MAHARASHTRA STATE BOARAD OF TECHNICAL EDUCATIOD
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(ISO/IEC-27001-2005 Certified)
SUMMER– 2014 Examinations
Subject Code: 12144
Model Answer
Page 1 of 26
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answer scheme.
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The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
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values may vary and there may be some difference in the candidate’s answers and model answer.
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based on candidate understands.
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concept.
Q.1 A) Attempt any three of the following: -----------------------------------------------------12 Marks
a) Describe the principle of working of 3 phase induction motor on the basis of the concept of
rotating magnetic field.
(4 Marks)
The principle of working of 3 phase induction motor on the basis of the concept of
rotating magnetic field can be explained as follows:
When 3-Ph AC supply is given to stator of three phase induction motor, rotating
magnetic field is produced in air gap, which starts to rotate around the stator frame with
synchronous speed (Ns = 120f/P). There is a relative motion between rotating magnetic field and
stationary rotor conductors which is (Ns-N). According to faradays laws of electromagnetic
induction, emf will be induced in the rotor conductors. As the rotor conductor are short
circuited on either sides by end rings, current flows through it. According to ‘Lenz Law’ the
rotor current should oppose the cause which produces it.
Here the cause is relative speed between rotating magnetic field and rotor conductors.
Now the rotor conductors will try to catch the rotating magnetic field to minimize the relative
speed. But due to inertia and friction of rotor, they never succeed. Hence due to relative speed i.e
Ns-N, rotor will be in continuous rotation with speed N, which is always less than Ns.
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Model Answer
b) State various methods to control the speed of 3 phase induction motor and explain any one
method.
Following methods to control the speed of 3 phase induction motor:
(2 Mark)
1. by varying applied voltage (voltage control)
2. By Varying applied frequency (frequency control)
3. by varying number of poles of the stator winding (Pole changing control)
4. by rotor rheostatic control
5. By V/f method
1. by varying applied voltage (voltage control): (Any one explanation expected)



(2 Mark)
This method is very easy but rarely used in commercial practice because a large
variation of voltage produces a very small change in speed and much energy is wasted.
In this method three resistances are inserted in series with the stator winding of the
motor and the value of these resistances is varied by a common handle, so that equal
resistances come in the stator circuit.
For a particular load when voltage increases, speed of the motor also increases and
vice-versa.
2. by varying applied Frequency (Frequency control):




120  f
.
P
It is clear from the equation that the speed of the induction motor can be changed by
changing the frequency of the supply.
The speed of the motor will increase if frequency increased and it will increase and it
will decrease if frequency decreased.
Changing the frequency of supply to the motor is not an easy job. Therefore this
method is only employed where the variable frequency alternator is available for the
above purpose.
The synchronous speed of an induction motor is given by N S 
3. by varying number of poles of the stator winding (pole changing control):
120  f
.
P

The synchronous speed of an induction motor is given by N S 

It is clear from the equation that if the number of poles of the stator is decreased, the
speed of the motor will increased.
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Model Answer
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
When the number of poles are increases, the speed of the motor decreases.

The poles of the stator winding can be changed by having two or more separate stator
windings of different pole combination housed in common stator frame. By selecting
proper number of pole combination, Ns can be varied and hence the speed.
4. by rotor rheostatic control:

In this method star connected external resistances (of continuous rating) are connected
in the rotor circuit.

The speed of the motor increases with the decrease of resistance in the rotor circuit.

The change in speed is approximately inversely proportional to the external resistance
connected in the rotor circuit.

This method of the speed control is applied where a small variation of speed is
required and the power wasted is of no great importance.
5. By Voltage/ frequency control (V/f) method:

If the ratio of voltage to frequency is kept constant, the flux remains constant.
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
Model Answer
Page 4 of 26
The maximum torque which is independent of frequency can be maintained
approximately constant.

However at a low frequency, the air gap flux is reduced due to drop in the stator
impedance and the voltage has to be increased to maintain the torque level.

This type of control is usually known as Volts/ Hertz or V/f control.

A simple circuit arrangement for obtaining variable voltage and frequency is as shown
in the above figure.
c) What is split phase technique of starting? Why is it necessary to adopt such a technique in
single phase induction motors?
Necessary to adopt split phase technique in single phase induction motors:
(2 Mark)
 1-phase induction motor is having only one main winding.
 When 1-ph A.C. Supply is given, only alternating magnetic field will be produced which
does not rotates.
 For induction principle, the basic requirement of the motor is the production of rotating
magnetic field. Due to absence of rotating magnetic field, single phase induction motor is
not having “self starting” property.
 To make single phase induction motor having self starting characteristics, split phase
technique of starting is used.
Split phase technique of starting:
(2 Mark)
 In this type of technique, another winding called as starting/auxiliary is used.
 Starting winding is having high resistance and low reactance and main/running winding is
having low resistance and high reactance.
 When current passes through these two windings, two separate fluxes having phase
difference about 600 to 90 0 elec. will be produced,
 The interaction between these two AC fluxes having phase difference creates rotating
magnetic field.
 The rotor rotates according to induction principle.
 The capacitor can also be adopted for Split phase technique.
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Model Answer
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Page 5 of 26
d) With the help of necessary graphs and phasor diagram, explain the procedure to calculate
voltage regulation of a 3 phase alternator by synchronous impedance method.
Figure:
or Equivalent fig
(2 Marks)
The procedural steps for synchronous impedance method are as follows: --
(2 Marks)
1) The Open Circuit Characteristics OCC is plotted from open circuit test
2) Short Circuit characteristics is plotted from short circuit test:
Short circuit characteristics are straight line through origin. Both characteristics
plotted for common field current base. Consider field current If and the corresponding OC
voltage E1.During short circuit, at the same field current, the whole E1 is being used to
circulate the short circuit current in armature Isc.
3) The synchronous impedance Zs can be calculated as,
E1  I SC Z S  Z S 
E1OCC
I SC
4) By performing resistance test, Effective armature resistance, Ra can be calculated
Synchronous reactance can be calculated as
X S  Z S2  Ra2
5) The regulation of the alternator at a particular load condition can be calculated as, the
generated EMF; E0 can be calculated as,
E 0  (V Cos  I a Ra ) 2  (V Sin  I a X S ) 2
The % regulation =
E0  V
100
V
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Model Answer
Subject Code: 12144
Page 6 of 26
Q.1 B) Attempt any one of the following: -------------------------------------------------------06 Marks
a) Explain any two methods of staring a synchronous motor.
Methods of starting a synchronous motor:-
(2 Mark)
i) Using a small DC Machine (Exciter)
ii) Using a small induction Motor (pony motor)
iii) Using a damper winding. and
Explanation:-
( explanation of any two method should be given, 02 marks each)
i) Using a small DC Machine (Exciter) :Many a time’s large size synchronous motors are provided with DC generator which acts
as exciter. It’s DC output is used to supply a field winding of synchronous motor.
At the beginning the exciter is fed with DC supply to run it as a motor. This motor brings
speed of rotor to synchronous speed. The field of synchronous motor is then excited and motor
gets synchronized.
ii) Using a small induction Motor (pony motor) :A small induction motor is used to bring the rotor to synchronous speed then DC field is
switched ON, motor gets synchronized and induction motor is decoupled.
iii) Using a Damper winding:It is an additional winding induced on the rotor side. Damper winding is in the form of
copper bars placed in pole shoes of rotor. These copper bars are connected to each other using
copper rings.(similar to squirrel cage rotor of I.M)
Explanation:When the rotor is rotating at synchronous speed, the relative velocity between RMF and rotor
is zero. Hence induced emf and current in the damper winding is zero. The damper winding is
ineffective when motor is running.
At the time of Starting, relative velocity between emf and rotor is equal to Synchronous
speed. This will induce emf into the damper winding and current will flow. Similar to I.M
principle
Copper rotor is accelerated in the direction same as that of emf. As soon as rotor reaches
near the synchronous speed, magnetic locking takes place between stator and rotor. The rotor
then start rotating synchronously and damper winding is now ineffective. Thus damper winding
is used for starting.
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Model Answer
Subject Code: 12144
Page 7 of 26
b) Derive the emf equation of an alternator.
Let, P = no. of rotor poles.  = Flux per pole
Z= Number of stator conductors
N = Speed in rpm
turns per phase (Tph) 
Z Ph

2
Frequency of induced emf is
f = Cycles per rotation x rotation per sec

P N

2 60
f 
PN
---------------------------------------------------------------- (1 Marks)
120
Consider one rotation of rotor then change in flux linkage is,
d  P.  Time required for one rotation is,
 dt 
1
1
60


          Sec. ----------------------------- (1 Marks)
n ( N 60) N
By faradays law of Electromagnetic induction
 Average emf per conductor =
ave / Conductor =
ave / Conductor =
d
dt
P.
( N 60)
P   N
       Volt ---------------------- (1 Marks)
60
ave / turn = 2 ave/ Conductor
ave / turn = 2
P   N
       Volt
60
P   N
       Volt
60
4 P N
       Volt
120

=

= 4(
PN
)
120
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Model Answer
Subject Code: 12144
ave / turn = 4 f   ( f 
PN
)
120
ave / Phs = ave / x Number of turns per phase
= 4 f   TPh -------------------------------------------------------------- (1 Marks)
RMS Value per phase is given by,
Eph = Eph (ave) x Form Factor
= 4 f   TPh 1.11 ---------------------------------------------- (1 Marks)
Eph = 4.44 f . . TPh
It is for full pitched concentrated winding. If winding is distributed & short pitched then
EPh = 4.44 f . TPh . kd.kc
---------------------------------- (1 Marks)
Kc = coil span factor or chording factor
Kd = Distribution factor
Q.2 Attempt any two of the following: ---------------------------------------------------------16 Marks
a) An 18.65 KW, 4Pole, 50 Hz, 3Phase induction motor has friction and windage losses 2.5
percent of the output. The full load slips 4% compute for full load.
i) The rotor Copper losses ii) The rotor input iii) The shaft torque
iv) The gross electromagnetic torque
Given Data:
3Ph, 4 Pole, 50 Hz I.M
Sf = slip at full load = 0.04
Full load Slip = 3.5 %
Net motor o/p = 18.65 kW I.M
Windage losses = 0.025
Windage & frictional losses (Mech. Losses) = 0.025  18.65 103
= 466.25 watts
Gross (Net) Rotor output = Net motor output + mech. losses
= 18650  466.25
= 19116.25 watts --------------------------------------------- (1 Marks)
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Model Answer
Subject Code: 12144
i) Rotor Input =
Rotor Input =
Gross Rotor output
----------------------------------------------------- (1 Marks)
(1  S )
19116.25
 19912.75 Watts
(1  0.04)
Rotor Input = 19912.75 Watts ----------------------------------------------------------- (1 Marks)
ii) Rotor Copper Losses = S  Rotor Input
Rotor Copper Losses = 0.04  19912.75
Rotor Copper Losses = 796.51 Watts -------------------------------------------------- (1 Marks)
NS 
120 f 120  50

 1500 RPM ------------------------------------------------------ (1 Marks)
P
4
Full Load Speed = Ns (1-S),
Full Load Speed = 1500 (1-0.04)
Full Load Speed = 1440 RPM
N in RPS = N 
1440
 24 RPS ------ ----------------------------------- (1 Marks)
60
iii) The shaft torque or Net torque =

Net output
2 N

19116.25
2  3.142  24
Where, N in RPS 
Tsh  126.75 N  m
----------------------------- (1 Marks)
iv) Gross electromagnetic torque =

Gross output
2 N

19912.75
2  3.142  24
Tg  132.03 N  m
Where, N in RPS 
------------------------------- (1 Marks)
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b) The power input to a 3 phase induction motor is 40 kW. The stator losses total are 1 kW
and frictional and windage losses total 2kW. If the Slip of motor is 4% find:
i) The mechanical power output ii) rotor copper loss per phase iii) The efficiency
Given Data:
Net motor input= 40 KW
Stator Losses = 1 kW Slip = 0.04
Rotor I/P = Stator Input – Stator Losses
= (40-1)
= 39 Kw
------------------------------------------------------------- (1 Mark)
i) Total Rotor Copper losses = S (Rotor I/P)
= 0.04 x 39000
= 1560 watts------------------------------------------------ (1 Mark)
rotor Cu losses per phase =
1560
 520 watts -------------------------------- (1 Mark)
3
Gross Rotor Output  Rotor input  Rotor Copper losses
Gross Rotor Output  (39000)  1560
Gross Rotor Output  37440 Watts ---------------------------------------------------- (1 Mark)
ii ) Net Mech. power o / p  Gross Rotor O / P  Mech losses
 37440  2000 W
----------------- (2 Mark)
 35440 Watt
iii) Efficiency %  
Net motor output
 100
Net Motor Input
35440
 100
40000
%   88.6 % -------------------------------------------------------------- (2 Mark)
% 
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Model Answer
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c) A 3 phase , 6600 volts, 50Hz, Star connected synchronous motor takes 50 Amp current. The
resistance and synchronous reactance per phase are 1 ohm and 20 ohm respectively. Find
the power supplied to the motor and induced emf for a power factor of i) 0.8 Lagging
ii) 0.8 leading
Given Data:
3ph, 6600V, 50 Hz, star connected, IL = 50A,
Phase Resistance, Ra = 1 ohm, Phase reactance, XS = 20 ohm
Power sup plied to the motor  3  VL I L  Cos ----------------------------------- (1/2Mark)
Power sup plied to the motor  3  6600  50  0.8
Power sup plied to the motor  457.2614 kW --------------------------------------- (1/2Mark)
Supply voltage per phase VTph 
  Cos 1 (0.8)  36.520 elec.
6600
 3810 volt
3
and   tan 1 ( X S Ra )  ( 20 / 1)  87.80 elec. ---- (1Mark)
Z S / ph  Ra2  X S2  (1) 2  ( 20) 2  20.02  ------------------------------ (1Mark)
 impedance drop  I a Z S  50  20.02  1001V per phase ----------------- (1Mark)
i) induced emf or Eb for 0.8 Lagging Power Factor;
2
emf Eb  ( Er ) 2  VTPh ) 2  2 (VTph ) ( Er ) CoS (51.08) --------------------------- (1/2Mark)
2
emf Eb  (1001) 2  (3810) 2  2 (3810) (1001) Cos (51.08)
emf Eb  3263 V / phase
Line induced e.m. f  3263  3  5651 V / phase ----------------------------- (1Mark)
ii) ) induced emf or Eb for 0.8 leading Power Factor:
  Cos 1 (0.8)  36.52 0 elec.
and   tan 1 ( X S Ra )  ( 20 / 1)  87.80 elec.
 Now (   )  87.80  36.520 1240
 Cos 1240   Cos 560 ----------------------------------------------------------- (1Mark)
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Model Answer
Subject Code: 12144
2
emf Eb  ( Er ) 2  VTPh ) 2  2 (VTph ) ( Er )  Cos 560 ------------------------------ (1/2Mark)
2
emf Eb  (1001) 2  (3810) 2  2  3810  1001  Cos 56 0
emf Eb  4447 V / phase
Line induced e.m. f  4447  3  7700 V / phase ---------------------------- (1Mark)
Q.3 Attempt any four of the following: ---------------------------------------------------------16 Marks
a) With the help of neat diagram, explain the working of shaded pole induction motor.
i) Shaded Pole Induction Motor :-
(Figure-2 Mark & Explanation: 2 Mark)
or Equivalent Fig.
Correct diagram of pole axis shifting from left to right
Working:When single phase supply is applied across the stator winding an alternating field is
created. The flux distribution is non uniform due to shading coils on the poles.
Now consider three different instants of time t1, t2, t3 of the flux wave to examine the
effect of shading coil as shown in the fig above. The magnetic neutral axis shifts from left to
right in every half cycle, from non shaded area of pole to the shaded area of the pole. This
gives to some extent a rotating field effect which may be sufficient to provide starting torque to
squirrel cage rotor.
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Model Answer
b) Give the constructional details of three phase induction motor.
Construction of three phase induction motor: (Explanation is not expected)
The following are three main parts of three phase induction motor
i) Stator:
(2 Mark)
or equivalent fig
 It is formed out of thin circular ring shaped steel laminations. Slots are cut around the
inner periphery as shown the figure
 A three phase distributed winding is placed in the slots. For two, four, six and eight poles.
 This stator is fixed inside the frame of machine
 Three phase supply is given to the stator winding, which produces rotating magnetic field
rotating at a synchronous speed N S 
120  f
.
P
 Stator winding can be connected in star or delta.
ii) Rotor: Squireel Cage rotor: (1 Mark)
Wound Rotor:
(Mark)
or equivalent fig
 For a squirrel cage motor, thick copper bars are fixed around the periphery & are short
circuited by the two end rings as shown in figure.
 For a slip-ring type motor, rotor has shaft as shown in figure-2.
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 A three phase winding is placed in the slots. The winding is connected in star & the
terminal is connected to the three copper slip-rings mounted on the shaft.
 By short circuiting these slip-rings with the help of brushes, rotor winding can be short
circuited.
c) Draw and explain the torque speed Slip characteristics of 3 phase induction motor.
(Figure-2 Marks & Effect- 2 Marks)
or Equivalent fig
Explanation: From the above characteristics:- for squirrel cage induction motor (1R)
 When Slip (S)  0 (i.e N  Ns) torque is almost zero at no load, hence characteristics start
from origin
 As load on motor increases Slip increases and therefore torques increases.
 For lower values of load, torque proportional to slip, and characteristics will having linear
nature.
 At a particular value of Slip, maximum torque conditions will be obtained which is R2 =
SX2
 For higher values of load i.e for higher values of slip, torque inversely proportional to slip
and characteristics will having hyperbolic nature. In short breakdown occurs due to over
load.
For squirrel cage induction motor (2R-4R-6R)
 When rotor resistance increases, maximum torque condition occurs at higher values of slip
and characteristics shifts towards left hand side.
 The maximum torque condition can be obtained at any required slip by changing rotor
resistance.
 The safe zone of operation of induction motor is shown by dark portion on characteristics.
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Model Answer
Subject Code: 12144
d) What are the causes and effects of hunting in synchronous motor? How it can be reduced?
Causes and Effects of hunting:-
(2 Mark)
Hunting produces oscillations of the rotor of synchronous motor .If the time period of these
oscillations happens to be equal to the natural time period of the machine than mechanical
resonance is set up. The amplitude of these oscillations is built up to a large value and may throw
the machine out of synchronism.
It can be reduced:
(2 Mark)
Hunting is reduced by placing damper windings on the rotor pole surface. Damper
windings are short circuited copper bars placed on rotor pole faces. During rotor equilibrium
condition, damper windings are inactive. Whenever the rotor oscillates, because of the relative
motion between stator and rotor, currents are induced in damper windings. This produces a torque
which restores the rotor back to its equilibrium position and hunting is prevented.
e) Draw and explain the working of capacitor start capacitor run induction motor.
(Figure- 2 Mark & Explanation- 2 Mark)
OR
or Equivalent fig
In these motors one capacitor is connected in series with the auxiliary winding. There is
no centrifugal switch. Thus this winding along with the capacitor remains energized for both
starting and running conditions. Capacitor used serves the purpose of obtaining necessary phase
displacement at the time of starting and also improves the power factor of the motor
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Q.4 A) Attempt any three of the following: ---------------------------------------------------12 Marks
a) Prove that rotor copper loss in induction motor is Slip times rotor input.
(4 Mark)
Consider 3-ph induction motor,
Gross rotor output  2  N Tg Where, N in rps & Tg in N-m
If rotor rotates with synchronous speed, there will not be copper losses in rotor and hence
rotor output will be equal rotor input
 Gross rotor input  2  N S Tg Where N S in rps
 Rotor copper losses  Gross rotor input  Gross rotor output
 Rotor copper losses  2 N S Tg  2 N Tg
 Rotor copper losses  2 Tg ( N S  N )

Rotor Cu losses 2 Tg ( N S  N )

Rotor input
2 Tg  N S

Rotor Cu losses
(NS  N )

Rotor input
NS

Rotor Cu losses
 Slip
Rotor input
 Rotor Cu losses  Slip  rotor input
b) Why single phase induction motor does not have a self starting torque?
(4 Marks)
 When single phase AC supply is given to main winding it produces alternating flux.
 According to double field revolving theory, alternating flux can be represented by two
opposite rotating flux of half magnitude.
 These oppositely rotating flux induce current in rotor & there interaction produces two
opposite torque hence the net torque is Zero and the rotor remains standstill.
 Hence Single-phase induction motor is not self starting. OR
When single phase A.C supply is applied across the single phase stator winding, an
alternating field is produced. The axis of this field is stationary in horizontal direction. The
alternating field will induce an emf in the rotor conductors by transformer action. Since the
rotor has closed circuit, current will flow through the rotor conductors. Due to induced emf
and current in the rotor conductors the force experienced by the upper conductors of the rotor
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will be downward and the force experienced by the lower conductors of the rotor will be
upward .The two sets of force will cancel each other and the rotor will experience no torque
.Therefore single phase motors are not self starting.
c) Derive the equation of torque for a polyphase induction motor at any value of Slip.
 Torque equation of polyphase induction motor at any value of slip:
(4 Mark)
Torque of 3-phase induction motor under running condition is directly
proportional to the product of working flux per pole, rotor current under running and P.f. of
motor circuit under running.
 T   I R cos R .......... .......... .1
Now, IR = Rotor current under running / ph

Er Rotor induced Emf / ph under running

Zr
Rotor impedance / ph under running

S E2
2
R2  S 2 X 2
S  Slip
Where,
R2  Rotor resis tan ce / ph
X 2  S tan dstill rortor reac tan ce / ph
E2  S tan dstill Rotor induced Emf / ph
cosr 
R2
Zr

R2
2
R2  S 2 X 2
Putting Value of Ir and cos r in equation.......1 of Torque
T


S E2


 R2 2  S 2 X 2 
SE R
 2 22 2 2
R2  S X
T


R2


 R2 2  S 2 X 2 
K  S E2 R2
2
R2  S 2 X 2
Where, K= proportionality constant
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d) What is chorded and distributed winding and how it affects coil emf, phase emf and line
emf
i) Chorded winding:
(1.5 Mark)
 A full pitch coil has a distance of 1800 elec. (i.e pole pitch) between its two coil sides.
 When a distance of less than 180 0 elec. is kept between the two coil sides of a coil then this
type of winding is said to be a chorded winding.
ii) Distributed winding:
(1.5 Mark)
 Generally the winding is a concentrated winding.
 When a space angle in electrical degrees between adjacent stator slots i.e. time angle of
phase difference between successive coils is present then the winding becomes a distributed
winding.
Effect on coil emf, phase emf and line emf:
(1 Mark)
 Due to these two types of windings the magnitude of induced emf slightly gets reduced, as
compared to concentrated winding and full pitch coils.
Q.4 B) Attempt any one of the following: -------------------------------------------------------06 Marks
a) With the help of necessary diagrams, explain the construction of Linear induction Motor
(LVM)
(Figure- 3 Marks & explanation of construction – 3 Marks)
or Equivalent fig.
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Model Answer
Subject Code: 12144
 Explanation construction of linear induction motor:Construction wise a LIM is similar to three phase induction motor . In a sector IM, if sector
is made flat and squirrel cage winding is brought to it we get linear I.M. In practice instead of a
flat squirrel cage winding, aluminum or copper or iron plate is used as rotor.
The flat stator produces a flux that moves in a straight line from its one end to other at a
linear synchronous speed given by Vs = 2 wf
Where, Vs = linear synchronous speed in m/sec,
w = width of one pitch in m.
f = supply frequency (Hz)
b) Compare salient pole and smooth cylindrical type rotor used in 3 phase alternator (Give at
least 6 points)
(Any 6 Point expected-1 Mark each)
S.No
Parameter/Machine
Salient pole type rotor
Smooth cylindrical type
rotor
1
Operating speed
Low medium
high
2
Number of poles
large
Small & medium
3
Rotor construction
4
5
6
7
8
Axial length
Diameter
Operation
Centrifugal stresses
Application
Projected type bulky & heavy
weight
short
large
noisy
Cylindrical poles type
comparatively moderate weight
large
small
Very smooth
Non uniform
uniform
In hydro power stations
Thermal power station
Q.5 Attempt any two of the following: ---------------------------------------------------------16 Marks
a) Develop and draw the equivalent circuit and vector diagram of 3 phase induction motor
when running at Slip S.
(4 Mark)
or equivalent fig.
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Model Answer
Subject Code: 12144
Vector diagram of 3 phase induction motor:
(4 Mark)
or equivalent fig.
b) The following data were obtained in the open circuit and short circuit tests on a 25 KVA,
440V, 50Hz, three phase, star connected alternator.
Field Current (Amp)
Terminal Voltage (V)
S.C. Current (Amp)
2
156
11
4
288
22
6
396
34
7
440
40
8
474
46
10
530
57
12
568
69
14
592
80
Effective resistance between terminals is 0.32 ohm. Calculate the full load regulation by
synchronous impedance method at 0.8 lag, 0.8 lead and unity p.f.
Solution:Phase Resistance, Ra = 0.32 ohm
Phase Voltage V=
440
 254 volts
3
Full load Line current I = Ia =
Field Current (Amp)
Terminal Voltage (V)/line
Terminal Voltage/phase
S.C. Current (Amp)
2
156
90.07
11
4
288
166.3
22
25 103
 32.8 A ----------------------- (1/Mark)
3  440
6
396
228.6
34
7
440
254
40
8
474
273.7
46
10
530
306
57
12
568
328
69
14
592
341.8
80
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Graph:
Page 21 of 26
(2 Mark)
From OCC and SCC:-
 Synchronous impedance per phase  Zs 
O.C Voltage / ph 222

S .C Current / ph 32.8
 Synchronous impedance per phase  Zs / ph  6.77  ------------------------ (1Mark)
Effective Resistance between terminals = 0.32 ohm
Synchronous reactance, X S / ph  Z S2  Ra2  (6.77)2  (0.16) 2  6.74  ---- (1/2Mark)
i) 0.8 Lagging Power Factor: cos  0.8, sin   0.6
E / ph  (VT Cos  I a R a ) 2  (VT sin   I a X S)2 ---------------------------------- (1/2Mark)
E ph  ( 254  0.8  32.8  0.16) 2  ( 254  0.6  32.8  6.74) 2
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Eph  ( 203.0  5.248) 2  (152.4  221.07) 2
Eph  43450.57  139479.84
Eph  427.7 Volt --------------------------------------------------------------- (1/2Mark)
Re gulation 
427.7  254
 100  68.33 %
254
% Regulation = 68.33 %----------------------------------------------------------------- (1/2Mark)
ii) % regulation at 0.8 leading Power Factor;
E / ph  (VT Cos  I a R a ) 2  (VT sin   I a X S) 2 ----------------------------------- (1/2Mark)
E ph  ( 254  0.8  32.8  0.16) 2  ( 254  0.6  32.8  6.74) 2
Eph  ( 203.0  5.248) 2  (152.4  221.07) 2
Eph  43450.57  4715.57
Eph  219.47 Volt ------------------------------------------------------------- (1/2Mark)
Re gulation 
219.47  254
100   13.78 %
254
% Regulation = -13.78 %---------------------------------------------------------------- (1/2Mark)
iii) Unity Power Factor;
E / ph  (VT Cos  I a R a ) 2  (VT sin   I a X S)2
E ph  (254  1  32.8  0.16) 2  (254  0  32.8  6.74) 2
Eph  43450.57  48872.83
Eph  303.85 Volt ------------------------------------------------------------- (1/2Mark)
Re gulation 
303.85  254
 100  19.63 %
254
% Regulation = 19.63 %----------------------------------------------------------------- (1/2Mark)
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Model Answer
Subject Code: 12144
c) Explain how resistance inserted in the rotor circuit of 3 phase induction motor acts as a
starter. How the speed can be controlled by this resistance?
Diagram for resistance inserted in the rotor circuit of 3 phase induction motor:
(4 Marks)
or equivalent figure
This method is only applicable to slip-ring motors. At the instant of starting, the external
rotor resistance can be kept at maximum value. Therefore heavy starting current can be
controlled. Thus for slip ring induction motor external resistance connected in rotor circuit can
acts as starter.
The speed can be controlled by external resistance:
(4 Marks)
For an induction motor working at or near full load slip, the torque and slip are related
with each other approximately as -
T
s
R2
Where, R2 is rotor resistance per phase in ohms,
This external resistance can be connected to rotor circuit as shown in the above figure.
If the effective resistance of the rotor circuit increases, for maintaining the torque to drive given
load, the slip increases and speed decreases and vice-versa. Thus by connecting external series
rheostat the speed of the induction motor can be changed effectively.
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Model Answer
Q.6 Attempt any four of the following: ---------------------------------------------------------16 Marks
a) Explain why three phase induction motor can never run on synchronous speed? (4 Marks)
The working principle of three phase induction motor is based on relative motion between
rotating magnetic field and rotor conductors i.e (NS - N), According to Lenz’s law rotor will try to
catch the synchronous speed of rotating magnetic field to oppose the ‘cause producing it’. But
rotor never succeeds due to frictional losses.
If rotor catches the synchronous speed of rotating magnetic field, (NS - N) i.e is relative
motion will be zero and rotor stops to rotate and therefore three phase induction motor can never
run on synchronous speed .
b) Draw a schematic diagram of slipring induction motor and name their parts.
Schematic diagram of slipring induction motor: (Figure:2 Mark & name of parts: 2 Mark)
or equivalent figure
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Model Answer
c) Explain with neat sketches the production of rotating magnetic field in 3 phase induction
motor.
or Equivalent fig.--------------------- (1 Mark)
ii) wt = 600
i) wt = 0
or Equivalent fig.
i) Wt = 00, 
r
3
 m
2
3
ii) Wt =600,  r  m
2
iii) wt = 1200
--------------------------------- (2 Marks)
3
iii) Wt =1200,  r  m .
2
From the above vector diagrams at different phase angles particularly at 00, 60 0 and
1200 referred in waveform diagram, it is clear that the resultant flux vector is not stationary but
it rotates with NS
(1Marks)
d) Explain the principle of operation of an Induction generator.
Figure:(Figure- 2 Marks & Principle – 2 Marks)
or Equivalent fig.
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 The principle of operation induction Generator:
When rotor of induction motor runs faster than synchronous speed (N>Ns), induction
motor runs as generator and called as induction generator. It converts mechanical energy it
receives from the shaft into electrical energy which is released by stator. However, for
creating its own magnetic field, it absorbs reactive power Q from the line to which it is
connected. The reactive power is supplied by a capacitor bank connected at the induction
generator output terminals.
e) Draw a schematic diagram of an A.C Series motor. How to change its speed and its
direction of rotation?
Schematic diagram of an A.C Series motor:
(2 Mark)
OR
Reason for change its speed:
(1 Mark)
 Since the speed of this motor is not imitated by the supply frequency hence the speed
control of this motor is best obtained by solid state devices.
Reason for change its direction of rotation:
(1 Mark)
 The direction of rotation can be changed by interchanging connection to the field with
respect to the armature.