Induction Motor in Running Conditions MCQ Quiz - Objective Question with Answer for Induction Motor in Running Conditions - Download Free PDF

Last updated on Apr 7, 2025

Latest Induction Motor in Running Conditions MCQ Objective Questions

Induction Motor in Running Conditions Question 1:

In a 3-phase induction motor, rotor resistance is equal to rotor reactance. What will be the starting torque?

  1. Moderate
  2. Negligible
  3. Zero
  4. Maximum

Answer (Detailed Solution Below)

Option 4 : Maximum

Induction Motor in Running Conditions Question 1 Detailed Solution

Torque in the 3-phase induction motor

The torque of a 3-phase induction motor is given by:

T=3ωsV2(R2s2)+X22R2s

where, V = Supply

ωs = Synchronous speed in radian/sec

R2 = Rotor resistance

X2 = Rotor reactance

s = Slip

Condition for maximum torque

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

s=R2X2

If the maximum torque has to occur in starting, the value of the slip (s)=1

1=R2X2

R2=X2

The maximum torque of a 3ϕ induction motor occurs when rotor resistance is equal to the rotor reactance.

Induction Motor in Running Conditions Question 2:

A 6-pole induction machine is working as an induction generator. If the supply frequency is 60 Hz and the rotor current frequency is 5 Hz, find the speed of the rotor.

  1. 1400 rpm
  2. 1300 rpm
  3. 1100 rpm
  4. 1200 rpm

Answer (Detailed Solution Below)

Option 2 : 1300 rpm

Induction Motor in Running Conditions Question 2 Detailed Solution

Concept

The frequency of EMF induced in an induction motor is given by:

fr=sfs

The speed of an 

where, fr = Rotor induced frequency

fs = Supply frequency

s = Slip

The speed of a 3ϕ induction generator is given by:

Nr=Ns(1+s)

Nr=120fsP(1+s)

Calculation

Given, fr = 5 Hz 

fs = 60 Hz

5=s×60

s=112

 

Nr=120×606(1+112)

Nr=1300 rpm

Induction Motor in Running Conditions Question 3:

The rotor of a 3-phase induction motor has 0.04 Ω resistance per phase and 0.2 Ω standstill reactance per phase. An external resistance is used in the rotor circuit in order to get half of the maximum torque at starting. Neglect stator impedance. By what percentage will this external resistance change the power factor at starting?

Answer (Detailed Solution Below) 30 - 35

Induction Motor in Running Conditions Question 3 Detailed Solution

TestTem=12TemTem=2smt1+1smt

smt24smt+1=0

⇒ smt = 3.73 (or) 0.27

smt = 3.72 is not valid.

So, smt = 0.27

smt=r2x2=0.27

r2=0.2×0.27=0.054Ω

External resistance that must be inserted in the rotor circuit =r2r2=0.0540.04=0.014Ω

Without external resistance,

Power factor =0.040.042+0.22=0.196

With external resistance,

Power factor =0.0540.0542+0.22=0.261

Percentage improvement in power factor

=0.2610.1960.196×100=33.16%

Induction Motor in Running Conditions Question 4:

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

Induction Motor in Running Conditions Question 4 Detailed Solution

Concept:

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

Slip (s) = =(Ns+Nr)Ns

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, Ns=120fsP 

=120×604=1800

We know that,

fr = (s) fs

⇒ 5 = (s) (60)

s=112

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

112=1800+Nr1800

Nr = 1950 rpm

Induction Motor in Running Conditions Question 5:

If the full load speed of a 3 phase 50 Hz 6 pole induction motor is 950 rpm. What is its half load speed nearly equal to

  1. 1000 rpm
  2. 450 rpm
  3. 1900 rpm
  4. 975 rpm

Answer (Detailed Solution Below)

Option 4 : 975 rpm

Induction Motor in Running Conditions Question 5 Detailed Solution

The synchronous speed of induction motor,

Ns=120fp=120×506=1000rpm

At full load, slip s=10009501000=0.05

We know that slip is directly proportional to load, hence at half load slip will be 0.025

At half load rotor speed, Nr=Ns(1s)=1000(10.025)=975rpm

Top Induction Motor in Running Conditions MCQ Objective Questions

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

Induction Motor in Running Conditions Question 6 Detailed Solution

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Concept:

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

Slip (s) = =(Ns+Nr)Ns

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, Ns=120fsP 

=120×604=1800

We know that,

fr = (s) fs

⇒ 5 = (s) (60)

s=112

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

112=1800+Nr1800

Nr = 1950 rpm

If the full load speed of a 3 phase 50 Hz 6 pole induction motor is 950 rpm. What is its half load speed nearly equal to

  1. 1000 rpm
  2. 450 rpm
  3. 1900 rpm
  4. 975 rpm

Answer (Detailed Solution Below)

Option 4 : 975 rpm

Induction Motor in Running Conditions Question 7 Detailed Solution

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The synchronous speed of induction motor,

Ns=120fp=120×506=1000rpm

At full load, slip s=10009501000=0.05

We know that slip is directly proportional to load, hence at half load slip will be 0.025

At half load rotor speed, Nr=Ns(1s)=1000(10.025)=975rpm

The figure shows the per-phase equivalent circuit of a two-pole three-phase induction motor operating at 50 Hz. The “air-gap” voltage, Vg across the magnetizing inductance, is 210 V rms, and the slip, is 0.05. The torque (in Nm) produced by the motor is _______.

F2 U.B Madhu 27.04.20 D5

Answer (Detailed Solution Below) 400 - 403

Induction Motor in Running Conditions Question 8 Detailed Solution

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Concept:

Exact Equivalent Circuit

GATE CS 39 25Q GATE 2019 Part1.docx 7

Here, R1 is the winding resistance of the stator

X1 is the inductance of the stator winding

R0 is the core loss component

XM is the magnetizing reaction of the winding

R2/s is the power of the rotor, which includes output mechanical power and loss of rotor

Important:

If we draw the circuit with refereed to the stator then the circuit will look like-

GATE CS 39 25Q GATE 2019 Part1.docx 8

Here all the other parameters are same except R2’ is the rotor winding resistance with referred to stator winding. R2(1-s) / s is the resistance which shows the power which is mechanical power output or useful power. The power dissipated in that resistor is the useful power output or shaft power.

Explanation:

 Gate EE 2015 paper 2 Images-Q20

We redraw the given circuit as

Gate EE 2015 paper 2 Images-Q20.1

Given Vg=210V,s=0.05

Air gap voltage is the voltage across AB, i.e.

VAB=Vg=210V(rms)I2=VABz=2101+j0.22=205.095312.40

Hence, torque =3ωsI22r2s

=3314.15×(205.0953)2×1=401.09Nm.

The starting line current of a 415V,3phase, delta connected induction motor is 120A, when the rated voltage is applied to its stator winding. The starting line current at a reduced voltage of 110V, in ampere, is _________.

Answer (Detailed Solution Below) 31 - 33

Induction Motor in Running Conditions Question 9 Detailed Solution

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At starting slip, S=1

CurrentdrawnαvoltageIst1Ist2=V1V2120Ist2=415110Ist2=110×120415=31.807A

A 6-pole induction machine is working as an induction generator. If the supply frequency is 60 Hz and the rotor current frequency is 5 Hz, find the speed of the rotor.

  1. 1400 rpm
  2. 1300 rpm
  3. 1100 rpm
  4. 1200 rpm

Answer (Detailed Solution Below)

Option 2 : 1300 rpm

Induction Motor in Running Conditions Question 10 Detailed Solution

Download Solution PDF

Concept

The frequency of EMF induced in an induction motor is given by:

fr=sfs

The speed of an 

where, fr = Rotor induced frequency

fs = Supply frequency

s = Slip

The speed of a 3ϕ induction generator is given by:

Nr=Ns(1+s)

Nr=120fsP(1+s)

Calculation

Given, fr = 5 Hz 

fs = 60 Hz

5=s×60

s=112

 

Nr=120×606(1+112)

Nr=1300 rpm

In a 3-phase induction motor, rotor resistance is equal to rotor reactance. What will be the starting torque?

  1. Moderate
  2. Negligible
  3. Zero
  4. Maximum

Answer (Detailed Solution Below)

Option 4 : Maximum

Induction Motor in Running Conditions Question 11 Detailed Solution

Download Solution PDF

Torque in the 3-phase induction motor

The torque of a 3-phase induction motor is given by:

T=3ωsV2(R2s2)+X22R2s

where, V = Supply

ωs = Synchronous speed in radian/sec

R2 = Rotor resistance

X2 = Rotor reactance

s = Slip

Condition for maximum torque

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

s=R2X2

If the maximum torque has to occur in starting, the value of the slip (s)=1

1=R2X2

R2=X2

The maximum torque of a 3ϕ induction motor occurs when rotor resistance is equal to the rotor reactance.

Induction Motor in Running Conditions Question 12:

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

Induction Motor in Running Conditions Question 12 Detailed Solution

Concept:

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

Slip (s) = =(Ns+Nr)Ns

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, Ns=120fsP 

=120×604=1800

We know that,

fr = (s) fs

⇒ 5 = (s) (60)

s=112

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

112=1800+Nr1800

Nr = 1950 rpm

Induction Motor in Running Conditions Question 13:

If the full load speed of a 3 phase 50 Hz 6 pole induction motor is 950 rpm. What is its half load speed nearly equal to

  1. 1000 rpm
  2. 450 rpm
  3. 1900 rpm
  4. 975 rpm

Answer (Detailed Solution Below)

Option 4 : 975 rpm

Induction Motor in Running Conditions Question 13 Detailed Solution

The synchronous speed of induction motor,

Ns=120fp=120×506=1000rpm

At full load, slip s=10009501000=0.05

We know that slip is directly proportional to load, hence at half load slip will be 0.025

At half load rotor speed, Nr=Ns(1s)=1000(10.025)=975rpm

Induction Motor in Running Conditions Question 14:

A 3 – phase, slip ring induction motor with delta connected rotor has an induced emf of 230 V between slip rings at standstill. Resistance and leakage reactance are 0.5 and 2.5 Ω respectively at standstill. What is the rotor current per phase at rotor developing maximum torque?

  1. 68.13 A
  2. 65.06 A
  3. 70.16 A
  4. 74.07 A

Answer (Detailed Solution Below)

Option 2 : 65.06 A

Induction Motor in Running Conditions Question 14 Detailed Solution

We have, R2 = 0.5 Ω, X2 = 2.5 Ω

At maximum torque, R2 = sX2  

Where s is the slip.

S=R2X2=0.52.5=0.2

Impedance at maximum torque,

Zr=R22+Xr2=(0.5)2+(0.5)2

= 0.707 Ω          (∵ Xr = sX2)

Standstill phase voltage in rotor is 

E2=230V ( ∵ Delta connection)

Now rotor induced emf is

Er=sE2=0.2×230

= 46 V

Rotor current per phase, I2=ErZr=460.707=65.06A

Induction Motor in Running Conditions Question 15:

The rotor of a 3-phase induction motor has 0.04 Ω resistance per phase and 0.2 Ω standstill reactance per phase. An external resistance is used in the rotor circuit in order to get half of the maximum torque at starting. Neglect stator impedance. By what percentage will this external resistance change the power factor at starting?

Answer (Detailed Solution Below) 30 - 35

Induction Motor in Running Conditions Question 15 Detailed Solution

TestTem=12TemTem=2smt1+1smt

smt24smt+1=0

⇒ smt = 3.73 (or) 0.27

smt = 3.72 is not valid.

So, smt = 0.27

smt=r2x2=0.27

r2=0.2×0.27=0.054Ω

External resistance that must be inserted in the rotor circuit =r2r2=0.0540.04=0.014Ω

Without external resistance,

Power factor =0.040.042+0.22=0.196

With external resistance,

Power factor =0.0540.0542+0.22=0.261

Percentage improvement in power factor

=0.2610.1960.196×100=33.16%

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