Transmission and Distribution MCQs 2026

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

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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 1 Questions 110 of 80
  1. Q1 Past Paper · PPSC/FPSC/NTS easy

    When ambient temperature rises, conductor sag generally

    1. A decreases always
    2. B remains unchanged always
    3. C increases because thermal expansion lengthens the conductor
    4. D becomes zero
    💡 Explanation:

    Thermal expansion reduces tension for fixed supports, increasing sag.

  2. Q2 Past Paper · PPSC/FPSC/NTS medium

    Maximum conductor tension under worst loading usually occurs at

    1. A hottest day only
    2. B minimum temperature with maximum wind and/or ice loading
    3. C no wind condition only
    4. D average annual temperature only
    💡 Explanation:

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

  3. Q3 hard

    Ruling span in sag-tension calculations is defined as

    1. A shortest span in the section only
    2. B average tower height
    3. C equivalent span that gives the same maximum tension as the actual supports
    4. D ground clearance distance
    💡 Explanation:

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

  4. Q4 Past Paper · PPSC/FPSC/NTS medium

    Wind pressure on conductors increases effective weight and therefore

    1. A reduces sag
    2. B increases sag and reduces ground clearance
    3. C eliminates tension variation
    4. D removes aeolian vibration
    💡 Explanation:

    Transverse wind load adds to resultant conductor loading.

  5. Q5 Past Paper · PPSC/FPSC/NTS easy

    Ice loading on transmission conductors

    1. A reduces conductor weight
    2. B eliminates need for dampers
    3. C lowers tower height requirements
    4. D increases mechanical load and sag significantly
    💡 Explanation:

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

  6. Q6 medium

    Stringing chart for overhead lines relates

    1. A sag and tension to temperature and span
    2. B only voltage regulation
    3. C only fault level
    4. D only relay settings
    💡 Explanation:

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

  7. Q7 hard

    For supports at different elevations, sag calculation must account for

    1. A only span length ignoring height difference
    2. B only insulator type
    3. C only bundle spacing
    4. D unequal support heights and horizontal tension component
    💡 Explanation:

    Unequal supports change the catenary shape and clearance profile.

  8. Q8 medium

    Aeolian vibration of conductors is caused by

    1. A vortex shedding in steady wind across the cylindrical conductor
    2. B only short-circuit forces
    3. C only ice melting
    4. D only transformer inrush
    💡 Explanation:

    Wind vortices excite conductor vibration at certain wind speeds.

  9. Q9 Past Paper · PPSC/FPSC/NTS medium

    Stockbridge dampers on transmission lines are used to mitigate

    1. A voltage regulation only
    2. B aeolian vibration and fatigue of conductors
    3. C fault current magnitude
    4. D transformer magnetizing inrush
    💡 Explanation:

    Dampers dissipate vibration energy to protect conductors and fittings.

  10. Q10 medium

    Spacer dampers in bundled conductors prevent

    1. A sub-conductor clashing and control bundle oscillation
    2. B increase of line voltage to infinity
    3. C elimination of series impedance
    4. D blocking of zero-sequence current
    💡 Explanation:

    Spacers maintain bundle geometry and reduce sub-conductor motion.

  11. Q11 Past Paper · PPSC/FPSC/NTS easy

    Ground clearance of an overhead line is minimum at

    1. A support/tower location
    2. B only during faults
    3. C transformer terminals
    4. D mid-span where sag is maximum
    💡 Explanation:

    Lowest point of catenary is typically mid-span.

  12. Q12 Past Paper · PPSC/FPSC/NTS easy

    Higher erection tension during stringing results in

    1. A higher sag always
    2. B zero tension at supports
    3. C lower sag at operating conditions
    4. D infinite span capability
    💡 Explanation:

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

  13. Q13 Past Paper · PPSC/FPSC/NTS easy

    Pin type insulators are commonly used on

    1. A 400 kV EHV long strings only
    2. B underground cables only
    3. C SF6 switchgear tanks only
    4. D distribution lines up to about 33 kV
    💡 Explanation:

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

  14. Q14 Past Paper · PPSC/FPSC/NTS easy

    Suspension insulator strings on transmission towers are used because

    1. A they provide flexibility and mechanical strength for higher voltages
    2. B they eliminate the need for conductors
    3. C they replace transformers
    4. D they conduct load current
    💡 Explanation:

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

  15. Q15 Past Paper · PPSC/FPSC/NTS medium

    Strain insulators are employed at

    1. A dead ends, sharp corners and long spans where tension is high
    2. B transformer bushings only
    3. C motor terminals only
    4. D neutral grounding resistors only
    💡 Explanation:

    Strain insulators withstand mechanical tension in addition to voltage stress.

  16. Q16 Past Paper · PPSC/FPSC/NTS medium

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

    1. A all discs share equal voltage always
    2. B line frequency is zero
    3. C conductor is perfectly smooth
    4. D capacitance to earth causes unequal voltage distribution across discs
    💡 Explanation:

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

  17. Q17 Past Paper · PPSC/FPSC/NTS easy

    Flashover of an insulator refers to

    1. A puncture through the solid insulator body only
    2. B melting of conductor only
    3. C tripping of distance relay only
    4. D discharge over the insulator surface between electrodes
    💡 Explanation:

    Flashover is external arc; puncture damages the insulator internally.

  18. Q18 Past Paper · PPSC/FPSC/NTS medium

    Puncture of a porcelain insulator means

    1. A recoverable surface flashover only
    2. B permanent breakdown through the insulator material
    3. C corona without discharge
    4. D harmonic resonance only
    💡 Explanation:

    Puncture destroys the insulator and requires replacement.

  19. Q19 hard

    Guard rings on insulator strings help by

    1. A increasing line inductance
    2. B carrying load current
    3. C improving voltage distribution and reducing stress on lower discs
    4. D replacing earth wire
    💡 Explanation:

    Grading rings capacitively shunt voltage for more uniform disc stress.

  20. Q20 Past Paper · PPSC/FPSC/NTS medium

    Number of insulator discs in a string depends on

    1. A conductor material only
    2. B power factor of load only
    3. C line voltage, insulation coordination and pollution level
    4. D motor slip only
    💡 Explanation:

    BIL, switching surges and contamination determine disc count.

  21. Q21 medium

    Composite polymeric insulators compared to porcelain offer

    1. A always lower creepage requirement regardless of site
    2. B zero need for maintenance ever
    3. C lighter weight and better pollution performance in many cases
    4. D inability to used outdoors
    💡 Explanation:

    Silicone composite insulators resist wet pollution flashover well.

  22. Q22 Past Paper · PPSC/FPSC/NTS medium

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

    1. A indoor switchboards only
    2. B coastal and polluted environments with longer creepage paths
    3. C DC machines only
    4. D underground joints only
    💡 Explanation:

    Extended creepage reduces flashover under contaminated conditions.

  23. Q23 Past Paper · PPSC/FPSC/NTS medium

    Transformer per unit impedance referred to a common MVA base is

    1. A always higher on LV by turns ratio squared only without MVA change
    2. B zero on one side always
    3. C the same on HV and LV sides
    4. D dependent only on frequency
    💡 Explanation:

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

  24. Q24 hard

    Per unit admittance Ypu equals

    1. A actual admittance divided by MVA only
    2. B line SIL only
    3. C actual admittance multiplied by base impedance
    4. D reciprocal of sag
    💡 Explanation:

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

  25. Q25 easy

    A common system study base MVA is

    1. A always 1 kVA only
    2. B 100 MVA or a value matching the largest plant rating
    3. C always equal to line length in km
    4. D always zero
    💡 Explanation:

    Engineers pick convenient MVA base for numerical simplicity.

  26. Q26 Past Paper · PPSC/FPSC/NTS easy

    Per unit active power Ppu is computed as

    1. A actual MW divided by base MVA
    2. B MW multiplied by kV only
    3. C MVAR divided by frequency
    4. D kA times ohm only
    💡 Explanation:

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

  27. Q27 Past Paper · PPSC/FPSC/NTS medium

    Per unit current Ipu equals

    1. A actual current times voltage only
    2. B fault level only
    3. C actual current divided by base current Ibase
    4. D charging current only
    💡 Explanation:

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

  28. Q28 easy

    Nominal voltage kV chosen as base is typically

    1. A minimum sag voltage only
    2. B corona onset voltage only
    3. C rotor bar voltage only
    4. D rated or nominal system voltage of the bus
    💡 Explanation:

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

  29. Q29 medium

    Advantage of per unit values near 1.0 is

    1. A elimination of all fault currents
    2. B improved numerical conditioning in calculations
    3. C zero need for symmetrical components
    4. D automatic relay coordination
    💡 Explanation:

    Quantities of similar magnitude simplify hand and computer analysis.

  30. Q30 hard

    Line charging susceptance in per unit is often small compared to

    1. A short-circuit impedance magnitudes in transmission studies
    2. B transformer nameplate MVA only
    3. C motor slip values
    4. D fuse cut-off ratings
    💡 Explanation:

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

  31. Q31 Past Paper · PPSC/FPSC/NTS easy

    Voltage regulation of a line approximated at lagging power factor is

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

    Regulation compares sending and receiving voltages at specified load pf.

  32. Q32 Past Paper · PPSC/FPSC/NTS medium

    Ferranti effect occurs when

    1. A heavy lagging load at short line only
    2. B fault at transformer only
    3. C receiving end voltage exceeds sending end voltage at light load on long lines
    4. D motor starting only
    💡 Explanation:

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

  33. Q33 Past Paper · PPSC/FPSC/NTS medium

    Ferranti effect is most pronounced on

    1. A short distribution feeders at full load
    2. B DC resistive loads only
    3. C long EHV lines at light or no load
    4. D synchronous motor full load
    💡 Explanation:

    Large shunt capacitance relative to load causes voltage rise.

  34. Q34 Past Paper · PPSC/FPSC/NTS medium

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

    1. A leading due to line charging
    2. B always unity only
    3. C always lagging only
    4. D zero always
    💡 Explanation:

    Capacitive charging supplies leading reactive power at the receiving end.

  35. Q35 hard

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

    1. A always zero
    2. B always equal to short-circuit level
    3. C equal to DC battery voltage only
    4. D higher than sending end due to capacitive Ferranti rise
    💡 Explanation:

    Open-end voltage rise is classic Ferranti behaviour.

  36. Q36 Past Paper · PPSC/FPSC/NTS easy

    Positive voltage regulation at lagging load means

    1. A receiving voltage is always higher
    2. B sending end voltage magnitude exceeds receiving end under load
    3. C line has zero impedance
    4. D power flows from load to source only
    💡 Explanation:

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

  37. Q37 Past Paper · PPSC/FPSC/NTS medium

    Load power factor strongly affects voltage regulation because

    1. A resistance becomes zero
    2. B frequency doubles
    3. C reactive component of current causes additional voltage drop in line impedance
    4. D SIL becomes negative
    💡 Explanation:

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

  38. Q38 Past Paper · PPSC/FPSC/NTS easy

    Shunt capacitors installed at load buses improve voltage profile by

    1. A increasing line charging on EHV only
    2. B increasing series impedance
    3. C supplying reactive power locally and reducing line VAR flow
    4. D blocking positive-sequence faults
    💡 Explanation:

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

  39. Q39 Past Paper · PPSC/FPSC/NTS medium

    Series capacitors on transmission lines reduce voltage regulation by

    1. A increasing Ferranti rise only
    2. B partially compensating line inductive reactance
    3. C eliminating the need for conductors
    4. D replacing distance protection
    💡 Explanation:

    Series compensation lowers effective X and improves stability and voltage.

  40. Q40 Past Paper · PPSC/FPSC/NTS medium

    On-load tap changers on transformers regulate voltage by

    1. A changing transformer turns ratio under load
    2. B changing line length physically
    3. C inserting fuses in series
    4. D opening circuit breaker contacts only
    💡 Explanation:

    OLTC adjusts secondary voltage without interrupting supply.

  41. Q41 Past Paper · PPSC/FPSC/NTS medium

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

    1. A increase Ferranti effect
    2. B raise receiving voltage further
    3. C replace series impedance completely
    4. D absorb excess reactive power and limit voltage rise
    💡 Explanation:

    Reactors compensate capacitive charging during low load periods.

  42. Q42 Past Paper · PPSC/FPSC/NTS hard

    ABCD parameters of a transmission line are used to compute

    1. A sending end voltage and current from receiving end conditions
    2. B only transformer copper loss
    3. C only motor slip
    4. D only fuse melting time only
    💡 Explanation:

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

  43. Q43 hard

    Critical loading for voltage support on uncompensated lines relates to

    1. A motor starting current only
    2. B SIL and line length where reactive power balance changes character
    3. C Buchholz trip level only
    4. D PID derivative time only
    💡 Explanation:

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

  44. Q44 Past Paper · PPSC/FPSC/NTS easy

    Characteristic impedance Zc of a lossless overhead line equals

    1. A R/L only
    2. B sqrt(L/C) where L is series inductance and C is shunt capacitance per unit length
    3. C C/R only
    4. D sqrt(RG) only
    💡 Explanation:

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

  45. Q45 Past Paper · PPSC/FPSC/NTS medium

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

    1. A V/Zc only
    2. B Zc/V²
    3. C V×Zc×√3 only
    4. D V²/Zc in MW for line-to-line voltage V
    💡 Explanation:

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

  46. Q46 medium

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

    1. A only at the receiving end
    2. B at both sending and receiving ends as half the total line capacitance each
    3. C only at the sending end
    4. D entirely in the middle as a lump
    💡 Explanation:

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

  47. Q47 Past Paper · PPSC/FPSC/NTS easy

    Series impedance per unit length of an overhead line includes

    1. A only shunt capacitance
    2. B only corona loss resistance alone
    3. C only transformer leakage
    4. D resistance and inductive reactance of conductors
    💡 Explanation:

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

  48. Q48 hard

    Geometric mean distance (GMD) in inductance formula refers to

    1. A resistance of earth wire only
    2. B insulator string length
    3. C tower footing resistance
    4. D equivalent spacing among bundled or multi-conductor arrangements
    💡 Explanation:

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

  49. Q49 hard

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

    1. A only corona onset voltage
    2. B equivalent inductance per phase lower than single conductor
    3. C only sag at maximum tension
    4. D only zero-sequence resistance only
    💡 Explanation:

    Bundling reduces effective GMR and hence line inductance per phase.

  50. Q50 Past Paper · PPSC/FPSC/NTS medium

    Skin effect in AC transmission conductors causes

    1. A uniform current across the cross-section always
    2. B reduction of resistance at higher frequency
    3. C current concentration near the surface increasing effective resistance
    4. D zero inductance change
    💡 Explanation:

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

  51. Q51 hard

    Proximity effect in parallel conductors results in

    1. A non-uniform current distribution and increased effective resistance
    2. B increased shunt capacitance only
    3. C elimination of inductive reactance
    4. D infinite surge impedance
    💡 Explanation:

    Mutual magnetic fields redistribute current within conductors, raising losses.

  52. Q52 Past Paper · PPSC/FPSC/NTS medium

    Bundled conductors on EHV lines primarily reduce

    1. A corona loss and line inductance
    2. B series resistance to zero
    3. C shunt capacitance to zero
    4. D surge impedance below 1 ohm always
    💡 Explanation:

    Sub-conductors lower surface voltage gradient and effective inductance.

  53. Q53 Past Paper · PPSC/FPSC/NTS easy

    A short transmission line model neglects

    1. A shunt capacitance and uses only series impedance
    2. B series resistance only
    3. C both series and shunt completely
    4. D transformer magnetizing branch
    💡 Explanation:

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

  54. Q54 hard

    Propagation constant γ of a uniform line is

    1. A only jωL
    2. B only R/L
    3. C only G/C without imaginary part
    4. D γ = sqrt((R+jωL)(G+jωC)) = α + jβ
    💡 Explanation:

    γ has attenuation constant α and phase constant β.

  55. Q55 medium

    Nominal T equivalent of a line places

    1. A all impedance at one end only
    2. B capacitance only at sending end
    3. C no shunt branch
    4. D half 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.

  56. Q56 Past Paper · PPSC/FPSC/NTS easy

    Charging current in an unloaded long overhead line is due to

    1. A shunt capacitance of the line
    2. B series resistance only
    3. C transformer no-load current only
    4. D fault current only
    💡 Explanation:

    Capacitive charging draws leading current even without load.

  57. Q57 easy

    Resistance of a conductor per km increases with temperature because

    1. A inductance falls to zero
    2. B resistivity of metal increases with temperature
    3. C capacitance doubles always
    4. D GMR becomes infinite
    💡 Explanation:

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

  58. Q58 Past Paper · PPSC/FPSC/NTS easy

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

    1. A s = wL/T
    2. B s = 8T/(wL²)
    3. C s = wL²/(8T) where w is weight per unit length and T is tension
    4. D s = w/T only
    💡 Explanation:

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

  59. Q59 hard

    Arcing horns on insulator fittings protect by

    1. A increasing series impedance of line
    2. B measuring fault distance
    3. C providing a path for flashover away from critical metal parts
    4. D turning on SCR
    💡 Explanation:

    Horns intercept arcs to reduce damage to hardware and conductors.

  60. Q60 Past Paper · PPSC/FPSC/NTS medium

    Corona on insulators in wet weather can lead to

    1. A improved string efficiency above 100%
    2. B power loss, audible noise and eventual flashover
    3. C zero charging current
    4. D automatic voltage boost without load
    💡 Explanation:

    Corona causes losses and can precede pollution flashover.

  61. Q61 Past Paper · PPSC/FPSC/NTS easy

    Creepage distance of an insulator is

    1. A direct straight-line distance through air only
    2. B surface path length between live and earth electrodes
    3. C conductor cross-section area
    4. D tower footing resistance
    💡 Explanation:

    Adequate creepage prevents pollution flashover along the surface.

  62. Q62 Past Paper · PPSC/FPSC/NTS easy

    Underground cables compared with overhead lines have

    1. A higher capacitance and charging current per km
    2. B negligible shunt capacitance always
    3. C no insulation requirement
    4. D lower voltage rating always
    💡 Explanation:

    Close conductor-sheath spacing gives large capacitance in cables.

  63. Q63 hard

    In belted type paper-insulated cables, insulation layers are

    1. A always individually metal screened
    2. B immersed in SF6 gas
    3. C replaced by air gaps only
    4. D laid helically without metallic screens between phases
    💡 Explanation:

    Belted construction is older; stress concentration limits voltage.

  64. Q64 Past Paper · PPSC/FPSC/NTS medium

    Screened or shielded cables reduce electric stress by

    1. A eliminating the need for conductor
    2. B using only one phase in a duct
    3. C removing the sheath
    4. D keeping uniform radial field within each core insulation
    💡 Explanation:

    Metallic screens at ground potential grade stress within insulation.

  65. Q65 hard

    Capacitance grading of high-voltage cables uses

    1. A layers of different permittivity to equalize radial stress
    2. B only thicker conductor without insulation change
    3. C air gaps between cores only
    4. D open delta connection
    💡 Explanation:

    Graded permittivity reduces maximum stress in insulation.

  66. Q66 Past Paper · PPSC/FPSC/NTS medium

    Charging current in a long underground cable at no load is

    1. A purely resistive always
    2. B leading current due to cable capacitance
    3. C zero because cable is short
    4. D same as fault current
    💡 Explanation:

    Large cable capacitance draws significant leading charging current.

  67. Q67 medium

    Dielectric loss in cable insulation contributes to

    1. A increased surge impedance only
    2. B heating of insulation and reduced transmission efficiency
    3. C negative sequence currents only
    4. D motor starting torque
    💡 Explanation:

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

  68. Q68 Past Paper · PPSC/FPSC/NTS easy

    XLPE insulated cables are preferred today because

    1. A they have good dielectric strength and easier jointing than PILC
    2. B they require only paper oil impregnation
    3. C they cannot be used underground
    4. D they have no thermal limit
    💡 Explanation:

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

  69. Q69 Past Paper · PPSC/FPSC/NTS easy

    Cable ampacity is primarily limited by

    1. A only skin effect at DC
    2. B only corona on sheath
    3. C only relay pickup current
    4. D conductor temperature rise under load
    💡 Explanation:

    Insulation and sheath temperature limits define continuous current rating.

  70. Q70 hard

    Sheath bonding in cable systems is done to

    1. A control sheath circulating currents and voltage rise
    2. B increase line inductance to infinity
    3. C replace the conductor
    4. D eliminate the need for earthing
    💡 Explanation:

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

  71. Q71 hard

    Murray loop test on cables is used for

    1. A locating earth faults in cable cores
    2. B measuring power factor of motor
    3. C setting distance relay zones
    4. D calibrating PID controller
    💡 Explanation:

    Wheatstone-type bridge methods locate fault distance in cables.

  72. Q72 easy

    Minimum bending radius during cable installation must be observed to avoid

    1. A increase of surge impedance only
    2. B damage to insulation and conductors
    3. C automatic voltage regulation
    4. D symmetrical component unbalance only
    💡 Explanation:

    Excessive bending cracks insulation and displaces conductors.

  73. Q73 medium

    Paper-oil impregnated (PILC) cables require

    1. A no sealing against moisture ever
    2. B careful handling of oil impregnation and jointing skill
    3. C operation only at DC voltage
    4. D absence of any metallic sheath
    💡 Explanation:

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

  74. Q74 Past Paper · PPSC/FPSC/NTS medium

    Direct burial of cables requires consideration of

    1. A only tower height
    2. B only insulator string length
    3. C soil thermal resistivity and mechanical protection
    4. D only excitation reactance of alternator
    💡 Explanation:

    Soil heat dissipation and mechanical damage govern burial design.

  75. Q75 medium

    Cable joints and terminations must control

    1. A electric stress concentration at insulation interfaces
    2. B only rotor slip
    3. C only prime mover governor gain
    4. D only Buchholz gas volume only
    💡 Explanation:

    Stress cones and proper kits prevent partial discharge at terminations.

  76. Q76 Past Paper · PPSC/FPSC/NTS easy

    Per unit impedance is defined as

    1. A actual impedance multiplied by base voltage only
    2. B fault MVA divided by load MVA only
    3. C line length in km divided by SIL
    4. D actual impedance divided by base impedance
    💡 Explanation:

    Zpu = ZΩ/Zbase normalizes impedances for system studies.

  77. Q77 Past Paper · PPSC/FPSC/NTS easy

    Base impedance on a three-phase system is Zbase =

    1. A MVA_base / kV
    2. B kV / MVA_base only
    3. C sqrt(kV×MVA) only
    4. D (kV_line-line)² / MVA_base
    💡 Explanation:

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

  78. Q78 Past Paper · PPSC/FPSC/NTS hard

    When changing per unit base, impedance transforms as

    1. A Zpu_new = Zpu_old only regardless of base
    2. B multiply by kV_new only
    3. C Zpu_new = Zpu_old × (MVA_new/MVA_old) × (kV_old/kV_new)²
    4. D divide by frequency
    💡 Explanation:

    Both MVA and kV bases affect per unit impedance values.

  79. Q79 Past Paper · PPSC/FPSC/NTS medium

    Per unit system simplifies fault calculations because

    1. A it eliminates all resistances
    2. B it removes the need for sequence networks
    3. C it sets all voltages to zero
    4. D transformer turns ratios need not be explicitly applied in the network
    💡 Explanation:

    Consistent bases make interconnected elements add directly in pu.

  80. Q80 Past Paper · PPSC/FPSC/NTS easy

    For three-phase analysis, voltage base is usually

    1. A phase-to-neutral only always
    2. B peak phase voltage only
    3. C line-to-line nominal kV
    4. D DC link voltage only
    💡 Explanation:

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