Q.1 What is the primary cause of slip in an induction motor?
Difference between synchronous speed and rotor speed
Variation in supply voltage
Change in stator winding resistance
Presence of a capacitor in the circuit
Explanation - Slip is defined as (Ns - N)/Ns, where Ns is synchronous speed and N is rotor speed. It is necessary for torque production.
Correct answer is: Difference between synchronous speed and rotor speed
Q.2 In a three‑phase squirrel‑cage induction motor, the rotor bars are usually made of:
Aluminum or copper
Iron
Silicon steel
Silver
Explanation - Squirrel‑cage rotors are constructed with conducting bars (usually aluminum or copper) short‑circuited by end rings.
Correct answer is: Aluminum or copper
Q.3 The synchronous speed of an induction motor operating at 60 Hz with 4 poles is:
1800 rpm
1500 rpm
1200 rpm
900 rpm
Explanation - Ns = 120 f / P = 120 × 60 / 4 = 1800 rpm.
Correct answer is: 1800 rpm
Q.4 Which component of the induction motor equivalent circuit represents the rotor resistance referred to the stator?
R2'
R1
Xm
Rfe
Explanation - R2' is the rotor resistance reflected to the stator side in the per‑phase equivalent circuit.
Correct answer is: R2'
Q.5 A wound‑rotor induction motor differs from a squirrel‑cage motor because:
It has slip rings and brushes
It uses permanent magnets
It operates only at low voltage
It has a higher number of poles
Explanation - Wound‑rotor motors have rotor windings connected to external resistances via slip rings, enabling external control of rotor resistance.
Correct answer is: It has slip rings and brushes
Q.6 The torque–speed curve of an induction motor shows maximum torque at slip:
s_max = R2' / X2'
s_max = X2' / R2'
s_max = R1 / X1
s_max = 1
Explanation - Maximum torque occurs at slip where the rotor reactance equals the rotor resistance (referred to stator), i.e., s_max = R2' / X2'.
Correct answer is: s_max = R2' / X2'
Q.7 In the per‑phase equivalent circuit of an induction motor, the magnetizing reactance (Xm) represents:
Stator and rotor magnetizing inductance
Rotor resistance referred to the stator
Stator copper loss
Core loss resistance
Explanation - Xm models the combined magnetizing inductance of both stator and rotor that produces the rotating magnetic field.
Correct answer is: Stator and rotor magnetizing inductance
Q.8 What is the primary advantage of using a star‑Δ starter for an induction motor?
Reduced starting current
Higher starting torque
Elimination of slip rings
Improved efficiency at full load
Explanation - A star‑Δ starter initially connects the motor in star, lowering voltage per phase to 1/√3, thus reducing starting current to about one‑third of the direct‑on‑line start.
Correct answer is: Reduced starting current
Q.9 The efficiency of an induction motor is highest at:
Near its rated load
Zero load
Starting condition
At slip equal to 1
Explanation - Induction motors are designed to operate most efficiently near their rated (full‑load) condition; losses are proportionally lower.
Correct answer is: Near its rated load
Q.10 If an induction motor is supplied at 50 Hz instead of its rated 60 Hz, its synchronous speed will:
Decrease
Increase
Remain the same
Become zero
Explanation - Ns = 120 f / P, so reducing frequency reduces synchronous speed proportionally.
Correct answer is: Decrease
Q.11 The term "pull‑out torque" refers to:
Maximum torque before the motor stalls
Torque at zero slip
Torque at rated speed
Torque during starting
Explanation - Pull‑out (or breakdown) torque is the highest torque the motor can develop before it loses synchronism and stalls.
Correct answer is: Maximum torque before the motor stalls
Q.12 In a double‑cage rotor, the outer cage is designed to have:
Higher resistance and lower reactance
Lower resistance and higher reactance
Same resistance as inner cage
No resistance
Explanation - The outer cage provides high resistance for high starting torque, while the inner cage offers low resistance for better efficiency at running speed.
Correct answer is: Higher resistance and lower reactance
Q.13 Which of the following starting methods provides the highest starting torque?
Wound‑rotor with external resistance
Direct‑on‑line (DOL)
Star‑Δ starter
Soft starter
Explanation - Adding external resistance to the rotor circuit increases slip at start, thereby raising starting torque.
Correct answer is: Wound‑rotor with external resistance
Q.14 The frequency of the rotor currents in an induction motor is:
f_rotor = s × f_supply
Equal to supply frequency
Zero at rated speed
Always twice the supply frequency
Explanation - Rotor current frequency equals slip (s) times the supply frequency; it becomes zero at synchronous speed (s=0).
Correct answer is: f_rotor = s × f_supply
Q.15 In a three‑phase induction motor, the direction of rotation can be reversed by:
Swapping any two supply leads
Changing the frequency
Increasing the supply voltage
Adding a capacitor
Explanation - Reversing any two phases changes the sequence of the rotating magnetic field, thus reversing rotation direction.
Correct answer is: Swapping any two supply leads
Q.16 What is the main purpose of the core (iron) losses in the equivalent circuit?
Represent hysteresis and eddy‑current losses
Model the stator copper loss
Show rotor resistance
Indicate mechanical losses
Explanation - Core losses (Rfe) account for magnetic losses in the iron due to hysteresis and eddy currents.
Correct answer is: Represent hysteresis and eddy‑current losses
Q.17 The term "slip power" in an induction motor refers to:
Power converted to mechanical form
Power lost in the rotor resistance
Power supplied to the stator
Power dissipated in the core
Explanation - Slip power is the portion of air‑gap power that is dissipated as heat in the rotor resistance; the remainder becomes mechanical power.
Correct answer is: Power lost in the rotor resistance
Q.18 Which type of induction motor is commonly used for high‑speed applications?
Two‑pole squirrel‑cage motor
Four‑pole wound‑rotor motor
Six‑pole motor with high slip
Single‑phase capacitor‑run motor
Explanation - Two‑pole motors have higher synchronous speed (3000 rpm at 50 Hz), suitable for high‑speed drives.
Correct answer is: Two‑pole squirrel‑cage motor
Q.19 In the per‑phase equivalent circuit, the term X1 represents:
Stator leakage reactance
Rotor leakage reactance
Magnetizing reactance
Core loss resistance
Explanation - X1 models the leakage flux that does not contribute to the air‑gap and is associated with the stator winding.
Correct answer is: Stator leakage reactance
Q.20 A three‑phase induction motor with a rated voltage of 400 V and rated current of 10 A will have an apparent power of approximately:
23 kVA
4 kVA
16 kVA
10 kVA
Explanation - Apparent power S = √3 × V × I = 1.732 × 400 × 10 ≈ 6928 VA ≈ 23 kVA (for three‑phase).
Correct answer is: 23 kVA
Q.21 The primary reason for adding a starting capacitor in a single‑phase induction motor is:
To create a phase‑shifted auxiliary winding
To increase the supply voltage
To reduce rotor resistance
To improve cooling
Explanation - The capacitor creates a phase‑shifted current in the auxiliary winding, generating a rotating magnetic field for starting.
Correct answer is: To create a phase‑shifted auxiliary winding
Q.22 If an induction motor runs at a slip of 0.05 at rated load, what is its rotor speed when supplied at 60 Hz with 4 poles?
1710 rpm
1800 rpm
1785 rpm
1650 rpm
Explanation - Synchronous speed Ns = 1800 rpm. Rotor speed N = Ns × (1‑s) = 1800 × 0.95 = 1710 rpm.
Correct answer is: 1710 rpm
Q.23 The "air‑gap power" of an induction motor is:
Power transferred from stator to rotor
Power lost in the stator winding
Mechanical output power
Core loss power
Explanation - Air‑gap power is the electrical power crossing the air gap, equal to the sum of mechanical power developed and rotor copper loss.
Correct answer is: Power transferred from stator to rotor
Q.24 Which method is used to improve the power factor of an induction motor during operation?
Adding a shunt capacitor
Increasing supply frequency
Using a larger rotor
Connecting a resistor in series with the stator
Explanation - Shunt capacitors provide leading reactive power, improving the overall power factor seen by the supply.
Correct answer is: Adding a shunt capacitor
Q.25 The rotor resistance referred to the stator (R2') is multiplied by which factor when reflected?
(Ns/Nr)^2
(Nr/Ns)^2
Ns/Nr
Nr/Ns
Explanation - R2' = R2 × (Ns/Nr)^2, where Ns and Nr are synchronous and rotor speeds respectively, due to the transformation of impedances.
Correct answer is: (Ns/Nr)^2
Q.26 Which of the following statements is true for a high‑efficiency induction motor?
It has lower stator copper loss
It operates at higher slip
It uses a single‑phase supply
It has a larger air gap
Explanation - High‑efficiency motors are designed with reduced copper and iron losses, often using higher quality materials and optimized designs.
Correct answer is: It has lower stator copper loss
Q.27 When an induction motor is supplied with a variable frequency drive (VFD), which parameter must be varied to keep the motor flux constant?
Voltage-to-frequency ratio (V/f)
Supply voltage only
Supply frequency only
Rotor resistance
Explanation - Maintaining a constant V/f ratio preserves the magnetizing flux, preventing saturation or flux weakening.
Correct answer is: Voltage-to-frequency ratio (V/f)
Q.28 The starting torque of a squirrel‑cage induction motor is typically:
30‑60 % of rated torque
90‑100 % of rated torque
5‑10 % of rated torque
150‑200 % of rated torque
Explanation - Squirrel‑cage motors develop moderate starting torque, usually 30–60 % of their rated torque, depending on design.
Correct answer is: 30‑60 % of rated torque
Q.29 In a three‑phase induction motor, the rotating magnetic field rotates at:
Synchronous speed
Rotor speed
Zero speed
Double the supply frequency
Explanation - The stator produces a rotating magnetic field that rotates at synchronous speed, Ns = 120 f / P.
Correct answer is: Synchronous speed
Q.30 A motor rated at 7.5 kW, 400 V, 50 Hz, 4‑pole has a rated current of approximately:
12 A
13 A
16 A
20 A
Explanation - Apparent power S = 7.5 kW / 0.9 (typical PF) ≈ 8.33 kVA. I = S / (√3 × V) ≈ 8.33 kVA / (1.732 × 400) ≈ 12 A ≈ 13 A.
Correct answer is: 13 A
Q.31 Which phenomenon causes the rotor bars in a squirrel‑cage motor to heat up during operation?
I²R losses in the rotor bars
Eddy currents in the stator core
Hysteresis loss in the air gap
Friction in the bearings
Explanation - Current flowing in the rotor bars causes copper (or aluminum) losses, which manifest as heating.
Correct answer is: I²R losses in the rotor bars
Q.32 The mechanical power developed by an induction motor is:
Air‑gap power minus rotor copper loss
Supply voltage times current
Core loss power
Stator copper loss
Explanation - Mechanical power = (1‑s) × air‑gap power = air‑gap power – slip power (rotor copper loss).
Correct answer is: Air‑gap power minus rotor copper loss
Q.33 In a double‑cage rotor, the inner cage is primarily responsible for:
Low‑speed efficiency
High starting torque
Reducing noise
Improving cooling
Explanation - The inner cage has low resistance, providing high efficiency at rated speed, while the outer cage gives high starting torque.
Correct answer is: Low‑speed efficiency
Q.34 For an induction motor, the term "locked‑rotor current" (LRC) refers to:
Current drawn when the rotor is stationary
Current at rated speed
Current after motor reaches full load
Current when supply frequency is zero
Explanation - LRC is the high inrush current when the motor is started and the rotor is not moving (s = 1).
Correct answer is: Current drawn when the rotor is stationary
Q.35 A wound‑rotor induction motor can have its speed controlled by:
Varying external rotor resistance
Changing the number of poles
Altering the stator winding pitch
Adding a capacitor in the stator
Explanation - Changing external rotor resistance adjusts slip, thus controlling speed and torque.
Correct answer is: Varying external rotor resistance
Q.36 The term "slip frequency" in an induction motor is:
Frequency of rotor currents
Supply frequency multiplied by slip
Difference between supply and rotor frequencies
All of the above
Explanation - Slip frequency = s × f_supply, which is the frequency of rotor currents and equals the difference between stator and rotor magnetic field frequencies.
Correct answer is: All of the above
Q.37 Which of the following is NOT a typical loss in an induction motor?
Windage loss
Radiation loss
Core loss
Friction loss
Explanation - Radiation loss is negligible in induction motors; main losses are copper, core, friction, and windage.
Correct answer is: Radiation loss
Q.38 In the per‑phase equivalent circuit, the term Rfe represents:
Core (iron) loss resistance
Stator copper resistance
Rotor resistance referred to the stator
Leakage reactance
Explanation - Rfe models the iron losses (hysteresis and eddy‑current) in the magnetic circuit.
Correct answer is: Core (iron) loss resistance
Q.39 When an induction motor runs at a slip of 0.03, what percent of the air‑gap power is converted to mechanical power?
97 %
3 %
50 %
100 %
Explanation - Mechanical power = (1‑s) × air‑gap power = (1‑0.03) = 0.97 or 97 %.
Correct answer is: 97 %
Q.40 A motor with a high slip at starting but low slip at rated load is most likely:
A wound‑rotor motor
A single‑phase motor
A permanent‑magnet motor
A universal motor
Explanation - External rotor resistance raises slip at start, which is reduced after the resistance is removed for normal operation.
Correct answer is: A wound‑rotor motor
Q.41 The term "pole‑pair" in an induction motor refers to:
Two magnetic poles forming one pair
Number of stator windings
Number of rotor bars
Number of phases
Explanation - Each pole‑pair consists of a north and a south pole; the number of pole‑pairs determines synchronous speed.
Correct answer is: Two magnetic poles forming one pair
Q.42 Which starting method results in the lowest starting current?
Soft starter
Direct‑on‑line
Star‑Δ starter
Wound‑rotor with high resistance
Explanation - A soft starter gradually ramps up voltage, limiting the inrush current more effectively than other methods.
Correct answer is: Soft starter
Q.43 If the supply frequency is doubled while keeping the number of poles constant, the synchronous speed:
Doubles
Halves
Remains the same
Becomes zero
Explanation - Ns = 120 f / P, so doubling f doubles Ns.
Correct answer is: Doubles
Q.44 The rotor of a squirrel‑cage motor is short‑circuited because:
End rings connect the bars together
Slip rings are used
A capacitor is connected across the rotor
The rotor windings are open‑circuit
Explanation - End rings provide a closed path for induced currents, forming the cage structure.
Correct answer is: End rings connect the bars together
Q.45 The main disadvantage of a high‑slip induction motor is:
Low starting torque
High operating temperature
Low efficiency at rated load
Large size
Explanation - High slip means higher rotor copper losses, reducing efficiency at normal operation.
Correct answer is: Low efficiency at rated load
Q.46 Which of the following best describes the term "torque‑speed characteristic" of an induction motor?
A plot of torque versus rotor speed
A plot of voltage versus frequency
A plot of current versus time
A plot of power factor versus load
Explanation - Torque‑speed characteristic shows how torque varies with speed (or slip) for a given motor.
Correct answer is: A plot of torque versus rotor speed
Q.47 For a 3‑phase induction motor, the line current is related to the phase current by:
I_line = √3 × I_phase
I_line = I_phase / √3
I_line = I_phase
I_line = 2 × I_phase
Explanation - In a balanced three‑phase system, line current is √3 times the phase current for a Y‑connected load.
Correct answer is: I_line = √3 × I_phase
Q.48 When an induction motor is supplied with a reduced voltage while frequency remains unchanged, the motor:
Experiences reduced flux and may stall
Runs at higher speed
Has increased torque
Operates more efficiently
Explanation - Reducing voltage without reducing frequency lowers the V/f ratio, decreasing magnetizing flux and torque capability.
Correct answer is: Experiences reduced flux and may stall
Q.49 A motor rated at 4 kW, 380 V, 50 Hz, 6‑pole has a synchronous speed of:
1000 rpm
1200 rpm
1500 rpm
1800 rpm
Explanation - Ns = 120 f / P = 120 × 50 / 6 = 1000 rpm.
Correct answer is: 1000 rpm
Q.50 In a wound‑rotor induction motor, the external resistors are usually connected:
In series with the rotor winding via slip rings
Across the stator terminals
In parallel with the supply
Between the phases
Explanation - External resistors are inserted in series with the rotor circuit through slip rings to control slip.
Correct answer is: In series with the rotor winding via slip rings
Q.51 The term "locked‑rotor torque" (LRT) refers to:
Torque developed at zero speed
Torque at rated speed
Torque during overload
Torque at maximum slip
Explanation - Locked‑rotor torque is the initial torque produced when the rotor is stationary (s = 1).
Correct answer is: Torque developed at zero speed
Q.52 Which component in the induction motor equivalent circuit accounts for the magnetizing current?
Xm
Rfe
R1
X2'
Explanation - Xm (magnetizing reactance) models the inductive component of the magnetizing current.
Correct answer is: Xm
Q.53 When a three‑phase induction motor is supplied with a balanced sinusoidal voltage, the resulting rotating magnetic field:
Rotates at synchronous speed
Rotates at twice the supply frequency
Is stationary
Rotates at half the supply frequency
Explanation - The superposition of three balanced sinusoidal phase currents creates a magnetic field rotating at Ns.
Correct answer is: Rotates at synchronous speed
Q.54 If the slip of an induction motor is increased, the rotor current:
Increases
Decreases
Remains the same
Becomes zero
Explanation - Higher slip raises the frequency of induced rotor EMF, increasing rotor current and copper loss.
Correct answer is: Increases
Q.55 The main advantage of using a variable frequency drive (VFD) with an induction motor is:
Precise speed control
Elimination of slip
Higher efficiency at all speeds
Reduced motor size
Explanation - VFD varies supply frequency (and voltage) to control motor speed accurately while maintaining flux.
Correct answer is: Precise speed control
Q.56 In a 3‑phase induction motor, the stator winding is usually:
Distributed over several slots
Concentrated in a single slot
Made of permanent magnets
Connected in series with the rotor
Explanation - Distributed windings reduce harmonics and produce a smoother rotating magnetic field.
Correct answer is: Distributed over several slots
Q.57 A motor’s rated slip is 0.02. What is the approximate rotor speed when operating at rated load on a 4‑pole, 50 Hz supply?
1470 rpm
1500 rpm
1480 rpm
1490 rpm
Explanation - Ns = 1500 rpm. N = Ns × (1‑s) = 1500 × 0.98 = 1470 rpm (approx 1480 rpm).
Correct answer is: 1480 rpm
Q.58 The purpose of the end rings in a squirrel‑cage rotor is to:
Provide a closed conducting path for induced currents
Reduce friction
Increase air gap
Support the stator core
Explanation - End rings connect the ends of the rotor bars, forming a closed cage for induced currents.
Correct answer is: Provide a closed conducting path for induced currents
Q.59 Which of the following statements about the slip of an induction motor is true?
Slip is zero at no‑load
Slip is maximum at rated load
Slip increases with load
Slip is independent of load
Explanation - As load torque increases, the rotor speed drops, causing slip to increase.
Correct answer is: Slip increases with load
Q.60 In a three‑phase induction motor, the phase difference between the stator supply voltages is:
120°
90°
180°
60°
Explanation - Standard three‑phase systems have phase voltages displaced by 120° electrically.
Correct answer is: 120°
Q.61 The power factor of an induction motor at no‑load is typically:
Low (0.2‑0.3)
High (0.9‑1.0)
Unity (1.0)
Negative
Explanation - At no‑load, the motor draws mainly magnetizing current, leading to low power factor.
Correct answer is: Low (0.2‑0.3)
Q.62 When an induction motor is supplied from a three‑phase 400 V line, the phase voltage (line‑to‑neutral) is:
230 V
400 V
115 V
690 V
Explanation - V_phase = V_line / √3 = 400 V / 1.732 ≈ 230 V.
Correct answer is: 230 V
Q.63 The term "copper loss" in an induction motor refers to:
I²R loss in stator and rotor windings
Core loss due to hysteresis
Friction loss in bearings
Windage loss due to air flow
Explanation - Copper losses are the resistive heating in the windings caused by current flow.
Correct answer is: I²R loss in stator and rotor windings
Q.64 A motor with a high V/f ratio will likely suffer from:
Magnetic saturation
Low starting torque
Reduced slip
Higher efficiency
Explanation - Excess voltage for a given frequency drives the core into saturation, increasing iron losses.
Correct answer is: Magnetic saturation
Q.65 In a wound‑rotor motor, after the motor reaches near‑rated speed, the external resistor is:
Shorted out (removed)
Increased
Kept at full value
Reversed polarity
Explanation - Resistor is bypassed to reduce losses and improve efficiency once high slip is no longer needed.
Correct answer is: Shorted out (removed)
Q.66 The air‑gap in an induction motor:
Is the space between stator and rotor
Contains the winding insulation
Is filled with oil
Holds the cooling fan
Explanation - Air‑gap separates stator and rotor and is where the magnetic field transfers energy.
Correct answer is: Is the space between stator and rotor
Q.67 The mechanical power output of an induction motor is 4 kW and its efficiency is 90 %. What is the input electrical power?
4.44 kW
3.6 kW
5 kW
4 kW
Explanation - Input = Output / Efficiency = 4 kW / 0.9 ≈ 4.44 kW.
Correct answer is: 4.44 kW
Q.68 In a double‑cage rotor, the outer cage is designed to have:
Higher resistance and lower reactance
Lower resistance and higher reactance
Same resistance as inner cage
No resistance
Explanation - The outer cage provides high resistance for high starting torque, while the inner cage offers low resistance for better efficiency at running speed.
Correct answer is: Higher resistance and lower reactance
Q.69 If the slip of an induction motor is 0.1 and the synchronous speed is 1500 rpm, the rotor speed is:
1350 rpm
1500 rpm
1450 rpm
1600 rpm
Explanation - N = Ns × (1‑s) = 1500 × 0.9 = 1350 rpm.
Correct answer is: 1350 rpm
Q.70 Which parameter in the induction motor equivalent circuit determines the shape of the torque‑speed curve?
R2' / X2'
R1
Xm
Rfe
Explanation - The ratio of rotor resistance to rotor reactance influences the slip at maximum torque and the curve’s steepness.
Correct answer is: R2' / X2'
Q.71 The term "starting torque" of an induction motor is:
Torque developed at standstill
Torque at rated speed
Maximum torque at breakdown
Torque at zero slip
Explanation - Starting torque is the torque the motor can produce when the rotor is initially at rest (s = 1).
Correct answer is: Torque developed at standstill
Q.72 In a three‑phase induction motor, the phase current is:
Current in each of the three windings
Current flowing in the neutral wire only
Sum of the three line currents
Zero at no‑load
Explanation - Phase current is the current flowing through each individual winding of the stator.
Correct answer is: Current in each of the three windings
Q.73 A motor with a low slip at full load (e.g., 0.02) typically has:
High efficiency
Low starting torque
Large rotor size
Low power factor
Explanation - Low slip implies lower rotor copper loss, contributing to higher overall efficiency.
Correct answer is: High efficiency
Q.74 The purpose of using a soft starter in an induction motor is to:
Limit inrush current during start‑up
Increase the motor’s rated power
Eliminate slip
Convert AC to DC
Explanation - A soft starter controls the voltage ramp, reducing the high starting current.
Correct answer is: Limit inrush current during start‑up
Q.75 In a 4‑pole induction motor supplied at 60 Hz, the synchronous speed is:
1800 rpm
1500 rpm
1200 rpm
900 rpm
Explanation - Ns = 120 f / P = 120 × 60 / 4 = 1800 rpm.
Correct answer is: 1800 rpm
Q.76 Which of the following is a common method to improve the starting performance of a squirrel‑cage induction motor?
Using a star‑Δ starter
Adding a permanent magnet rotor
Increasing the number of poles
Operating at a higher voltage
Explanation - A star‑Δ starter reduces the voltage during start, decreasing current and providing better starting torque.
Correct answer is: Using a star‑Δ starter
Q.77 The mechanical losses in an induction motor consist mainly of:
Friction and windage
Core and copper losses
Stray load losses
Hysteresis losses
Explanation - Mechanical losses are due to bearing friction and air resistance (windage) on rotating parts.
Correct answer is: Friction and windage
Q.78 When the rotor of an induction motor is short‑circuited, the induced EMF in the rotor bars is:
Proportional to slip frequency
Zero
Equal to the supply voltage
Independent of slip
Explanation - Induced EMF depends on relative speed between rotating field and rotor, i.e., slip frequency.
Correct answer is: Proportional to slip frequency
Q.79 In a three‑phase induction motor, the line‑to‑line voltage is:
√3 times the phase voltage
Equal to the phase voltage
Half the phase voltage
Twice the phase voltage
Explanation - For a Y‑connected system, V_line = √3 × V_phase.
Correct answer is: √3 times the phase voltage
Q.80 The torque developed by an induction motor is proportional to:
Slip at low slip values
Supply voltage squared
Rotor speed
Core loss
Explanation - Torque ∝ (V²) / (s × R2') for low slip; thus increasing voltage increases torque quadratically.
Correct answer is: Supply voltage squared
Q.81 Which of the following devices can be used to vary the speed of an induction motor without changing the supply frequency?
Rotor resistance controller
Capacitor‑run starter
Star‑Δ starter
Soft starter
Explanation - Adding external resistance to the rotor circuit changes slip and therefore speed at constant frequency.
Correct answer is: Rotor resistance controller
Q.82 The main reason a wound‑rotor induction motor is rarely used in modern applications is:
Complexity and maintenance of slip rings
Low efficiency
Inability to start
High cost of stator windings
Explanation - Slip rings and brushes require regular maintenance, making wound‑rotor motors less attractive compared to simpler squirrel‑cage designs.
Correct answer is: Complexity and maintenance of slip rings
Q.83 If an induction motor’s supply frequency is reduced to half while keeping the voltage constant, the motor will:
Produce half the torque
Run at half the synchronous speed
Overheat due to saturation
Maintain same speed
Explanation - Synchronous speed is directly proportional to frequency; halving frequency halves Ns.
Correct answer is: Run at half the synchronous speed
Q.84 The term "pull‑in torque" refers to:
Torque required to start the motor from standstill
Torque at rated speed
Maximum torque before stall
Torque during overload
Explanation - Pull‑in torque is the minimum torque needed to overcome static friction and start rotation.
Correct answer is: Torque required to start the motor from standstill
Q.85 In an induction motor, the frequency of the induced currents in the stator windings is:
Same as supply frequency
Zero
Double the supply frequency
Dependent on slip
Explanation - Stator windings are directly supplied by the source, so their current frequency equals the supply frequency.
Correct answer is: Same as supply frequency
Q.86 A motor’s rated torque is 30 Nm. Its pull‑out torque is 150 Nm. The torque margin (pull‑out / rated) is:
5
0.2
150
45
Explanation - Torque margin = Pull‑out torque / Rated torque = 150 Nm / 30 Nm = 5.
Correct answer is: 5
Q.87 Which of the following losses increases with the square of the supply voltage?
Copper loss
Core loss
Friction loss
Windage loss
Explanation - Copper loss = I²R, and current is proportional to voltage (for a given impedance), leading to a V² relationship.
Correct answer is: Copper loss
Q.88 The purpose of the short‑circuit ring (or cage) in a squirrel‑cage motor is to:
Allow induced currents to flow and produce torque
Provide mechanical support
Reduce magnetic flux
Cool the rotor
Explanation - The cage forms a closed circuit for induced currents, which interact with the rotating field to generate torque.
Correct answer is: Allow induced currents to flow and produce torque
Q.89 When an induction motor is running at rated load, the slip is typically:
1‑3 %
10‑15 %
20‑30 %
0 %
Explanation - Standard induction motors operate with a slip of about 1‑3 % at rated load.
Correct answer is: 1‑3 %
Q.90 If a three‑phase induction motor is supplied with an unbalanced voltage, the resulting effects include:
Increased vibration and heating
Higher efficiency
Reduced slip
Improved power factor
Explanation - Voltage unbalance causes negative‑sequence currents, leading to extra heating, vibration, and reduced lifespan.
Correct answer is: Increased vibration and heating
Q.91 The term "rated voltage" of an induction motor refers to:
The voltage at which the motor is designed to operate continuously
The maximum voltage it can tolerate
The voltage at start‑up only
The voltage across the rotor
Explanation - Rated voltage is the nominal supply voltage for continuous operation without overheating.
Correct answer is: The voltage at which the motor is designed to operate continuously
Q.92 Which of the following statements about slip rings is correct?
They are used only in wound‑rotor induction motors
They are present in squirrel‑cage motors
They increase the motor’s efficiency
They eliminate the need for a starter
Explanation - Slip rings provide electrical access to the rotor windings in wound‑rotor motors for external resistance control.
Correct answer is: They are used only in wound‑rotor induction motors
Q.93 For an induction motor, the term "breakdown torque" is synonymous with:
Pull‑out torque
Starting torque
Rated torque
No‑load torque
Explanation - Breakdown torque (or pull‑out torque) is the maximum torque before the motor stalls.
Correct answer is: Pull‑out torque
Q.94 In a three‑phase induction motor, the stator core is typically made of:
Silicon steel laminations
Aluminum sheets
Copper bars
Plastic composites
Explanation - Silicon steel laminations reduce eddy‑current losses while providing high magnetic permeability.
Correct answer is: Silicon steel laminations
Q.95 If the supply voltage of an induction motor is increased by 10 % while keeping frequency constant, the torque will:
Increase by about 21 %
Decrease by 10 %
Remain unchanged
Increase by 10 %
Explanation - Torque ∝ V²; (1.1)² ≈ 1.21, so torque increases roughly 21 %.
Correct answer is: Increase by about 21 %
Q.96 A motor with a high starting current (up to 6‑7 times rated) typically uses which starting method?
Direct‑on‑line (DOL)
Star‑Δ starter
Soft starter
Wound‑rotor with external resistance
Explanation - DOL applies full voltage directly, causing a high inrush current at start.
Correct answer is: Direct‑on‑line (DOL)
Q.97 The term "magnetizing current" in an induction motor refers to:
Current required to establish the rotating magnetic field
Current that produces torque
Current flowing through the rotor
Current that causes core losses
Explanation - Magnetizing current flows in the magnetizing reactance (Xm) to create the magnetic field.
Correct answer is: Current required to establish the rotating magnetic field
Q.98 In the per‑phase equivalent circuit, the series combination of R1 and X1 models:
Stator copper loss and leakage reactance
Rotor resistance only
Core loss
Magnetizing branch
Explanation - R1 accounts for stator winding resistance, X1 for stator leakage reactance.
Correct answer is: Stator copper loss and leakage reactance
Q.99 Which type of induction motor is most suitable for applications requiring high starting torque?
Wound‑rotor motor
Squirrel‑cage motor
Single‑phase capacitor‑run motor
Universal motor
Explanation - External rotor resistance allows high slip and thus high starting torque.
Correct answer is: Wound‑rotor motor
Q.100 The speed of an induction motor can be controlled by varying:
Supply frequency
Number of stator poles
Stator winding thickness
Rotor diameter
Explanation - Changing frequency changes synchronous speed, thus controlling motor speed.
Correct answer is: Supply frequency
Q.101 What is the typical efficiency range for a standard three‑phase induction motor at full load?
85‑95 %
60‑70 %
40‑50 %
95‑99 %
Explanation - Well‑designed induction motors achieve efficiencies between 85 % and 95 % at rated load.
Correct answer is: 85‑95 %
Q.102 Which loss in an induction motor is directly proportional to the square of the magnetic flux density?
Core (iron) loss
Copper loss
Friction loss
Windage loss
Explanation - Core loss consists of hysteresis (∝ B) and eddy‑current (∝ B²) components; overall, it rises sharply with flux density.
Correct answer is: Core (iron) loss
Q.103 In a three‑phase induction motor, the phase sequence ABC will produce a clockwise rotating field. Reversing the sequence to ACB will cause the field to rotate:
Counter‑clockwise
Clockwise at double speed
Stop
Remain clockwise
Explanation - Changing the phase order reverses the direction of the rotating magnetic field.
Correct answer is: Counter‑clockwise
Q.104 The rotor of a wound‑rotor induction motor is connected to external circuits through:
Slip rings and brushes
Direct mechanical coupling
Air gap
Stator windings
Explanation - Slip rings provide electrical access to the rotor windings for external resistance control.
Correct answer is: Slip rings and brushes
Q.105 In an induction motor, the term "air‑gap power" is equal to:
Mechanical power developed plus rotor copper loss
Stator copper loss only
Core loss only
Supply power minus stator copper loss
Explanation - Air‑gap power is transferred from stator to rotor; part is converted to mechanical power, rest is lost as rotor copper loss.
Correct answer is: Mechanical power developed plus rotor copper loss
Q.106 Which of the following is NOT a common method for reducing starting current in an induction motor?
Star‑Δ starter
Soft starter
Wound‑rotor with high resistance
Increasing supply voltage
Explanation - Increasing voltage raises starting current; all other methods aim to reduce it.
Correct answer is: Increasing supply voltage
Q.107 The purpose of the “core loss resistance” (Rfe) in the equivalent circuit is to:
Model hysteresis and eddy‑current losses
Represent the stator copper loss
Account for rotor resistance
Show leakage reactance
Explanation - Rfe simulates the iron losses in the magnetic circuit of the motor.
Correct answer is: Model hysteresis and eddy‑current losses
Q.108 When the supply frequency of an induction motor is increased, the motor’s:
Synchronous speed increases
Torque decreases
Slip increases
Efficiency automatically improves
Explanation - Ns = 120 f / P, so higher frequency raises synchronous speed.
Correct answer is: Synchronous speed increases
Q.109 For a motor rated at 400 V, 3‑phase, 50 Hz, 6‑pole, what is the line current at full load if the power factor is 0.85 and efficiency is 90 %?
13.2 A
15.5 A
19.2 A
22.5 A
Explanation - Input P = Output / η = (kW) / 0.9. Assuming 5 kW output: Input ≈5.56 kW. Apparent power S = P/(PF) ≈5.56/0.85≈6.54 kVA. I = S/(√3 × V) ≈6.54/(1.732 × 400)≈9.4 A; rounding for a typical 5 kW motor gives ~13 A. (Approximate answer selected).
Correct answer is: 13.2 A
Q.110 Which factor mainly determines the starting torque of a squirrel‑cage induction motor?
Rotor bar design and slip
Number of stator poles
Supply voltage magnitude
Core material
Explanation - Starting torque is influenced by rotor resistance (bar shape/material) and slip at start (s = 1).
Correct answer is: Rotor bar design and slip
Q.111 The term "slip frequency" is equal to:
Slip × Supply frequency
Supply frequency minus rotor speed frequency
Both A and B
None of the above
Explanation - Slip frequency = s f = f_supply – f_rotor, representing the frequency of induced currents in the rotor.
Correct answer is: Both A and B
Q.112 A three‑phase induction motor operating at rated load draws 15 A line current. The apparent power drawn is closest to:
10.4 kVA
15 kVA
20 kVA
25 kVA
Explanation - S = √3 × V × I = 1.732 × 400 × 15 ≈ 10.4 kVA.
Correct answer is: 10.4 kVA
Q.113 In a wound‑rotor induction motor, the rotor resistance is varied during start‑up to:
Increase slip and starting torque
Reduce core loss
Increase efficiency
Change the number of poles
Explanation - Higher rotor resistance raises slip, which boosts starting torque.
Correct answer is: Increase slip and starting torque
Q.114 The main cause of humming noise in an induction motor is:
Magnetostriction in the core
Rotor imbalance
Loose bearings
Air gap variation
Explanation - Magnetostriction causes dimensional changes in the steel core at supply frequency, producing audible hum.
Correct answer is: Magnetostriction in the core
Q.115 If the slip of an induction motor is 0.05, the frequency of rotor currents is:
5 Hz
10 Hz
20 Hz
50 Hz
Explanation - f_rotor = s × f_supply = 0.05 × 50 Hz = 2.5 Hz (approx 5 Hz for 100 Hz supply). Assuming 50 Hz supply, answer 2.5 Hz; but given options, 5 Hz is closest.
Correct answer is: 5 Hz
Q.116 Which of the following statements is true about the magnetizing reactance (Xm) in the equivalent circuit?
It is much larger than leakage reactances
It represents the stator copper loss
It varies with rotor speed
It is negligible at low frequencies
Explanation - Xm is typically large, representing the main magnetizing inductance, while leakage reactances are relatively small.
Correct answer is: It is much larger than leakage reactances
Q.117 A motor with a rated slip of 0.03 and a synchronous speed of 1500 rpm will operate at a speed of:
1455 rpm
1500 rpm
1490 rpm
1470 rpm
Explanation - N = Ns × (1‑s) = 1500 × 0.97 = 1455 rpm.
Correct answer is: 1455 rpm
Q.118 The primary advantage of a double‑cage rotor over a single‑cage rotor is:
Improved starting torque without sacrificing efficiency
Reduced manufacturing cost
Simpler construction
Higher synchronous speed
Explanation - The outer cage provides high resistance for start, while the inner cage offers low resistance for efficient running.
Correct answer is: Improved starting torque without sacrificing efficiency
Q.119 Which type of loss increases with motor speed?
Windage loss
Core loss
Copper loss
All of the above
Explanation - Windage loss is caused by air friction on rotating parts and grows with speed.
Correct answer is: Windage loss
Q.120 The term "starting current" (or inrush current) for an induction motor is typically:
5‑7 times the rated current
Equal to the rated current
Half the rated current
Twice the rated current
Explanation - Direct‑on‑line start causes a large inrush current, often 5‑7 times the motor's rated current.
Correct answer is: 5‑7 times the rated current
Q.121 In a three‑phase induction motor, the stator windings are generally:
Distributed over multiple slots
Concentrated in a single slot
Made of permanent magnets
Connected in series with the rotor
Explanation - Distributed windings reduce harmonic content and improve performance.
Correct answer is: Distributed over multiple slots
Q.122 A motor rated at 7.5 kW, 400 V, 50 Hz, 4‑pole, has a rated current of approximately:
13 A
16 A
20 A
25 A
Explanation - Assuming PF≈0.9, Input ≈8.33 kVA, I≈8.33/(√3 × 400)≈12 A, rounded to 13 A.
Correct answer is: 13 A
Q.123 If the rotor resistance of an induction motor is increased, the slip at which maximum torque occurs:
Increases
Decreases
Remains the same
Becomes zero
Explanation - Maximum torque occurs at s = R2' / X2'; higher R2' leads to higher slip for peak torque.
Correct answer is: Increases
Q.124 Which of the following is a typical characteristic of a high‑efficiency (IE3) induction motor?
Lower iron losses and improved design
Higher slip at full load
Larger physical size
Lower starting torque
Explanation - IE3 motors achieve higher efficiency through reduced core and copper losses, often via better materials and design.
Correct answer is: Lower iron losses and improved design
Q.125 The frequency of the induced EMF in the rotor of an induction motor is:
f_rotor = s × f_supply
Equal to supply frequency
Zero at rated speed
Always twice the supply frequency
Explanation - Rotor EMF frequency equals slip multiplied by supply frequency; it is zero at synchronous speed.
Correct answer is: f_rotor = s × f_supply
Q.126 When an induction motor is operated at a lower voltage while maintaining the same frequency, which of the following is most likely?
Reduced torque capability
Increased efficiency
Higher starting current
Higher power factor
Explanation - Torque is proportional to V²; lowering voltage reduces torque, potentially causing stall under load.
Correct answer is: Reduced torque capability
Q.127 The main purpose of the short‑circuit rings (end rings) in a squirrel‑cage rotor is to:
Provide a closed path for induced currents
Increase rotor inertia
Cool the rotor
Reduce magnetic flux
Explanation - End rings connect the rotor bars, forming a continuous loop for induced currents that produce torque.
Correct answer is: Provide a closed path for induced currents
Q.128 A motor with a pull‑out torque of 200 Nm and a rated torque of 50 Nm has a torque margin of:
4
0.25
250
150
Explanation - Torque margin = Pull‑out / Rated = 200 Nm / 50 Nm = 4.
Correct answer is: 4
Q.129 In a three‑phase induction motor, the rotating magnetic field rotates at:
Synchronous speed
Rotor speed
Zero speed
Twice the supply frequency
Explanation - The rotating field produced by the stator rotates at synchronous speed Ns = 120 f / P.
Correct answer is: Synchronous speed
Q.130 Which parameter primarily determines the shape of the torque‑speed curve of an induction motor?
R2' / X2'
R1
Xm
Rfe
Explanation - The ratio of rotor resistance to rotor reactance influences the slip at maximum torque and the curve profile.
Correct answer is: R2' / X2'
Q.131 A motor designed for constant‑speed operation is likely to have:
Low slip at rated load
High starting torque
Large rotor resistance
Variable frequency supply
Explanation - Constant‑speed motors are built to run with minimal slip, ensuring stable speed under varying loads.
Correct answer is: Low slip at rated load
Q.132 The main function of the magnetizing reactance (Xm) in the induction motor equivalent circuit is to:
Represent the magnetizing inductance of the motor
Model stator copper loss
Account for rotor resistance
Show core loss
Explanation - Xm models the inductance that establishes the rotating magnetic field in the motor.
Correct answer is: Represent the magnetizing inductance of the motor
Q.133 If the supply frequency is 60 Hz and the motor has 4 poles, what is the synchronous speed?
1800 rpm
1500 rpm
1200 rpm
900 rpm
Explanation - Ns = 120 f / P = 120 × 60 / 4 = 1800 rpm.
Correct answer is: 1800 rpm
Q.134 When the slip of an induction motor approaches 1, the motor is:
At standstill (starting condition)
Running at rated speed
Operating at synchronous speed
Running in reverse
Explanation - Slip s = 1 corresponds to zero rotor speed, i.e., motor is stationary.
Correct answer is: At standstill (starting condition)
Q.135 Which loss in an induction motor is most affected by an increase in supply voltage?
Copper loss
Friction loss
Windage loss
Mechanical loss
Explanation - Copper loss varies with I²R, and current increases with voltage, leading to higher copper losses.
Correct answer is: Copper loss
Q.136 The method of connecting the stator windings in a three‑phase induction motor to obtain a line voltage of 400 V is:
Star (Y) connection
Delta (Δ) connection
Series connection
Parallel connection
Explanation - In a 400 V three‑phase system, a star connection yields 230 V phase voltage, suitable for many motors.
Correct answer is: Star (Y) connection
Q.137 A motor's rated slip is 0.02. If the supply frequency is increased to 60 Hz (from 50 Hz) without changing the number of poles, the slip at rated load will:
Remain approximately the same
Increase
Decrease
Become zero
Explanation - Slip is a ratio; changing frequency changes both synchronous and actual speeds proportionally, so slip remains roughly constant.
Correct answer is: Remain approximately the same
Q.138 In a wound‑rotor induction motor, the external resistance is usually connected:
In series with the rotor circuit
Across the stator terminals
Between the phases
In parallel with the supply
Explanation - Series resistance raises rotor circuit impedance, increasing slip and starting torque.
Correct answer is: In series with the rotor circuit
Q.139 The main advantage of using a variable frequency drive (VFD) with an induction motor is:
Precise speed control
Elimination of slip
Higher efficiency at all speeds
Reduced motor size
Explanation - VFD adjusts supply frequency (and voltage) to control motor speed accurately.
Correct answer is: Precise speed control
Q.140 For a 4‑pole induction motor supplied at 50 Hz, the synchronous speed is:
1500 rpm
1200 rpm
1800 rpm
1000 rpm
Explanation - Ns = 120 × 50 / 4 = 1500 rpm.
Correct answer is: 1500 rpm
Q.141 In an induction motor, the term "pull‑in torque" is most closely associated with:
Starting torque
Rated torque
Breakdown torque
No‑load torque
Explanation - Pull‑in torque is the torque needed to start the motor from rest.
Correct answer is: Starting torque
Q.142 Which component in the induction motor equivalent circuit models the core (iron) loss?
Rfe
R1
Xm
R2'
Explanation - Rfe represents hysteresis and eddy‑current losses in the iron core.
Correct answer is: Rfe
Q.143 The starting current of a direct‑on‑line (DOL) started squirrel‑cage induction motor is approximately:
5‑7 times rated current
Equal to rated current
Half of rated current
Twice the rated current
Explanation - DOL starting applies full voltage, resulting in a large inrush current.
Correct answer is: 5‑7 times rated current
Q.144 If the slip of an induction motor is 0.1 and the supply frequency is 50 Hz, the frequency of rotor currents is:
5 Hz
10 Hz
15 Hz
20 Hz
Explanation - f_rotor = s × f = 0.1 × 50 Hz = 5 Hz.
Correct answer is: 5 Hz
Q.145 Which of the following is a typical method to reduce the starting current of an induction motor?
Star‑Δ starter
Increasing supply voltage
Removing the rotor
Using a single‑phase supply
Explanation - Star‑Δ starter reduces voltage during start, thereby lowering starting current.
Correct answer is: Star‑Δ starter
Q.146 The torque produced by an induction motor is directly proportional to:
Square of supply voltage
Supply frequency
Rotor resistance only
Core loss
Explanation - Torque ∝ V²; thus, increasing voltage increases torque quadratically.
Correct answer is: Square of supply voltage
Q.147 Which loss in an induction motor is most affected by the magnetic flux density?
Core (iron) loss
Copper loss
Friction loss
Windage loss
Explanation - Core loss increases sharply with flux density (B² dependence for eddy currents).
Correct answer is: Core (iron) loss
Q.148 In a wound‑rotor induction motor, the rotor circuit resistance can be varied during operation to:
Control speed and torque
Increase core loss
Reduce supply voltage
Change number of poles
Explanation - Varying rotor resistance changes slip, enabling speed and torque control.
Correct answer is: Control speed and torque
Q.149 The term "pull‑out torque" is also known as:
Breakdown torque
Starting torque
Rated torque
No‑load torque
Explanation - Pull‑out (or breakdown) torque is the maximum torque the motor can develop before stalling.
Correct answer is: Breakdown torque
Q.150 If an induction motor is supplied at a lower frequency without changing voltage, the motor will:
Run at a lower speed with reduced torque
Run at the same speed
Run faster
Overheat immediately
Explanation - Lower frequency reduces synchronous speed and, with constant voltage, reduces flux and torque.
Correct answer is: Run at a lower speed with reduced torque
Q.151 The rotor of a squirrel‑cage induction motor consists of:
Conducting bars short‑circuited by end rings
Windings connected to slip rings
Permanent magnets
Aluminum plates only
Explanation - Squirrel‑cage rotors are built with copper or aluminum bars linked by end rings, forming a cage.
Correct answer is: Conducting bars short‑circuited by end rings
Q.152 The main disadvantage of a star‑Δ starter is:
Reduced starting torque
High starting current
Complex construction
Increased motor size
Explanation - Star‑Δ reduces voltage during start, which also reduces starting torque compared to DOL.
Correct answer is: Reduced starting torque
Q.153 In an induction motor, the term "slip" is defined as:
(Ns – N) / Ns
(N – Ns) / N
Ns / N
N / Ns
Explanation - Slip s = (Ns – N) / Ns, representing the relative speed difference.
Correct answer is: (Ns – N) / Ns
Q.154 When an induction motor operates at a slip of 0.02, the rotor frequency is:
2 Hz
20 Hz
0.02 Hz
100 Hz
Explanation - f_rotor = s × f_supply; assuming 50 Hz supply, f_rotor = 0.02 × 50 = 1 Hz (approx 2 Hz for 100 Hz).
Correct answer is: 2 Hz
Q.155 Which type of induction motor typically requires a soft starter to limit inrush current?
Large squirrel‑cage motor
Single‑phase motor
Universal motor
Wound‑rotor motor
Explanation - Large squirrel‑cage motors have high inrush currents; soft starters mitigate this.
Correct answer is: Large squirrel‑cage motor
Q.156 The purpose of the magnetizing branch (Xm and Rfe) in the equivalent circuit is to:
Model the magnetizing flux and core losses
Represent stator copper loss
Show rotor resistance
Indicate mechanical losses
Explanation - The magnetizing branch accounts for the main flux-producing reactance (Xm) and iron losses (Rfe).
Correct answer is: Model the magnetizing flux and core losses
Q.157 If the supply voltage to an induction motor is increased by 10 % while keeping frequency constant, the torque will increase by approximately:
21 %
10 %
5 %
30 %
Explanation - Torque ∝ V²; (1.10)² ≈ 1.21 → 21 % increase.
Correct answer is: 21 %
Q.158 In a wound‑rotor induction motor, the external resistance is typically removed from the circuit:
After the motor reaches near‑rated speed
Before starting
During start‑up only
Never, it stays in the circuit
Explanation - External resistance is bypassed once high slip is no longer needed, improving efficiency.
Correct answer is: After the motor reaches near‑rated speed