Three Phase Induction Motor MCQ Quiz - Objective Question with Answer for Three Phase Induction Motor - Download Free PDF

Last updated on Jun 26, 2025

Latest Three Phase Induction Motor MCQ Objective Questions

Three Phase Induction Motor Question 1:

Which of the following statements describe applications of three-phase squirrel cage induction motors?

I. Driving escalators and elevators.

II. Powering household refrigerators.

III. Operating centrifugal pumps in industrial plants.

IV. Used in electric vehicles for propulsion. 

  1. II, and IV only correct
  2. I, II, and III only correct 
  3. I and III only correct
  4. I and IV only correct

Answer (Detailed Solution Below)

Option 3 : I and III only correct

Three Phase Induction Motor Question 1 Detailed Solution

Explanation:

Applications of Three-Phase Squirrel Cage Induction Motors

Introduction: Three-phase squirrel cage induction motors are widely used in various industrial and commercial applications due to their robust construction, simplicity, and efficiency. These motors operate on the principle of electromagnetic induction and are characterized by their ability to handle heavy loads effectively, making them suitable for applications involving continuous operation and high reliability.

Correct Option Analysis:

The correct option is:

Option 3: I and III only correct

Explanation:

Let us analyze the statements provided:

  • Statement I: Driving escalators and elevators.
  • Statement III: Operating centrifugal pumps in industrial plants.

Statement I - Driving escalators and elevators: Three-phase squirrel cage induction motors are commonly used for driving escalators and elevators. These applications require high torque and reliability to ensure smooth and safe operation. The robust design and ability to handle variable loads make these motors an ideal choice for such applications. Additionally, these motors can be easily controlled using variable frequency drives (VFDs) to adjust the speed and torque as needed for elevators and escalators.

Statement III - Operating centrifugal pumps in industrial plants: Centrifugal pumps are widely used in industries for pumping fluids such as water, chemicals, and other liquids. Three-phase squirrel cage induction motors are preferred for driving these pumps due to their ability to deliver consistent performance under varying load conditions. Their high efficiency and reliability ensure continuous operation, which is critical in industrial settings.

Hence, both Statement I and Statement III correctly describe applications of three-phase squirrel cage induction motors, making Option 3 the correct choice.

Important Information

To further understand the analysis, let’s evaluate the other statements:

  • Statement II - Powering household refrigerators: Household refrigerators typically use single-phase induction motors rather than three-phase squirrel cage induction motors. Single-phase motors are more suitable for low-power applications like refrigerators, where the required power and torque are comparatively lower than industrial applications.
  • Statement IV - Used in electric vehicles for propulsion: Electric vehicles generally use specialized motors such as permanent magnet synchronous motors (PMSMs) or brushless DC motors (BLDCs) for propulsion due to their high efficiency, compact design, and ability to deliver precise torque control. Three-phase squirrel cage induction motors are not commonly used in electric vehicles.

Conclusion:

Three-phase squirrel cage induction motors are versatile and widely used in applications requiring high reliability, efficiency, and the ability to handle heavy loads. While they are ideal for driving escalators, elevators, and centrifugal pumps in industrial plants, they are not suitable for powering household refrigerators or electric vehicles. Understanding the specific requirements of different applications helps in selecting the appropriate motor type, ensuring optimal performance and efficiency.

Three Phase Induction Motor Question 2:

Why is WRIG preferred over Squirrel Cage Induction Generator (SCIG) in some wind turbines?

  1. WRIG does not require maintenance due to the absence of slip rings
  2. WRIG allows external rotor resistance control, enabling better efficiency and power factor control
  3. WRIG works only with constant-speed wind turbines
  4. WRIG has no moving parts, making it more durable

Answer (Detailed Solution Below)

Option 2 : WRIG allows external rotor resistance control, enabling better efficiency and power factor control

Three Phase Induction Motor Question 2 Detailed Solution

Explanation:

Why is WRIG preferred over Squirrel Cage Induction Generator (SCIG) in some wind turbines?

Correct Option: Option 2 - WRIG allows external rotor resistance control, enabling better efficiency and power factor control.

Introduction:

The Wound Rotor Induction Generator (WRIG) and the Squirrel Cage Induction Generator (SCIG) are two common types of induction generators used in wind turbines. The choice between these two depends on several factors, including efficiency, controllability, maintenance requirements, and cost. While SCIGs are widely used due to their simplicity and robustness, WRIGs are often preferred in certain applications due to their enhanced controllability and adaptability to varying operational conditions.

Explanation of the Correct Option:

WRIGs have a unique design feature that sets them apart from SCIGs: their rotors are equipped with slip rings and external resistors. This configuration allows external rotor resistance control, which is key to their advantages in wind turbine applications.

1. External Rotor Resistance Control:

  • In a WRIG, the rotor windings are connected to external resistors through slip rings. By adjusting the external resistance, the performance characteristics of the generator can be modified as per the operational requirements.
  • This capability allows for better control of the generator's torque-speed characteristics, which is particularly important in wind turbines where wind speeds can vary significantly.

2. Efficiency and Power Factor Control:

  • The external resistance control in WRIGs helps optimize the efficiency of the generator. By adjusting the resistance, energy losses can be minimized, and the generator can operate closer to its optimal efficiency point under varying wind conditions.
  • Power factor control is another significant advantage. WRIGs can adjust their reactive power output by varying the external resistance, which helps improve the overall power factor of the wind turbine system. This is especially beneficial in grid-connected systems where maintaining a good power factor is essential for reducing losses and complying with grid codes.

3. Application in Wind Turbines:

  • WRIGs are often used in variable-speed wind turbines, where their ability to adapt to changing wind conditions is a significant advantage. This adaptability is achieved through the external resistance control and the generator's inherent design flexibility.
  • In contrast, SCIGs are typically limited to constant-speed applications due to their fixed rotor design, which does not allow for external resistance control.

4. Limitations of WRIGs:

  • Despite their advantages, WRIGs have some drawbacks, such as higher maintenance requirements due to the presence of slip rings and brushes.
  • However, in many cases, the benefits of enhanced controllability and efficiency outweigh these maintenance concerns, making WRIGs a preferred choice for specific wind turbine applications.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: WRIG does not require maintenance due to the absence of slip rings.

This option is incorrect because WRIGs use slip rings and brushes, which require periodic maintenance. In fact, the presence of slip rings is one of the main distinguishing features of WRIGs compared to SCIGs, which do not have slip rings and are therefore maintenance-free in this regard.

Option 3: WRIG works only with constant-speed wind turbines.

This statement is incorrect as WRIGs are particularly suited for variable-speed wind turbines. Their ability to control external rotor resistance makes them adaptable to varying wind speeds, unlike SCIGs, which are typically limited to constant-speed applications.

Option 4: WRIG has no moving parts, making it more durable.

This option is also incorrect because WRIGs, like all rotating electrical machines, have moving parts, including the rotor, slip rings, and brushes. The presence of these components means WRIGs are not as maintenance-free or durable as SCIGs, which have a simpler and more robust construction.

Option 5: Statement is missing or does not apply to WRIGs.

This option does not provide any valid information relevant to the comparison between WRIGs and SCIGs, and therefore, it does not contribute to the discussion.

Conclusion:

The key advantage of WRIGs over SCIGs lies in their ability to control external rotor resistance, which enables better efficiency and power factor control. This feature makes WRIGs highly suitable for variable-speed wind turbine applications, where adaptability to changing wind conditions is crucial. While WRIGs have higher maintenance requirements due to the presence of slip rings, their enhanced performance and controllability often justify their use in specific scenarios. Understanding the operational characteristics and limitations of different types of induction generators is essential for selecting the most appropriate technology for wind turbine applications.

Three Phase Induction Motor Question 3:

The rotor current frequency in an induction motor operating at 50 Hz, with 4% slip is

  1. 48 Hz
  2. 4 Hz
  3. 2 Hz
  4. 16 Hz

Answer (Detailed Solution Below)

Option 3 : 2 Hz

Three Phase Induction Motor Question 3 Detailed Solution

Explanation:

Rotor Current Frequency in an Induction Motor

Definition: The rotor current frequency in an induction motor is the frequency of the current induced in the rotor circuit, which depends on the slip of the motor and the supply frequency. It is an essential parameter in understanding the performance of an induction motor, as it determines the relative motion between the stator's rotating magnetic field and the rotor.

Slip Formula:

S = ((Ns - Nr) / Ns) × 100

Where:

  • Ns = Synchronous speed of the stator's magnetic field (in revolutions per minute, RPM).
  • Nr = Rotor speed (in RPM).

The slip determines the frequency of the rotor current (fr), which is given by the relationship:

Rotor Current Frequency Formula:

fr = S × fs

Where:

  • fr = Rotor current frequency (in Hz).
  • S = Slip (expressed as a decimal).
  • fs = Supply frequency (in Hz).

Calculation:

Given the supply frequency (fs) is 50 Hz and the slip (S) is 4% (or 0.04 in decimal form), the rotor current frequency (fr) can be calculated as follows:

fr = S × fs

fr = 0.04 × 50

fr = 2 Hz

Therefore, the rotor current frequency is 2 Hz.

Three Phase Induction Motor Question 4:

The maximum starting torque of a 3-phase induction motor occurs when:

  1. rotor resistance is 3/4th of the rotor reactance
  2. rotor resistance is 1/4th of rotor reactance 
  3. rotor resistance is 1/2th of rotor reactance
  4. rotor resistance is equal to rotor reactance

Answer (Detailed Solution Below)

Option 4 : rotor resistance is equal to rotor reactance

Three Phase Induction Motor Question 4 Detailed Solution

Explanation:

The Maximum Starting Torque of a 3-Phase Induction Motor

Definition: The starting torque of a 3-phase induction motor is the torque produced by the motor when it starts from a stationary position. It is an important parameter for applications where the motor needs to overcome high initial load inertia or resistance at startup.

The maximum starting torque occurs under specific conditions related to the relationship between the rotor resistance (Rr) and the rotor reactance (Xr). This phenomenon can be explained using the torque equation of an induction motor.

Torque Equation of an Induction Motor:

The torque (T) developed in an induction motor is given by the following equation:

T ∝ (s * E22 * Rr) / [(Rr2 + (s * Xr)2)]

Where:

  • s = slip of the motor
  • E2 = rotor induced EMF
  • Rr = rotor resistance
  • Xr = rotor reactance

Key Point: The starting torque is obtained when the slip (s) is equal to 1, as the motor is stationary at startup.

Under these conditions, the torque equation simplifies to:

T ∝ (E22 * Rr) / [(Rr2 + Xr2)]

To maximize the starting torque, the denominator of the equation should be minimized. This occurs when:

Rr = Xr

Explanation of the Correct Option:

From the above analysis, it is clear that the maximum starting torque of a 3-phase induction motor occurs when the rotor resistance is equal to the rotor reactance. This is the condition under which the torque equation reaches its peak value, ensuring maximum torque at startup.

Advantages of Maximum Starting Torque:

  • Helps the motor overcome high initial loads or inertia.
  • Ensures smooth and reliable startup, especially in applications like conveyors, cranes, and elevators.

Correct Option:

The correct answer is:

Option 4: Rotor resistance is equal to rotor reactance.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Rotor resistance is 3/4th of the rotor reactance.

This option is incorrect because if the rotor resistance is less than the rotor reactance, the starting torque will not reach its maximum value. The torque equation depends on the balance between Rr and Xr, and a mismatch in this ratio reduces the torque.

Option 2: Rotor resistance is 1/4th of rotor reactance.

This option is also incorrect. If the rotor resistance is significantly smaller than the rotor reactance, the denominator in the torque equation becomes larger, leading to a reduction in torque. This condition is far from the optimal Rr = Xr ratio required for maximum torque.

Option 3: Rotor resistance is 1/2th of rotor reactance.

This option is incorrect as well. While it comes closer to the optimal condition compared to options 1 and 2, the starting torque will still not be maximized unless Rr equals Xr. A ratio of 1/2 implies that the resistance is still smaller than the reactance, which affects the torque negatively.

Conclusion:

Understanding the relationship between rotor resistance and reactance is critical for analyzing the performance of a 3-phase induction motor. For maximum starting torque, the rotor resistance must be equal to the rotor reactance, as this minimizes the denominator in the torque equation and allows the motor to produce the highest possible torque at startup. This condition is particularly beneficial in applications requiring high initial torque to overcome load inertia or resistance.

Three Phase Induction Motor Question 5:

A 3-phase, 4 pole, 55 hp squirrel cage induction motor has the following result from the no load test: supply frequency = 50 Hz, line voltage = 2 kV, line current = 4.5 A and input power = 1600 W. Assuming an average DC resistance per stator phase being 2.8 Ω, determine the no load rotational loss.

  1. 1429.9 W
  2. 1562.2 W
  3. 1829.9 W
  4. 1494.16 W

Answer (Detailed Solution Below)

Option 1 : 1429.9 W

Three Phase Induction Motor Question 5 Detailed Solution

Explanation:

Calculation of No Load Rotational Loss:

To determine the no load rotational loss of the given motor, we need to analyze the results provided in the no-load test and calculate the power losses. The no-load rotational loss primarily consists of friction, windage, and core losses.

Given Data:

  • Supply frequency (f) = 50 Hz
  • Line voltage (VL) = 2 kV = 2000 V
  • Line current (IL) = 4.5 A
  • Input power (Pin) = 1600 W
  • Stator phase DC resistance (Rs) = 2.8 Ω

Step-by-Step Calculation:

1. Convert Line Voltage to Phase Voltage:

The motor is a three-phase system, and the relationship between line voltage and phase voltage for a star connection is:

Vph = VL / √3

Substituting the values:

Vph = 2000 / √3 = 1154.7 V

2. Calculate Phase Current:

For a three-phase system, the line current is equal to the phase current in a star connection:

Iph = IL = 4.5 A

3. Determine Stator Copper Loss:

The stator copper loss is the power loss due to the resistance of the stator winding. It is calculated using:

PCu = 3 × Iph2 × Rs

Substituting the values:

PCu = 3 × (4.5)2 × 2.8

PCu = 3 × 20.25 × 2.8 = 170.1 W

4. Calculate No Load Rotational Loss:

The no-load input power is the total power supplied to the motor under no-load conditions. This power is used to overcome the rotational losses and stator copper losses. The rotational loss can be determined by subtracting the stator copper loss from the input power:

Protational = Pin - PCu

Substituting the values:

Protational = 1600 - 170.1 = 1429.9 W

Conclusion:

The no load rotational loss is 1429.9 W, which corresponds to Option 1.

Important Information

To further analyze the other options, let's evaluate their values:

Option 2: 1562.2 W - This value is incorrect because it does not account for the proper calculation of stator copper loss. It seems to be an overestimation.

Option 3: 1829.9 W - This option is incorrect as it assumes higher losses than actually calculated, indicating a significant deviation from the correct procedure.

Option 4: 1494.16 W - While closer to the correct answer, this option is still incorrect due to a slight overestimation of losses, possibly from rounding errors or incorrect assumptions.

Conclusion:

By accurately calculating the stator copper loss and subtracting it from the input power, the correct value of no load rotational loss is determined to be 1429.9 W, as explained above. This corresponds to Option 1, reaffirming its correctness based on the given data and proper analysis.

Top Three Phase Induction Motor MCQ Objective Questions

In the method of speed control of induction motor by inducing emf in the rotor circuit, if the injected voltage is in phase opposition to the induced rotor emf, then:

  1. the rotor resistance decreases
  2. the rotor resistance increases
  3. the rotor reactance decreases
  4. the rotor reactance increases

Answer (Detailed Solution Below)

Option 2 : the rotor resistance increases

Three Phase Induction Motor Question 6 Detailed Solution

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Rotor emf injection method:

For below-rated speeds: In this method, injected emf has the same frequency of rotor slip frequency and that emf is 180° out of phase with rotor emf.

F1 U.B N.J 31-07-2019 D 5

E2R is resultant emf in the rotor

E2R = E2 – E1

\(T \propto \frac{{sE_{2R}^2}}{{{R_2}}}\)

R2 is rotor resistance

T is the torque

s is the slip

Here, the value of rotor emf becomes less. To maintain constant torque, the value of slip will increase. Therefore, the speed will be decreased.

At this condition, effective rotor resistance increases.

For above-rated speeds: In this method, injected emf has the same frequency of rotor slip frequency and that emf is in phase with rotor emf.

F1 U.B N.J 31-07-2019 D 6

E2R is resultant emf in the rotor

E2R = E2 + E1

\(T \propto \frac{{sE_{2R}^2}}{{{R_2}}}\)

R2 is rotor resistance

T is the torque

s is the slip

Here, the value of rotor emf becomes more. To maintain constant torque, the value of slip will decrease. Therefore, the speed will be increased.

At this condition, effective rotor resistance decreases.

Induction generators deliver power at ______ power factor

  1. Lagging
  2. Leading
  3. Unity
  4. Zero

Answer (Detailed Solution Below)

Option 2 : Leading

Three Phase Induction Motor Question 7 Detailed Solution

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Induction generator always works with leading power factor since it will take large amount of reactive power to produce sufficient amount of working flux so that armature reaction is always magnetizing hence it will work always with leading pf.

F3 Savita Engineering 21-7-22 D4

Important Point

  • induction generator is basically an induction motor, which runs above the synchronous speed
  • when it acts as a generator it will supply the active power back to source, but for this supply of active power it needs reactive power as input to keep its winding excited
  • in case if the induction motor is connected to the grid, it will draw the required reactive power for the excitation of windings, but if it is standalone system (ie. not connected to grid) then a capacitor bank will be always connected, and this will provide leading reactive power to keep the winding excited for the process of mechanical to electrical energy conversion
  • since the reactive power is supplied by the capacitor, the induction generator is operating in leading power factor

The given symbols show a/an:

F35 Neha B 12-4-2021 Swati D44

  1. rheostat
  2. capacitor
  3. inductor
  4. diode

Answer (Detailed Solution Below)

Option 1 : rheostat

Three Phase Induction Motor Question 8 Detailed Solution

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Rheostat:

A rheostat is a type of variable resistor, whose resistance can be changed for varying the amount of electric current flowing through an electrical circuit. The word rheostat is made of two words (‘rheo’ meaning flow of current in Greek and ‘stat’ meaning stationary instrument). When placed in an electric circuit, the flow of electricity changed through two terminals: one terminal near the slider/adjustable contact and the other connected near the bottom.

A rheostat is internationally denoted by the following symbol:

The rheostat is generally used in applications where high voltage or current is required such as:

  1. Changing the light intensity of a light bulb. An increase in the resistance of the rheostat decreases the flow of electric current, leading to the dimming of lights and vice-versa.
  2. Generators
  3. Motor speed
  4. Heater and oven temperature control
  5. Volume control

 

Additional Information

Some important electronic components symbols given below-

F1 Neha B 20.4.21 Pallavi D8

The parameter of an equivalent circuit of a three-phase induction motor affected by reducing the rms value of the supply voltage at the rated frequency is

  1. rotor resistance
  2. rotor leakage reactance
  3. magnetizing reactance
  4. stator resistance

Answer (Detailed Solution Below)

Option 3 : magnetizing reactance

Three Phase Induction Motor Question 9 Detailed Solution

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Magnetic reactance (Xm) is depends on airgap flux and the flux is depends on V/f.

Xm ∝ ϕ ∝ V/f

Hence, magnetizing reactance gets affected by reducing the rms value of the supply at the rated frequency.

Additional Information

The total resistance at rotor is represented as \(\frac{{{r_2}}}{s}\), where \(s\) is slip. Now, if we create an equivalent transformer circuit for induction motor, the secondary resistance will be \({r_2}\). Thus the load resistance will be \({r_2}\left( {\frac{1}{s} - 1} \right)\).

Induction motor modelled as a transformer

F1 U.B Deepak 24.10.2019 D 3

When all the rotor parameters are shifted to stator side induction motor circuit is given by

F1 U.B Deepak 24.10.2019 D 4

So, \(r_{2}^{'}\left( \frac{1}{s}-1 \right)\) is the resistance which shows the power which is converted to mechanical power output or useful power.

In an induction motor, if the rotor is locked, then the rotor frequency of induction motor will be:

  1. Equal to the supply frequency
  2. Less than the supply frequency
  3. More than the supply frequency
  4. Zero

Answer (Detailed Solution Below)

Option 1 : Equal to the supply frequency

Three Phase Induction Motor Question 10 Detailed Solution

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Concept:

Slip in Induction Motor (s)

\(s=\frac{N_s\ -\ N_r}{N_s}\)   ---(1)

Where,

Ns is the stator frequency

Nr is the rotor frequency

Also, fr = s fs     ---(2)

Where,

fr is the rotor frequency

fis the supply frequency

Explanation:

The rotor is locked means Nr = 0

So, from equation (1), s = 1

From equation (2),

f= fs

Therefore, the rotor frequency of the induction motor = supply frequency.

A 4 pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in RPM is

  1. 1350
  2. 1650
  3. 1950
  4. 2250

Answer (Detailed Solution Below)

Option 3 : 1950

Three Phase Induction Motor Question 11 Detailed Solution

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Concept:

When 3-ϕ induction machine working as an induction generator, then 

Slip (s) = \(= \frac{{\left( { - {N_s} + {N_r}} \right)}}{{{N_s}}}\)

Where, 

N= Synchronous speed

Nr = Rotor speed

Frequency of rotor current = s × f

Where s is the slip

f is the supply frequency

Calculation:

Given that,

Supply frequency (fs) = 60 Hz

Rotor current frequency (fr) = 5 Hz

Number of poles = 4

Synchronous speed, \({{\rm{N}}_{\rm{s}}} = \frac{{120{{\rm{f}}_{\rm{s}}}}}{{\rm{P}}}\) 

\(= \frac{{120 × 60}}{4} = 1800\)

We know that,

fr = (s) fs

⇒ 5 = (s) (60)

\(\Rightarrow s = \frac{1}{{12}}\)

To work as an induction generator, rotor speed should be slip speed greater than synchronous speed, therefore 

\(\Rightarrow \frac{1}{{12}} = \frac{{ - 1800 + {N_r}}}{{1800}}\)

Nr = 1950 rpm

The magnetic field produced in the stator of a three phase induction motor travels at _______

  1. Rotating speed
  2. Asynchronous speed
  3. Synchronous speed
  4. Slip speed

Answer (Detailed Solution Below)

Option 3 : Synchronous speed

Three Phase Induction Motor Question 12 Detailed Solution

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In a three phase induction motor,

a) The speed of stator is zero

b) The speed of stator magnetic field is synchronous speed

c) The speed of rotor magnetic field is synchronous speed

d) The speed of rotor is rotating speed

e) The difference between the rotor magnetic field and rotor is slip speed

A three-phase induction motor is running at 4% slip. If the input to the rotor is 1,000 W, then the mechanical power developed by the motor will be _______.

  1. 960 W
  2. 9,600 W
  3. 96 W
  4. 0.96 W

Answer (Detailed Solution Below)

Option 1 : 960 W

Three Phase Induction Motor Question 13 Detailed Solution

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Power Stages in an Induction Motor:

Stator iron loss (eddy current loss and hysteresis losses) considered as constant loss and it depends on the supply frequency and magnetic flux density in the iron core.

The iron loss of the rotor is negligible because the frequency of rotor currents under normal running conditions is always small.

F1 Shweta Anil 02.03.21 D4

\( {{P}_{2}}:{{P}_{cu}}:{{P}_{m}}=1:s:\left( 1-s \right)\)

Where,

s is the slip of motor and Pcu is rotor copper loss.

Calculation:

Given,

s = 4% = 0.04

P2 = 1000 W

From above concept,

P2 : Pm = 1 : (1 - s)

1000 : Pm = 1 : (1 - 0.04)

1000 : Pm = 1 : 0.96

\(\frac{1000}{P_m}=\frac{1}{0.96}\)

Pm = 1000 × 0.96 = 960 W

For heavy loads, the relation between torque (T) and slip (S) in induction motor is given by ______.

  1. \(T \propto \frac{S}{{1 - S}}\)
  2. T ∝ S
  3. T ∝ (1 - S)
  4. \(T \propto \frac{1}{S}\)

Answer (Detailed Solution Below)

Option 4 : \(T \propto \frac{1}{S}\)

Three Phase Induction Motor Question 14 Detailed Solution

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  • Induction motor is a type of electric motor in which alternating current from a power source is fed through a primary winding and induces a current in a secondary winding, with the parts arranged so that the resulting magnetic field causes a movable rotor to rotate with respect to a fixed stator.
  • The torque-slip characteristics is represented by a rectangular hyperbola.
  • For the immediate value of slip, the graph changes from one form to another.
  • The torque equation of the induction motor is:

\(T = \frac{{Ks{R_2}E_{20}^2}}{{R_2^2 + {{\left( {s{X_{20}}} \right)}^2}}}\)
F1 Vinanti Engineering 12.04.23 D1
The torque slip characteristic curve is divided into three regions:

  • Low slip region
  • Medium slip region
  • High slip region


Low slip region: (Near full load)

At synchronous speed, slip = 0, therefore the torque is zero.

When there is a light load, the speed is very near to synchronous speed.

The slip is very low and (sX20)2 is negligible in comparison with R2. Therefore

\(T = \frac{{{K_1}s}}{{{R_2}}}\)

i.e. T S

High slip region: (After starting to maximum load)

As the slip increases, the speed of the motor decreases with the increase in load.

The term (sX20)2 becomes large.

The term R22 may be neglected in comparison with the term (sx20)2 and the torque equation becomes

\(T = \frac{{{K_3}{R_2}}}{{sX_{20}^2}}\)

i.e. \(T \propto \frac{1}{s}\) 

Hence for heavy loads, torque is inversely proportional to slip.

In order to______ semi-closed slots or totally closed slots are used in induction motors.

  1. improve starting torque
  2. improve power factor
  3. increase efficiency
  4. increase pull-one torque

Answer (Detailed Solution Below)

Option 2 : improve power factor

Three Phase Induction Motor Question 15 Detailed Solution

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Slots in Induction motor:

  • The speed of the induction motor is inversely proportional to the load torque. In semi-closed and closed slots, the air gap between the stator and rotor is small as compared to open slots. As the air gap is small, the requirement of magnetizing current to establish the flux in the air gap is less.
  • It results in improved power factor In order to semi-closed slots or totally closed slots are used in induction motors.
  • Among all the three types of slots, semi-closed type slots are preferred for induction machines as semi-closed slots having the partial advantages of open type and partial advantages of closed type slots.
  • Open-type slots are generally preferred for synchronous and dc machines.
  • In general, closed type slots are used in low hp motors, to control the starting current, as the leakage reactance offered by closed type slots is very high compared to other types of slots.
  • Large size induction motors use open slots so that already prepared and properly insulated coils can be easily inserted in open slots.
  • In order to improve the power factor semi-closed slots or totally closed slots are used in induction motors.
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