Wire Line – 1 and Answers

Power Systems Multiple Choice Questions on “Inductance of a Single-Phase Two-Wire Line – 1”.

1. Effect of earth is considered insignificant while calculating the inductance of a single phase transmission line.
A. True
B. False

Answer: A
Clarification: The earth’s magnetic symmetry does not affect the inductance of the line.

2. Effect of earth’s presence on the magnetic field geometry is _________________
A. significant
B. insignificant
C. cannot be said
D. none of the mentioned

Answer: A
Clarification: This is because the relative permeability of earth is about the same as that of air and its electrical conductivity is relatively small.

3. The inductance of the transmission line is do not contain effect of earth electric field geometry while the capacitances contain.
A. True
B. False

Answer: A
Clarification: It is true as the earth’s magnetic field symmetry is insignificant to the inductance while the electric field symmetry is.

4. The end effect information is included in the inductance of a transmission line.
A. True
B. False

Answer: B
Clarification: Since TL is infinitely long, the end effects are ignored and they are not a part of the inductance formula.

5. The number of the strands included in the stranded cables while implementing a similar cable equivalent of single strand should be ________
A. 3x²-3x+1
B. 2x²-3x+1
C. 3x²-2x+1
D. 3x²-x+1

Answer: A
Clarification: It is 3x²-3x+1.

6. The overall diameter of a stranded conductor if ‘x’ is the number of layers and each layer has a diameter ‘d’ will be
A. (2x-1)d
B. (3x-1)d
C. (6x-3)d
D. (2x+1)d

Answer: A
Clarification: It is given by (2x-1)d.

7. _______ conductors are the latest trend of the extra high voltage lines.
A. Bundled
B. Parallel
C. ACSR
D. Bundled or ACSR

Answer: A
Clarification: The bundled ACSR conductors have corona free characteristics and they have good tensile strength.

8. A 10 MVA, 11 kV alternator has positive, negative and zero sequence reactance of 25%, 40%,10% respectively. The resistance that must be placed in the neutral such that zero sequence of fault current for a single phase to ground fault will not exceed the rated line current in ohms is
A. 2.66 Ω
B. 1.66 Ω
C. 12.1 Ω
D. 0.22 Ω

Answer: A

250+ TOP MCQs on Geometric Mean Distance and Geometric Mean Radius – 1 and Answers

Power Systems Multiple Choice Questions on “Geometric Mean Distance and Geometric Mean Radius – 1”.

1. Which of the system would require more insulation, a 400 kV ac line or 400 kV dc line?
A. 400 kV ac
B. 400 kV dc
C. Both will need same level of insulation
D. Data is insufficient to decide

Answer: A
Clarification: In ac system, the peak value attained will be 400*√2= 564 kV.

2. The amount of dc compensation needed by a 400 kV dc line for transmitting the data over 300 km having resistance as 5 ohms?
A. 400 var
B. 564 var
C. 600 VAR
D. None of the mentioned

Answer: A
Clarification: A dc line does not require any reactive compensation.

3. Steady state problem is not very serious problem in dc transmission line system due to which of the following systems?
A. No reactive compensation
B. Only resistance drop
C. Stable voltage regulation
D. No reactive compensation, resistance drop, voltage regulation

Answer: A
Clarification: Due to all the above reasons, mainly the reactive power compensation there is very less variation in stability.

4. DC line transmission can be used to provide _______ link.
A. asynchronous
B. synchronous
C. lossless
D. none of the mentioned

Answer: A
Clarification: As dc link provides to connect to different frequencies of the operating buses. It can transfer power frequency independent.

5. (i) Low short circuit current is required in dc system
(ii) Fast control of the converter switches cannot effectively control the damping of the oscillations.
Choose the most appropriate.
A. (i) and (ii) both are true
B. Only (ii) is false
C. Only (i) is false
D. Both statements are false

Answer: B
Clarification: It’s an advantage the using converters in dc link can control the fast damping of the oscillations which occur in ac systems.

6. The reactive power demand of the converters in HVDC can be provided by ___________
A. AC filters
B. Shunt Capacitors
C. SVC
D. AC Filters, Shunt Capacitors and SVC

Answer: D
Clarification: Reactive power demand by the HVDC system can be fulfilled by all the mentioned equipment’s.

7. As compared to the HVDC line, the corona and radio interference in a HVAC line are ____
A. more
B. lesser
C. same
D. cannot be determined

Answer: A
Clarification: The corona and radio interference losses are more in ac system.

8. What of the below mentioned statements are incorrect as compared to the HVDC system?
A. Back to back connection is possible
B. More corona losses
C. Distance limitation
D. Extra reactive power compensation

Answer: B
Clarification: Out of all the corona losses are less in HVDC system.

9. It is required to implement the HVDC system using the thyristors in large numbers. Why it becomes necessary to use a large number of thyristor switches?
A. Due to low voltage rating of thyristor switches
B. Due to low current rating of thyristor switches
C. Due to lack of response for open circuit fault
D. Cannot be determined

Answer: A
Clarification: Thyristors have low voltage rating and to achieve high voltage, a series of switches that have to used.

10. The corona loss on a particular system at 50 Hz is 1 kW/phase per km. The corona loss on the same system in case of DC transmission will be ____________
A. Zero
B. 0.5 kW/phase/km
C. 0.33 kW/phase/km
D. 0.6 kW/phase/km

Answer: C

250+ TOP MCQs on Comparison of Faults in Three Phase System – 2 and Answers

Power Systems Question Bank on “Comparison of Faults in Three Phase System – 2”.

1. The zero sequence depiction of the unbalancing occurring at the terminals of the induction motor will be most likely _____________

Answer: A
Clarification: The zero sequence components will not have any phase displacement and they are co- phasal in nature.

2. Complete the given phasor diagram by assuming that operator ‘a’ is unity magnitude and counter clockwise rotation of 120 degrees.
A. 1-a
B. a-1
C. 1
D. –a³

Answer: A
Clarification: Based on the phasor calculations the unknown vector is found using balanced condition of the three vectors A+B+C = 0 and 1+a+a²=0. Then the third vector is 1-a.

3. The unknown vector in the given figure is ______ if we assume the system is balanced with a as unity magnitude and counter clockwise rotation of 120 degrees.
A. a
B. 0
C. a³
D. a⁴

Answer: A
Clarification: 1+a+ a² = 0.

4. For a balanced three phase system having a vector representation as mentioned in the figure, complete the vector X and Y.
A. X=1-a² ; Y=a²-a
B. bX=a² -a; Y=1-a²
C. X=a² -1; Y=1-a
D. X=a² -1; Y=a-1

Answer: A
Clarification: This question can be attempted by using the options. For a balanced system the net voltage will be zero. So option a and b are the shortlisted options. Out of these applying the rotation criteria, option a is the best match.

5. For a balanced three phase system having a vector representation as mentioned in the figure, complete the vector X and Y.
A. X=√3∠30° ; Y = √3∠-90°
B. X=√3∠-30° ; Y = √3∠90°
C. X=√3∠-30° ; Y = √3∠-150°
D. X=√3∠30° ; Y = √3∠150°

6. The unknown vector in the given figure is ______ if we assume the system is balanced with a as unity magnitude and counter clockwise rotation of 120 degrees.
A. 1∠120°
B. 1∠30°
C. 1∠0°
D. 1∠-120°

Answer: A
Clarification: 1+a+ a² = 0; a= 1∠120°.

7. Complete the given phasor diagram by assuming that operator ‘a’ is unity magnitude and counter clockwise rotation of 120 degrees.
A. √3∠-30°
B. √3∠30°
C. √3∠-60°
D. √3∠90°

Answer: A
Clarification: Based on the phasor calculations the unknown vector is found using balanced condition of the three vectors A+B+C = 0 and 1+a+a²=0. Then the third vector is 1-a. Converting it in the polar form, it is √3∠-30°.

8. In the cylindrical rotor alternator, the sub transient and negative sequence reactances are same.
A. True
B. False

Answer: A
Clarification: It is true because the saliency offered is zero in the cylindrical pole machines.

9. The zero sequence impedance of a synchronous machine is independent of the pitch of the armature coils.
A. True
B. False

Answer: B
Clarification: The zero sequence impedance does depend on the pitch of the armature coils of the synchronous machine.