Transmission and Distribution MCQs 2026

80 questions with detailed answers · 50 from past papers · 8 quiz batches available

📚 Electrical Engineering📄 50 Past-Paper Qs✓ Free · No Login Needed
🎯 Mock Test

Read each question, think about the answer, then click Show Answer to reveal the correct option and explanation. Load 10 at a time so it stays manageable — perfect for one-topic study sessions on the bus or during a break.

Page 1 of 1Questions 110 of 80
  1. Q1Past Paper · PPSC/FPSC/NTSmedium

    Ferranti effect occurs when

    1. Aheavy lagging load at short line only
    2. Bfault at transformer only
    3. Creceiving end voltage exceeds sending end voltage at light load on long lines
    4. Dmotor starting only
    💡 Explanation:

    Capacitive charging of long lines raises receiving voltage under light load.

  2. Q2Past Paper · PPSC/FPSC/NTSmedium

    Ferranti effect is most pronounced on

    1. Ashort distribution feeders at full load
    2. BDC resistive loads only
    3. Clong EHV lines at light or no load
    4. Dsynchronous motor full load
    💡 Explanation:

    Large shunt capacitance relative to load causes voltage rise.

  3. Q3Past Paper · PPSC/FPSC/NTSmedium

    At light load on a long overhead line, receiving end power factor tends to be

    1. Aleading due to line charging
    2. Balways unity only
    3. Calways lagging only
    4. Dzero always
    💡 Explanation:

    Capacitive charging supplies leading reactive power at the receiving end.

  4. Q4hard

    Open-circuit receiving end of a long line may show voltage

    1. Aalways zero
    2. Balways equal to short-circuit level
    3. Cequal to DC battery voltage only
    4. Dhigher than sending end due to capacitive Ferranti rise
    💡 Explanation:

    Open-end voltage rise is classic Ferranti behaviour.

  5. Q5Past Paper · PPSC/FPSC/NTSeasy

    Positive voltage regulation at lagging load means

    1. Areceiving voltage is always higher
    2. Bsending end voltage magnitude exceeds receiving end under load
    3. Cline has zero impedance
    4. Dpower flows from load to source only
    💡 Explanation:

    Loaded lagging lines typically show voltage drop from sending to receiving.

  6. Q6Past Paper · PPSC/FPSC/NTSmedium

    Load power factor strongly affects voltage regulation because

    1. Aresistance becomes zero
    2. Bfrequency doubles
    3. Creactive component of current causes additional voltage drop in line impedance
    4. DSIL becomes negative
    💡 Explanation:

    I×X drop depends on reactive current magnitude and angle.

  7. Q7Past Paper · PPSC/FPSC/NTSeasy

    Shunt capacitors installed at load buses improve voltage profile by

    1. Aincreasing line charging on EHV only
    2. Bincreasing series impedance
    3. Csupplying reactive power locally and reducing line VAR flow
    4. Dblocking positive-sequence faults
    💡 Explanation:

    Local VAR support raises voltage and reduces I×X drops.

  8. Q8Past Paper · PPSC/FPSC/NTSmedium

    Series capacitors on transmission lines reduce voltage regulation by

    1. Aincreasing Ferranti rise only
    2. Bpartially compensating line inductive reactance
    3. Celiminating the need for conductors
    4. Dreplacing distance protection
    💡 Explanation:

    Series compensation lowers effective X and improves stability and voltage.

  9. Q9Past Paper · PPSC/FPSC/NTSmedium

    On-load tap changers on transformers regulate voltage by

    1. Achanging transformer turns ratio under load
    2. Bchanging line length physically
    3. Cinserting fuses in series
    4. Dopening circuit breaker contacts only
    💡 Explanation:

    OLTC adjusts secondary voltage without interrupting supply.

  10. Q10Past Paper · PPSC/FPSC/NTSmedium

    Shunt reactors on long EHV lines at light load are used to

    1. Aincrease Ferranti effect
    2. Braise receiving voltage further
    3. Creplace series impedance completely
    4. Dabsorb excess reactive power and limit voltage rise
    💡 Explanation:

    Reactors compensate capacitive charging during low load periods.

  11. Q11Past Paper · PPSC/FPSC/NTShard

    ABCD parameters of a transmission line are used to compute

    1. Asending end voltage and current from receiving end conditions
    2. Bonly transformer copper loss
    3. Conly motor slip
    4. Donly fuse melting time only
    💡 Explanation:

    Two-port ABCD equations relate Vs, Is to Vr, Ir.

  12. Q12hard

    Critical loading for voltage support on uncompensated lines relates to

    1. Amotor starting current only
    2. BSIL and line length where reactive power balance changes character
    3. CBuchholz trip level only
    4. DPID derivative time only
    💡 Explanation:

    Around SIL, line reactive generation and consumption tend to balance.

  13. Q13Past Paper · PPSC/FPSC/NTSeasy

    Characteristic impedance Zc of a lossless overhead line equals

    1. AR/L only
    2. Bsqrt(L/C) where L is series inductance and C is shunt capacitance per unit length
    3. CC/R only
    4. Dsqrt(RG) only
    💡 Explanation:

    For a lossless line, Zc = sqrt(L/C) and is also called surge impedance.

  14. Q14Past Paper · PPSC/FPSC/NTSmedium

    Surge impedance loading (SIL) of a transmission line is given by

    1. AV/Zc only
    2. BZc/V²
    3. CV×Zc×√3 only
    4. DV²/Zc in MW for line-to-line voltage V
    💡 Explanation:

    SIL = V²/Zc represents the natural loading at which reactive power balance is favourable.

  15. Q15medium

    In the nominal π model of a medium-length line, shunt capacitance is placed

    1. Aonly at the receiving end
    2. Bat both sending and receiving ends as half the total line capacitance each
    3. Conly at the sending end
    4. Dentirely in the middle as a lump
    💡 Explanation:

    The π model splits total shunt admittance equally at both ends for medium lines.

  16. Q16Past Paper · PPSC/FPSC/NTSeasy

    Series impedance per unit length of an overhead line includes

    1. Aonly shunt capacitance
    2. Bonly corona loss resistance alone
    3. Conly transformer leakage
    4. Dresistance and inductive reactance of conductors
    💡 Explanation:

    Per-km series Z = R + jωL accounts for conductor resistance and inductance.

  17. Q17hard

    Geometric mean distance (GMD) in inductance formula refers to

    1. Aresistance of earth wire only
    2. Binsulator string length
    3. Ctower footing resistance
    4. Dequivalent spacing among bundled or multi-conductor arrangements
    💡 Explanation:

    GMD replaces actual spacing in inductance calculations for complex conductor geometry.

  18. Q18hard

    Geometric mean radius (GMR) of a conductor bundle is used to compute

    1. Aonly corona onset voltage
    2. Bequivalent inductance per phase lower than single conductor
    3. Conly sag at maximum tension
    4. Donly zero-sequence resistance only
    💡 Explanation:

    Bundling reduces effective GMR and hence line inductance per phase.

  19. Q19Past Paper · PPSC/FPSC/NTSmedium

    Skin effect in AC transmission conductors causes

    1. Auniform current across the cross-section always
    2. Breduction of resistance at higher frequency
    3. Ccurrent concentration near the surface increasing effective resistance
    4. Dzero inductance change
    💡 Explanation:

    Higher frequency or large conductors show increased AC resistance due to skin effect.

  20. Q20hard

    Proximity effect in parallel conductors results in

    1. Anon-uniform current distribution and increased effective resistance
    2. Bincreased shunt capacitance only
    3. Celimination of inductive reactance
    4. Dinfinite surge impedance
    💡 Explanation:

    Mutual magnetic fields redistribute current within conductors, raising losses.

  21. Q21Past Paper · PPSC/FPSC/NTSmedium

    Bundled conductors on EHV lines primarily reduce

    1. Acorona loss and line inductance
    2. Bseries resistance to zero
    3. Cshunt capacitance to zero
    4. Dsurge impedance below 1 ohm always
    💡 Explanation:

    Sub-conductors lower surface voltage gradient and effective inductance.

  22. Q22Past Paper · PPSC/FPSC/NTSeasy

    A short transmission line model neglects

    1. Ashunt capacitance and uses only series impedance
    2. Bseries resistance only
    3. Cboth series and shunt completely
    4. Dtransformer magnetizing branch
    💡 Explanation:

    Lines shorter than about 80 km (50 Hz) are often treated as short lines.

  23. Q23hard

    Propagation constant γ of a uniform line is

    1. Aonly jωL
    2. Bonly R/L
    3. Conly G/C without imaginary part
    4. Dγ = sqrt((R+jωL)(G+jωC)) = α + jβ
    💡 Explanation:

    γ has attenuation constant α and phase constant β.

  24. Q24medium

    Nominal T equivalent of a line places

    1. Aall impedance at one end only
    2. Bcapacitance only at sending end
    3. Cno shunt branch
    4. Dhalf the series impedance at each end and total shunt admittance at the centre
    💡 Explanation:

    T model is an alternative to π for medium-length line analysis.

  25. Q25Past Paper · PPSC/FPSC/NTSeasy

    Charging current in an unloaded long overhead line is due to

    1. Ashunt capacitance of the line
    2. Bseries resistance only
    3. Ctransformer no-load current only
    4. Dfault current only
    💡 Explanation:

    Capacitive charging draws leading current even without load.

  26. Q26easy

    Resistance of a conductor per km increases with temperature because

    1. Ainductance falls to zero
    2. Bresistivity of metal increases with temperature
    3. Ccapacitance doubles always
    4. DGMR becomes infinite
    💡 Explanation:

    Copper and aluminium resistivity rises with temperature, increasing line losses.

  27. Q27Past Paper · PPSC/FPSC/NTSeasy

    Sag of a conductor on a level span with uniform loading is approximately

    1. As = wL/T
    2. Bs = 8T/(wL²)
    3. Cs = wL²/(8T) where w is weight per unit length and T is tension
    4. Ds = w/T only
    💡 Explanation:

    Parabolic approximation gives sag proportional to span squared and inversely to tension.

  28. Q28Past Paper · PPSC/FPSC/NTSeasy

    When ambient temperature rises, conductor sag generally

    1. Adecreases always
    2. Bremains unchanged always
    3. Cincreases because thermal expansion lengthens the conductor
    4. Dbecomes zero
    💡 Explanation:

    Thermal expansion reduces tension for fixed supports, increasing sag.

  29. Q29Past Paper · PPSC/FPSC/NTSmedium

    Maximum conductor tension under worst loading usually occurs at

    1. Ahottest day only
    2. Bminimum temperature with maximum wind and/or ice loading
    3. Cno wind condition only
    4. Daverage annual temperature only
    💡 Explanation:

    Cold weather contracts the conductor while ice/wind increase mechanical loading.

  30. Q30hard

    Ruling span in sag-tension calculations is defined as

    1. Ashortest span in the section only
    2. Baverage tower height
    3. Cequivalent span that gives the same maximum tension as the actual supports
    4. Dground clearance distance
    💡 Explanation:

    Ruling span links actual unequal spans to equivalent uniform sag-tension behaviour.

  31. Q31Past Paper · PPSC/FPSC/NTSmedium

    Wind pressure on conductors increases effective weight and therefore

    1. Areduces sag
    2. Bincreases sag and reduces ground clearance
    3. Celiminates tension variation
    4. Dremoves aeolian vibration
    💡 Explanation:

    Transverse wind load adds to resultant conductor loading.

  32. Q32Past Paper · PPSC/FPSC/NTSeasy

    Ice loading on transmission conductors

    1. Areduces conductor weight
    2. Beliminates need for dampers
    3. Clowers tower height requirements
    4. Dincreases mechanical load and sag significantly
    💡 Explanation:

    Ice accretion increases weight and diameter, worsening sag and wind loading.

  33. Q33medium

    Stringing chart for overhead lines relates

    1. Asag and tension to temperature and span
    2. Bonly voltage regulation
    3. Conly fault level
    4. Donly relay settings
    💡 Explanation:

    Stringing charts guide erection tension for desired clearance at design conditions.

  34. Q34hard

    For supports at different elevations, sag calculation must account for

    1. Aonly span length ignoring height difference
    2. Bonly insulator type
    3. Conly bundle spacing
    4. Dunequal support heights and horizontal tension component
    💡 Explanation:

    Unequal supports change the catenary shape and clearance profile.

  35. Q35medium

    Aeolian vibration of conductors is caused by

    1. Avortex shedding in steady wind across the cylindrical conductor
    2. Bonly short-circuit forces
    3. Conly ice melting
    4. Donly transformer inrush
    💡 Explanation:

    Wind vortices excite conductor vibration at certain wind speeds.

  36. Q36Past Paper · PPSC/FPSC/NTSmedium

    Stockbridge dampers on transmission lines are used to mitigate

    1. Avoltage regulation only
    2. Baeolian vibration and fatigue of conductors
    3. Cfault current magnitude
    4. Dtransformer magnetizing inrush
    💡 Explanation:

    Dampers dissipate vibration energy to protect conductors and fittings.

  37. Q37medium

    Spacer dampers in bundled conductors prevent

    1. Asub-conductor clashing and control bundle oscillation
    2. Bincrease of line voltage to infinity
    3. Celimination of series impedance
    4. Dblocking of zero-sequence current
    💡 Explanation:

    Spacers maintain bundle geometry and reduce sub-conductor motion.

  38. Q38Past Paper · PPSC/FPSC/NTSeasy

    Ground clearance of an overhead line is minimum at

    1. Asupport/tower location
    2. Bonly during faults
    3. Ctransformer terminals
    4. Dmid-span where sag is maximum
    💡 Explanation:

    Lowest point of catenary is typically mid-span.

  39. Q39Past Paper · PPSC/FPSC/NTSeasy

    Higher erection tension during stringing results in

    1. Ahigher sag always
    2. Bzero tension at supports
    3. Clower sag at operating conditions
    4. Dinfinite span capability
    💡 Explanation:

    Greater initial tension reduces subsequent sag for given span and loading.

  40. Q40Past Paper · PPSC/FPSC/NTSeasy

    Pin type insulators are commonly used on

    1. A400 kV EHV long strings only
    2. Bunderground cables only
    3. CSF6 switchgear tanks only
    4. Ddistribution lines up to about 33 kV
    💡 Explanation:

    Pin insulators mount directly on cross-arms for lower voltages.

  41. Q41Past Paper · PPSC/FPSC/NTSeasy

    Suspension insulator strings on transmission towers are used because

    1. Athey provide flexibility and mechanical strength for higher voltages
    2. Bthey eliminate the need for conductors
    3. Cthey replace transformers
    4. Dthey conduct load current
    💡 Explanation:

    Disc strings hang from tower cross-arms and support conductor weight.

  42. Q42Past Paper · PPSC/FPSC/NTSmedium

    Strain insulators are employed at

    1. Adead ends, sharp corners and long spans where tension is high
    2. Btransformer bushings only
    3. Cmotor terminals only
    4. Dneutral grounding resistors only
    💡 Explanation:

    Strain insulators withstand mechanical tension in addition to voltage stress.

  43. Q43Past Paper · PPSC/FPSC/NTSmedium

    String efficiency of an insulator disc string is less than 100% when

    1. Aall discs share equal voltage always
    2. Bline frequency is zero
    3. Cconductor is perfectly smooth
    4. Dcapacitance to earth causes unequal voltage distribution across discs
    💡 Explanation:

    Voltage grading across discs is non-uniform without correction measures.

  44. Q44Past Paper · PPSC/FPSC/NTSeasy

    Flashover of an insulator refers to

    1. Apuncture through the solid insulator body only
    2. Bmelting of conductor only
    3. Ctripping of distance relay only
    4. Ddischarge over the insulator surface between electrodes
    💡 Explanation:

    Flashover is external arc; puncture damages the insulator internally.

  45. Q45Past Paper · PPSC/FPSC/NTSmedium

    Puncture of a porcelain insulator means

    1. Arecoverable surface flashover only
    2. Bpermanent breakdown through the insulator material
    3. Ccorona without discharge
    4. Dharmonic resonance only
    💡 Explanation:

    Puncture destroys the insulator and requires replacement.

  46. Q46hard

    Guard rings on insulator strings help by

    1. Aincreasing line inductance
    2. Bcarrying load current
    3. Cimproving voltage distribution and reducing stress on lower discs
    4. Dreplacing earth wire
    💡 Explanation:

    Grading rings capacitively shunt voltage for more uniform disc stress.

  47. Q47Past Paper · PPSC/FPSC/NTSmedium

    Number of insulator discs in a string depends on

    1. Aconductor material only
    2. Bpower factor of load only
    3. Cline voltage, insulation coordination and pollution level
    4. Dmotor slip only
    💡 Explanation:

    BIL, switching surges and contamination determine disc count.

  48. Q48medium

    Composite polymeric insulators compared to porcelain offer

    1. Aalways lower creepage requirement regardless of site
    2. Bzero need for maintenance ever
    3. Clighter weight and better pollution performance in many cases
    4. Dinability to used outdoors
    💡 Explanation:

    Silicone composite insulators resist wet pollution flashover well.

  49. Q49Past Paper · PPSC/FPSC/NTSmedium

    Fog-type or anti-fog insulator profiles are designed for

    1. Aindoor switchboards only
    2. Bcoastal and polluted environments with longer creepage paths
    3. CDC machines only
    4. Dunderground joints only
    💡 Explanation:

    Extended creepage reduces flashover under contaminated conditions.

  50. Q50hard

    Arcing horns on insulator fittings protect by

    1. Aincreasing series impedance of line
    2. Bmeasuring fault distance
    3. Cproviding a path for flashover away from critical metal parts
    4. Dturning on SCR
    💡 Explanation:

    Horns intercept arcs to reduce damage to hardware and conductors.

  51. Q51Past Paper · PPSC/FPSC/NTSmedium

    Corona on insulators in wet weather can lead to

    1. Aimproved string efficiency above 100%
    2. Bpower loss, audible noise and eventual flashover
    3. Czero charging current
    4. Dautomatic voltage boost without load
    💡 Explanation:

    Corona causes losses and can precede pollution flashover.

  52. Q52Past Paper · PPSC/FPSC/NTSeasy

    Creepage distance of an insulator is

    1. Adirect straight-line distance through air only
    2. Bsurface path length between live and earth electrodes
    3. Cconductor cross-section area
    4. Dtower footing resistance
    💡 Explanation:

    Adequate creepage prevents pollution flashover along the surface.

  53. Q53Past Paper · PPSC/FPSC/NTSeasy

    Underground cables compared with overhead lines have

    1. Ahigher capacitance and charging current per km
    2. Bnegligible shunt capacitance always
    3. Cno insulation requirement
    4. Dlower voltage rating always
    💡 Explanation:

    Close conductor-sheath spacing gives large capacitance in cables.

  54. Q54hard

    In belted type paper-insulated cables, insulation layers are

    1. Aalways individually metal screened
    2. Bimmersed in SF6 gas
    3. Creplaced by air gaps only
    4. Dlaid helically without metallic screens between phases
    💡 Explanation:

    Belted construction is older; stress concentration limits voltage.

  55. Q55Past Paper · PPSC/FPSC/NTSmedium

    Screened or shielded cables reduce electric stress by

    1. Aeliminating the need for conductor
    2. Busing only one phase in a duct
    3. Cremoving the sheath
    4. Dkeeping uniform radial field within each core insulation
    💡 Explanation:

    Metallic screens at ground potential grade stress within insulation.

  56. Q56hard

    Capacitance grading of high-voltage cables uses

    1. Alayers of different permittivity to equalize radial stress
    2. Bonly thicker conductor without insulation change
    3. Cair gaps between cores only
    4. Dopen delta connection
    💡 Explanation:

    Graded permittivity reduces maximum stress in insulation.

  57. Q57Past Paper · PPSC/FPSC/NTSmedium

    Charging current in a long underground cable at no load is

    1. Apurely resistive always
    2. Bleading current due to cable capacitance
    3. Czero because cable is short
    4. Dsame as fault current
    💡 Explanation:

    Large cable capacitance draws significant leading charging current.

  58. Q58medium

    Dielectric loss in cable insulation contributes to

    1. Aincreased surge impedance only
    2. Bheating of insulation and reduced transmission efficiency
    3. Cnegative sequence currents only
    4. Dmotor starting torque
    💡 Explanation:

    Loss tangent of insulation causes I²R type heating under voltage stress.

  59. Q59Past Paper · PPSC/FPSC/NTSeasy

    XLPE insulated cables are preferred today because

    1. Athey have good dielectric strength and easier jointing than PILC
    2. Bthey require only paper oil impregnation
    3. Cthey cannot be used underground
    4. Dthey have no thermal limit
    💡 Explanation:

    Cross-linked polyethylene is widely used in modern MV/HV cables.

  60. Q60Past Paper · PPSC/FPSC/NTSeasy

    Cable ampacity is primarily limited by

    1. Aonly skin effect at DC
    2. Bonly corona on sheath
    3. Conly relay pickup current
    4. Dconductor temperature rise under load
    💡 Explanation:

    Insulation and sheath temperature limits define continuous current rating.

  61. Q61hard

    Sheath bonding in cable systems is done to

    1. Acontrol sheath circulating currents and voltage rise
    2. Bincrease line inductance to infinity
    3. Creplace the conductor
    4. Deliminate the need for earthing
    💡 Explanation:

    Single-point or cross bonding reduces sheath losses and voltage.

  62. Q62hard

    Murray loop test on cables is used for

    1. Alocating earth faults in cable cores
    2. Bmeasuring power factor of motor
    3. Csetting distance relay zones
    4. Dcalibrating PID controller
    💡 Explanation:

    Wheatstone-type bridge methods locate fault distance in cables.

  63. Q63easy

    Minimum bending radius during cable installation must be observed to avoid

    1. Aincrease of surge impedance only
    2. Bdamage to insulation and conductors
    3. Cautomatic voltage regulation
    4. Dsymmetrical component unbalance only
    💡 Explanation:

    Excessive bending cracks insulation and displaces conductors.

  64. Q64medium

    Paper-oil impregnated (PILC) cables require

    1. Ano sealing against moisture ever
    2. Bcareful handling of oil impregnation and jointing skill
    3. Coperation only at DC voltage
    4. Dabsence of any metallic sheath
    💡 Explanation:

    PILC systems depend on oil-impregnated paper and skilled joints.

  65. Q65Past Paper · PPSC/FPSC/NTSmedium

    Direct burial of cables requires consideration of

    1. Aonly tower height
    2. Bonly insulator string length
    3. Csoil thermal resistivity and mechanical protection
    4. Donly excitation reactance of alternator
    💡 Explanation:

    Soil heat dissipation and mechanical damage govern burial design.

  66. Q66medium

    Cable joints and terminations must control

    1. Aelectric stress concentration at insulation interfaces
    2. Bonly rotor slip
    3. Conly prime mover governor gain
    4. Donly Buchholz gas volume only
    💡 Explanation:

    Stress cones and proper kits prevent partial discharge at terminations.

  67. Q67Past Paper · PPSC/FPSC/NTSeasy

    Per unit impedance is defined as

    1. Aactual impedance multiplied by base voltage only
    2. Bfault MVA divided by load MVA only
    3. Cline length in km divided by SIL
    4. Dactual impedance divided by base impedance
    💡 Explanation:

    Zpu = ZΩ/Zbase normalizes impedances for system studies.

  68. Q68Past Paper · PPSC/FPSC/NTSeasy

    Base impedance on a three-phase system is Zbase =

    1. AMVA_base / kV
    2. BkV / MVA_base only
    3. Csqrt(kV×MVA) only
    4. D(kV_line-line)² / MVA_base
    💡 Explanation:

    Zbase uses line-to-line kV squared over three-phase MVA base.

  69. Q69Past Paper · PPSC/FPSC/NTShard

    When changing per unit base, impedance transforms as

    1. AZpu_new = Zpu_old only regardless of base
    2. Bmultiply by kV_new only
    3. CZpu_new = Zpu_old × (MVA_new/MVA_old) × (kV_old/kV_new)²
    4. Ddivide by frequency
    💡 Explanation:

    Both MVA and kV bases affect per unit impedance values.

  70. Q70Past Paper · PPSC/FPSC/NTSmedium

    Per unit system simplifies fault calculations because

    1. Ait eliminates all resistances
    2. Bit removes the need for sequence networks
    3. Cit sets all voltages to zero
    4. Dtransformer turns ratios need not be explicitly applied in the network
    💡 Explanation:

    Consistent bases make interconnected elements add directly in pu.

  71. Q71Past Paper · PPSC/FPSC/NTSeasy

    For three-phase analysis, voltage base is usually

    1. Aphase-to-neutral only always
    2. Bpeak phase voltage only
    3. Cline-to-line nominal kV
    4. DDC link voltage only
    💡 Explanation:

    Industry practice uses line-to-line voltage for three-phase pu bases.

  72. Q72Past Paper · PPSC/FPSC/NTSmedium

    Transformer per unit impedance referred to a common MVA base is

    1. Aalways higher on LV by turns ratio squared only without MVA change
    2. Bzero on one side always
    3. Cthe same on HV and LV sides
    4. Ddependent only on frequency
    💡 Explanation:

    On a common MVA base, pu Z is identical on both windings.

  73. Q73hard

    Per unit admittance Ypu equals

    1. Aactual admittance divided by MVA only
    2. Bline SIL only
    3. Cactual admittance multiplied by base impedance
    4. Dreciprocal of sag
    💡 Explanation:

    Ypu = YΩ × Zbase is the dual of impedance conversion.

  74. Q74easy

    A common system study base MVA is

    1. Aalways 1 kVA only
    2. B100 MVA or a value matching the largest plant rating
    3. Calways equal to line length in km
    4. Dalways zero
    💡 Explanation:

    Engineers pick convenient MVA base for numerical simplicity.

  75. Q75Past Paper · PPSC/FPSC/NTSeasy

    Per unit active power Ppu is computed as

    1. Aactual MW divided by base MVA
    2. BMW multiplied by kV only
    3. CMVAR divided by frequency
    4. DkA times ohm only
    💡 Explanation:

    Real power in pu = P(MW)/Sbase(MVA).

  76. Q76Past Paper · PPSC/FPSC/NTSmedium

    Per unit current Ipu equals

    1. Aactual current times voltage only
    2. Bfault level only
    3. Cactual current divided by base current Ibase
    4. Dcharging current only
    💡 Explanation:

    Ibase = MVA_base / (√3 × kV_base) for three-phase systems.

  77. Q77easy

    Nominal voltage kV chosen as base is typically

    1. Aminimum sag voltage only
    2. Bcorona onset voltage only
    3. Crotor bar voltage only
    4. Drated or nominal system voltage of the bus
    💡 Explanation:

    Rated nominal kV defines the reference for pu voltages near 1.0 pu.

  78. Q78medium

    Advantage of per unit values near 1.0 is

    1. Aelimination of all fault currents
    2. Bimproved numerical conditioning in calculations
    3. Czero need for symmetrical components
    4. Dautomatic relay coordination
    💡 Explanation:

    Quantities of similar magnitude simplify hand and computer analysis.

  79. Q79hard

    Line charging susceptance in per unit is often small compared to

    1. Ashort-circuit impedance magnitudes in transmission studies
    2. Btransformer nameplate MVA only
    3. Cmotor slip values
    4. Dfuse cut-off ratings
    💡 Explanation:

    Shunt B is secondary to series Z in many fault and load-flow studies.

  80. Q80Past Paper · PPSC/FPSC/NTSeasy

    Voltage regulation of a line approximated at lagging power factor is

    1. A(|Vr| − |Vs|)/|Vs| only at leading pf always
    2. Bzero for all loads always
    3. C(|Vs| − |Vr|) / |Vr| × 100% for comparable angles
    4. Dequal to SIL always
    💡 Explanation:

    Regulation compares sending and receiving voltages at specified load pf.