Three-Phase Induction Motor .pdf

NAGARAJU KATTA Assistant professor EEE Dept. GCET
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4.1 Three phase induction introduction
4.2 Three phase induction construction
4.3 Three phase induction working principle
and operation
4.4 Rotating magnetic field
4.6 Torque equation
UNIT-IV
THREE-PHASE
INDUCTION MOTOR

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4.1 Three phase induction motor introduction
❖ The induction motor especially three phase induction motors are widely used
AC motor to produce mechanical power in industrial applications.
❖ Almost 80% of the motor is a three-phase induction motor among all motors used
in industries.
❖ Therefore, the induction motor is the most important motor among all other types
of motor.
❖ A three phase induction motor is a type of AC induction motors which operates on
three phase supply as compared to the single phase induction motor where single
phase supply is needed to operate it.
❖ The three-phase supply current produces an electromagnetic field in the stator
winding, which leads to generate the torque in the rotor winding of three phase
induction motor having magnetic field.
4.1.1 Advantages of three phase induction motor
❖ Induction motors are simple and rugged in construction. Advantage of induction
motors are that they are robust and can operate in any environmental condition
❖ Induction motors are cheaper in cost due to the absence of brushes, commutators,
and slip rings
❖ They are maintenance free motors unlike dc motors and synchronous motors due to
the absence of brushes, commutators and slip rings.
❖ Induction motors can be operated in polluted and explosive environments as they
do not have brushes which can cause sparks
❖ 3 phase induction motors will have self-starting torque unlike synchronous motors;
hence no starting methods are employed unlike synchronous motor. However,
single-phase induction motors do not have self-starting torque, and are made to
rotate using some auxiliaries.

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4.1.1 Disadvantages of three phase induction motor
❖ The main disadvantage of an induction motor is, it draws a very high inrush current
during its starting, for this reason, it makes a dip in voltage during starting which
can affect other electrical loads connected to the same power source. So, the
induction motor must require a starter to control the current during its starting.
❖ Induction motor provides a very poor starting torque, especially the squirrel cage
induction motor provides low starting torque. And the speed control of the
induction motor is more complicated than other motors like DC Motors. Nowadays
High-Cost device VFD (Variable Frequency Drives) are used to control the speed of
the induction motors.
❖ Although the induction motor creates a very low I^2R loss but it creates a high
magnetizing loss in its air gap. The induction motor always causes the lagging power
factor. So, when we use induction motors, power factor improvement must be used
to avoid penalty.
❖ Due to poor starting torque, induction motors are cannot be used for lifting, forcing
applications. Induction motor must need a starting arrangement such as a starter
when connected to heavy load.

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4.2 Constructional details of Three-Phase induction Motor
The rotor and stator are separated by a small air gap ranges from 0.5 mm to 4 mm
depending on the power rating of the motor.

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4.2.1 Stator of Three-Phase induction Motor
.
Stator Core
❖ The function of the stator
core is to carry the
alternating magnetic flux
which produces
hysteresis and eddy
current loss.
❖ To minimize these losses,
the core is laminated by
high-grade steel
stampings thickness of
0.3 to 0.6 mm.
❖ These stampings are
insulated from each other
by varnish.
❖ All stampings stamp
together in the shape of
the stator core and fixed
it with the stator frame.
❖ An inner layer of the
stator core has a number
of slots.
Stator Frame
❖ The stator frame is the
outer part of the motor.
❖ The function of the stator
frame is to provide support
to the stator core and
stator winding.
❖ It provides mechanical
strength to the inner parts
of the motor.
❖ The frame has fins on the
outer surface for heat
dissipation and cooling of
the motor.
❖ The frame is casted for
small machines and it is
fabricated for a large
machine.
❖ According to the
applications, the frame is
made up of die-cast or
fabricated steel,
aluminum/ aluminum
alloys, or stainless steel.
Stator Winding
❖ The stator winding is placed inside
the stator slots available inside the
stator core.
❖ Three-phase winding is placed as a
stator winding. And three-phase
supply is given to the stator
winding.
❖ The number of poles of a motor
depends on the internal connection
of the stator winding and it will
decide the speed of the motor.
❖ If the number of poles is greater,
the speed will less and if the
number of poles is lesser than the
speed will high.
❖ The poles are always in pairs.
Therefore, the total number of poles
always an even number.
❖ The relation between synchronous
speed and number poles is as shown
in the below equation,
❖ 𝑁𝑠 =
120𝑓
𝑃
RPM
❖ The stator is the stationary part of the
motor.
❖ It consists of a steel frame which encloses
a hollow cylindrical core.
❖ The core of the three-phase induction
motor is made up of thin laminations of
silicon steel to reduce the eddy current
and hysteresis losses.
Fig (a): Stator

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4.2.2 Rotor of Three-Phase induction Motor
The rotor is a rotating part of induction motor. The rotor is connected to the mechanical
load through the shaft.
The rotor of the three-phase induction motor is further classified as,
1. Squirrel cage rotor
2. Slip ring rotor or wound rotor or phase wound rotor.
Depending upon the type of rotor construction used the three-phase induction motor are
classified as:
1. Squirrel cage induction motor
2. Slip ring induction motor or wound induction motor or phase wound
induction motor.
4.2.2.1 Squirrel cage induction motor
❖ The squirrel cage rotor consists of a
cylindrical laminated core having slots on
its outer periphery which are nearly parallel
to the shaft axis or skewed.
❖ Conductors are heavy copper or aluminium
bars which fit in each slot. These conductors
are brazed to the short-circuiting end rings.
❖ The slots are not exactly made parallel to
the axis of the shaft but are slotted a little
skewed for the following reasons:
➢ They reduce magnetic hum or noise
➢ They avoid stalling of the motor

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4.2.2.1(i) Advantages and Disadvantages
4.2.2.1(ii) Applications
These motors are commonly used in:
❖ Centrifugal pumps
❖ Industrial drives (e.g. to run conveyor belts)
❖ Large blowers and fans
❖ Machine tools
❖ Lathes and other turning equipment
4.2.2.2 Slip-Ring Induction Motor
❖ It consists of a laminated cylindrical core and carries a 3-phase winding, similar to
the one on the stator.
❖ The rotor winding is uniformly distributed in the slots and is usually star-connected.
Advantages
❖ They are low cost.
❖ Require less maintenance (as there are no slip
rings or brushes).
❖ Good speed regulation (they are able to
maintain a constant speed).
❖ High efficiency in converting electrical
energy to mechanical energy (while running,
not during startup).
Disadvantages
❖ Very poor speed control.
❖ They are more sensitive to fluctuations in
the supply voltage.
❖ They have high starting current and poor
starting
torque (the starting current can be 5-9
times the full load current; the starting
torque can be 1.5-2 times the full load
torque)

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❖ The open ends of the rotor winding are brought out and joined to three insulated
slip rings mounted on the rotor shaft with one brush resting on each slip ring.
❖ The three brushes are connected to a 3-phase star-connected rheostat as shown in
fig.
❖ At starting, the external resistances are included in the rotor circuit to give a large
starting torque.
❖ These resistances are gradually reduced to zero as the motor runs up to speed.
❖ The external resistances are used during starting period only.
❖ When the motor attains normal speed, the three brushes are short-circuited so that
the wound rotor runs like a squirrel cage rotor.
4.2.2.2(i) Advantages and Disadvantages
4.2.2.2(ii) Applications
❖ Can be used where high torque and low starting current is required.
❖ Used in elevator, and compressors.
❖ Used in cranes, conveyors, hoists, and more.
Advantages
❖ Having high starting torque.
❖ Having the low starting current.
❖ Can take full load current up to 6-7
times higher.
Disadvantages
❖ High maintenance cost.
❖ More copper loss.
❖ Less efficiency.
❖ Less power factor.
❖ More expensive.

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4.2.2 Comparison between squirrel cage rotor and slip ring rotor
Parameter of
Comparison
Squirrel-Cage
Induction Motor
Slip-Ring Induction
Motor
Rotor Type
Squirrel cage type rotor with
permanently short-circuited
rotor bars.
Wound type rotor with 3-phase
windings similar to stator
winding.
Construction Simple Complicated
External
Rotor
Resistance
It is not possible to add
external resistance because
the rotor bars are permanently
shorted with the help of end
rings.
External resistance is to be added
during starting.
Starting
Torque
Low High
Brushes Not used Used
Maintenance Less maintenance is required.
Frequent maintenance is
required.
Copper Loss Low High
Efficiency High Low
Power Factor High Low
Speed Control Not possible Possible
Starting
Current
High Low
Cost Cheap Expensive
Application
In lath machine, fans, pumps,
blowers, etc.
In cranes, hoists, elevators,
conveyors and loads which
require high starting torque, etc.

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4.3 Working principle and operation of Three Phase Induction Motor
Fig(a): working of three phase induction motor

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❖ A 3 Phase Induction Motor works on the principle of electromagnetic induction.
❖ The three-phase induction motor has a stator part and a rotor part.
❖ The stator part consists of a stator coil that is designed to operate with three phase
power supply. And the rotor part also consists of a rotor coil that is generally short-
circuited.
❖ The rotor part is cylindrical in shape and it is free to rotate. When there is a torque
applied to the rotor it will start rotating.
❖ When a 3-phase supply is given to stator winding, a rotating magnetic field (RMF) is
set up which rotates around the stator at synchronous speed NS.
❖ The rotating field passes through the air gap and cuts the rotor conductors which
are in a stationary position during starting.
❖ Due to the relative speed between the rotating flux and the stationary rotor, EMF is
induced in the rotor conductors.
❖ Since the rotor circuit is short-circuited, currents start flowing in the rotor
conductors.

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❖ The current-carrying rotor conductors are placed in the magnetic field produced by
the stator and mechanical force acts on the rotor conductors.
❖ The sum of the mechanical forces on all the rotor conductors produces a torque that
tends to move the rotor in the same direction as the rotating field.
❖ The fact that the rotor is urged(strongly) to follow the stator field by means of Lenz’s
law.
❖ Now, the cause producing the rotor currents is the relative speed between the
rotating field and the stationary rotor conductors.
❖ Hence to reduce this relative speed, the rotor starts running in the same direction as
that of the stator field and tries to catch it.
NOTE
❖ If the speed of the rotor is equal to the synchronous speed, no relative motion occurs
between the rotating magnetic field of the stator and the conductors of the rotor.
❖ Thus, the EMF is not induced on the conductor, and zero current develops on it.
❖ Without current, the torque is also not produced, that's why the induction motor
can never run on synchronous speed, Induction motor runs always speed lesser than
the speed of the magnetic field (synchronous speed) hence they are Asynchronous.

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4.4 Generation of Rotating Magnetic field (RMF)
❖ Consider three phase windings displaced in space by 120°, supplied by a three-phase
a.c supply.
❖ The three phase currents are also displaced from each other by 120°.
❖ The flux produced by each phase current is also sinusoidal in nature and all three
fluxes are separated from each other by 120°.
❖ If the phase sequence of the windings is R-Y-B, then the mathematical equation for
the instantaneous values of the fluxes ΦR, ΦY and ΦB can be given as,
Fig: Three phase wave forms Fig: Vector diagram
ΦR = Φm Sin (𝜔𝑡 − 0°) (1)
ΦY = Φm Sin (𝜔𝑡 – 120°) (2)
ΦB = Φm Sin (𝜔𝑡 – 240°) (3)
❖ The effective or total flux (ɸT) in the air gap is equal to the phasor sum of the three
components of fluxes ɸR, ɸY and, ɸB.
❖ Therefore, ɸT = ɸR + ɸY + ɸB
❖ Step 1: We will find the values of total flux ɸT for different values of 𝜔𝑡 such as 0,
60, 120, 180 …. 360o.
❖ Step 2: For every value of θ in step 1, we will draw the phasor diagrams.

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❖ By observing these phasor diagrams, you can understand easily that ɸT keep shifting
its position from 90o to 30o to – 30o to – 90o and so on. In other words, ɸT rotates
in the clock wise direction.
❖ We will use the phasor ɸR as the reference phasor i.e. all the angles are drawn with
respect to this phasor.
4.4. Case(i) At 𝝎𝒕 = 0o
Substituting 𝜔𝑡 = 00
in the equations (1), (2) and (3) we get,
ΦR = Φm sin 0o
= 0
ΦY = Φm sin(-120o
) = -
√3
2
Φm = -0.866 Φm
ΦB = Φm sin (-240o
) =
√3
2
Φm = 0.866 Φm
The resultant flux ΦT is the
phasor addition of
ΦR, ΦY and ΦB.
ΦT = ΦY + ΦB
By Parallelogram law of vector addition,
when two vectors having equal in
magnitude and displaced by same
phase angle, then the resultant vector is given by,
ΦT =√(−𝜙𝑦)
2
+ (𝜙𝐵)2 − 2(−𝜙𝑦)𝜙𝐵 𝑐𝑜𝑠 𝜃
ΦT = √(−
√3
2
Φm )
2
+ (
√3
2
Φm)
2
− 2 (−
√3
2
Φm )
√3
2
Φm 𝑐𝑜𝑠 60
ΦT = √
3
4
𝜙𝑚
2 +
3
4
𝜙𝑚
2 + 2
3
4
𝜙𝑚
2 ×
1
2
=
3
2
Φm

Fig. Vector diagram of 𝜔𝑡 = 00
ΦT = 1.5 Φm

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4.4. Case(ii) At 𝝎𝒕 = 60o
ΦR = Φm sin 60o
=
√3
2
Φm = 0.866 Φm
ΦY = Φm sin (-60o
) = -
√3
2
Φm = -0.866 Φm
ΦB = Φm sin (-180o
) = 0 Φm
phasor addition of
ΦR, ΦY and ΦB.
ΦT = ΦR + ΦY
ΦT =√(𝜙𝑅)2 + (−𝜙𝑦)
2
− 2(−𝜙𝑦)𝜙𝑅 𝑐𝑜𝑠 𝜃
ΦT = √(−
√3
2
Φm )
2
+ (
√3
2
Φm)
2
− 2 (−
√3
2
Φm )
√3
2
Φm 𝑐𝑜𝑠 60
ΦT = √
3
4
𝜙𝑚
2 +
3
4
𝜙𝑚
2 + 2
3
4
𝜙𝑚
2 ×
1
2
=
3
2
Φm 
ΦT = 1.5 Φm

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4.4. Case(iii) At 𝝎𝒕 = 120o
ΦR = Φm sin 120o
=
√3
2
Φm = 0.866 Φm
ΦY = Φm sin (0o
) = 0
ΦB = Φm sin (-60o
) = -
√3
2
Φm = -0.866 Φm
phasor addition of
ΦR, ΦY and ΦB.
ΦT = ΦR + ΦB
ΦT =√(𝜙𝑅)2 + (−𝜙𝐵)2 − 2(−𝜙𝐵)𝜙𝑅 𝑐𝑜𝑠 𝜃
ΦT = √(−
√3
2
Φm )
2
+ (
√3
2
Φm)
2
− 2 (−
√3
2
Φm )
√3
2
Φm 𝑐𝑜𝑠 60
ΦT = √
3
4
𝜙𝑚
2 +
3
4
𝜙𝑚
2 + 2
3
4
𝜙𝑚
2 ×
1
2
=
3
2
Φm 
ΦT = 1.5 Φm

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4.4. Case(iv) At 𝝎𝒕 = 180o
ΦR = Φm sin 180o
= 0
ΦY = Φm sin (60o
) =
√3
2
Φm = 0.866 Φm
ΦB = Φm sin (-60o
) = -
√3
2
Φm = -0.866 Φm
phasor addition of
ΦR, ΦY and ΦB.
ΦT = ΦY + ΦB
ΦT =√(𝜙𝑌)2 + (−𝜙𝐵)2 − 2(−𝜙𝐵)𝜙𝑌 𝑐𝑜𝑠 𝜃
ΦT = √(
√3
2
Φm )
2
+ (−
√3
2
Φm)
2
− 2 (−
√3
2
Φm )
√3
2
Φm 𝑐𝑜𝑠 60
ΦT = √
3
4
𝜙𝑚
2 +
3
4
𝜙𝑚
2 + 2
3
4
𝜙𝑚
2 ×
1
2
=
3
2
Φm 
Note: Observe that at each and every instant the magnitude of resultant flux is constant
i.e ΦT = 1.5 Φm
ΦT = 1.5 Φm

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Similarly, at 𝝎𝒕 = 240o
, at 𝝎𝒕 = 300o
, and at 𝝎𝒕 = 360o
the magnitude of resultant flux
is constant i.e ΦT = 1.5 Φm .
Fig: 𝝎𝒕 = 240o
ΦR = Φm sin 240o
= Φm sin(1800
+600
)
= - Φm sin(600
)
ΦY = Φm sin (240o
-1200
)
= Φm sin (1200
)
ΦB = Φm sin (240o
-240o
)
ΦR = Φm sin 360o
ΦY = Φm sin (360o
-1200
)
= Φm sin (2400
)
= - Φm sin (600
)
ΦB = Φm sin (360o
-240o
)
ΦB = Φm sin (120o
)
ΦR = Φm sin 300o
= Φm sin(2400
+600
)
= Φm sin(2400
)
= - Φm sin(600
)
ΦY = Φm sin (300o
-1200
)
= Φm sin (1800
)
ΦB = Φm sin (300o
-240o
)
ΦB = Φm sin (60o
)
ΦR = -
√𝟑
𝟐
Φm = -0.866 Φm
ΦY =
√𝟑
𝟐
Φm = 0.866 Φm
ΦB = 0
ΦY = 0
ΦB =
√3
2
Φm = 0.866 Φm
ΦR = 0
ΦR = -
√𝟑
𝟐
Φm = -0.866 Φm
ΦY = -
√3
2
Φm = -0.866 Φm
ΦB =
√3
2
Φm = 0.866 Φm

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Fig. 𝝎𝒕 = 0o
Fig. 𝝎𝒕 = 60o
Fig. 𝝎𝒕 = 120o
Note: From the above vector diagrams, the direction of RMF is clock-wise

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4.5.1 Speed of Rotating magnetic field
❖ The speed of rotation of the RMF, is called synchronous speed.
❖ It mainly depends upon the supply frequency and number of poles on the stator.
❖ We can measure synchronous speed by using this formula,
Where,
𝑵𝒔 = synchronous speed ------- RMP
f = supply frequency ---------Hz
P = number of poles on the stator
4.5.2 Rotor Speed (Nr)
❖ Generally, the speed of the induction motor (running condition) considered as rotor
speed, denoted by Nr.
❖ The induction motor rotates at a speed (Nr) close to but less than the synchronous
speed.
4.5.3 Slip Speed
❖ An induction motor cannot run at synchronous speed.
❖ If it runs at synchronous speed, there would be no cutting of the flux by the rotor
conductors and there would be no induced EMF, no current and no torque.
❖ Therefore, the rotor of the induction motor rotates at a speed slightly less than the
synchronous speed. For this reason, an induction motor is also known
as asynchronous motor.
The difference between the synchronous speed and the actual rotor speed is known as slip
speed, i.e.,
Slip speed = NS - Nr
𝑵𝒔 =
𝟏𝟐𝟎𝒇
𝑷

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4.5.4 Slip of three phase induction motor
❖ A rotor of a three-phase induction motor rotates at a speed close to the synchronous
speed but not equal to the synchronous speed.
❖ The difference between the synchronous speed and the rotor speed is known as slip
speed.
❖ Slip is defined as the ratio of slip speed to the rotor speed, denoted by S.
❖ Therefore slip,
Rotor speed
where,
𝑵𝒔 = synchronous speed ------- RPM
Nr = rotor speed ------------- RPM
S = slip
S =
𝑵𝑺−𝑵𝒓
𝑵𝑺
 S =
𝑵𝑺−𝑵𝒓
𝑵𝑺
x 100
Nr = (1-S) NS
Important Points:
❖ In a three-phase induction motor, the stator is stationary, and the rotor is a
rotating part.
❖ Both the stator field and rotor field rotate with synchronous speed.
❖ The rotor rotates with a speed less than synchronous speed.
❖ When the rotor is stationary, i.e., Nr = 0, then the slip, s = 1 or 100 %.
❖ In an induction motor, change in slip from no load to full load is hardly 1% to 5%
so that it is essentially a constant speed motor.
❖ Three phase induction motor is a particular form of transformer which has
secondary winding rotating.
❖ In transformer, the frequency of primary and secondary winding is same but
in induction motor the frequency of emf induced in rotor depends on slip.

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Numerical examples on Synchronous speed and Slip
1. A 3-phase induction motor has 26 poles and is connected to 50 Hz AC supply. Find
the synchronous speed of the RMF in the motor.
Solution: Given data,
No. of poles on the stator, P = 26
Supply frequency, f = 50 Hz
Calculate,
synchronous speed, NS =?
synchronous speed, NS =
𝟏𝟐𝟎𝒇
𝑷
=
120𝑥50
26
2. A 3-phase induction motor is wound for 8 poles and is supplied from 50 Hz source.
Calculate (1) synchronous speed (2) slip of the motor when speed is 720 RPM.
Calculate,
(1) Synchronous speed, NS =?
(2) Slip, S =? When rotor is running at 720 RPM
(1) Synchronous speed (NS):
𝟏𝟐𝟎𝒇
𝑷
=
120𝑥50
8
(2) Slip S=? when Nr=720 RPM
Slip S =
𝑵𝑺−𝑵𝒓
𝑵𝑺
=
𝟕𝟓𝟎−𝟕𝟐𝟎
𝟕𝟓𝟎
 Slip S =
𝑵𝑺−𝑵𝒓
𝑵𝑺
x 100 =
𝟕𝟓𝟎−𝟕𝟐𝟎
𝟕𝟓𝟎
x 100
NS = 230.769 RPM
NS = 750 RPM
S=0.04 per unit
%slip = 4%

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3. A 3-phase ,4-pole,50Hz induction motor works a full load slip of 30%.
Find (i) Synchronous speed, (ii) Actual speed of the motor
Full load slip, S= 30%
i.e., S=
𝟑𝟎
𝟏𝟎𝟎
=0.03 per unit
Calculate,
(1) Synchronous speed, NS =?
(2) Actual speed (Rotor speed) Nr=?
(1) Synchronous speed, NS:
𝟏𝟐𝟎𝒇
𝑷
=
120𝑥50
4
(2) Actual speed (Rotor speed) Nr:
Actual speed (Rotor speed) Nr = (1-S) NS
Nr = (1-0.03) x 1500
NS = 1500 RPM
Nr=1455 RPM

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4.4.5 Effect of slip on rotor parameters
❖ In case of a transformer, frequency of the induced e.m.f. in the secondary is same
as the voltage applied to primary.
❖ Now in case of induction motor at starting Nr = 0 and slip s = 1. Under this condition
as long as s = 1, the frequency (fr) of induced e.m.f. in rotor is same as the voltage
applied to the stator i.e., f=fr.
❖ But as motor gathers speed, induction motor has some slip corresponding to speed
Nr. In such case, the frequency (fr) of induced e.m.f. in rotor is no longer same as that
of stator voltage.
❖ Slip affects the frequency (fr) of rotor induced e.m.f.
❖ Due to these some other rotor parameters also get affected.
❖ Let us study the effect of slip on the following rotor parameters.
1.Rotor frequency 2. Rotor Resistance
3. Rotor reactance 3. Impedance
5. Rotor current 6. Rotor power factor
7. Magnitude of rotor induced e.m.f
4.4.5.i Rotor frequency (f2r)
Dividing (2) by (1) we get,
𝑵𝑺−𝑵𝒓
𝑵𝒓
=
𝟏𝟐𝟎𝒇𝟐𝒓
𝒑
x
𝒑
𝟏𝟐𝟎𝒇
S =
𝒇𝟐𝒓
𝒇
Therefore, rotor frequency
i. At standstill Nr = 0 hence s = 1
ii. At Nr = Ns, s = 0 (imaginary condition).
Case(i): At start when Nr = 0, s = 1 and
stationary rotor has maximum relative
motion with respect to R.M.F. Hence
maximum e.m.f. gets induced in the rotor
at start.
The frequency of this induced e.m.f. at
start is same as that of supply frequency.
NS =
𝟏𝟐𝟎𝒇
𝑷
(1)
Case(ii): As motor actually rotates with speed
Nr, the relative speed of rotor with respect R.M.F.
decreases and becomes equal to slip speed of
Ns – Nr. The induced e.m.f. in rotor depends on
rate of cutting flux i.e. relative speed Ns – Nr.
Hence in running condition magnitude of
induced e.m.f. decreases so as to its frequency.
Relative speed (Ns – Nr) =
𝟏𝟐𝟎𝒇𝒓
𝑷
(2)
f2r= Sf

25P a g e
Thus, frequency of rotor induced e.m.f. in running condition (fr) is slip times the supply
frequency (f).
At start we have s = 1 hence rotor frequency is same as supply frequency. As slip of the
induction motor is in the range 0.01 to 0.05, rotor frequency is very small in the running
condition.
Example: A 4 pole, 3 phase, 50 Hz induction motor runs at a speed of 1470 p.m. Find the
frequency of the induced e.m.f in the rotor under this condition.
Rotor speed, Nr = 1470 r.p.m
Find,
frequency of the induced e.m.f under running condition i.e., fr =?
fr = Sf
Slip S =
𝑵𝑺−𝑵𝒓
𝑵𝒓
Synchronous speed NS =
𝟏𝟐𝟎𝒙𝟓𝟎
𝟒
S =
𝟏𝟓𝟎𝟎−𝟏𝟒𝟕𝟎
𝟏𝟒𝟕𝟎
NS = 1500 r.p.m
S = 0.02
 fr = s f = 0.02 x 50 = 1 Hz
NOTE: It can be seen that in running condition, frequency of rotor induced e.m.f. is
very small.

26P a g e
4.4.5.ii Rotor Resistance (R2)
❖ There is no effect of rotor induced EMF frequency on its resistance.
❖ Hence the rotor resistance remains constant irrespective of the speed of the
induction motor.
4.4.5.iii Rotor Reactance (X2)
❖ Let X2 be the rotor reactance per phase at standstill. The rotor frequency at
standstill is fr = f. Therefore, X2 = 2πfL2 ohm/phase.
❖ In the running condition the frequency of rotor voltage is fr = sf. Hence rotor
reactance in the running condition X2r is given by,
X2r = 2πfrL2
= 2π(sf)L2
= s (2πfL2)
ohm/phase.
4.4.5.iv Rotor induced (Z2)
❖ The rotor impedance per phase at standstill is condition given by,
Z2 = √𝑅2
2
+ 𝑋2
2
ohm/phase
❖ The rotor impedance per phase at running condition is given by,
Z2r =√𝑅2
2
+ (𝑆𝑋2)2 ohm/phase
❖ Where R2 = Rotor resistance per phase
X2 = Reactance of the rotor winding per phase at standstill.
4.4.5.v Magnitude of rotor induced E.M.F (E2)
❖ The rotor induced E.M.F per phase at standstill condition is given by,
𝐸2 =
𝐼2
𝑧2
=
𝐼2
√𝑅2
2+𝑋2
2
volts/phase
❖ The rotor induced E.M.F per phase at running condition is given by,
𝐸2𝑟 =
𝐼2𝑟
𝑧2𝑟
=
𝑠𝐼2
√𝑅2
2+(𝑆𝑋2)2
volts/phase
X2r = sX2.

27P a g e
4.4.5.vi Rotor current (I2)
❖ The current per phase at standstill condition is given by,
𝐼2 =
𝐸2
𝑧2
=
𝐸2
√𝑅2
2+(𝑋2)2
ampere/phase
❖ The rotor current per phase at running condition is given by,
𝐼2𝑟 =
𝐸2𝑟
𝑧2𝑟
=
𝑠𝐸2
√𝑅2
2+(𝑆𝑋2)2
ampere/phase
4.4.5.vii Power factor (cos 2)
❖ The rotor power factor at standstill condition is given by,
𝑐𝑜𝑠 𝜙2 =
𝑅2
𝑧2
=
𝑅2
√𝑅2
2+𝑋2
2
❖ The rotor power factor at running condition is given by,
𝑐𝑜𝑠 𝜙2𝑟 =
𝑅2
𝑧2𝑟
=
𝑅2
√𝑅2
2
+ (𝑆𝑋2)2
4.4.5.viii Rotor circuit
Case 1 – When the rotor is stationary
When the rotor is stationary, the motor
is at standstill (slip, s = 1)
Case 2 – When the motor is running at slip ‘s’

28P a g e
Slip effect on rotor parameters sample problem
1. A 3-phase, 4-pole induction motor is connected to a 50 Hz supply. The voltage induced
in the rotor bar conductors is 5 V when the rotor is at standstill. Calculate the voltage and
frequency induced in the rotor conductors at 500 RPM.
Rotor voltage at standstill condition, E2 = 5 V
Rotor speed Nr = 500 RPM
At running condition Calculate, (i) Rotor induced voltage, E2r =?
(ii) Rotor frequency, fr =?
Synchronous speed NS =
𝟏𝟐𝟎𝒇
𝑷
=
𝟏𝟐𝟎𝒙𝟓𝟎
𝟒
Slip S =
𝑵𝑺−𝑵𝒓
𝑵𝑺
=
𝟏𝟓𝟎𝟎−𝟓𝟎𝟎
𝟏𝟓𝟎𝟎
(i) Rotor induced voltage, E2r = SE2 = 0.67 x 5
(ii) Rotor frequency f2r = Sf = 0.67 x 50
NS =1500 r.p.m.
Slip S = 0.66
E2r = 3.35 V
f2r = 33.5 Hz

29P a g e
4.6 Torque equation of three phase induction motor
The torque (τ) developed by the rotor of a 3-phase induction motor is directly
proportional to −
• Rotor current (I2)
• Rotor EMF (E2)
• Rotor circuit power factor (cos ϕ2)
Therefore,   E2I2cos2 (1)
Where, K is the constant of proportionality.
➢ 4.6. i Starting Torque of 3-Phase Induction Motor
Let,
• Rotor resistance/Phase = R2
• Rotor reactance/Phase at standstill = X2
• Rotor EMF/Phase at standstill = E2
∴ Rotor impedance/Phase at standstill, Z2 = √𝑅2
2
+ 𝑋2
2
Rotor current/Phase at standstill, 𝐼2 =
𝐸2
𝑧2
=
𝐸2
√𝑅2
2+(𝑋2)2
And, Rotor power factor at standstill, 𝑐𝑜𝑠 𝜙2 =
𝑅2
𝑧2
=
𝑅2
√𝑅2
2+𝑋2
2
∴ Starting torque, S =K E2I2cos2 = K E2
𝐸2
√𝑅2
2+(𝑋2)2
𝑅2
√𝑅2
2+𝑋2
2
Starting torque, S =
𝑘𝐸2
2𝑅2
(𝑅2
2+𝑋2
2)
(2)
In general, the stator supply voltage V is constant so that the magnetic flux per pole set
up by the stator is also constant. Therefore, the EMF E2 in the rotor will be constant.
Starting torque, S =
𝑘1𝑅2
(𝑅2
2+𝑥2
2)
(3)
Where K1 is another constant = KE2
2
 =K E2I2cos2

30P a g e
➢ Condition for Maximum Starting Torque of 3-Phase Induction Motor
As the starting torque is given by, Starting torque,
S =
𝑘1𝑅2
(𝑅2
2+𝑋2
2)
(4) Divide equation(4) by R2,we get
=
𝑘1
(𝑅2+
𝑋2
2
𝑅2
)
(5)
To be the maximum of the stating torque, the denominator of the eqn. (5) should be
minimum, i.e.,
ⅆ
ⅆ𝑅2
(𝑅2 +
𝑋2
2
𝑅2
)=0 1-
𝑋2
2
𝑅2
2 =0
Hence, the stating will be maximum when,
➢ 4.6. ii Torque of 3-Phase Induction Motor under Running Condition
Let,
• Rotor resistance/Phase = R2
• Rotor reactance/Phase at standstill = X2
• Rotor EMF/Phase at standstill = E2
If ‘s’ is the slip under running condition of the motor, then,
Fig: Rotor circuit under running condition Fig: Impedance triangle
R2 = X2
Rotor Resistance/Phase=Rotor Reactance/Phase at standstill

31P a g e
Rotor reactance/Phase at running, X2r = SX2
Rotor impedance/Phase at standstill, Z2r = √𝑅2
2
+ (𝑆𝑋2)2
Rotor current/Phase at standstill, 𝐼2𝑟 =
𝐸2𝑟
𝑧2𝑟
=
𝑆𝐸2
√𝑅2
2+(𝑆𝑋2)2
And, Rotor power factor at standstill, 𝑐𝑜𝑠 𝜙2𝑟 =
𝑅2
𝑧2
=
𝑅2
√𝑅2
2+(𝑆𝑋2)2
∴ Running torque, r  E2rI2rcos2r
∵ E2r  stator flux ()
∴ Running torque, r  I2rcos2r
r =K I2rcos2r = K 
𝑆𝐸2
√𝑅2
2+(𝑆𝑋2)2
𝑅2
√𝑅2
2+(𝑆𝑋2)2
r =
𝑘𝑆𝐸2𝑅2
(𝑅2
2+(𝑆𝑋2)2)
(1)
∵ stator flux ()  𝐸2
∴ Running torque, r =
𝑘𝑆𝐸2
2𝑅2
(𝑅2
2+(𝑆𝑋2)2)
(2)
Eqn. (1) gives the value of running torque. It can be seen that
• The running torque is directly proportional to slip, i.e., if slip increases,
the torque will increase and vice-versa.
• The running torque is directly proportional to square of supply voltage
since (E2 ∝ V).
As the running torque of a 3-phase induction motor is given by,
r =
𝑘𝑆𝐸2𝑅2
(𝑅2
2+(𝑆𝑋2)2)
Since the supply voltage (V) is constant, then stator flux () and hence the EMF
E2 will be constant.
r =
𝐾1𝑆𝑅2
(𝑅2
2+(𝑆𝑋2)2)
(3) Where, K1 = K ϕ E2 is a constant.

32P a g e
Divide equation (3) ‘S’ by, we get
r =
𝑘1𝑅2
𝑅2
2
𝑆
+𝑠𝑋2
2
(4)
➢ Condition for Maximum Running Torque of 3-Phase Induction Motor
To be the maximum of the running torque, the denominator of eqn. (4) should be
minimum. Hence, differentiating denominator of eqn. (3) with respect to slip 's'
and equating it to zero, i.e.,
ⅆ
ⅆ𝑠
(
𝑅2
2
𝑠
+ 𝑠𝑋2
2
) = 0
−
𝑅2
2
𝑠2
+ 𝑋2
2
= 0
𝑅2
2
= 𝑠2
𝑋2
2
Thus, for maximum torque under running conditions,
Now,
r 
𝑆𝑅2
(𝑅2
2+(𝑆𝑋2)2)
Putting R2 = s X2 in the above expression, the maximum torque under running
condition is given by,
r 
1
2𝑋2
The slip corresponding to maximum torque is given by,
𝑅2 = 𝑆𝑋2
Rotor Resistance/Phase=Per unit slip × standstill rotor reactance/phase
𝒔𝒎 =
𝑹𝟐
𝒙𝟐

33P a g e
4.7 Torque slip characteristics of three phase induction motor
❖ Slip of induction motor plays an important role in the operation of
the induction motor.
❖ The torque produced by the induction motor is directly proportional to induction
motor slip.
❖ At no-load induction motor requires small torque to meet with the frictional, iron
and other losses, therefore slip is small.
❖ When the motor is loaded, greater torque is required to drive the load, therefore,
slip increases and rotor speed decreases slightly.
❖ Thus, induction motor slip adjusts itself to such a value so as to meet the required
driving torque.
The torque of a 3-phase induction motor under running conditions is given by,
r =
𝑘𝑆𝐸2
2𝑅2
(𝑅2
2+(𝑆𝑋2)2)
We have seen that for a constant supply voltage, E2 is also constant. So, we can write
torque equations as, r 
𝑆𝑅2
(𝑅2
2+(𝑆𝑋2)2)
(1)
From the eqn. (1), it can be seen that if R2 and X2 are kept constant, the torque depends
upon the slip 's'. The torque-slip characteristics curve can be divided into two regions, viz.
• Low-slip region
• High-slip region
i. Low-slip region
❖ At synchronous speed, the slip s = 0, thus, the torque is 0.
❖ When the speed is very near to the synchronous speed, the slip is very low and the
term (𝑠𝑋2)2
in denominator is negligible in comparison with R2. Therefore,
❖ ∴ r 
𝑆𝑅2
𝑅2
2 r 
𝑆
𝑅2
∵ R2=constant
r  s

34P a g e
❖ Hence 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.
ii. High slip region
❖ In this region, slip is high i.e., slip value is approaching to 1.
❖ Here it can be assumed that the term R22 is very very small as compared to (s X2)2
.
❖ Hence neglecting from the denominator, we get
❖ r 
𝑆𝑅2
((𝑆𝑋2)2)
r 
𝑅2
𝒔𝑋2
2 ∵ R2 and X2 are constant
❖ So, in high slip region torque is inversely proportional to the slip. Hence its nature
is like rectangular hyperbola.
❖ Now when load increases, load demand increases but speed decreases. As speed
decreases, slip increases. In high slip region as T α1/s, torque decreases as slip
increases.
r 
𝟏
𝑺

35P a g e
Fig: Torque slip characteristics
4.8 Objective questions
1. The frame of an induction motor is usually made of [ B ]
A. silicon steel B. cast iron C. aluminium D. bronze
2. The shaft of an induction motor is made of [ C ]
A. high speed steel B. stainless steel C. carbon steel D. cast iron
3. In an induction motor, no-load the slip is generally [ A ]
A. less than 1% B.1.5% C.2% D.4%
4. In squirrel cage induction motors, the rotor slots are usually given slight skew in order to [ D ]
A. reduce windage losses B. reduce eddy currents
C. reduce accumulation of dirt and dust D. reduce magnetic hum
5. Slip rings are usually made of [ C ]
A. copper B. carbon C. phospor bronze D. aluminium
6. A 3-phase 440 V, 50 Hz induction motor has 4% slip. The frequency of rotor e.m.f. will be [ C ]
A.200 Hz B.50 Hz C.2 Hz D.0.2 Hz
7. In Ns is the synchronous speed and s the slip, then actual running speed of an induction motor
will be
A. Ns B. sN C. (l-s)Ns D.(Ns-l)s [ C ]
8. The efficiency of an induction motor can be expected to be nearly [ B ]
A.60 to 90% B.80 to 90% C.95 to 98% D.99%
9. The starting torque of a squirrel-cage induction motor is [ A ]
A. low B. negligible
C. same as full-load torque D. slightly more than full-load torque
10. An induction motor with 1000 r.p.m. speed will have [ B ]
A.8 poles B.6 poles C.4 poles D.2 poles
11. An induction motor is identical to [ D ]
A.D.C. compound motor B. D.C. series motor
C. synchronous motor D. asynchronous motor
12.In order to control the speed of a 3-phase slip ring induction motor through injected voltage in
its rotor circuit, this voltage and the rotor voltage should essentially be [ D ]
A. in same phase B. in quadrature phase
C. in opposition phase D. of same frequency
Explanation: To control the speed of a 3-phase slip ring induction motor,
the injected voltage in its rotor circuit must have the same frequency as the
rotor voltage. This matching frequency allows for the adjustment of slip and
enables speed control.

36P a g e
13. In a three-phase induction motor, the starting torque will be maximum when [ B ]
A. R2X2 = 1 B.R2 /X2 = 1 C. R2 = X2
2
D. R2 = √X
14. A 3-phase 400 V, 50 Hz induction motor requires a current of 80 A. Calculate its HP at a P.F.
of 0.8 and efficiency 85%. [ A ]
A. 50.5 B.30 C. 40.5 D. 60
15.A 4-pole 50 Hz induction motor is running at 1460 rpm. The frequency of rotor current is
[ A ]
A. 1.33 Hz B. 1.0 Hz C. 15.0 Hz D. 50 Hz
16. In an induction motor, if the rotor is locked, then the rotor frequency of induction motor will
be: [ A ]
A. Equal to the supply frequency B. Less than supply frequency
Explanation: 3 phase IM V = 400 Volts, f = 50 Hz, I = 80 A, pf = 0.8
Efficiency = 85% = 0.85
Output power in HP (Horse Power) =?
We know that Efficiency is the ratio of output and Input = Output/ Input
Output = 0.85 × Input ----(1)
Input = 1.732 × 400 × 80 × 0.8 = 44340 W ----(2)
Putting the value of (2) in (1) we get
Output= 0.85 × 44340 = 37689 W
We know that 1 HP = 746 W
So, the output power in HP = 37689/746 = 50.5 HP
The correct answer is option (a).
Explanation: Number of poles (P) = 4, Frequency (f) = 50 Hz
Rotor speed (Nr) = 1460 rpm, rotor current frequency f2r = sf =?
Synchronous speed Ns =
120𝑓
𝑝
=
120 𝑥 50
4
= 1500 rpm
Slip, S=
𝑁𝑠−𝑁𝑟
𝑁𝑠
=
1500−1460
1500
= 0.0266
rotor current frequency f2r = sf = 0.0266 x 50 = 1.33

37P a g e
C. More than supply frequency D. Zero
17. When a 3-phase induction motor is supplied with balanced 3-phase supply, it produces a
rotating magnetic field of magnitude: [ C ]
A. Twice the peak value of the flux due to any individual phase
B. 0.5 times the peak value of the flux due to any individual phase
C. 1.5 times the peak value of the flux due to any individual phase
D. None of the above
18. The magnetic field produced in the stator of a three phase induction motor travels at _______
A. Rotating speed B. Asynchronous speed C. Synchronous speed D. Slip speed [ C ]
19. For heavy loads, the relation between torque (T) and slip (S) in induction motor is given by
______. [ B ]
A. T
𝑠
1−𝑠
B. Ts C. T(1-s) D. T
1
𝑠
20. Rotor rheostat control method of speed control is used for [ B ]
A. Squirrel-cage induction motors only B. Slip ring induction motors only
C. both (a) and (b) D. None of the above
21. If any two phases for an induction motor are interchanged [ A ]
A. The motor will run in reverse direction
B. The motor will run at reduced speed
C. The motor will not run
D. The motor will burn
22. An induction motor is [ C ]
A. Self-starting with zero torque B. Self-starting with high torque
C. Self-starting with low torque D. Non-self-starting
23. The maximum torque in an induction motor depends on [ D ]
A. Frequency B. Rotor inductive reactance
C. Square of supply voltage D. All of the above

38P a g e
24. In three-phase squirrel-cage induction motors [ B ]
A. Rotor conductor ends are short-circuited through slip rings
B. Rotor conductors are short-circuited through end rings
C. Rotor conductors are kept open
D. Rotor conductors are connected to insulation
25. In a three-phase induction motor, the number of poles in the rotor winding is always [ D ]
A. Zero B. More than the number of poles in stator
C. Less than number of poles in stator D. Equal to number of poles in stator
26. The speed of a squirrel-cage induction motor can be controlled by all of the following except
A. Changing supply frequency B. Changing number of poles [ C ]
C. Changing winding resistance D. Changing supply voltage
27. Slip ring motor is recommended where [ D ]
A. Speed control is required B. Frequent starting, stopping and reversing is required
C. High starting torque is needed D. All above features are required
28. As load on an induction motor goes on increasing [ D ]
A. its power factor goes on decreasing
B. its power factor remains constant
C. its power factor goes on increasing even after full load
D. its power factor goes on increasing upto full load and then it falls again
29. If a 3-phase supply is given to the stator and rotor is short circuited rotor will move [ B ]
A.in the opposite direction as the direction of the rotating field
B.in the same direction as the direction of the field
C.in any direction depending upon phase squence of supply
D. none
30. A 500 kW, 3-phase, 440 volts, 50 Hz, A.C. induction motor has a speed of 960 r.p.m. on full
load. The machine has 6 poles. The slip of the machine will be [ D ]
A.0.01 B.0.02 C.0.03 D.0.04

39P a g e
4.9 Review questions
1. What is the Three Phase induction motor Working Principle? Explain with neat sketch.
2. What are the advantages and disadvantages of three phase induction motor?
3. Describe the construction of a three-phase motor? And explain each part with neat
diagram.
4. What is the effect of no. of poles on speed of motor?
5. What are the major types of Rotor Winding placed in 3 phase induction motor?
6. Differentiate between Squirrel cage type rotor and Wound type rotor?
7. What do you mean by squirrel cage induction motors?
8. How Rotating Magnetic Field is produced in three phase induction motor?
9. What is Synchronous speed?
10. What are the advantages of slip-ring induction motors over the squirrel cage motors?
11. Is there any disadvantage of slip-ring motors?
12. What do you mean by synchronous speed of a 3-phase induction motor?
13. why does the rotor of a 3-phase induction motor rotate in the same direction as the
rotating field?
14. Why cannot 3-phase induction motor run at synchronous ‘speed?
15. What is the importance of slip in a 3-phase induction motor?
16. What are the essential differences between 3-phase induction motor and a transformer?
17. Derive the torque equation of three-phase induction motor under running condition.
18. Draw the torque-speed characteristics of three-phase induction motor.

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