Faculty of Degree Engineering - 083 Department of ...

Faculty of Degree Engineering - 083

Department of Electrical Engineering - 09 MCQ Question Bank EPS-II 2160908

MCQ Questions & Answers

Electrical Power System



QUESTIONS

Unit: 1 Transmission line

1. Which of the following transmission line can be considered as short transmission line? a) Transmission line of length upto 600 Km b) Transmission line of length upto 500 Km c) Transmission line of length upto 200 Km d) Transmission line of length upto 80 Km 2. Which of the following is correct operating voltage range for short transmission lines. a) Less than 456 KV b) Less than 132 KV c) Less than 20 KV d) Less than 100 KV 3. What is the line to earth capacitance value of the short transmission line. a) Very high b) Medium c) Low d) Negligible 4. Performance of short transmission lines depends on which of the following? a) Resistance and Capacitance b) Resistance and Inductance c) Inductance and Capacitance d) Resistance, Inductance and Capacitance 5. Performance analysis of short transmission line is done __________ a) By symmetrical component analysis method b) By reactance diagram c) On per phase basis d) By neglecting line inductance 6. What is the value of shunt conductance of short transmission line? a) Very high b) Medium c) Low d) Negligable 7. In short transmission line the reciving end voltage may be greater than sending end voltage due to Ferranti effect.



a) True b) False 8. Series inductance and series resistance of short transmission lines are taken as ___________ a) Lumped and Distributed b) Distributed and Lumped c) Both Lumped d) Both Distributed 9. What is the value of charging current in short transmission lines? a) Less than medium transmission lines b) Equal to medium transmission lines c) More than medium transmission lines d) More than long transmission lines 10. What is the value of Inductive reactance in short transmission lines? a) More at reciving end b) More at sending end c) Uniformly distributed over entire length d) More in middle of sending end and reciving end 11. Regulation of short transmission lines depends: a) Only on line resistance b) Only on line inductance c) Only on line capacitance d) On line inductance and line resistance 12. What is the percentage voltage regulation of short transmission line if its sending end and reciving end voltages are 160 KV and 132 KV respectively? a) 30 % b) 21.21 % c) 12.12 % d) 38.22 % 13. A single phase transmission line is transmitting 1,100 KW power at 11 KV and at unity power fector. If it has a total resistance of 5 Ω, what is the efficiency of the transmission line? a) 80 % b) 89.65 % c) 97.24 % d) 99.54 %



14. Voltage regulation of a transmission line should be ________ a) Minimum b) Maximum c) Greater than 50 % d) Less than 50 % 15. Which of the following is like equivalent circuit of short transmission line? a) Series RLC circuit b) Parallel RLC circuit c) Series RL circuit d) Parallel RL circuit 16. In single phase transmission lines resistance and inductance are considered only up to neutral. a) True b) False 17. Which of the following transmission line can be considered as medium transmission line? a) Transmission line of length upto 600 Km b) Transmission line of length upto 500 Km c) Transmission line of length upto 200 Km d) Transmission line of length upto 80 Km 18. Which of the following is correct operating voltage range for medium transmission lines. a) More than 765 KV b) More than 400 KV c) More than 20 KV d) More than 132 KV 19. What is the value of the charging current flowing to earth in medium transmission line. a) Very high b) Medium c) Negligible d) No capacitance 20. Performance of short transmission lines depends on which of the following? a) Resistance and Capacitance b) Resistance and Inductance c) Inductance and Capacitance d) Resistance, Inductance and Capacitance 21. Performance analysis of medium transmission line is done ________ a) By reactance diagram b) By symmetrical component analysis method



c) By neglecting line inductance d) On per phase basis 22. What is the value of shunt capacitance of medium transmission line? a) Very high b) Medium c) Zero d) Negligable 23. In medium transmission line the reciving end voltage may be greater than sending end voltage due to Ferranti effect. a) True b) False 24. Series inductance and series resistance of medium transmission lines are taken as: a) Distributed and Lumped b) Lumped and Distributed c) Distributed d) Lumped 25. Which of the following is like equivalent circuit of medium transmission line? a) Series RLC circuit b) RLC circuit in pie form c) Series RL circuit d) Parallel RL circuit 26. In medium transmission lines shunt capacitances are considered as lumped. a) true b) false 27. In medium transmission lines capacitive current is ________________ a) Less than short transmission lines b) Equal to short transmission lines c) More than short transmission lines d) More than long transmission lines 28. In actual the capacitance of line is ____________ a) More at receiving end b) More at sending end c) Uniformly distributed over entire length d) More in middle of sending end and receiving end 29. Charging current in medium transmission line is _________________ a) Maximum at receiving end b) Maximum at sending end



c) More in between sending and receiving end d) Equal throughout the line 30. In end condenser method of solution of medium transmission line, capacitance of the line is assumed to be _____________ a) Distributed uniformly from sending end to receiving end b) Lumped at the load end c) Lumped at the generation end d) Lumped at middle of generation and load end 31. In nominal-T method of solution of medium transmission line, capacitance is assumed to be __________ a) Distributed uniformly from sending end to receiving end b) Lumped at the load end c) Lumped at the generation end d) Lumped at middle of generation and load end 32. In nominal-π method of solution of medium transmission line, capacitance is assumed to be: a) Distributed uniformly from sending end to receiving end b) Divided into two halves c) Lumped at the generation end d) Lumped at middle of generation and load end 33. Medium transmission line operates below voltage level of 20 KV. a) True b) False 34. Which of the following is not a method for solution of medium transmission line? a) End condenser method b) Nominal-T method c) Nominal-π method d) Nominal-H method 35. In a nominal-π method the resistance of line is assumed to be __________ a) Distributed uniformly from sending end to receiving end b) Divided into two halves c) Lumped at the generation end d) Lumped at middle of the line 36. In nominal-T method of solution of medium transmission line, resistance and inductive reactance of the line is divided into two half’s. a) True b) False



37. Length of long transmission line is more than ___________ a) 80 Km b) 50 Km c) 120 Km d) 200 Km 38. In long transmission lines Resistance and Capacitance parameters of lines are connected in __________ a) Series, shunt b) Series, series c) Shunt, shunt d) Shunt, parallel 39. Range of surge impedance for an overhead transmission line is ____________ a) 12 Ω – 144 Ω b) 40 Ω – 60 Ω c) 400 Ω – 600 Ω d) 300 Ω – 900 Ω 40. Range of surge impedance for an underground cables is _____________ a) 12 Ω – 144 Ω b) 40 Ω – 60 Ω c) 400 Ω – 600 Ω d) 300 Ω – 900 Ω 41. Synchronous phase modifiers are installed at which of the following position of the transmission line? a) Reciving end b) Sending end c) Between reciving end and sending end d) Near reciving end 42. The voltage rating of long transmission line is _________ a) 20 KV to 100 KV b) Upto 20 KV c) Above 100 KV d) 60 KV to 80 KV 43. The shunt capacitive susceptance in long transmission line is greater than that in medium and short transmission line. a) True b) False 44. What is the value of characteristics impedance for loss free transmission line? a) √(L/C)



b) √(R/C) c) √(LC) d) √(C/L) 45. The leakage current through the shunt admittance is __________ a) Maximum at sending end b) Maximum at receiving end c) Uniform over length of line d) Maximum at centre of line 46. Value of leakage current at reciving end of transmission line is zero. a) True b) False

UNIT: 2 Symmetrical faults Short Circuit of a Synchronous Machine (No Load)

1. Transient in synchronous generator is similar to which of the following circuit? a) Parallel RLC circuit b) Series RLC circuit c) Series RL circuit d) Parallel RL circuit 2. When all three phases of a synchronous generator on no load are suddenly short circuited then symmetry of the short circuit current depends on which of the following? a) Position of fault b) Symmetry of fault c) Instantaneous Voltage at which fault occurs d) Resistance of armature winding 3. In a synchronous generator for how much time subtransient period of symmetrical short circuit current lasts? a) For 200 Cycles b) For 500 Cycles c) For 30 Cycles d) For 2 Cycles 4. In a synchronous generator for how much time transient period of symmetrical short circuit current lasts? a) For 200 Cycles



b) For 500 Cycles c) For 30 Cycles d) For 2 Cycles 5. After how many cycles in a synchronous generator symmetrical short circuit current reaches to its steady state value? a) After 200 Cycles b) After 500 Cycles c) After 30 Cycles d) After 2 Cycles 6. What is the phase current in phase Y and in phase B out of R-Y-B when sudden short circuit to all three phases occurs at no load? It is given that after fault current in phase R is equal (Vm Sin ωt) a) Y = Vm Sin(ωt +60°) & B = VmSin(ωt +120°) b) Y = Vm Sin(ωt +120°) & B = VmSin(ωt+240°) c) Y = Vm Sin(ωt +240°) & B = Vm Sin(ωt +120°) d) Y = Vm Sin(ωt +90°) & B = Vm Sin(ωt -90°) 7. In case of synchronous generator the reactants is constant. a) True b) False 8. If Xl is leakage reactance, Xf is reactance due to field winding, Xdw is reactance due to damper winding and Xa is armature winding reactance then reactance of synchronous generator in subtransient state is due to which of the following? a) Xl ,Xdw and Xa b) Xl and Xa c) Xl , Xf ,Xdw and Xa d) Xl and Xdw 9. Which of the following reactance is eliminated first in synchronous generator just after the symmetrical fault? a) Leakage reactance b) Damper winding reactance c) Armature winding reactance d) Field winding reactance 10. Steady state direct axis reactance is greater than subtransient direct axis reactance and transient direct axis reactance. a) True b) False



UNIT:3 Symmetrical Components

Symmetrical Component Transformation

1. The receiving end active power for a short transmission line is (where the angles have their usual meanings)

2. The receiving end reactive power for a short transmission line is (where the angles have their usual meanings)

3. The sending end active power for a 20 km transmission line with Vs as the sending end voltage and Vr as receiving end voltage, can be given by most appropriately

4. The sending end reactive power for a 20 km transmission line with Vs as the sending end voltage and Vr as receiving end voltage, can be given by most appropriately

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5. The simplified ABCD representation of a 40 km transmission line is best given as

6. For a 35 km transmission line having a lumped impedance of the line as 20 ohms, is required to be shown in the ABCD form, it is given as

7. If it is tried to represent the active and reactive power on a circle, then the radius would be

8. The maximum power delivered to the load for short transmission line is at a) β=α b) β>α c) β=δ d) β>δ







9. The maximum real active power delivered to the load is defined most accurately by

10. For a given power system, its zero and maximum regulation will occur at the impedance angle of a) 45 b) 90 c) 0 d) 60 11. The charging currents due to shunt admittance can be neglected for ______ transmission line? a) short b) long c) medium d) all of the mentioned 12. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify the missing term marked as ’?’. Vs = ?*Vr+B*Ir Is = C*Vr+D*Ir

a) 1+YZ b) Z c) Y d) 1 13. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify the missing term marked as ’?’. Vs = A*Vr+?*Ir





Is = C*Vr+D*Ir

a) 1+YZ b) Z c) Y d) 1 14. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify the missing term marked as ’?’. Vs = A*Vr+B*Ir Is = ?*Vr+D*Ir

a) 1+YZ b) Z c) Y d) 1 15. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify the missing term marked as ’?’. Vs = A*Vr+B*Ir Is = C*Vr+?*Ir





a) 1+YZ b) Z c) Y d) 1 16. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir

a) 1+YZ, Z b) Z,Y c) Y, 1+YZ d) 1+YZ, Y-1+Z 17. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and C parameters. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir


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a) 1+YZ, Z b) Z,Y c) 1+YZ, Y d) 1+YZ, Y-1+Z 18. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and D parameters. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir

a) 1+YZ, 1 b) 1+YZ, Z c) Y, 1+YZ d) 1+YZ, Y-1+Z 19. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify B and A parameters. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir






a) 1+YZ, Z b) Z,YZ+1 c) Z, 1+YZ d) 1+YZ, Y-1+Z 20. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify C and A parameters. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir

a) 1+YZ, Z b) Z,Y c) Y, 1+YZ d) 1+YZ, Y-1+Z 21. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify D and A parameters. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir

a) 1+YZ, Z b) 1+ZY, 1 c) Y, 1+YZ d) 1, 1+YZ 22. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find AB value. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir





a) (1+YZ)* Z b) 1+YZ* Z c) Y*( 1+YZ) d) (1+YZ)*Y-1+Z 23. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find ‘A*C’ value. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir

a) (1+YZ)* Y b) (1+YZ)* Z c) Y*( 1+YZ) d) (1+YZ)*Y-1+Z 24. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find ‘A*D’ value. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir






a) (1+YZ) b) (1+YZ)* Z c) 1 d) 0 25. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find ‘B*D’ value. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir

a) (1+YZ) b) (1+YZ)* Z c) Z d) YZ 26. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find ‘B*C’ value. Vs = A*Vr + B*Ir Is = C*Vr + D*Ir

a) (1+YZ) b) (1+YZ)* Z c) 1 d) YZ 27. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find ‘C*D’ value. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir





a) (1+YZ) b) (1+YZ)* Z c) Z d) Y 28. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find ‘AB-CD’ value. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir

a) Z-Y+YZ2 b) (1+YZ)* Z c) 1+YZ+YZ2 d) 1-YZ+ZY2 30. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find ‘AD-BC’ value. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir






a) Z-Y+YZ2 b) (1+YZ)* Z c) YZ+YZ2 d) 1 30. The transmission line equations are given by the below set of equations based on the line diagram as given. Identify A and B parameters and find ‘A*C*D’ value. Vs = A*Vr+B*Ir Is = C*Vr+D*Ir

a) ZY+(YZ)2 b) (1+YZ)* Z c) YZ+YZ2 d) 1-YZ+ZY2

Comparison of Faults in Three Phase System

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 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 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 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 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 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 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



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

a)

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b)

c)

d) 13. 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 14. 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.

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a) a b) 0 c) a3 d) a4 15. 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 16. 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° 17. 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° 18. 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° 19. In the cylindrical rotor alternator, the sub transient and negative sequence reactances are same. a) True b) False 20. The zero sequence impedance of a synchronous machine is independent of the pitch of the armature coils. a) True b) False

UNIT: 4 Unsymmetrical Faults

Per Unit (PU) System

1. A power system network is connected as shown in the figure.

Sd1=15+j5 pu


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Sd2=25+j15 pu Zcable = j0.05pu |V1|=|V2|=1 pu. The torue angle for the system will be __________ a) 14.4 b) 22.1 c) 16.2 d) 18.2 2. A single phase distributor of 1 km long has resistance and reactance per conductor of 0.1Ω and 0.15 Ωrespectively. If the far end voltage Vb=200V and current is at 100A at 0.8 lag. At the midpoint a current of 100A is tapped at a pf of 0.6 pf with ref to voltage Vm at mid point. The voltage magnitude at M is ________ a) 218V b) 200V c) 232V d) 220V 3. A single phase motor is connected to 400V, 50Hz supply. The motor draws a current of 31.7A at a power factor 0.7 lag. The capacitance required in parallel with motor to raise the power factor of 0.9 lag (in micro farads) is __________ a) 94.62 b) 282.81 c) 108.24 d) 46.87 4. A single phase motor is connected to 400V, 50Hz supply. The motor draws a current of 31.7A at a power factor 0.7 lag. The additional reactive power (in VAR) to be supplied by the capacitor bank will be ___________ a) 4756 b) 4873 c) 4299 d) 9055.3 5. A 275 kV TL has following line constants A=0.85ʟ5o, B=200ʟ75o. The active power received if the voltage to be maintained is 275kV will be __________ a) 117.63 b) 220 c) 120 d) 115.25 6. A 275 kV TL has following line constants A=0.85ʟ5o, B=200ʟ75o. The active power angle such that the voltage to be maintained at the other end will be 275 kV ____________ a) 22 b) 16



c) 18 d) 24 7. A power system has a maximum load of 15 MW. Annual load factor is 50%. The reserve capacity of plant is _____ if the plant capacity factor is 40%. a) 3.75 MW b) 4.75 MW c) 18.75 MW d) 5.75 MW 8. A 100 MVA synchronous generator operates on full load at a frequency of 50 Hz. The load is suddenly reduced to 50 MW. Due to time lag in governor system, the steam valve begins to close after 0.4s. The change in the frequency is ________(H=5 kW-s/KVA). a) 1 b) 0.5 c) -1.5 d) 0.8 9. A 50 Hz four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8 MJ/MVA. If the mechanical input is suddenly raised to 80 MW for an electrical load of 50MW, then the rotor acceleration is ____________ a) 337.5 b) 3.375 c) 457.5 d) 4.57 10. A single phase TL has copper conductor of 0.775 cm2 cross section through which 200 kW at UPF at 330 V is to be maintained. If the efficiency of transmission line is 90%, then the minimum length of TL is ___________(in km and take specific resistance to be 1.785 μΩ/cm). a) 13.6 km b) 14 km c) 136 km d) 16.4 km 11. A three phase transformer has a nameplate rating of 30 MVA, 230Y/69Y kV with a leakage -reactance of 10% and the transformer connection is wye-wye. Choosing a base of 30MVA and 230 kV on high voltage side, the reactance of transformer in per units is __________ a) 0.1 b) 0.3 c) 0.03 d) 1.5



12. A three phase transformer has a nameplate rating of 30 MVA, 230Y/69Y kV with a leakage -reactance of 10% and the transformer connection is wye-wye. Choosing a base of 30MVA and 230 kV on high voltage side, the high voltage side impedance ____________ a) 1763.3 Ω b) 158.7 Ω c) 15.87 Ω d) 176.3 Ω 13. A three phase transformer has a nameplate rating of 30 MVA, 230Y/69Y kV with a leakage -reactance of 10% and the transformer connection is wye-wye. Choosing a base of 30MVA and 230 kV on high voltage side, the low voltage side impedance is ___________ a) 158.7 Ω b) 176.3 Ω c) 1763.3 Ω d) 15.87 Ω 14. A three phase transformer has a nameplate rating of 30 MVA, 230Y/69Y kV with a leakage -reactance of 10% and the transformer connection is wye-wye. Choosing a base of 30MVA and 230 kV on high voltage side, the transformer reactance referred to the high voltage side will be _________(in ohms). a) 176.33 Ω b) 17.67 Ω c) 158.7 Ω d) 15.87 Ω

Ferranti Effect & Methods of Voltage Control

1. Ferranti effect is not a problem for ________ a) Long Transmission lines b) Medium Transmission lines c) Short Transmission lines d) Transmission line having high capacitance 2. What happens during the Ferranti effect? a) Reciving end current becomes more than sending end current b) Reciving end voltage becomes more than sending end voltage c) Reciving end power becomes more than sending end power d) Reciving end frequency becomes more than sending end frequency 3. During Ferranti effect the voltage drop across line resistance ___________ a) In phase with reciving end voltage b) lags behind reciving end voltage



c) lead the reciving end voltage d) lags behind sending end voltage 4. A transmission line of 200 Km is supplying at 50Hz frequency. What is the percentage rise in voltage at reciving end? a) 20% b) 1.2% c) 2.19% d) 20.8% 5. Which of the following equipment or element can reduce Ferranti effect? a) Relay b) Circuit Breaker c) Resistors d) Current limiting reactors 6. Under no load condition inductance of line predominant the line to earth capacitances. a) True b) False 7. What is the limit within which the consumer’s end terminal voltage should be maintained? a) ± 9 % b) ± 10 % c) ± 5 % d) ± 6 % 8. What is the main reason for maintaining consumer end voltage within prescribed limit? a) Because it is declared by the supplier b) For satisfactory operation of electrical equipments c) For easy calculation of units supplied d) To reduce the line losses 9. Which of the following equipment is not used for voltage control? a) Tap changing transformer b) Induction generators c) Series compensators d) Synchronous phase modifiers 10. What is the full form of AVR? This term is related to voltage control? a) Automatic Voltage Rectifier b) Actuated Voltage Rectifier c) Automatic Voltage Rectifier d) Automatic Voltage Regulator



11. On which side of the transformer tap changer is provided? a) On High Voltage side b) On low voltage side c) On high voltage and low voltage side d) In core

Skin Effect and Proximity Effect

1. Ripple of the DC-output at power converter is ____________ a) maximum value of instantaneous difference between average and instantaneous value b) minimum value of instantaneous difference between average and instantaneous value c) average value of output d) maximum value of instantaneous value 2. Smooth reactor is connected between _____________ a) DC output of converter and load b) load and ground c) source and input of converter d) any of the mentioned 3. Which of the above is not a DC link? a) Monopolar link b) Bipolar link c) Homopolar link d) None of the mentioned 4. (i) Monopolar DC link has one energised conductor of positive polarity. (ii) It was ground as return path Opt the most appropriate. a) (i) is False & (ii) is True b) (i) is True & (ii) is True c) Both are false d) (i) is True & (ii) is False 5. (i) Bipolar link has two conductors (ii) Both conductors have same magnitude of voltage. a) (i) is False & (ii) is True b) (i) is True & (ii) is True c) Both are false d) (i) is True & (ii) is False



6. If the currents in the two conductors of bipolar HDVC link is same, then _______ a) ground current is zero b) two poles work independently c) ground current zero and two poles work independently d) none of the mentioned 7. Homopolar link is employed instead of bipolar link to ___________ a) reduce corona loss b) reduce radio interference c) improve reliability d) all of the mentioned 8. Which of the DC links can be operated when a fault occurs in the transmission line? a) Bipolar link and Homopolar link b) Monopolar link and Bipolar link c) Monopolar link d) Bipolar link 9. The overvoltage that occur mainly under converter stations are due to _________ (i) Lightening and switching overvoltage (ii) Transients produced in conductors (iii) Overvoltage due to fault clearing Choose the correct option. a) (i), (ii), (iii) b) (ii), (iii) c) (iii) d) (i), (ii) 10. Insulator capacitance is does not play significant role in HVDC lines a) True b) False

UNIT:6 Corona

1. Higher the frequency, _____________. a. Lower the corona loss. b. Higher is the corona loss. c. Does not effect. d. Depends on the physical conditions. 2. Which harmonics are generated during the corona, which leads to the increase in corona losses?



a. Third harmonics. b. Fifth harmonics. c. Seventh harmonics. d. None of these. 3. What is the use of bundled conductors? a. Reduces surface electric stress of conductor. b. Increases the line reactance. c. Decreases the line capacitance. d. None of these. 4. On which factor is the corona loss dependent on? a. Material of the conductor. b. Diameter of the conductor. c. Height of the conductor. d. None of these. 5. In which climate does the chances of occurrence of corona is maximum a. Dry b. Hot summer. c. Winter. d. Humid. 6. Corona loss can be reduced by using 1. Solid conductor. 2. Hollow conductor. 3. Bundle conductor. a. 1 only b. 1 and 2 only. c. 1, 2 and 3 only. d. 2 and 3 only. 7. What is the effect on corona, if the spacing between the conductors is increased? a. Corona increases. b. Corona is absent. c. Corona decreases. d. None of these.



8. Why are the hollow conductors used? a. Reduce the weight of copper. b. Improve stability. c. Reduce corona d. Increase power transmission capacity. 9. Which of these given statements is wrong in consideration with bundled conductors? a. Control of voltage gradient. b. Reduction in corona loss. c. Reduction in the radio interference. d. Increase in interference with communication lines. 10. Why are bundled conductors employed? a. Appearance of the transmission line is improved. b. Mechanical stability of the line is improved. c. Improves current carrying capacity. d. Improves the corona performance of the line. 11. The effect of dirt on the surface of the conductor is to _____________ irregularity and thereby ________________ the break down voltage. a. Decreases, reduces. b. Increases, increases. c. Increases, reduces. d. Decreases, increases. 12. Find the spacing between the conductors a 132 kV 3 phase line with 1.956 cm diameter conductors is built so that corona takes place, if the line voltage exceeds 210 kV (rms). With go = 30 kV/cm. a. 1.213 m b. 2.315 m. c. 3.451 m. d. 4.256 m.



ANSWERS

UNIT: 1 Transmission lines 1. Answer: d Explanation: The transmission lines having length less than 80 Km are considered as short transmission lines. Their operating voltage is also less than medium and long transmission lines. 2. Answer: c Explanation: Transmission lines having length lesser than 80 Km and operating voltage lower than 20 KV fall in the category of short transmission lines. 3. Answer: d Explanation: Due to smaller distance and lower line voltage, the capacitance effects are extremely small and, so are neglected. 4. Answer: b Explanation: Due to smaller distance and lower line voltage, the capacitance effects are extremely small and, so are neglected. So, performance of short transmission lines depends only on inductance and resistance of the line. 5. Answer: c Explanation: Performance analysis of balance transmission line is done only. And is done on per phase basis. No symmetrical component method or reactance diagram method is required for performance analysis. 6. Answer: d Explanation: In short transmission lines the shunt conductance and shunt capacitance are neglected and so only series resistance and inductance are to be considered. 7. Answer: b Explanation: Ferranti effect is caused due to the high capacitance of the transmission lines. In short transmission lines capacitance effects are negligible. So, Ferranti effect doesn’t happens in short transmission lines. 8. Answer: c Explanation: The series inductance and resistance of short transmission lines are taken as lumped. This makes the performance analysis easy 9. Answer: c Explanation: Line to Earth capacitance of short transmission line is less than that in medium



and long transmission line and is negligable. Because of this the charging current of short transmission line is less than that in medium and long transmission line and is negligable. 10. Answer: c Explanation: Actually the inductive reactance of line is uniformly distributed over its entire length. To make calculations simple this reactance is assumed to be lumped and connected in series with the line.

11. Answer: d Explanation: The expression of voltage regulation of short transmission line shows that, its voltage regulation depends on line resistance and line inductive reactance. Capacitance of short transmission line is negligible so it does not cause its effecting it.

12. Answer: b Explanation: % Voltage Regulation = (Sending end voltage -Reciving end voltage) ÷ Reciving end voltage = [( 160 – 132 ) × 100] ÷ 132 = 21.21 %.

13 Answer: c Explanation: Efficiency (η) = Power delivered ÷ (Power Delivered + line losses) Line current (I) = P/V = (1100 × 1000/11000) = 100 Amp. Line losses = I2 R = (100)2 5 = 50 KW η = [11000/(11000+50)] × 100 = 99.54 %. 14. Answer: a Explanation: More value of voltage regulation means more voltage fluctuations and this is undesirable. So the value of voltage regulation should be as low as possible. 15.Answer: c Explanation: Due to smaller distance and lower line voltage, the shunt capacitance effects are extremely small and also the shunt conductance effects are very low. So both of them are neglected and equivalent circuit contains only series inductance(L) and resistance(R). 16. Answer: b Explanation: In case of single phase transmission lines the total loop resistance and inductance is to be taken into account. Resistance and inductance only up to neutral are taken into account for 3 phase transmission lines.



17. Answer: c Explanation: The transmission lines having length more than 80 Km and less than 200 Km are considered as medium transmission lines.Their operating voltage is more than short transmission lines but less than long transmission lines.

18. Answer: c Explanation: Transmission lines having length more than 80 Km and less than 200 Km, operating voltage between 20 KV and 100 KV fall in the category of medium transmission lines.

19. Answer: b Explanation: Owing to appreciable length and voltage of the line the charging current is appreciable in medium transmission lines. Also the capacitance effect cannot be ignored in medium transmission lines.

20. Answer: d Explanation: Medium transmission lines have enough value of resistance and inductance. Length and operating voltage of the medium transmission lines are sufficiently large to cause unavoidable charging current and therefore capacitance effect can not be ignored. Hence, performance of Medium transmission lines depends on R, L and C.

21. Answer: d Explanation: Performance analysis of balance transmission line is done only. And is done on per phase basis. No symmetrical component method or reactance diagram method is required for performance analysis.

22. Answer: b Explanation: Length and operating voltage of the medium transmission lines are sufficiently large to cause unavoidable charging current and therefore capacitance effect can not be ignored. Capacitance effect in medium transmission line is less than that in long transmission line.

23. Answer: a Explanation: Ferranti effect happens when capacitance effect in medium or long transmission lines becomes greater than inductance effect. This happens during low or no load period.

24. Answer: d Explanation: The series inductance and resistance of medium transmission lines are taken as lumped. This makes the performance analysis easier.

25. Answer: b Explanation: Medium transmission lines have considerable amount of inductance, resistance and shunt capacitance. Medium transmission lines are represented as either pie form or as T form.

26. Answer: a Explanation: In case of medium transmission lines the charging current is unavoidable but



less than that in long transmission lines. So it does not cause much error in calculations if shunt capacitances are taken as lumped

27. Answer: c Explanation: Line to Earth capacitance of medium transmission line is more than that in short transmission line and less than that in long transmission line. Because of this the charging current of medium transmission line is more than that in short transmission line and less than that in long transmission line.

28. Answer: c Explanation: Actually the capacitance of line is uniformly distributed over its entire length. To make calculations simple the capacitance of the system is assumed to be divided up and lumped in the form of capacitors shunted across the line at one or more points.

29. Answer: a Explanation: The magnitude of capacitive current flowing at any point along the line is that required to charge the section of the line between the given point and receiving end. Hence, it has maximum value at the sending in and diminished at practically uniform rate down to zero at the receiving end.

30. Answer: b Explanation: In end condenser method the capacitance of the line is assumed to be lumped at the load end. This method overestimates the effect of capacitance.

31.Answer: d Explanation: In nominal- T method or middle condenser method of analysis of medium transmission line the whole of the line capacitance is assumed to be concentrated at the middle point of the line and half the line resistance and reactance to be lamp on either side.

32. Answer: b Explanation: In nominal-π method of solution of medium transmission line the capacitance of each line conductor is assumed to be divided into two halves. One half being shunted between line conductor and neutral at the receiving in and other half at the sending end.

33. Answer: b Explanation: The voltage limit of 20 KV is for short transmission lines. Medium transmission lines can operate for voltage more than 20 KV.

34. Answer: d Explanation: Actual method used for solution of medium transmission lines are end condenser method, nominal-T method and nominal π method.

35. Answer: d Explanation: In nominal-π method the capacitance of each line conductor is assumed to be divided into two halves and the resistance of the line is assumed to be lumped in between them at the middle of the line.

36. Answer: a Explanation: In nominal-T method of solution of medium transmission line, the whole of the



line capacitance is assumed to be concentrated at the middle point of the line and half the line resistance and reactance to be lumped on either side of it.

37. Answer: d Explanation: Transmission lines having length more than 200 kilometres are categorised as long transmission lines. Transmission lines having length less than 200 kilometres are medium transmission lines.

38. Answer: d Explanation: The resistance and inductance parameters are series elements of long transmission line. Capacitive susceptance and conductance parameters are shunt elements of transmission lines.

39. Answer: c Explanation: Surge impedance is the characteristic impedance of a loss free transmission line. It is the square root of ratio of line impedance(Z) and shunt admittance (Y). Its value varies between 400 Ω and 600 Ω.

40. Answer: b Explanation: Surge impedance is the characteristic impedance of a loss free transmission line. It is the square root of ratio of line impedance(Z) and shunt admittance (Y). Its value varies between 40 Ω and 60 Ω.

41. Answer: b Explanation: For constant voltage transmission, specially designed synchronous motors, called synchronous phase modifiers. Synchronous phase modifiers are installed at receiving end to maintain the voltage drop along the line constant.

42. Answer: c Explanation: The transmission line having length above time 200 kilometre and line voltage above 100 KV fall into the category of long transmission line.

43. Answer: a Explanation: Due to long length and high voltage of line the charging current is very high for the long transmission line and therefore the capacitance in long transmission line is also very high and is more than that in short and medium transmission line.

44. Answer: a Explanation: Characteristic impedance of a long transmission line is given by Zc = √[(R+jX) / ( G+jB)] For loss less transmission line resiatance (R) and shunt conductance (G) are equal to zero. And hence, Z0 ( Characteristic impedance for loss-free transmission line) is given by Z0 = √(L/C).

45. Answer: a Explanation: The leakage current through the shunt admittance is maximum at the sending



end of the transmission line. This current decreases continuously as it move towards the receiving end of the line and becomes zero at the receiving end.

46. Answer: a Explanation: Value of leakage current through the shunt admittance is maximum at the sending end of the transmission line. This current decreases continuously as we move towards the receiving end of the line and becomes zero at the receiving end.

UNIT:2 Symmetrical Faults

Short Circuit of a Synchronous Machine (No Load)

1. Answer: c Explanation: The current flowing in armature of synchronous generator when its terminals are short circuited is similar to that flowing when sinusoidal voltage is suddenly applied to an RL series circuit.

2. Answer: c Explanation: Symmetry of the short circuit current about time axis and DC offset depends upon the value of phase voltage at which short circuit occurs.

3. Answer: d Explanation: Subtransient period lasts for only about 2 cycles. During this period the subtransient current decay is very rapidly.

4. Answer: c Explanation: In a synchronous generator transient period of symmetrical short circuit current lasts for about 30 cycles. During this period current decreases somewhat slower than the current in subtransient period.

5. Answer: c Explanation: In a synchronous generator subtransient period of short circuit current lasts for up to 2 cycles the transient period of that lies for about 30 cycles. After that the current reaches to its steady state value.

6. Answer: b Explanation: The given phase sequence is R-Y-B . When all three phases are short circuited then the fault is symmetrical that means fault current flowing through each phases will be 120° apart from each other.

7. Answer: b Explanation: In case of synchronous generator the reactance is not a constant one but is a function of time. The reactance of synchronous generator is different for subtransient state, transient state and steady state.



8. Answer: c Explanation: During the initial part of short circuit or during subtransient period equivalent circuit of alternator contains the field winding reactance damper winding reactance armature winding reactance in parallel and leakage reactance in series.

9. Answer: b Explanation: As the damper winding currents are first to die out, damper winding reactance effectively becomes open circuited. After that Approximate circuit model of synchronous generator contains only leakage reactance, field winding reactance and armature winding reactance.

10. Answer: a Explanation: During transient and subtransient period the armature winding reactance contains other reactances like damper winding reactance and field winding reactances in parallel with it. So, the net reactance during subtransient and transient period are less than the net reactance during steady state period.

UNIT:3 Symmetrical Components

Symmetrical Component Transformation

1. Answer: a Explanation: Receiving end active power for a short transmission line is

2. Answer: c Explanation: Receiving end reactive power for a short transmission line is

3. Answer: a

Explanation: The sending end active power is

4. Answer: d

Explanation: The sending end reactive power is

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5. Answer: a Explanation: From the given length, it is a short TL. Hence,

6. Answer: a Explanation: From the given length, it is a short TL. Hence,

7. Answer: a Explanation:

8. Answer: c Explanation: Maximum power occurs for β=δ.

9. Answer: a Explanation: The maximum real active power delivered to the load

is







10. Answer: a Explanation: At θ=45°, ZVR and maximum VR coincide.

11. Answer: a Explanation: The shun admittance for lines more than 100 km become very prominent and can not be neglected.

12. Answer: a Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir.

13. Answer: b Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir.

14. Answer: c Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir.

15. Answer: d Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir.


17. Answer: c Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir.


19. Answer: b Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir.

20. Answer: c Explanation: Using KVL to the line diagram,



Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir.




24. Answer: a Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+ Z*Ir Is = Y*Vr + Ir AD = (1+YZ).

25. Answer: c Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr + Z*Ir Is = Y*Vr + Ir BD = Z.

26. Answer: a Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir BC = YZ.

27. Answer: a Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir CD = Y*1 = Y.

28. Answer: a Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir AB-CD = (1+YZ)*Z-Y.



29. Answer: d Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir AD-BC = (1+YZ)-YZ = 1.

30 .Answer: a Explanation: Using KVL to the line diagram, Vs = (1+YZ)*Vr+Z*Ir Is = Y*Vr+Ir ACD = (1+YZ)*ZY.

Comparison of Faults in Three Phase System – 1

1. Answer: a Explanation: n= VHV/VLV = (230/√3)/69=1.9245.

2. Answer: a Explanation: Zb(HV) = kVb(HV))2/MVAb Zb(HV) = 2302/30=1763.33 Ω XΩ =0.1*1763.33 = 176.33 Ω.

3. Answer: a Explanation: Zb(HV) = (kVb(HV)

2/MVAb = 2302/30=1763.33 Ω XΩ(HV) =0.1*1763.33 = 176.33 Ω XΩ(LV) =176.33/(1.92452) = 47.61 Ω.

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

5. Answer: a Explanation: 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. Answer: a Explanation: 1 α15 ; P2 α 20;

P1 = 0.75 P2 …(1)




P1 +P2 = 12 …(2) Solving above, P2 = 6.85 MW.

7. Answer: a Explanation: 1 α15 ; P2 α 20;

P1 = 0.75 P2 …(1) P1 +P2 = 12 …(2) Solving above equations, P1 = 5.14 MW.

8. Answer: a Explanation: Qc = P[tanθ1 – tanθ2] =300[(4/3) – cos-10.9] = 254.72 kVAR.

9. Answer: a Explanation: cosФ = 3/5 = 0.6.

10. Answer: a Explanation: In an unbalanced system, all the symmetrical components will be present.

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

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

13. Answer: a Explanation: 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.

14. Answer: a Explanation: 1+a+ a2 = 0.

15. Answer: a Explanation: 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.

16. Answer: a Explanation: 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 in polar form.

17. Answer: a Explanation: 1+a+ a2 = 0; a= 1∠120°.




18. Answer: a Explanation: 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°.

19. Answer: a Explanation: It is true because the saliency offered is zero in the cylindrical pole machines.

20. Answer: b Explanation: The zero sequence impedance does depend on the pitch of the armature coils of the synchronous machine.

UNIT: 4 Unsymmetrical Faults

Per Unit (PU) System

1. Answer: a Explanation: As the resistance is zero, losses will be zero. PG1=PD1+PD2=40 pu For the equal sharing of load at the station PG1=PG2=20pu Real power flow from bus 1 to 2

2. Answer: a Explanation: Drop in MB= (100ʟ-36.67)(0.1+j0.15)= 18.027ʟ19.44 VM= 200+18.027 = 218.027 V.

3. Answer: a Explanation: Active power drawn by the motor=VIcosФ = 400*31.7*0.7 = 8876 W Reactive power = VIsinФ=400*31.7*sin(45.57.29) = 9055.3 VAR

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New power factor=cosθ2 = 0.9 θ2=cos-1(0.9) Q2=8876*tan(25.84) = 4298.855 VAR Change in reactive power=9055.3-4298.855 = 4756.4 VAR Qc = V2/Xc = V2*2πfC C=4756.4/(4002*2π*50) = 94.62μF.

4. Answer: a Explanation: Active power drawn by the motor=VIcosФ = 400*31.7*0.7 = 8876 W Reactive power = VIsinФ=400*31.7*sin(45.57.29) = 9055.3 VAR New power factor=cosθ2 = 0.9 θ2=cos-1(0.9) Q2=8876*tan(25.84) = 4298.855 VAR Change in reactive power=9055.3-4298.855 = 4756.4 VAR.

5. Answer: a Explanation: |Vs|=|Vr|=275 kV 6. Answer: a Explanation: |Vs|=|Vr|=275 kV α= 5o , β= 75o

7. Answer: a Explanation: LF = (Average Demand)/(Maximum Demand)=0.5 Plant capacity factor =(Average Load)/(Plant Capacity)= 0.5/0.4





Plant capacity= (0.5/0.4)*15 = 18.75 MW Reserve Capacity = 18.75-15 = 3.75 MW.

8. Answer: a Explanation: Energy transferred in 0.4 sec = 50*0.4= 20 J

9. Answer: a Explanation: Energy stored = 100*8 =800MJ Accelerating power = Md2δ/dt2 M=GH/180f = 800/(180*50) = 4/45 MJs/elect. Deg

10. Answer: a Explanation: Pr=200 kW, efficiency=0.9 Ps= 200/0.9=222.22 kW Losses=22.22 kW Current, I=200000/3300 = 60.60 A Line losses=2I2R ( for a 2 wire line) R=22.22/(2*60.602)=3.02 Ω R=ρl/a Length, l = (3.025*0.775)/(1.785*10-6) = 13.6 km.





11. Answer: a Explanation: The pu value of a transformer does not change.

12. Answer: a Explanation: On the high voltage side, Zb=kVb2/MVAb(3-ph) = 2302/30 = 1763.33 Ω.

13. Answer: a Explanation: On the low voltage side, Zb=kVb

2/MVAb(3-ph) =692/30 = 158.7 Ω.

14. Answer: a Explanation: Zb=kVb2/MVAb(3-ph) = 2302/30 = 1763.33 Ω X Ω = Xpu*Xb(HV) = 0.1*1763.33 = 176.33 Ω.

Ferranti Effect & Methods of Voltage Control

1. Answer: c Explanation: Ferranti effect is happens when line to earth capacitance becomes more than the inductance parameter. This effect is more predominant in medium and long transmission lines but is not a problem for short transmission lines. This is because line to earth capacitance for short transmission line is negligible.

2. Answer: b Explanation: In a medium or long transmission line when open circuited or lightly loaded the receiving end voltage is found more than the standing in voltage. This phenomenon of rise in voltage at receiving end of open circuited or lightly loaded line is called Ferranti effect.

3. Answer: c Explanation: During Ferranti effect the leading charging current is more than lagging component of current. Net line current leads the sending end and reciving end voltage. Hence the iR drop leads these voltages.

4. Answer: c Explanation: Percentage rise in voltage = (ω2 l2 10-8 )/18 Solving this % Vrise = (2 × Ω × 50) × 2002 × 10-8/18 = 2.19%.

5. Answer: d Explanation: For high voltage and extra high voltage lines shunt reactors are provided to absorb part of charging current. As leading charging current is reduced. Ferranti effect is also reduced.

6. Answer: b Explanation: Under no load condition capacitance between line and earth predominant the inductance of long and medium transmission lines.



7. Answer: d Explanation: The voltage fluctuations at consumers end may cause malfunction or event damage of consumer’s equipments. So, the voltage at the consumers end must be maintained within prescribed limit upto ± 6% of declared supply voltage so that the consumers apparatus may operate satisfactorily.

8. Answer: b Explanation: All electrical equipments are designed to work at a certain voltage mentioned as voltage rating of that equipment. Change of supply voltage beyond a certain limit may cause malfunction or even damage to the equipment. So, the consumer end voltage must be maintained to a prescribed limit for satisfactory operation of consumer’s equipments.

9. Answer: b Explanation: Induction generators are not used for voltage control. Tap changing transformer is used to control distribution end voltage for upto limited voltage variations. Series compensators and Synchronous phase modifiers controls line voltage by supplying required leading or lagging VAR.

10. Answer: d Explanation: AVR stands for Automatic Voltage Regulator. It operates on the principle of detection of error and is used to control the line voltage.

11. Answer: a Explanation: In tap changing transformer there is always a typing on the high voltage winding which when connected to the rated voltage gives rated voltage on the low voltage side. This tapping is called principal typing.

Skin Effect and Proximity Effect 1. Answer: a Explanation: Ripple at a power converter is maximum value of instantaneous difference between average and instantaneous value.

2. Answer: a Explanation: Smoothening reactor is connected between the dc output and load to provide sinusoidal source current.

3. Answer: d Explanation: All are types of DC link.

4. Answer: a Explanation: Negative polarity is preferred on OH lines due to less radio interference.

5. Answer: b Explanation: Both the statements are correct.

6. Answer: b Explanation: When the two conductors in bipolar link are same then ground current is zero and they have no interaction.



7. Answer: a Explanation: Homopolar link reduces corona loss.

8. Answer: a Explanation: 50% of rated power can be supplied.

9. Answer: a Explanation: All the statements are correct.

10. Answer: a Explanation: There is no capacitive current which flows in HVDC link.

UNIT: 06 Corona

1. ANSWER: b. Higher is the corona loss.

2. ANSWER: a. Third harmonics.

3. ANSWER: a. Reduces surface electric stress of conductor.

4. ANSWER: b. Diameter of the conductor.

5. ANSWER: d. Humid.

6. ANSWER: d. 2 and 3 only.

7. ANSWER: b. Corona is absent.

8. ANSWER: c. Reduce corona

9. ANSWER: d. Increase in interference with communication lines.

10 .ANSWER: d. Improves the corona performance of the line.

11. ANSWER: c. Increases, reduces.

12. ANSWER: c. 3.451 m.

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