Three-Phase Induction Motors MCQs 2026

90 questions with detailed answers · 32 from past papers · 9 quiz batches available

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Page 1 of 1 Questions 110 of 90
  1. Q1 Past Paper · PPSC/FPSC/CSS easy

    The slip of a three-phase induction motor is defined as

    1. A (Nr - Ns) / Nr
    2. B Nr / Ns
    3. C Ns / Nr
    4. D (Ns - Nr) / Ns
    💡 Explanation:

    Slip s expresses how far rotor speed Nr lags synchronous speed Ns.

  2. Q2 easy

    Synchronous speed of a 50 Hz, 4-pole induction motor is

    1. A 3000 rpm
    2. B 750 rpm
    3. C 1500 rpm
    4. D 1000 rpm
    💡 Explanation:

    Ns = 120f/P = 120×50/4 = 1500 rpm.

  3. Q3 Past Paper · PPSC/FPSC/CSS easy

    At standstill, slip of an induction motor is

    1. A 0
    2. B 0.5
    3. C infinity
    4. D 1
    💡 Explanation:

    When Nr = 0, s = (Ns - 0)/Ns = 1.

  4. Q4 easy

    At synchronous speed, slip of an induction motor is

    1. A 1
    2. B 0.5
    3. C 0
    4. D negative
    💡 Explanation:

    When Nr = Ns, s = 0 and no torque is developed.

  5. Q5 Past Paper · PPSC/FPSC/CSS medium

    Rotor frequency in a running induction motor equals

    1. A f / s
    2. B f + s
    3. C 2 × f
    4. D s × f
    💡 Explanation:

    Rotor induced EMF and current frequency is sf where s is slip.

  6. Q6 medium

    For a 50 Hz motor at 4% slip, rotor frequency is

    1. A 50 Hz
    2. B 4 Hz
    3. C 0.5 Hz
    4. D 2 Hz
    💡 Explanation:

    sf = 0.04 × 50 = 2 Hz.

  7. Q7 Past Paper · PPSC/FPSC/CSS medium

    The torque-slip curve of an induction motor is approximately linear in the region

    1. A low slip (stable operating region)
    2. B high slip near breakdown
    3. C only at synchronous speed
    4. D only at standstill
    💡 Explanation:

    In normal load range, torque rises nearly linearly with slip.

  8. Q8 medium

    Breakdown torque of an induction motor occurs at

    1. A zero slip
    2. B synchronous speed
    3. C a specific slip beyond which torque falls sharply
    4. D infinite slip only
    💡 Explanation:

    Maximum torque point on the torque-slip curve is breakdown or pull-out torque.

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

    Starting torque of a squirrel-cage motor is typically

    1. A 1.5 to 2.5 times rated torque
    2. B equal to breakdown torque always
    3. C less than 0.5 times rated torque
    4. D zero until rated speed
    💡 Explanation:

    Design of rotor bars sets starting torque relative to full-load torque.

  10. Q10 easy

    Direct-on-line (DOL) starting applies

    1. A reduced voltage through rotor resistance
    2. B only half the poles initially
    3. C DC excitation to rotor
    4. D full line voltage to stator terminals at start
    💡 Explanation:

    DOL gives highest starting current and torque for a given motor.

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

    Main disadvantage of DOL starting for large motors is

    1. A no starting torque
    2. B permanent speed reduction
    3. C need for slip rings
    4. D very high starting current (5-7 times FLC)
    💡 Explanation:

    High inrush current causes voltage dip and thermal stress.

  12. Q12 medium

    Star-delta starting reduces starting current to approximately

    1. A one-third of DOL value
    2. B one-half
    3. C one-tenth
    4. D same as DOL
    💡 Explanation:

    Line current in star is 1/√3 of delta; power and torque scale similarly at start.

  13. Q13 Past Paper · PPSC/FPSC/CSS medium

    Star-delta starter is applicable only when

    1. A motor is designed for delta running with six leads accessible
    2. B motor has wound rotor
    3. C motor always runs in star
    4. D supply is single phase
    💡 Explanation:

    Stator must be reconnectable between star start and delta run.

  14. Q14 medium

    Autotransformer starting reduces voltage to motor by

    1. A inserting resistance in rotor circuit
    2. B changing number of poles
    3. C reversing two phases
    4. D tapping a fraction of line voltage
    💡 Explanation:

    Reduced stator voltage lowers starting current and torque proportionally to V².

  15. Q15 Past Paper · PPSC/FPSC/CSS easy

    Rotor resistance starting is used with

    1. A wound-rotor (slip-ring) induction motors
    2. B squirrel-cage motors only
    3. C synchronous motors only
    4. D single-phase shaded-pole motors
    💡 Explanation:

    External resistance in rotor circuit increases starting torque and limits current.

  16. Q16 medium

    Adding external resistance in wound-rotor circuit at start

    1. A increases starting torque and reduces starting current
    2. B reduces both torque and current
    3. C increases speed above synchronous
    4. D eliminates slip
    💡 Explanation:

    Higher rotor resistance shifts breakdown torque toward higher slip, aiding start.

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

    Squirrel-cage rotor construction uses

    1. A slip rings and brush gear
    2. B copper or aluminum bars short-circuited by end rings
    3. C separate DC winding
    4. D permanent magnets on rotor
    💡 Explanation:

    Cage rotor is rugged, maintenance-free, and low cost.

  18. Q18 easy

    Wound-rotor motors differ from cage motors mainly by

    1. A higher synchronous speed
    2. B absence of stator winding
    3. C accessible rotor winding via slip rings for external resistance
    4. D operation only at unity slip
    💡 Explanation:

    Slip rings allow resistance control and slip energy recovery schemes.

  19. Q19 Past Paper · PPSC/FPSC/CSS hard

    Crawling in induction motors is caused by

    1. A harmonics (especially 7th) producing forward torque at sub-synchronous speeds
    2. B excessive load only
    3. C unity power factor
    4. D open circuit in one phase
    💡 Explanation:

    Harmonic torques can make motor run stably at about 1/7th of synchronous speed.

  20. Q20 hard

    Cogging in induction motors occurs when

    1. A rotor runs above synchronous speed
    2. B rotor and stator slot numbers create locking tendency at low voltage
    3. C frequency is doubled
    4. D motor is overloaded
    💡 Explanation:

    Magnetic locking between stator and rotor teeth prevents smooth starting.

  21. Q21 Past Paper · PPSC/FPSC/CSS medium

    Ratio of starting torque to full-load torque is called

    1. A starting torque ratio or Tst/TL
    2. B slip ratio
    3. C breakdown factor only
    4. D service factor
    💡 Explanation:

    Tst defines ability to accelerate load from rest.

  22. Q22 medium

    High-resistance rotor bars increase

    1. A synchronous speed
    2. B starting torque but may reduce running efficiency
    3. C breakdown torque always to infinity
    4. D power factor at full load only
    💡 Explanation:

    Trade-off between start performance and running losses.

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

    DOL starting torque is proportional to

    1. A square of slip only
    2. B inverse of frequency cubed
    3. C square of applied voltage
    4. D rotor external resistance only
    💡 Explanation:

    Torque ∝ V² for induction motor at fixed frequency and slip.

  24. Q24 medium

    Star connection at start compared to delta run gives starting torque

    1. A three times delta torque
    2. B equal torque
    3. C nine times delta torque
    4. D one-third of delta-start torque
    💡 Explanation:

    Voltage per phase is 1/√3, so torque ∝ V² gives 1/3 torque.

  25. Q25 Past Paper · PPSC/FPSC/CSS hard

    Autotransformer starter set at 70% tap gives approximately

    1. A 70% torque
    2. B 49% of full-voltage starting torque
    3. C 30% torque
    4. D 100% torque
    💡 Explanation:

    Torque scales with V²: 0.7² ≈ 0.49 of DOL torque.

  26. Q26 hard

    For maximum starting torque with rotor resistance, total rotor circuit resistance should equal

    1. A stator resistance R1 only
    2. B zero ohms
    3. C infinite ohms
    4. D rotor standstill reactance X2
    💡 Explanation:

    Condition R2 = X2 places breakdown at s=1 for maximum start torque.

  27. Q27 Past Paper · PPSC/FPSC/CSS easy

    Slip rings on wound-rotor motors are used to

    1. A feed DC to stator
    2. B connect external resistance or recovery equipment to rotor winding
    3. C measure air-gap flux only
    4. D cool the bearings
    💡 Explanation:

    Brushes ride on rings to access three-phase rotor winding.

  28. Q28 easy

    Maintenance of squirrel-cage motors is generally lower because

    1. A rotor spins without any magnetic field
    2. B stator has no winding
    3. C rotor has no brushes or external connections
    4. D they never heat up
    💡 Explanation:

    Eliminating slip-ring assembly reduces wear and inspection.

  29. Q29 Past Paper · PPSC/FPSC/CSS hard

    To reduce cogging, designers often choose

    1. A equal stator and rotor slot counts
    2. B rotor slots skewed and slot combinations avoiding equal numbers
    3. C zero air gap
    4. D single-phase supply
    💡 Explanation:

    Skew and slot ratio selection minimizes locking torque.

  30. Q30 medium

    Skewing rotor bars helps reduce

    1. A synchronous speed
    2. B rated voltage
    3. C number of poles
    4. D cogging and noise
    💡 Explanation:

    Skew smooths torque variation as bars traverse stator field.

  31. Q31 Past Paper · PPSC/FPSC/CSS hard

    Outer cage in double-cage rotor has

    1. A no function at start
    2. B lower resistance than inner cage always
    3. C direct DC connection
    4. D higher resistance and lower leakage inductance for starting
    💡 Explanation:

    Outer cage carries high-frequency start current due to leakage flux path.

  32. Q32 hard

    Inner cage of double-cage rotor primarily carries

    1. A only harmonics at crawl speed
    2. B running current at low rotor frequency
    3. C stator current directly
    4. D excitation for synchronous speed
    💡 Explanation:

    Low frequency at run allows inner low-resistance cage to dominate.

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

    Below base speed in V/f drive, increasing frequency without raising voltage causes

    1. A flux increase
    2. B speed above synchronous mechanically
    3. C zero slip operation
    4. D flux weakening and reduced torque capability
    💡 Explanation:

    Constant V/f keeps air-gap flux near rated value.

  34. Q34 hard

    Above base speed, V/f drives often operate in

    1. A constant torque with rising voltage unlimited
    2. B constant power region with weakened flux
    3. C blocked rotor mode
    4. D pure DOL connection
    💡 Explanation:

    Voltage ceiling limits flux; torque falls with speed above base.

  35. Q35 Past Paper · PPSC/FPSC/CSS medium

    Pole-changing from 4 poles to 8 poles at same 50 Hz halves

    1. A supply frequency
    2. B rotor resistance
    3. C synchronous speed
    4. D stator slots
    💡 Explanation:

    Ns inversely proportional to pole count P.

  36. Q36 medium

    Two separate windings for two speeds is

    1. A impossible on induction machines
    2. B same as rotor resistance start
    3. C only for DC motors
    4. D less economical than Dahlander but flexible in ratio
    💡 Explanation:

    Dual-winding motors cost more but allow non-2:1 speed ratios.

  37. Q37 Past Paper · PPSC/FPSC/CSS hard

    Static Kramer drive inserts

    1. A resistance only permanently
    2. B controlled converter in rotor circuit to recover slip energy
    3. C DOL contactor in stator
    4. D permanent magnet excitation
    💡 Explanation:

    Controlled slip power return improves efficiency at variable speed.

  38. Q38 medium

    Air-gap power Pg in an induction motor is approximately

    1. A only stator copper loss
    2. B only windage loss
    3. C power transferred from stator to rotor across air gap
    4. D output shaft power directly
    💡 Explanation:

    Pg = T×ωs where ωs is synchronous angular speed.

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

    Efficiency of induction motor is highest near

    1. A no-load always
    2. B rated load where copper and iron losses are balanced
    3. C locked rotor always
    4. D slip equal to 1
    💡 Explanation:

    Stray and copper losses dominate away from rated point.

  40. Q40 medium

    Stator copper loss in equivalent circuit is represented by

    1. A I1²R1
    2. B I2²R2/s only
    3. C mechanical output
    4. D s×Pg term alone
    💡 Explanation:

    R1 carries stator winding I²R independent of slip representation.

  41. Q41 hard

    Referred rotor leakage reactance X2 in equivalent circuit is

    1. A stationary-rotor reactance scaled to stator side
    2. B variable with mechanical output only
    3. C equal to synchronous reactance of supply
    4. D zero at all slips
    💡 Explanation:

    Magnetic referral combines turns ratio with frequency referral.

  42. Q42 hard

    No-load current in circle diagram is mainly

    1. A magnetizing component with small real loss part
    2. B purely resistive short-circuit current
    3. C rotor blocked-rotor current
    4. D DC bias current
    💡 Explanation:

    No-load point lies near top of current locus with low power factor.

  43. Q43 hard

    Blocked-rotor point on circle diagram corresponds to

    1. A s = 0 operation
    2. B maximum efficiency point
    3. C field weakening region
    4. D s = 1 and starting impedance condition
    💡 Explanation:

    Short-circuit rotor test maps to standstill slip.

  44. Q44 medium

    NEMA design A motors have

    1. A low breakdown torque
    2. B external rotor resistance required
    3. C no cage rotor
    4. D high breakdown torque and normal starting current
    💡 Explanation:

    Design A suits applications needing high overload capability.

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

    IP23 enclosure typically indicates

    1. A dust-tight and jet-proof
    2. B no protection at all
    3. C oil-immersed construction
    4. D protected against fingers and dripping water up to 15° from vertical
    💡 Explanation:

    IP23 suits indoor industrial environments with limited moisture.

  46. Q46 Past Paper · PPSC/FPSC/CSS hard

    Double-cage rotor design improves

    1. A synchronous speed
    2. B power factor at no load only
    3. C need for slip rings
    4. D starting performance while maintaining good running efficiency
    💡 Explanation:

    Outer high-resistance cage aids start; inner low-resistance cage carries running flux.

  47. Q47 hard

    Deep-bar or double-cage rotors increase starting torque because

    1. A effective rotor resistance is higher at low rotor frequency
    2. B stator poles are doubled
    3. C supply frequency is increased at start
    4. D slip is forced to zero
    💡 Explanation:

    Skin effect raises bar resistance when rotor frequency is high at start.

  48. Q48 Past Paper · PPSC/FPSC/CSS medium

    V/f control of induction motors maintains

    1. A approximately constant flux below base speed
    2. B constant rotor resistance
    3. C synchronous speed fixed
    4. D unity slip always
    💡 Explanation:

    Stator voltage is varied proportionally with frequency to avoid saturation or weak flux.

  49. Q49 medium

    Pole-changing speed control alters

    1. A rotor resistance only
    2. B line frequency only
    3. C synchronous speed by changing effective stator poles
    4. D air-gap length
    💡 Explanation:

    Ns = 120f/P; doubling poles halves synchronous speed.

  50. Q50 Past Paper · PPSC/FPSC/CSS hard

    Dahlander winding arrangement allows

    1. A only wound-rotor starting
    2. B synchronous speed above 3000 rpm always
    3. C two-speed operation with single winding
    4. D DC injection braking only
    💡 Explanation:

    Dahlander connection switches pole groups for 1:2 speed ratio commonly.

  51. Q51 hard

    Slip energy recovery (Scherbius/Kramer system) returns

    1. A slip power from wound-rotor circuit to supply or load
    2. B stator copper loss to rotor
    3. C mechanical power to slip rings
    4. D all losses as heat only
    💡 Explanation:

    Recovering sPg improves efficiency when operating with large slip.

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

    Rotor copper loss in an induction motor equals

    1. A s × Pg
    2. B Pg / s
    3. C (1-s) × Pg
    4. D s × Pout
    💡 Explanation:

    Air-gap power Pg splits into rotor copper loss sPg and mechanical power (1-s)Pg.

  53. Q53 medium

    If air-gap power is 10 kW and slip is 0.04, rotor copper loss is

    1. A 4 kW
    2. B 40 W
    3. C 400 W
    4. D 9.6 kW
    💡 Explanation:

    Rotor I²R loss = sPg = 0.04 × 10000 = 400 W.

  54. Q54 Past Paper · PPSC/FPSC/CSS medium

    Mechanical power developed by rotor equals

    1. A s × Pg
    2. B (1 - s) × Pg
    3. C Pg / (1-s)
    4. D s × Pout
    💡 Explanation:

    Fraction (1-s) of air-gap power converts to shaft power minus friction/windage.

  55. Q55 hard

    The per-phase approximate equivalent circuit of an induction motor includes

    1. A only stator resistance
    2. B stator resistance, leakage reactance, and referred rotor branch R2/s + jX2
    3. C DC field winding
    4. D commutator segments
    💡 Explanation:

    Referred rotor impedance R2/s captures slip-dependent rotor behavior.

  56. Q56 Past Paper · PPSC/FPSC/CSS hard

    In the equivalent circuit, R2/s represents

    1. A rotor resistance and electromechanical power conversion
    2. B fixed stator leakage only
    3. C friction loss alone
    4. D supply frequency
    💡 Explanation:

    As s decreases at higher speed, R2/s rises, modeling reduced rotor current effects.

  57. Q57 hard

    Circle diagram of an induction motor graphically shows

    1. A only hysteresis loop of core
    2. B commutation sparking
    3. C open-circuit stator curve only
    4. D current locus and torque/power relations under varying slip
    💡 Explanation:

    Locus of stator current with slip is circular for approximate equivalent circuit.

  58. Q58 Past Paper · PPSC/FPSC/CSS hard

    From circle diagram, maximum torque corresponds to

    1. A zero stator current
    2. B infinite slip only visually
    3. C a specific current and power factor point on the locus
    4. D synchronous speed point
    💡 Explanation:

    Breakdown torque is read where tangent from origin meets the circle.

  59. Q59 medium

    NEMA design B squirrel-cage motors typically have

    1. A very high starting torque, high current
    2. B normal starting torque and normal starting current
    3. C low starting torque, low current
    4. D no starting torque
    💡 Explanation:

    Design B is general-purpose with moderate start characteristics.

  60. Q60 Past Paper · PPSC/FPSC/CSS hard

    NEMA design D motors are characterized by

    1. A low breakdown torque
    2. B high starting torque with high slip at rated load
    3. C synchronous operation
    4. D wound-rotor construction
    💡 Explanation:

    Design D uses high-resistance rotors for punch-press type loads.

  61. Q61 medium

    IP55 motor enclosure rating means

    1. A dust-protected and water jets from any direction will not harm motor
    2. B totally submersible without limit
    3. C open drip-proof only
    4. D explosion-proof for mines
    💡 Explanation:

    First digit 5 = dust; second digit 5 = water jet protection per IEC 60529.

  62. Q62 easy

    Induction motors are preferred for constant-speed industrial drives because

    1. A they always run above synchronous speed
    2. B they require commutators
    3. C they are rugged, low cost, and need little maintenance
    4. D rotor must be fed with DC
    💡 Explanation:

    Squirrel-cage machines dominate pumps, fans, and conveyors.

  63. Q63 Past Paper · PPSC/FPSC/CSS easy

    A 6-pole, 50 Hz motor has synchronous speed of

    1. A 1000 rpm
    2. B 1500 rpm
    3. C 750 rpm
    4. D 3000 rpm
    💡 Explanation:

    Ns = 120×50/6 = 1000 rpm.

  64. Q64 easy

    If a 4-pole, 50 Hz motor runs at 1455 rpm, slip is

    1. A 0.05
    2. B 0.97
    3. C 0.03
    4. D 0.15
    💡 Explanation:

    s = (1500-1455)/1500 = 45/1500 = 0.03.

  65. Q65 Past Paper · PPSC/FPSC/CSS easy

    Frequency of rotor currents at standstill equals

    1. A zero always
    2. B twice supply frequency
    3. C mechanical speed in Hz
    4. D supply stator frequency
    💡 Explanation:

    At s=1, sf = 1×f = f.

  66. Q66 easy

    As motor accelerates, rotor frequency

    1. A decreases proportionally with slip
    2. B increases linearly with speed
    3. C remains equal to supply frequency
    4. D becomes negative
    💡 Explanation:

    sf = s×f drops toward zero near full speed.

  67. Q67 Past Paper · PPSC/FPSC/CSS medium

    Operating point of motor on torque-slip curve at rated load is

    1. A beyond breakdown torque
    2. B at zero torque always
    3. C on the stable side left of breakdown torque peak
    4. D only at slip equal to 1
    💡 Explanation:

    Normal loads lie on rising portion before maximum torque.

  68. Q68 medium

    If load torque exceeds breakdown torque, the motor

    1. A speeds up above Ns
    2. B stalls or slows drastically with excessive slip
    3. C maintains rated speed automatically
    4. D becomes a generator without change
    💡 Explanation:

    Beyond pull-out torque, equilibrium is lost on unstable side.

  69. Q69 medium

    NEMA design C motors provide

    1. A low starting torque
    2. B operation only as generator
    3. C wound rotor only
    4. D high starting torque with moderate starting current
    💡 Explanation:

    Design C uses higher rotor resistance than design B.

  70. Q70 medium

    Service factor of 1.15 on a motor nameplate means

    1. A efficiency is 15%
    2. B slip is fixed at 0.15
    3. C motor can operate at 15% overload for short periods under specified conditions
    4. D speed is 15% above Ns
    💡 Explanation:

    SF allows temporary overload without insulation damage.

  71. Q71 medium

    IP44 enclosure offers protection against

    1. A complete submersion
    2. B no dust entry ever without test
    3. C solid objects over 1 mm and splashing water from any direction
    4. D internal explosions
    💡 Explanation:

    Common for washdown or outdoor protected installations.

  72. Q72 easy

    Totally enclosed fan-cooled (TEFC) motors dissipate heat by

    1. A open ventilation through windings
    2. B external shaft-mounted fan blowing over ribbed frame
    3. C oil circulation only
    4. D rotor slip rings cooling water
    💡 Explanation:

    TEFC prevents internal contamination while forcing air over casing.

  73. Q73 medium

    Induction motors driving centrifugal pumps are often selected with

    1. A constant heavy overload at start only
    2. B cubic torque-speed load characteristic consideration
    3. C synchronous speed operation required
    4. D commutator maintenance schedule
    💡 Explanation:

    Pump torque rises with speed squared; power with speed cubed.

  74. Q74 medium

    Conveyor belt applications may require motors with

    1. A zero breakdown torque
    2. B high starting torque to overcome static friction
    3. C synchronous speed above 6000 rpm
    4. D single-phase capacitor start only
    💡 Explanation:

    High Tst prevents prolonged high-slip starting on heavy belts.

  75. Q75 medium

    Crane hoist duty often uses wound-rotor motors because

    1. A they cannot reverse
    2. B they eliminate slip
    3. C cage rotors are banned
    4. D rotor resistance control gives high starting torque and speed control
    💡 Explanation:

    Hoisting needs controlled torque and plugging/braking options.

  76. Q76 medium

    Large compressor motors may use soft starters to

    1. A increase synchronous speed
    2. B limit inrush current while providing adequate starting torque
    3. C remove need for bearings
    4. D operate as synchronous machines
    💡 Explanation:

    Soft start ramps voltage, reducing mechanical and electrical stress.

  77. Q77 easy

    A 2-pole, 60 Hz motor synchronous speed is

    1. A 3600 rpm
    2. B 1800 rpm
    3. C 1200 rpm
    4. D 7200 rpm
    💡 Explanation:

    Ns = 120×60/2 = 3600 rpm.

  78. Q78 easy

    At 3% slip, rotor speed of 4-pole 50 Hz motor is approximately

    1. A 1500 rpm
    2. B 1410 rpm
    3. C 1455 rpm
    4. D 1470 rpm
    💡 Explanation:

    Nr = Ns(1-s) = 1500×0.97 = 1455 rpm.

  79. Q79 hard

    Negative slip in induction motor operation indicates

    1. A motor is off
    2. B regenerative braking with rotor faster than synchronous speed
    3. C normal motoring at full load
    4. D locked rotor test
    💡 Explanation:

    Super-synchronous operation returns power to supply.

  80. Q80 easy

    In motoring mode, rotor always turns

    1. A faster than synchronous speed
    2. B at exactly Ns under load
    3. C slower than synchronous speed (positive slip)
    4. D backward relative to field always
    💡 Explanation:

    Forward motoring requires 0 < s < 1 typically.

  81. Q81 medium

    Torque is maximum on stable operating region when

    1. A load demand matches motor torque at intersection before breakdown
    2. B slip is unity at rated speed
    3. C voltage is zero
    4. D rotor is open circuited
    💡 Explanation:

    Equilibrium at torque-slip and load-speed curve intersection defines speed.

  82. Q82 easy

    Pull-out torque is also known as

    1. A starting torque
    2. B full-load torque only
    3. C breakdown or maximum torque
    4. D cogging torque
    💡 Explanation:

    Terminology interchangeably describes torque peak on curve.

  83. Q83 medium

    Increasing supply voltage to a running induction motor generally

    1. A increases torque and may increase speed slightly (lower slip)
    2. B decreases flux always
    3. C forces synchronous speed change
    4. D opens rotor circuit
    💡 Explanation:

    Higher V raises developed torque for same slip, reducing slip for same load.

  84. Q84 medium

    Inserting too much rotor resistance at full speed causes

    1. A speed above Ns in motoring
    2. B excessive slip, poor efficiency, and reduced speed
    3. C unity power factor automatically
    4. D elimination of heating
    💡 Explanation:

    Resistance should be cut out as motor accelerates.

  85. Q85 medium

    Thermal limit during repeated DOL starts is critical because

    1. A rotor and stator heat accumulates from high I² during low-speed start
    2. B starting uses no current
    3. C slip is zero during start
    4. D no losses occur in rotor
    💡 Explanation:

    Limited starts per hour protect insulation from overtemperature.

  86. Q86 easy

    Phase sequence reversal of three-phase supply to motor

    1. A changes number of poles
    2. B doubles synchronous speed
    3. C reverses direction of rotating field and rotor rotation
    4. D converts motor to generator only
    💡 Explanation:

    Swapping any two line connections reverses rotation.

  87. Q87 medium

    Single phasing during operation can

    1. A improve efficiency
    2. B damage motor due to unequal currents and loss of torque
    3. C increase breakdown torque safely
    4. D raise power factor to unity
    💡 Explanation:

    One open supply line causes severe heating in remaining phases.

  88. Q88 medium

    Core loss in induction motor depends mainly on

    1. A rotor slip only
    2. B shaft diameter only
    3. C supply voltage and frequency (flux density)
    4. D external rotor resistance
    💡 Explanation:

    Hysteresis and eddy losses in stator core follow flux level.

  89. Q89 medium

    Friction and windage loss component increases approximately with

    1. A slip only at start
    2. B square of voltage alone
    3. C rotor resistance setting
    4. D speed (higher at higher rotor speed)
    💡 Explanation:

    Mechanical losses rise with running speed near full load.

  90. Q90 medium

    Power factor of induction motor at light load is usually

    1. A unity leading
    2. B low lagging due to dominant magnetizing current
    3. C unity lagging always
    4. D zero
    💡 Explanation:

    Magnetizing VARs keep PF poor until load current adds in-phase component.