Class lectur on Electrical Machine

BY

Prof. Mirza Golam Rabbani

Dean, School of Science and Engineering

Uttara University, Dhaka

Topic: Three phase induction motors

1
Number Syllabus Detailed Syllabus

Lecture Three phase induction motors: Principle of Operation of

principle of operation, types, Three Phase Induction
performance, torque-speed Motor, Torque Equation,
characteristics. Slip,
Condition for maximum
Torque.

2
Constructional details of 3-Phase Induction Motor
Stator
 The stator of induction motor
consists of stator frame, stator
core, distributed winding, two
end covers, bearings etc.
 The stator core is a stack of
cylindrical laminations which
are slotted along their inner
periphery for housing the 3-
phase winding.
 The stator core fits closely in
the cast iron stator frame.
 The two end convers are made
of cast-iron. 3
 Constructional details of 3-Phase Induction Motor
Rotor
 The induction motor has two types of rotors: Squirrel cage rotor and Slip ring or
wound rotor.
Squirrel cage rotor
 The rotor core is cylindrical and slotted on its periphery.
 The rotor consists of uninsulated copper or aluminium bars called rotor
conductors.
 The bars are placed in the slots.
 These bars are permanently shorted at each end with the help of conducting copper
ring called end ring.

4
Cage type structure of rotor
 Constructional details of 3-Phase Induction Motor
Squirrel cage rotor
 As rotor itself is short circuited, no external resistance can be introduced in the
rotor circuit.
 Fan blades are generally provided at the ends of the rotor core.
 This circulates the air through the machine while operation , providing the
necessary cooling .
 The air gap between stator and rotor is kept uniform and as small as possible.
 In this type of rotor the slots are not arranged parallel to the shaft axis but are
skewed.

5
Fig. Skewing in rotor construction
 Constructional details of 3-Phase Induction Motor

Squirrel cage rotor

 The advantages of skewing;
1. A magnetic hum gets reduced.
2. Smooth motor operation.
3. Magnetic locking gets reduced.

6
 Constructional details of 3-Phase Induction Motor
Slip ring rotor or wound rotor
 In this type of construction, rotor winding is similar to the stator.
 The rotor construction is laminated and slotted. The slots contain the rotor winding.
 The three ends of the three phase winding are permanently connected to the slip
rings. The slip rings are mounted on the same shaft.
 The external resistances can be added with the help of brushes and slip ring
arrangement in series with each phase of the rotor winding.

7
 Principle of Operation of 3-Phase Induction Motor

 The stator of three phase induction motor is connected to three phase supply.
 The flow of 3-phase currents in the 3-phase stator winding produces a
rotating magnetic field.
 The speed of rotating field is the synchronous speed.
 The rotating flux wave cuts the stationary rotor conductors and therefore
emfs are induced in the rotor conductor.
 As the rotor circuit is short circuited, these induced emfs give rise to current
in the rotor conductors.
 The interaction of these rotor currents with rotating flux wave produces
torque in the rotor of a 3-phase induction motor and as a consequence, rotor
begins to rotate.

8
 Principle of Operation of 3-Phase Induction Motor

 According to Lenz‟ s law the developed torque must oppose the flux cutting
action.
 This is possible only if the developed torque forces the rotor to rotate in the
direction of rotating field.
 The relative speed between rotating flux and rotor conductors is reduced and
therefore flux cutting action also gets reduced.
 This shows that rotor must rotate in the direction of rotating magnetic field.
 If the rotor is assumed to run at synchronous speed in the direction of rotating
field, then there would be no flux cutting action, no emf in rotor conductors,
no current in rotor bars and therefore no developed torque.
 Thus, the rotor of 3-phase induction motor can never attain synchronous
speed.
9
 Rotating Field of Induction Motor
Let the three phase supply to inductor motor produce threephase currents which in turn
produces three magnetic fluxes. Suppose they are:
𝝋𝒂 = 𝝋𝒎 𝒔𝒊𝒏𝝎𝒕 c
𝝋𝒃 = 𝝋𝒎 𝒔𝒊𝒏(𝝎𝒕 − 𝟏𝟐𝟎 )
𝝋𝑪 = 𝝋𝒎 𝒔𝒊𝒏(𝝎𝒕 + 𝟏𝟐𝟎 )

The phasor diagram of the fluxes is shown as:

a
At t = 0,
𝝋𝒂 = 𝟎
𝝋𝒃 = 𝝋𝒎 𝒔𝒊𝒏(𝟎 − 𝟏𝟐𝟎 ) = −𝟎. 𝟖𝟔𝟔𝝋𝒎
b
𝝋𝑪 = 𝝋𝒎 𝒔𝒊𝒏(𝟎 + 𝟏𝟐𝟎 ) = 𝟎. 𝟖𝟔𝟔𝝋𝒎
t
t2 = b2 + c2 + 2b c cos(60)
c 30 b = (0.866m)2 +(0.866m)2 + 2(0.866m)2  0.5
30 =2 (3/4)m2 + (3/4)m2 = (9/4) m2
t = (3/2)m = 1.5m
At t = 90,
𝝋𝒂 = 𝝋𝒎
𝝋𝒃 = 𝝋𝒎 𝒔𝒊𝒏(𝟗𝟎 − 𝟏𝟐𝟎 ) = −𝟎. 𝟓𝝋𝒎
𝝋𝑪 = 𝝋𝒎 𝒔𝒊𝒏(𝟗𝟎 + 𝟏𝟐𝟎 ) = −𝟎. 𝟓𝝋𝒎
 b r2 = b2 + c2 + 2b c cos(120)
= (0.5m)2 +(0.5m)2 + 2(0.5m)2  (-0.5)
a
120 t =2.25m2 - .25m2 = .25 m2
t =r + a = 0.5m + m = 1.5m
c
At t = 180,
𝝋𝒂 = 𝟎
𝝋𝒃 = 𝝋𝒎 𝒔𝒊𝒏(𝟏𝟖𝟎 − 𝟏𝟐𝟎 ) = 𝟎. 𝟖𝟔𝟔𝝋𝒎
𝝋𝑪 = 𝝋𝒎 𝒔𝒊𝒏(𝟏𝟖𝟎 + 𝟏𝟐𝟎 ) = −𝟎. 𝟖𝟔𝟔𝝋𝒎

t2 = b2 + c2 + 2b c cos(60)

c = (0.866m)2 +(0.866m)2 + 2(0.866m)2  0.5
b =2 (3/4)m2 + (3/4)m2 = (9/4) m2
t t = (3/2)m = 1.5m

t=0
t=90 It is seen that as the time increases (t changes from 0 to 180), the
resultant flux t also changes 0 to 180, that means it is rotating
with an angular speed of  and constant magnitude of 1.5m
t=180
t moving
 Slip of induction motor

 Slip: It is defined as the difference between the synchronous speed (Ns) and actual
speed of rotor (N) expressed as the percentage of the synchronous speed.

𝑵𝒔 − 𝑵
∴𝒔=
𝑵𝒔
The percentage slip is expressed as
𝑵𝒔 − 𝑵
𝒔= × 𝟏𝟎𝟎
𝑵𝒔

 The actual speed of motor (N) can be expressed as

𝑵 = 𝑵𝒔(𝟏 − 𝒔)

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 Performance parameters
 Rotor Frequency:
The of rotating magnetic field is
𝟏𝟐𝟎𝒇
𝑵𝒔 = (1)
𝑷

Where f is the supply frequency and P is the number of poles.

If fr is the frequency of rotor induced emf in running condition at slip speed 𝑵𝒔 − 𝑵

then there exists a fixed relation between (𝑵𝒔 − 𝑵).
From eqn.(1),
𝟏𝟐𝟎𝒇𝒓
(𝑵𝒔−𝑵) = (2)
𝑷
Dividing Eqn. (2) by Eqn. (1) we get,
𝟏𝟐𝟎𝒇𝒓
(𝑵𝒔−𝑵) = 𝒔 = 𝑷
𝑵𝒔 𝟏𝟐𝟎𝒇
𝑷
𝒇𝒓
𝒔=
𝒇

∴ 𝒇𝒓 = 𝒔𝒇 11
 Performance parameters

 Rotor induced EMF:

Let,

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 Performance parameters

 Rotor resistance and reactance:

Let,

13
 Performance parameters

 Rotor resistance and reactance:

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 Performance parameters

 Rotor Power Factor:

Let,

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 Performance parameters

 Rotor Current:

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 Torque Equation
 The torque produced in the induction motor can be expressed as

𝑻 ∝ ∅𝑰𝟐𝒓 cos ∅2𝑟 (1)

Where, ∅ = Flux,
𝑰𝟐𝒓 = Rotor current in running condition,
cos ∅2𝑟= Power factor of rotor in running condition
𝑬𝟐 ∝ ∅ (2)
𝑬 𝒔𝑬𝟐
𝑰𝟐𝒓 = 𝒁𝟐𝒓 = 𝟐 𝟐
(3)
𝟐𝒓 𝑹𝟐 + 𝒔𝑿𝟐
𝑹 𝑹𝟐
cos ∅2𝑟 = 𝒁 𝟐 = (4)
𝟐𝒓 𝑹𝟐 𝟐+ 𝒔𝑿𝟐 𝟐

Substitute equations (2), (3) and (4) in Eqn. (1)

𝒔𝑬𝟐 𝑹𝟐
𝑻 ∝ 𝑬𝟐 × 𝟐 𝟐
× (5)
𝑹𝟐 + 𝒔𝑿𝟐 𝑹𝟐 𝟐+ 𝒔𝑿𝟐 𝟐

𝟐
𝒔. 𝑬𝟐 .𝑹𝟐
𝑻∝ 𝑹𝟐 𝟐+ 𝒔𝑿𝟐 𝟐
(6)

𝟐
𝒌𝒔. 𝑬𝟐 .𝑹𝟐
𝑻= 𝑹𝟐 𝟐+ 𝒔𝑿𝟐 𝟐
(7)

Where, k is the proportionality constant = 𝟑

𝟐𝝅𝒏𝒔
Starting torque
 The torque produced by an induction motor at start is called the starting torque.
 At start, Speed (N) =0 and Slip (S)=1,
 Substituting S=1 in torque equation, the starting torque is given by

𝒌𝒔. 𝑬𝟐 𝟐.𝑹𝟐
𝑻𝒔𝒕 =
𝑹𝟐 𝟐+ 𝒔𝑿𝟐 𝟐

Maximum torque
 Condition for the maximum torque,

𝒅𝑻
=𝟎
𝒅𝒔

𝒌𝒔. 𝑬𝟐 𝟐.𝑹𝟐
Where, 𝑻 =
𝑹𝟐 𝟐+ 𝒔𝑿𝟐 𝟐

As both numerator and denominator contains „s‟ terms, differentiate Torque w.r.t slip
using the rule of differentiation (u/v)
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Maximum torque

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Maximum torque
 The maximum torque (Tm) can be obtained by substituting 𝐬 = 𝑹𝟐
in the torque
𝑿𝟐

equation,

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 Torque Ratios
Full load and Maximum torque ratio

21
 Torque Ratios
Starting and Maximum torque ratio

22
 Torque Ratios
Starting and Maximum torque ratio

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 Torque-Slip Characteristics

 Torque-Slip characteristics has two parts,

1. Straight line called stable region of operation.
2. Rectangular hyperbola called unstable region of operation.
 There are two slip regions;
1. Low slip region
2. High slip region
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 Torque-Slip Characteristics
 Low slip region:

 High slip region:

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 Torque-Speed Characteristics

 At N=Ns, the motor stops as it cannot produce any torque.

 At N=0, the starting condition, motor produces a torque called starting torque.
 For low slip region, i.e speeds near Ns the region is stable and the characteristics is
straight line in nature.
 Fall in speed from no-load to full load is about 4 to 6%.

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 Multiple Choice Questions
1. A 3-phase 440 V, 50 Hz induction motor has 4% slip. The
frequency of rotor current will be
(a) 50 Hz (b) 25 Hz
(c) 5 Hz (d) 2 Hz

Ans: (d)

Explanation:

Frequency of the rotor current is given as fr = sf,

fr= 0.4 x 50
= 2 Hz

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2. In which of the following region, induction motor operation is
stable?
(a) Low Slip region (b) High slip region
(c) Any of the two (d) None of the above

Ans: (a)

Explanation:

 For Low slip region “S” is very small therefore the term S.X2 can
be neglected and R2 is constant therefore T ∝ S.
 Hence the graph is in the straight line in nature and this reason is
a stable region.

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 3. A 4 pole 50 Hz induction motor is running at 1300 rpm. Find
the speed of stator magnetic field with respect to the rotor?

(a) 1500 rpm (b)200 rpm

(c) 1300 rpm (d)3000 rpm
Ans: (b)

Explanation:
 The relative speed of an induction motor stator magnetic field with
respect to rotor is given as Ns – Nr.
Ns = 120f/p=120 x 50 / 4 =1500.
 Hence speed of stator magnetic field with respect to rotor is
Ns – Nr =1500-1300 = 200 rpm
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4. If the applied rated voltage per phase is reduced to one-half, then
the starting torque of squirrel cage induction motor becomes
(a) 4 times the initial value (b) 2 times the initial value
(c) 1/4 of the initial value (d) 1/2 times the initial value

Ans: (c)

Explanation:

 In an induction, motor torque is proportional to square of the

applied voltage, T ∝ V2 , T = (V/2)2,
T = 1/4 x V2
⇒ 1/4 of the initial value

30
 5. An 8 pole, three phase induction motor is supplied from 50 Hz,
a.c. supply. On full load, the frequency of induced EMF in the
rotor is 2 Hz. Then the full load slip and the corresponding speed.
(a) 4% & 750 (b) 4% & 720
(c) 5% & 1000 (d) 5% & 1500
Ans: (b)
Explanation:

 Rotor frequency is given as fr = sf, s = 2/50, s = 0.04,

%s = 0.04 x 100 = 4%
 Now the corresponding speed i.e rotor speed, N = Ns(1 – s),
Where Ns = 120f/P = 120 x 50/8 = 750 R.P.M, N = 750(1 – 0.04),
31
N = 720 R.P.M
6. A 4 pole, 3 phase, 50 Hz induction motor runs at a speed of 1470
r.p.m. speed. Then the frequency of the induced e.m.f. in the
rotor under this condition.
(a) 1 Hz (b) 2 Hz
(c) 5 Hz (d) 6 Hz

Ans: (a)

Explanation:

 Ns = 120f/P = 120 x 50/4 = 1500 RPM,

 Slip of an induction motor,
s = (Ns – Nr)/Ns = (1500 – 1470)/1500 = 0.02
 Now Rotor frequency fr = sf = 0.02 x 50 = 1 Hz
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7. The rotor current in a 3-phase induction motor is ……. slip.
(a) Directly proportional to
(b) Inversely Proportional to
(c) Independent
(d) None of the above

Ans: (a)

Explanation:

 At normal speed, close to synchronism, sX2 is very small and may

be neglected. Since E2 and R2 are fixed, I2r ∝ s.

33
8. If the slip of a 3-phase induction motor increases, the p.f. of the
rotor circuit is
(a) Decreased (b) Remain unchanged
(c) Increased (d) None of the above

Ans: (a)

Explanation:

 Rotor reactance is a function of rotor frequency an increase in slip

causes an increase in rotor frequency and hence the total reactance.
 This in turn increases the total impedance Z2.
 Since rotor resistance (R2) is constant, rotor p.f. is decreased.
34
9. If the supply voltage of a 3-phase induction motor is increased
two times, then, torque is
(a) Increased 4 times (b) Decreased 4 Times
(c) Increased twice (d) Remain the same

Ans: (a)
Explanation:

 The torque of the induction motor is directly proportional to the

square of the voltage T ∝ V2.
 If the supply voltage is increased by two times then the torque
will increase four times T = 22 or T = 4.

35
10. For a slip of 0.05, find the ratio of rotor speeds with the motor
operating with 4 and 6 poles respectively.
(a) 2.8 (b) 1.5 (c) 3.2 (d) 4.5

Ans: (b)

Explanation:

 Rotor speed with 4 poles and slip of 0.05 is

Nr4= (1 – 0.05)×1500= 1425 RPM
 Rotor speed with 6 poles and slip of 0.05 is
Nr6=(1 – 0.05) ×1000= 950 RPM
 The ratio of rotor speeds of motor operating with 4 and 6 poles
is Nr4/Nr6 = 1425/950 = 1.5 36
 11. If the stator voltage of an induction motor is reduced to 50%
of its rated value, the torque developed is reduced by
% of its full load value.
(a) 50 (b) 25
(c) 75 (d) 57.7
Ans: (c)

Explanation:

 In an induction, motor torque is proportional to square of the

applied voltage , T ∝ V2 , T = (0.5V)2, T = 0.25 V2
⇒ 1/4 of the initial value or it is reduced by 75% of its full load
value
37
12. A three phase, 50Hz, 8-pole squirrel cage induction motor run
at a speed
(a) <750rpm
(b) >750 rpm but < 1000rpm
(c) >1000rpm but <1500rpm
(d) >1500rpm

Ans: (a)

Explanation:

 Ns=120f/P, Ns=(120*50)/8=750rpm
 Induction motor speed is always less than synchronous speed.

38
13. The slip ring or wound rotor induction motor find its
applications in
(a) lifts, cranes (b) lathes
(c) drilling machines (d) printing machines
Ans: (a)
Explanation:

 High braking torque can also be obtained ( slip > 1) easily. Its
application therefore is where high starting and braking torque
are required as in lifts and cranes. Slip ring motor induction
motor can deliver higher starting torque as compared to squirrel
cage induction motor.

39
14. Slip speed of the motor decides the magnitude of the induced emf
and the rotor current, which in turn decides the torque produced.
If Ns is the synchronous speed and N is the motor speed in rpm,
then the slip speed is given by
(a) Ns (b) Ns – N (c) Ns + N (d) N – Ns

Ans: (b)
Explanation:
 Slip speed is defined as the difference between the synchronous
speed (Ns) and actual speed (N).

40
15. At start, the slip of the induction motor is
(a) 1 (b) 0
(c) 0.5 (d) None of these
Ans: (a)

Explanation:

 At start, Speed (N) =0

 𝑵𝒔−𝑵 𝑵𝒔−𝟎
Slip, 𝐬 = =
𝑵𝒔 𝑵𝒔
∴𝒔=𝟏

41
16. A 50 Hz, 3 phase slip ring induction motor, has 6 poles on stator
and 4 poles on rotor. Then the machine will run at
(a) 1000 rpm (b) 1500 rpm
(c) 1400 rpm (d) Machine will not run at all
Ans: (d)

Explanation:

 If stator poles and rotor poles are not equal then resultant
torque will be zero and motor will not run.

42
17. Nature of the rotor power factor in running condition is always
(a) Leading (b) Lagging
(c) Both (a) and (b) (d) None of these

Ans: (b)
Explanation:

 The power factor of the induction motor is always lagging because the rotor
and the stator winding have inductive impedance.

43
18. If a voltmeter when connected to the rotor of an induction motor
gives 150 oscillations per minute and stator frequency is 50 Hz.
Then the slip of induction motor will be
(a) 3% (b) 4.5%
(c) 5% (d) 5.5%
Ans: (c)
Explanation:

 Rotor frequency, fr =150/60=2.5Hz

 fr=sf, s=fr/f=2.5/50=0.05
 The slip of induction motor is 0.05*100=5%

44
19. The nature of the graph in the low slip region and in the high
slip region is
(a) Rising exponential, decaying exponential
(b) Both will be straight line
(c) Straight line, rectangular hyperbola
(d) Straight line, decaying exponential
Ans: (c)
Explanation:
 In low slip region torque is directly proportional to slip. So as load increases, speed
decreases, increasing the slip. This increases the torque which satisfies the load
demand. Hence the graph is straight line in nature.

 In high slip region torque is inversely proportional to the slip. Hence its nature is
like rectangular hyperbola.

45
 20. If an induction motor is operating at a point in the high slip
region, then the motor will be
(a) Stable
(b) Unstable
(c) Either stable or unstable depending on the torque
(d) None of these
Ans: (b)
Explanation:
 In high slip region as T α1/s, torque decreases as slip increases.
 But torque must increases to satisfy the load demand. As torque decreases, due to extra
loading effect, speed further decreases and slip further increases.
 Hence speed further drops. Eventually motor comes to standstill condition. The motor
can not continue to rotate at any point in this high slip region. Hence this region is called
unstable region of operation.

46
21. The value of slip for motoring, generating and braking region
respectively are
(a) S > 1, 0 = s = 1, s < 0 (b) S < 0, 0 = s = 1, s > 1
(c) 0 = s = 1, s > 1, s < 1 (d) 0 = s = 1, s < 0, s > 1

Ans: (d)
Explanation:
Range s = 0 to s = 1 is called
motoring region and s<0 is called
generating region. The slip s>1 is
called the braking region.

47
22. A 16 pole, 50 Hz, star connected three phase induction motor has
rotor resistance of 0.012 ohm per phase and rotor reactance of
0.220 ohm per phase at standstill. Its full load torque is at speed
of 220 rpm. Ratio of its starting torque to maximum torque is
(a) 0.1208 (b) 0.1087 (c) 0.1102 (d) 0.1012
Ans: (b)
Explanation:
𝑻𝒔𝒕 𝟐𝒔𝒎 𝑹𝟐 𝟎.𝟎𝟏𝟐
 𝑻𝒎 = 𝟐
, Where 𝒔𝒎 is the slip at maximum torque,𝒔𝒎 = = = 𝟎. 𝟎𝟓𝟒𝟓
𝟏+ 𝒔𝒎 𝑿𝟐 𝟎.𝟐𝟐𝟎

𝑻𝒔𝒕 𝟐×𝟎.𝟎𝟓𝟒𝟓
 𝑻𝒎 = = 𝟎. 𝟏𝟎𝟖𝟔𝟕𝟕
𝟏+(𝟎.𝟎𝟓𝟒𝟓)𝟐

48
23. The torque developed in three phase induction motor depends
on
(a) stator flux and rotor current
(b) stator flux and stator current
(c) stator current and rotor flux
(d) rotor current and rotor flux
Ans: (a)
Explanation:
 𝑻 ∝ ∅𝑰𝟐𝒓 cos ∅2𝑟

 Torque of a three phase induction motor is proportional to flux

per stator pole, rotor current and the power factor of the rotor.

49
 24. An induction motor when loaded from no load to full load, its slip
also varies from s = 1 to s = 0. Then the torques in the low slip
region and in high slip region is
(a) Both are directly proportional to the slip
(b) Both are inversely proportional to the slip
(c) Directly proportional to the slip, inversely proportional to the slip
(d) Independent of slip, proportional to the slip
Ans: (c)

Explanation:

50
25. A 400V, 4 pole, 3 phase, 50 Hz star connected induction motor
has a rotor resistance and reactance per phase equal to 0.02 ohm
and 0.2 ohm respectively. Then the slip at which maximum
torque occurs is
(a) 10% (b) 20% (c) 12.5% (d) 15%

Ans: (a)
Explanation:
𝑹𝟐 𝟎. 𝟎𝟐
𝑺𝒎 = = = 𝟎. 𝟏 = 𝟎. 𝟏 ∗ 𝟏𝟎𝟎 = 𝟏𝟎%
𝑿𝟐 𝟎. 𝟐

51
 26. When the torque produced by the induction motor is at its
maximum, the slip is given by
(a) X2 / R2 (b) (R2 + X2) / X2
(c) R2 / X2 (d) R2 / (R2 + X2)
Ans: (c)
Explanation:

52
 27. Stable operation of induction motor can be expected between
(a) 0 to 100% slip (b) 0 to 50% slip
(c) 0 to maximum torque slip (d) 0 to 1% slip

Ans: (c)
Explanation:
Slip range s = 0 to s = sm is known as
stable region of operation. Motor
always operates at a point in this region.

53
 28. The rotor of an induction motor cannot run with synchronous
speed because
(a) rotor torque would then become zero
(b) Lenz‟s law would be violated
(c) induction motor would then become synchronous
(d) air friction prevents it from doing so

Ans: (a)
Explanation:
 The torque developed in the rotor is proportional to the Slip.
 If the rotor run at synchronous speed then there will be no difference between the
synchronous speed and rotor speed, so the slip will be zero, hence no torque will be
developed and the rotor will slow down.

54
 29. A 6-pole induction motor is running from 50Hz supply. The emf
in its rotor is of frequency 2.5 Hz. The speed of the motor is
(a) 50 rpm (b) 1000 rpm
(c) 950 rpm (d) 1050 rpm
Ans: (c)

Explanation:
𝟏𝟐𝟎 ∗ 𝒇 𝟏𝟐𝟎 × 𝟓𝟎
𝑵𝒔 = = = 𝟏𝟎𝟎𝟎𝒓𝒑𝒎
𝑷 𝟔

 fr=sf, Slip, s=fr/f=2.5/50=0.05

 𝑵 = 𝑵𝒔
𝟏 − 𝒔 , 𝑵 = 𝟏𝟎𝟎𝟎 𝟏 − 𝟎. 𝟎𝟓 = 𝟗𝟓𝟎𝒓𝒑𝒎

55
 30. For a three phase, 50Hz, squirrel cage induction motor, rotor
leakage reactance at standstill is twice of its resistance. The
frequency of the supply at which maximum torque is obtained at
starting is
(a) 50 Hz (b) 25 Hz (c) 100 Hz (d) 75 Hz

Ans: (b)
Explanation:
 Slip at maximum torque Sm=R2/X2, rotor leakage reactance at
standstill is twice of its resistance i.e X2 = 2R2
 Sm=R2/2R2 =1/2=0.5, f=50*0.5=25Hz

56
 31. A three-phase, 50 Hz, 4-pole induction motor runs at no-load
with a slip of 1%. With full load, the slip increases to 5%. The
% speed regulation of the motor (rounded off to 2 decimal
places) is
GATE 2020
Ans: (4.20 – 4.22)
Explanation:
57
 32. The parameter of an equivalent circuit of a three-phase
induction motor affected by reducing the rms value of the
supply voltage at the rate frequency is GATE 2019

(a) magnetizing reactance (b) rotor leakage reactance

(c) rotor resistance (d) stator resistance
Ans: (b)
Explanation:
58
 33. A 3-phase, 4-pole, 400 V, 50 Hz squirrel-cage induction motor
is operating at a slip of 0.02. The speed of the rotor flux in
mechanical rad/sec, sensed by a stationary observer,
is closest to GATE 2017

(a) 1500 (b) 1470 (c) 157 (d) 154

Ans: (c)

Explanation:

59
 34. A 220 V, 3-phase, 4-pole, 50 Hz inductor motor of wound
rotor type is supplied at rated voltage and frequency. The
stator resistance, magnetizing reactance, and core loss are
negligible. The maximum torque produced by the rotor is
225% of full load torque and it occurs at 15% slip. The actual
rotor resistance is 0.03 /phase. The value of external
resistance (in Ohm) which must be inserted in a rotor phase
If the maximum torque is to occur at start is
Ans: 0.17Ω GATE 2015
Explanation:

60
61
 35. An 8-pole, 3-phase, 50 Hz induction motor is operating at a
speed of 700 rpm. The frequency of the rotor current of
the motor in Hz is . GATE 2014
Ans: 3.33Hz
Explanation:

62
 36. A 3-phase, 50 Hz, six pole induction motor has a rotor
resistance of 0.1W and reactance of 0.92W . Neglect the
voltage drop in stator and assume that the rotor resistance is
constant. Given that the full load slip is 3%, the ratio of
maximum torque to full load torque is GATE 2014

(a) 1.567 (b) 1.712 (c) 1.948 (d) 2.134

Ans: (c)

Explanation:

63
 37. A 4-pole induction motor, supplied by a slightly unbalanced
three-phase 50 Hz source, is rotating at 1440 rpm. The
electrical frequency in Hz of the induced negative sequence
current in the rotor is GATE 2013

(a) 100 (b) 98 (c) 52 (d) 48

Explanation: Ans: (b)

64
 38. The slip of an induction motor normally does not depend on
(a) rotor speed (b) synchronous speed
(c) shaft torque (d) core-loss component
GATE 2012
Ans: (d)

Explanation:

65
 39. A 3-phase squirrel cage induction motor supplied from a
balanced 3-phase source drives a mechanical load. The
torque-speed characteristics of the motor(solid curve) and of
the load(dotted curve) are shown. Of the two equilibrium
points A and B, which of the following options correctly
describes the stability of A and B ? GATE 2009

(a) A is stable, B is unstable (b) A is unstable, B is stable

(c) Both are stable (d) Both are unstable
Explanation:
Ans: (a)

66
 40. A three-phase squirrel cage induction motor has a starting
torque of 150% and a maximum torque of 300% with respect
to rated torque at rated voltage and rated frequency. Neglect
the stator resistance and rotational losses. The value of slip
for maximum torque is
(a) 13.48% (b) 16.42% (c) 18.92% (d) 26.79%
Ans: (d) GATE 2007
Explanation:

67
 41. On the torque/speed curve of the induction motor shown in
the figure four points of operation are marked as W, X, Y and
Z. Which one of them represents the operation at a slip
greater than 1 ?
GATE 2005

(a) W (b) X
(c) Y (d) Z
Ans: (a)
Explanation:

68
 42. The direction of rotation of a 3-phase induction motor is
clockwise when it is supplied with 3-phase sinusoidal voltage
having phase sequence A-B-C. For counter clockwise rotation
of the motor, the phase sequence of the power
supply should be GATE 2004
(a) B-C-A (b) C-A-B

(c) A-C-B (d) B-C-A or C-A-B

Ans: (c)
Explanation:

 Given that if 3-f induction motor is rotated in clockwise then the phase
sequence of supply voltage is A-B-C. In counter clock wise rotation of
the motor the phase sequence is change so in the counter clockwise
rotation the phase sequence is A-C-B.

69
 43. A 400 V, 15 kW, 4-pole, 50Hz, Y-connected induction motor
has full load slip of 4%. The output torque of the machine at
full load is GATE 2004
(a) 1.66 Nm (b) 95.50 Nm
(c) 99.47 Nm (d) 624.73 Nm
Ans: (c)
Explanation:

70
 44. If a 400V, 50Hz, star connected, 3-phase squirrel cage induction motor is
operated from a 400V, 75Hz, the torque that the motor can now provide
while drawing rated current from the supply?
(a) reduces (b) increases (c) remains the same
(d) increase or reduces depending upon the rotor resistance
Ans: (a) GATE 2002

Explanation:

71
 45. If an induction machine is run at above synchronous speed it acts as
(a) a synchronous motor (b) an induction generator
(c) an induction motor (d) none of the above

GATE 1997
Ans: (b)

Explanation:

 When an induction machine is made to run above synchronous

speed by using a prime mover, it can works as induction generator.

72
46. Induction motor can be regarded as a generalized transformer
due to certain similarities except rated IES/ESE 2020
(a) Frequency (b) Flux
(c) Speed (d) Induced e.m.f
Ans: (c)

 Induction motor is a generalized transformer. Difference is that

transformer is an alternating flux machine while induction motor is
rotating flux machine.

73
 47. A 3-phase, 400/200 V, Y-Y connected wound-rotor induction
motor has 0.06 Ω rotor resistance and 0.3 Ω standstill reactance
per phase. To make the starting torque equal to the
maximum torque, the additional resistance required in the
rotor circuit will be IES/ESE 2020

(a) 0.24 Ω/phase (b) 0.34 Ω/phase

(c) 0.42 Ω/phase (d) 0.52 Ω/phase

Ans: (a)
Explanation:
For, max starting torque, R2 = X2, R2 + Rext = X2, ∴ Rext = 0.3 – 0.06 = 0.24 Ω/phase

74
 48. A 500hp, 6-pole, 3-phase, 440V, 50Hz induction motor has a
speed of 950 rpm on full load. The full load slip and the number
of cycles the rotor voltage makes per minutes will be
respectively IES/ESE 2019

(a) 10% and 150 (b) 10% and 125

(c) 5% and 150 (d) 5% and 125
Ans: (c)

Explanation:

75
 49. A 3-phase induction motor drives a blower where load torque is directly
proportional to speed squared. If the motor operates at 1450 rpm, the
maximum current in terms of rated current will be nearly
(a) 2.2 (b) 3.4 IES/ESE 2019
(c) 4.6 (d) 6.8
Ans: (a)
Explanation:

76
 50. When the value of slip of an induction motor approaches zero,
the effective resistance IES/ESE 2018

(a) is very low and the motor is under no-load

(b) of the rotor circuit is very high and the motor is under no-load
(c) is zero
(d) of the rotor circuit is infinity and the motor is equivalent to short-
circuited two-winding transformer
Explanation: Ans: (b)

77
 51. A 4-pole, 50 Hz, 3-phase induction motor with a rotor resistance of 0.25Ω
develops a maximum torque of 25 N.m at 1400 rpm. The rotor reactance x2
and slip at maximum torque smax,T respectively would be
𝟏 𝟏
(a) 𝟐. 𝟎 𝒂𝒏𝒅 (b) 𝟑. 𝟕𝟓 𝒂𝒏𝒅
𝟏𝟓 𝟏𝟐
𝟏 𝟏
(c) 𝟐. 𝟎 𝒂𝒏𝒅 (d) 𝟑. 𝟕𝟓 𝒂𝒏𝒅
𝟏𝟐 𝟏𝟓

Ans: (d) IES/ESE 2018

Explanation:

78
 52. In an induction motor for a fixed speed at constant frequency

(a) Both line current and torque are proportional to voltage

(b) Both line current and torque are proportional to the square of voltage
(c) Line current is proportional to voltage and torque is proportional to the
square of voltage
(d) Line current is constant and torque is proportional to voltage
IES/ESE 2018
Ans: (c)
Explanation:

79
 53. On the Torque/speed curve of an induction motor shown in the figure, four
points of operation are marked as A,B,C and D. Which one of them
represents the operation at a slip greater than 1?
IES/ESE 2017

(a) A (b) B (c) C (d) D

Ans: (a)
Explanation:

80
54. Increasing the air-gap of a squirrel-cage induction motor would
result in IES/ESE 2017

(a) Increase in no-load speed

(b) Increase in full-load power-factor
(c) Increase in magnetizing current

(d) Maximum available torque Ans: (c)

Explanation:
 Increase in air gap results in more reluctance and hence the
motor draws large Magnetising current to set up a constant
Rotating Magnetic field.

81
55. What is the material of slip-rings in an induction machine?
(a) Carbon (b) Nickel IES/ESE 2016
(c) Phosphor bronze (d) Manganese

Ans: (c)
Explanation:
„Phosphor bronze‟ is used to make slip rings. Slip rings will provide
the path for current from stationary device to dynamic device and
Vice – Versa.

82
56. If a 3-phase slip ring induction motor is fed from the rotor side
with stator winding short circuited, then frequency of currents
flowing in the short circuited stator is IES/ESE 2016

(a) Slip x frequency (b) Supply frequency

(c) Frequency corresponding to rotor speed (d) Zero

Ans: (a)
Explanation:

By means of Mutual induction and to satisfy lenz‟s law the

frequency of e.m.f. in stator will be 𝒔 × 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚.

83
57. The frequency of rotor emf of an 8-pole induction motor is 2 Hz.
If the supply frequency is 50 Hz, then the motor speed is
(a) 1500 rpm (b) 750 rpm IES/ESE 2015
(c) 375 rpm (d) 720 rpm

Ans: (d)
Explanation:

84
58. A 15 kW, 400 V, 4-pole, 50Hz, star connected 3-phase induction
motor has full load slip of 4%. The output torque of the
IES/ESE 2012
machine at full load is

(a) 1.66 Nm (b) 95.5 Nm (c) 99.47 Nm (d) 624.73 Nm

Ans: (c)
Explanation:

85
59. The rotor frequency of a 3-phase, 5 kW, 400V, 50Hz, 4-pole
slip ring induction motor is 25 Hz. The speed of the motor
when connected to a 400 volt, 50 Hz supply will be
(a) 1500 rpm (b) 1000 rpm IES/ESE 2012

(c) 750 rpm (d) zero

Ans: (c)
Explanation:

𝟏𝟐𝟎×𝒇 𝟏𝟐𝟎×𝟓𝟎
 𝑵𝒔 = 𝑷
=
𝟒
= 𝟏𝟓𝟎𝟎𝒓𝒑𝒎

𝟏 − 𝟎. 𝟓
 S=fr/f=25/50=0.5, 𝑵 = 𝑵𝒔 𝟏 − 𝒔 , 𝑵 = 𝟏𝟓𝟎𝟎 = 𝟕𝟓𝟎𝒓𝒑𝒎

86
 60. The power factor of an induction motor operating at no load
wi11 have a value around IES/ESE 2010
(a) 0.9 lag (b) 0.2 lead
(c) 0.2 lag (d) 0.9 lead

Ans: (c)
Explanation:
 At no load, an induction motor draws a large magnetizing current and a small active
component to meet the no-load losses. Therefore, the induction motor takes a high
no-load current lagging the applied voltage by a large angle. Hence the power factor
of an induction motor on no load is low i.e., from 0.1 to 0.2 lagging

87
61. What is the ratio of starting torque and maximum torque of a
3 phase, 50 Hz, 4 pole induction motor for a maximum torque
IES/ESE 2009
at 1200 rpm?

(a) 0.421 (b) 0.384 (c) 0.6 (d) 0.5

Ans: (b)

Explanation:

𝟏𝟐𝟎×𝒇 𝟏𝟐𝟎×𝟓𝟎 𝑵𝒔−𝑵 𝟏𝟓𝟎𝟎−𝟏𝟐𝟎𝟎

 𝑵𝒔
= = = 𝟏𝟓𝟎𝟎𝒓𝒑𝒎 , 𝒔 = = = 𝟎. 𝟐
𝑷 𝟒 𝑵𝒔 𝟏𝟓𝟎𝟎

𝑻𝒔𝒕 𝟐×𝒔 𝟐×𝟎.𝟐

 
𝑻𝒎 = 𝟏+(𝒔)𝟐 = = 𝟎. 𝟑𝟖𝟒𝟔
𝟏+(𝟎.𝟐)𝟐

88
 62. An induction motor having 8-poles runs at 727.5 rpm. If the
supply frequency is 50 Hz, the emf in the rotor will have a
frequency IES/ESE 2003

(a) 1.5Hz (b) 48.5 Hz (c) 51.5 Hz (d) 75 Hz

Ans: (a)
Explanation:

𝟏𝟐𝟎×𝟓𝟎 𝑵𝒔−𝑵 𝟕𝟓𝟎−𝟕𝟐𝟕.𝟓

 𝑵𝒔
= = 𝟕𝟓𝟎𝒓𝒑𝒎, 𝒔 = = = 𝟎. 𝟎𝟑
𝟖 𝑵𝒔 𝟕𝟓𝟎

 𝒇𝒓 = 𝒔𝒇 = 𝟎. 𝟎𝟑 ∗ 𝟓𝟎 = 𝟏. 𝟓 𝑯𝒛

89
 63. The absolute speed of the magnetic field in space of 3-phase
rotor fed induction motor is IES/ESE 1996
(a) synchronous speed, Ns (b) rotor speed, Nr
(c) (Ns-Nr) (d) Ns+Nr

Ans: (c)

Explanation:

 Slip speed = Ns-Nr

90
 64. A voltmeter gives 120 oscillations per minute when connected to
the rotor of an induction motor. The stator frequency is 50Hz.
The slip of the motor is IES/ESE 1995
(a) 2% (b) 2.5%
(c) 4% (d) 5%
Ans: (c)
Explanation:

𝟏𝟐𝟎 𝒇𝒓 𝟐
 𝒇𝒓 = 𝟔𝟎
= 𝟐𝑯𝒛, 𝒇𝒓 = 𝒔𝒇, 𝒔 = =
𝟓𝟎
= 𝟎. 𝟎𝟒
𝒇

 The slip of the motor is 0.04*100=4%

91
 Assignment problems
1. In a 3-phase induction machine, motoring, generating and
braking operations takes place in the range of slip “S” is
(a) 1>S>0, 0>S>-2 and S>1
(b) S>1, 1>S>-1 and 0>S>-1
(c) S>1, 0>S>-1 and 1>S>0
(d) 0>S>-1, S>1 and 1>S>0

92
2. A 4-pole, 3-phase induction motor is supplied from 50Hz supply.
Determine its synchronous speed. On full load, its speed is
observed to be 1440 rpm. Its full load slip is.
(a) 0.01 (b) 0.04
(c) 0.02 (d) 0.15

93
3. A 4-pole, 3-phase, 50Hz, star connected induction motor has full
load slip of 5%. Full load speed of the motor is
(a) 1440 rpm (b) 1425 rpm
(c) 1500 rpm (d) 950 rpm

94
4. A 4-pole, 3-phase, 50Hz, induction motor runs at a speed of
1440rpm. The frequency of the induced e.m.f in the rotor under
this condition is
(a) 2 Hz (b) 1.5Hz
(c) 50Hz (d) 1Hz

95
5. A 24 pole, 50Hz, star connected induction motor has rotor
resistance of 0.016Ω per phase and rotor reactance of 0.265Ω per
phase at standstill. It is achieving its full load torque at speed of
247 rpm. The ratio of full load torque to maximum torque is
(a) 0.3824 (b) 0.1203
(c) 0.7648 (d) 0.2406

96
6. Rotor resistance and standstill reactance per phase of a 3-phase
induction motor are 0.04Ω and 0.2Ω respectively. What would be
the external resistance required at start in rotor circuit to obtain
maximum torque at start
(a) 0.16Ω (b) 1.6Ω
(c) 0.0135Ω (d) 0.008Ω

97
For any queries/clarifications/suggestions...feel free to
contact through...
rabbani@uttara.ac.bd

98
 Solutions for Assignment problems

1. In a 3-phase induction machine, motoring, generating and

braking operations takes place in the range of slip “S” is
(a) 1>S>0, 0>S>-2 and S>1
(b) S>1, 1>S>-1 and 0>S>-1

Ans: (a) (c) S>1, 0>S>-1 and 1>S>0

(d) 0>S>-1, S>1 and 1>S>0

Range s = 0 to s = 1 is called
motoring region and s<0 is called
generating region. The slip s>1 is
called the braking region. 101
2. A 4-pole, 3-phase induction motor is supplied from 50Hz supply. Determine
its synchronous speed. On full load, its speed is observed to be 1440 rpm. Its
full load slip is.

(a) 0.01 (b) 0.04 (c) 0.02 (d) 0.15

Ans: (b)
Solution:
 Given values: P=4, f=50Hz, N=1440rpm
𝟏𝟐𝟎𝒇 𝟏𝟐𝟎×𝟓𝟎
 𝑵𝒔 = 𝑷
=
𝟒
= 𝟏𝟓𝟎𝟎𝒓𝒑𝒎

 𝑵𝒔−𝑵 𝟏𝟓𝟎𝟎−𝟏𝟒𝟒𝟎
Full load slip, 𝒔 = = = 𝟎. 𝟎𝟒
𝑵𝒔 𝟏𝟓𝟎𝟎

102
3. A 4-pole, 3-phase, 50Hz, star connected induction motor has full
load slip of 5%. Full load speed of the motor is
(a) 1440 rpm (b) 1425 rpm
(c) 1500 rpm (d) 950 rpm

Ans: (b)

Solution:
Given values: P=4, f=50Hz, slip=5%
𝟏𝟐𝟎𝒇 𝟏𝟐𝟎×𝟓𝟎
𝑵𝒔 = 𝑷
=
𝟒
= 𝟏𝟓𝟎𝟎𝒓𝒑𝒎

Full load speed, 𝑵 = 𝑵𝒔(−𝒔), 𝑵 = 𝟏𝟓𝟎𝟎(𝟏 − 𝟎. 𝟎5)

𝑵 = 𝟏𝟒𝟐𝟓𝒓𝒑𝒎

103
4. A 4-pole, 3-phase, 50Hz, induction motor runs at a speed of
1440rpm. The frequency of the induced e.m.f in the rotor under
this condition is
(a) 2 Hz (b) 1.5Hz (c) 50Hz (d) 1Hz
Ans: (a)
Solution:
𝟏𝟐𝟎𝒇 𝟏𝟐𝟎×𝟓𝟎
𝑵𝒔 = = = 𝟏𝟓𝟎𝟎𝒓𝒑𝒎
𝑷 𝟒

𝑵𝒔−𝑵 𝟏𝟓𝟎𝟎−𝟏𝟒𝟒𝟎
𝒔= = = 𝟎. 𝟎𝟒 ∴ 𝒇𝒓 = 𝒔𝒇 = 𝟎. 𝟎𝟒 × 𝟓𝟎 = 𝟐𝑯𝒛
𝑵𝒔 𝟏𝟓𝟎𝟎

104
5. A 24 pole, 50Hz, star connected induction motor has rotor
resistance of 0.016Ω per phase and rotor reactance of 0.265Ω per
phase at standstill. It is achieving its full load torque at speed of
247 rpm. The ratio of full load torque to maximum torque is

(a) 0.3824 (b) 0.1203 (c) 0.7648 (d) 0.2406

Ans: (a)

Solution:

105
6. Rotor resistance and standstill reactance per phase of a 3-phase
induction motor are 0.04Ω and 0.2Ω respectively. What would be
the external resistance required at start in rotor circuit to obtain
maximum torque at start
(a) 0.16Ω (b) 1.6Ω (c) 0.0135Ω (d) 0.008Ω

Ans: (a)

Solution:

106