V.S.B.

ENGINEERING COLLEGE, KARUR

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

ACADEMIC YEAR 2020-2021 (EVEN SEMESTER)

III YEAR/VI SEMESTER

2 MARK AND 13 MARK QUESTION BANK

SUBJECT PAGE
S.No SUBJECT NAME
CODE NO.

1 EE8601 SOLID STATE DRIVES 2

PROTECTION AND
2 EE8602 25
SWITCHGEAR
3 EE8691 EMBEDDED SYSTEMS 51
DESIGN OF ELECTRICAL
4 EE8002 64
APPARATUS
SPECIAL ELECTRICAL
5 EE8005 95
MACHINES

1
 EE8601- SOLID STATE DRIVES

UNIT – I
DRIVE CHARACTERISTICS
PART-A
1. What is meant by electrical drives? (Nov/Dec 2014)
Systems employed for motion control are called drives and they employ any of the prime
movers such as diesel or petrol engines, gas or steam turbines, hydraulic motors and electric
motors for supplying mathematical energy for motion control. Drives employing electric motion
are called electric drives.

2. What are the requirements of an electric drive?

Stable operation should be assured.
The drive should have good transient response

3. Specify the functions of power modulator.

Power modulator performs one or more of the following four functions.

a. Modulates flow of power from the source to the motor in such a manner that motor is
imparted speed-torque characteristics required by the load.

b. During transient operations, such as starting, braking and speed reversal, it restricts
source and motor currents within permissible values; excessive current drawn from
source may overload it or may cause a voltage dip.

4. Mention the different types of drives.

1) Group drive 2) Individual drive 3) Multi motor drive

5. List the different types of electrical drives.

1) DC drives 2)AC drives

6. What are the advantages of electric drives?

They have flexible control characteristics. The steady state and dynamic characteristics of
electrical drives can be shaped to satisfy load requirements.
Drives can be provided with automatic fault detection systems, programmable logic controllers
and computers can be employed to automatically ctrl the drive operations in a desired sequence.
They are available in which range of torque, speed and power.
It can operate in all the four quadrants of speed-torque plane. Electric braking gives smooth
deceleration and increases life of the equipment compared to other forms of braking.
Control gear required for speed control, starting and braking is usually simple and easy to operate.

7. What are the functions performed by electric drives?

Various functions performed by electric drives include the following.
a. Driving fans, ventilators, compressors and pumps etc.
b. Lifting goods by hoists and cranes
c. Imparting motion to conveyors in factories, mines and warehouses and
 d. Running excavators and escalators, electric locomotives, trains, cars, trolley buses,
lifts and drums winders etc.

8. What are the disadvantages of electric drives?

The disadvantages of electric drives are
a. Electric drives system is tied only up to the electrified area.

b. The condition arising under the short circuits, leakage from conductors and
breakdown of overhead conductor may lead to fatal accidents.

c. Failure in supply for a few minutes may paralyses the whole system.

9. What are the advantages of group drive over individual drive?

The advantages of group drive over individual drive are

a. Initial cost: Initial cost of group drive is less as compared to that of the individual drive.

b. Sequence of operation: Group drive system is useful because all the operations are
stopped simultaneously.

c. Space requirement: Less space is required in group drive as compared to

individual drive.

d. Low maintenance cost: It requires little maintenance as compared to individual drive.

10. What the group drive is not used extensively.

Although the initial cost of group drive is less but yet this system is not used extensively
because of following disadvantages.

 Power factor: Group drive has low power factor

 Efficiency: Group drive system when used and if all the machines are not working
together the main motor shall work at very much reduced load.

 Reliability: In group drive if the main motor fails whole industry will come to stand
still.

 Flexibility: Such arrangement is not possible in group drive i.e., this arrangement is not
suitable for the place where flexibility is the prime factor.
 Speed: Group drive does not provide constant speed.

 Types of machines: Group drive is not suitable fro driving heavy machines such as
cranes, lifts and hoists etc.

11. Write short notes on individual electric drives.

In individual drive, each individual machine is driven by a separate motor. This motor also
imparts motion to various other parts of the machine. Examples of such machines are single
spindle drilling machines (Universal motor is used) and lathes. In a lathe, the motor rotates the
spindle, moves the feed and also with the help of gears, transmits motion to lubricating and
cooling pumps. A three phase squirrel cage induction motor is used as the drive. In many such
applications the electric motor forms an integral part of the machine.

12. Mention the different factors for the selection of electric drives?
1) Steady state operation requirements.
2) Transient operation requirements.
3) Requirements related to the source.
4) Capital and running cost, maintenance needs life.
5) Space and weight restriction.
6) Environment and location.
7) Reliability.

13. Mention the parts of electrical drives.

1) Electrical motors and load.
2) Power modulator
3) Sources
4) Control unit
5) Sensing unit

14. Mention the applications of electrical drives(Nov/Dec 2016)

• Paper mills
• Electric traction Cement mills
• Steel mills

15. Mention the types of enclosures

Screen projected type
Drip proof type
Totally enclosed type

16. Mention the different types of classes of duty

Continuous duty, Discontinuous duty, Short time duty, intermittent duty.

17. What is meant by regenerative braking?

Regenerative braking occurs when the motor speed exceeds the synchronous speed. In this
case the IM runs as the induction m\c is converting the mechanical power into electrical power
which is delivered back to the electrical system. This method of braking is known as regenerative
braking.

18. What is meant by dynamic braking?

Dynamic braking of electric motors occurs when the energy stored in the rotating mass is
dissipated in an electrical resistance. This requires a motor to operate as a gen. to convert the
stored energy into electrical.

19. What is meant by plugging?

It is one method of braking of IM. When phase sequence of supply of the motor running at
the speed is reversed by interchanging connections of any two phases of stator with respect to
supply terminals, operation shifts from motoring to plugging region.
20. What is critical speed?
It is the speed that separates continuous conduction from discontinuous conduction mode.

21. Which braking is suitable for reversing the motor?

Plugging is suitable for reversing the motor.

22. Define equivalent current method

The motor selected should have a current rating more than or equal to the current. It is also
necessary to check the overload of the motor. This method of determining the power rating of the
motor is known as equivalent current method.

23. Define cooling time constant

It is defined as the ratio between C and A. Cooling time constant is denoted as Tau. Tau =
C/A Where C=amount of heat required to raise the temp of the motor body by 1 degree Celsius
A=amount of heat dissipated by the motor per unit time per degree Celsius.

24. What are the methods of operation of electric drives?

Steady state
Acceleration including starting
Deceleration including starting

25. Define four quadrant operations.

The motor operates in two mode: motoring and braking. In motoring, it converts electrical
energy into mechanical energy which supports its motion. In braking, it works as a generator,
converting mathematical energy into electrical energy and thus opposes the motion. Motor can
provide motoring and braking operations for both forward and reverse directions.

26. What is meant by mechanical characteristics?

The curve is drawn between speed and torque. This characteristic is called mechanical
characteristics.

27. Mention the types of braking

Regenerative braking
Dynamic braking Plugging

28. What are the advantage and disadvantages of D.C. drives?

The advantages of D.C. drives are,
 Adjustable speed
 Good speed regulation
 Frequent starting, braking and reversing.

The disadvantage of D.C. drives is the presence of a mechanical commutator which limits the
maximum power rating and the speed.

29. Give some applications of D.C. drives.

The applications of D.C. drives are,
a. Rolling mills b. Paper mills c. Mine winders d. Hoists
e. Machine tools f. Traction g. Printing presses h. Excavators
i. Textile mils j. Cranes
30. Why the variable speed applications are dominated by D.C. drives?
The variable speed applications are dominated by D.C. drives because of lower cost,
reliability and simple control.

31. List out the examples of active load torque in drive system(Apr/May 2017)
Load torque which have the potential to drive the motor under equilibrium conditions are
called active load torques. Ex. Torque due to force of gravity, torque due tension torque, due to
comparession and torsion, Torque due to friction, cutting etc.

32.State the condition of steady state stability of motor load system? (May/Jun 2016)
i) A decrease in speed, the motor torque is greater that the load torque.
ii) An increase in speed, the load torque is greater that the motor torque.

33. What are active and passive load torques? Give examples.(Apr/May 2015)
A passive load is a load consisting of only a resistor, capacitor or inductor, or a
combination of them. Ex. Torque due to friction, cutting.
An active load is a load which includes something which is current or voltage controlled,
particularly a semiconductor device. Ex. Torque due to force of gravity, torque due tension torque.

PART-B
1. Label the essential parts of electric drive. Explain its function.
2. (i) Discuss and Draw the speed-torque characteristics of various types of loads. (ii) Discuss in
detail about the multi quadrant dynamics of electric drives.
3. Define how the following speed transitions are carried out : (i) Increase in speed in same
direction. (ii) Decrease in speed in same direction. (iii) Speed reversal.

4. (i) Show a motor is coupled to a load having the following characteristics: Motor: Tm = 15 –
0.6 ωα Load: TL = 0.5 ω α 2 Find out the stable operating point for this condition.(ii) Explain in
detail about steady state stability in electrical drive system.
5. (i) Discuss in detail the multi quadrant dynamics in the speed – torque plane. (BTL2) (8) (ii)
Discuss the different modes of operation of an electrical drive.
6. (i) Explain the four quadrant operation of low speed hoist in detail. (ii) Explain and derive an
equation to find out equivalent load torque in a motor load system with translational and rotational
motion?
7. Compose the mathematical condition to obtain steady state stability of equilibrium point?

8. Explain in detail the multi quadrant operation of low speed hoist in speed torque plane.
9. Solve a motor drives two loads. One has rotational motion. It is coupled to the motor through a
reduction gear with a = 0.1 and efficiency of 90%. The load has a moment of inertia of 10 kg-m2
and a torque of 10 N-m. Other load has translational motion and consists of 1000kg weight to be
lifted up at a uniform speed of 1.5 m/s. coupling between this load and the motor has an efficiency
of 85%. Motor has inertia of 0.2 kg-m2 and runs at a constant speed of 1420 rpm. Determine
equivalent inertia referred to the motor shaft and power developed by the motor.10. Define in
detail about the braking of DC and AC drive.

UNIT – II
CONVERTER/CHOPPER FED DC MOTOR DRIVE
PART-A

1. What is the use of flywheel? Where it is used?

It is used for load equalization. It is mounted on the motor shaft in compound motor.

2. What are the advantages of series motor?

The advantages of series motors are,
a) High starting torque
3. Define and mention different types of braking in a dc motor?
In breaking the motor works as a generator developing a negative torque which opposes
the motion. Types are regenerative braking, dynamic or rheostat braking and plugging or reverse
voltage braking.

4. How the D.C. motor is affected at the time of starting?

A D.C. motor is started with full supply voltage across its terminals, a very high current
will flow, which may damage the motor due to heavy sparking at commuter and heating of the
winding. Therefore, it is necessary top limit the current to a safe value during starting.

5. List the drawbacks of armature resistance control?

In armature resistance control speed is varied by wasting power in external resistors that
are connected in series with the armature. since it is an inefficient method of speed control it was
used in intermittent load applications where the duration of low speed operations forms only a
small proportion of total running time.

6. What is static Ward-Leonard drive?

Controlled rectifiers are used to get variable d.c. voltage from an a.c. source of fixed
voltage controlled rectifier fed dc drives are also known as static Ward-Leonard drive.

7. What is a line commutated inverter?

Full converter with firing angle delay greater than 90 deg. is called line commutated
inverter. Such an operation is used in regenerative braking mode of a dc motor in which case a
back emf is greater than applied voltage.

8. Mention the methods of armature voltage controlled dc motor? When the supplied voltage
is ac?
Ward-Leonard schemes
Transformer with taps and un controlled rectifier bridge Static Ward-Leonard scheme or
controlled rectifiers when the supply is dc: Chopper control

9. How is the stator winding changed during constant torque and constant horsepower
operations?
 For constant torque operation, the change of stator winding is made form series - star to
parallel - star, while for constant horsepower operation the change is made from series-delta to
parallel-star. Regenerative braking takes place during changeover from higher to lower speeds.

10. Define positive and negative motor torque.

Positive motor torque is defined as the torque which produces acceleration or the positive
rate of change of speed in forward direction. Positive load torque is negative if it produces
deceleration.

11. Write the expression for average o/p voltage of full converter fed dc drives?
Vm=(2Vm/pi)cospi ........ continuous conduction
Vm=[Vm(cos alpha-cos beta)+(pi+alpha+beta)]/pi] discontinuous conduction

12. What are the disadvantages of conventional Ward-Leonard schemes?

Higher initial cost due to use of two additional m\cs. Heavy weight and size.

Needs more floor space and proper foundation. Required frequent maintenance

13. Mention the drawbacks of rectifier fed dc drives?

Distortion of supply. Low power factor. Ripple in motor current

14. What are the advantages in operating choppers at high frequency?

The operation at a high frequency improves motor performance by reducing current ripple
and eliminating discontinuous conduction.

15. Why self commutated devices are preferred over thyristors for chopper circuits?
Self commutated devices such as power MOSFETs power transistors, IGBTs, GTOs and
IGCTs are preferred over thyristors for building choppers because they can be commutated by a
low power control signal and don't need commutation circuit.

16. State the advantages of dc chopper drives?

DC chopper device has the advantages of high efficiency, flexibility in control, light
weight, small size, quick response and regeneration down to very low speed.

17. What are the advantages of closed loop c of dc drives?

Closed loop control system has the adv. of improved accuracy, fast dynamic response and
reduced effects of disturbance and system non-linearities.

18. What are the types of control strategies in dc chopper?

• Time ratio control.
• Current limit control.

19. What are the adv. of using PI controller in closed loop ctrl. of dc drive?
 Stabilize the drive
 Adjust the damping ratio at the desired value
  Makes the steady state speed error close to zero by integral action and filters out noise
again due to the integral action.

19. What are the different methods of braking applied to the induction motor?
Regenerative braking, Plugging, Dynamic braking.

21. What are the different methods of speed control of IM?

Stator voltage control, Supply frequency control, Rotor resistance control, Slip power
recovery control.

20. What is meant by stator voltage control?

The speed of the IM can be changed by changing the stator voltage. Because the torque is
proportional to the square of the voltage.

23. Mention the application of stator voltage control.

This method is suitable for applications where torque demand reduced with speed, which
points towards its suitability for fan and pump drives.

24. Mention the applications of ac drives.

AC drives are used in a no. of applications such as fans, blowers, mill run-out tables,
cranes, conveyors, traction etc.

25. What are the three regions in the speed-torque characteristics in the IM?
Motoring region (0<=s<=1) Generating region(s<0)
Plugging region (1<=s<=2) where s is the slip.

26. What are the advantages of stator voltage control method?

 The control circuitry is simple
 Compact size
 Quick response time
 There is considerable savings in energy and thus it is economical method as
compared to other methods of speed control.

27. What is meant by soft start?

The ac voltage controllers show a step less control of supply voltage from zero to rated voltage
they are used for soft start for motors.

28. List the adv of squirrel cage IM?

• Cheaper
• light in weight
• Rugged in construction
• More efficient
• Require less maintenance
• It can be operated in dirty and explosive environment

29. Define slip

 The difference between the synchronous speed (Ns)and actual speed(N)of the rotor is
known as slip speed. the % of slip is gn by,

%slip s=[(Ns-N)/Ns]x 100

30. Define base speed.

The synchronous speed corresponding to the rated freq is called the base speed.

PART-B
1. Explain the steady state analysis of the single phase fully controlled converter fed separately
excited DC motor drive for continuous current mode. Also explain its operation in motoring and
regenerative braking mode.
2. Solve a 250V separately excited dc motor has an armature resistance of 2.5Ω when driving a
load at 600 r.p.m. with constant torque, the armature takes 20 A. This motor is controlled by a
chopper circuit with a frequency of 400 Hz and an input voltage of 250 V. (i) What should be the
value of the duty ratio if one desires to reduce the speed from 600 to 540 r.p.m. with the load
torque maintained constant? (ii) Find out the value of duty ratio for which the per unit ripple
current will be maximum.
3. Describe about Electrical –mechanical characteristics of commonly used electric motors.

4. (i) Explain the operation of four quadrant dc chopper drive.(BTL4) (8) (ii) Solve a 220 V, 20
A, 1000 rpm separately excited dc motor has an armature resistance of 2.5 Ω. The motor is
controlled by a step-down chopper with a frequency of 1 kHz. The input dc voltage to the chopper
is 250V. Identify what will be the duty cycle of the chopper for the motor to operate at a speed of
600 rpm delivering the rated torque?
5. (i)Explain in detail the single phase fully controlled rectifier control of dc separately excited
motor with heat waveforms. (ii) Solve a 220 V, 1500 rpm, 10 A separately excited DC motor has
an armature resistance of fed from a single phase fully controlled rectifier with a source voltage
of 230 V 50 Hz. Assuming continuous load current. Compute (1) Motor speed at the firing angle
of 30◦ and Torque of 5 Nm. (2) Developed Torque at the firing angle of 45◦ and speed of 1000
rpm.
6. (i) Define in detail about the regenerative operation of three phase fully controlled rectifier
control of separately excited DC motor. (ii) Define in detail about the four quadrant operation of
chopper fed drive.

7. Compose the operation of single phase controlled converter fed separately excited DC motor in
continuous and discontinuous modes with neat diagram, waveforms and comment the steady state
analysis?

8. Discuss the four quadrant operation of chopper fed DC drive.

9. Define in detail about the operation of single phase fully-controlled converter fed dc separately
excited motor in continuous and discontinuous modes of operation with necessary waveforms and
steady state analysis.
10. (i) Discuss the different control techniques of chopper in detail. (ii) Discuss the four quadrant
operation of DC-DC converter.
 UNIT – III
INDUCTION MOTOR DRIVES
PART-A

1. What is meant by frequency control of IM?

The speed of IM can be controlled by changing the supply freq because the speed is
directly proportional to supply frequency. This method of speed ctrl is called freq control.

2. What is meant by V/F control l?

When the freq is reduced the i/p voltage must be reduced proportionally so as to maintain
constant flux otherwise the core will get saturated resulting in excessive iron loss and magnetizing
current. This type of IM behavior is similar to the working of dc series motor.

3. What are the advantages of V/F control?

 Smooth speed ctrl
 Small i/p current and improved power factor at low freq. start
 Higher starting torque for low case resistance
3. What is meant by stator current control?
The 3 phase IM speed can be controlled by stator current control. The stator current can be
varied by using current source inverter.
4. What are the 3 modes of region in the adjustable-freq IM drives characteristics?
• Constant torque region
• Constant power region
• High speed series motoring region

5. What are the two modes of operation in the motor?

The two modes of operation in the motor are, motoring and braking. In motoring, it
converts electrical energy to mechanical energy, which supports its motion. In braking, it works as
a generator converting mechanical energy to electrical energy and thus opposes the motion.

6. How will you select the motor rating for a specific application?
When operating for a specific application motor rating should be carefully chosen that the
insulation temperature never exceed the prescribed limit. Otherwise either it will lead to its
immediate thermal breakdown causing short circuit and damage to winding, or it will lead to
deterioration of its quality resulting into thermal breakdown in near future.

7. What is braking? Mention its types.

The motor works as a generator developing a negative torque which opposes the motion is
called barking.
It is of three types. They are,
a. Regenerative braking.
b. Dynamic or rheostat braking.
c. Plugging or reverse voltage braking.

8. What are the three types of speed control?

The three types of speed control as,
 a. Armature voltage control
b. Field flux control
c. Armature resistance control.

9. What are the advantages of armature voltage control?

The advantages of armature voltage control are,
a. High efficiency
b. Good transient response
c. Good speed regulation.

10. What are the methods involved in armature voltage control? When the supply in A.C.
b. Ward-Leonard schemes
b. Transformer with taps and an uncontrolled rectifier bridge.
c. Static ward Leonard scheme or controlled rectifiers when the supply in D.C.
d. Chopper control.

11. Give some drawbacks and uses of Ward-Leonard drive.

The drawbacks of Ward Leonard drive are.
a. High initial cost
b. Low efficiency
c. The Ward-Leonard drive is used in rolling mills, mine winders, paper mills, elevators,
machine tools etc.

12. Give some advantages of Ward-Leonard drive.

The advantages of Ward-Leonard drive are,
a. Inherent regenerative braking capability
b. Power factor improvement.

13. What is the use of controlled rectifiers?

Controlled rectifiers are used to get variable D.C. Voltage form an A.C. Source of fixed
voltage.

14. What is known as half-controlled rectifier and fully controlled rectifier?

The rectifiers provide control of D.C. voltage in either direction and therefore, allow motor
control in quadrants I and IV. They are known as fully-controlled rectifiers.
The rectifiers allow D.C. Voltage control only in one direction and motor control in quadrant I
only. They are known as half-controlled rectifiers.

15. What is called continuous and discontinuous conduction?

A D.C. motor is fed from a phase controlled converter the current in the armature may
flow in discrete pulses in called continuous conduction.
A D.C. motor is fed from a phase controlled converter the current in the armature may
flow continuously with an average value superimposed on by a ripple is called discontinuous
conduction.

16. What are the three intervals present in discontinuous conduction mode of single phase
half and fully controlled rectifier?
The three intervals present in half controlled rectifier are,
a. Duty interval b. Free, wheeling interval c. Zero current intervals.
 The two intervals present in fully controlled rectifier are
a. Duty interval
b. Zero current intervals.

17. What is called inversion?

Rectifier takes power from D.C. terminals and transfers it to A.C. mains is called
inversion.

18. What are the limitations of series motor? Why series motor is not used in traction
applications now days?

1. The field of series cannot be easily controlled. If field control is not

employed, the series motor must be designed with its base speed equal to
the highest desired speed of the drive.

2. Further, there are a number of problems with regenerative braking of a

series motor. Because of the limitations of series motors, separately excited
motors are now preferred even for traction applications.

20. What are the advantages of induction motors over D.C. motors?
The main drawback of D.C. motors is the presence of commutate and brushes, which require
frequent maintenance and make them unsuitable for explosive and dirty environments. On the
other hand, induction motors, particularly squirrel-cage are rugged, cheaper, lighter, smaller, more
efficient, require lower maintenance and can operate in dirty and explosive environments.

21. Give the applications of induction motors drives.

Although variable speed induction motor drives are generally expensive than D.C. drives, they are
used in a number of applications such as fans, blowers, mill run-out tables, cranes, conveyors,
traction etc., because of the advantages of induction motors. Other applications involved are
underground and underwater installations, and explosive and dirty environments.

22. How is the speed controlled in induction motor?

The induction motor speed can be controlled by supplying the stator a variable voltage,
variable frequency supply using static frequency converters. Speed control is also possible by
feeding the slip power to the supply system using converters in the rotor circuit, basically one
distinguishes two different methods of speed control.

Speed control by varying the slip frequency when the stator is fed from a constant voltage,
constant frequency mains.
Speed control of the motor using a variable frequency variable voltage motor operating a
constant rotor frequency.

23. How is the speed control by variation of slip frequency obtained?

Speed control by variation of slip frequency is obtained by the following ways.
a. Stator voltage control using a three-phase voltage controller.
b. Rotor resistance control using a chopper controlled resistance in the rotor circuit.
c. Using a converter cascade in the rotor circuit to recover slip energy.
 d. Using a cyclconverter in the rotor circuit.

24. Mention the effects of variable voltage supply in a cage induction motor.
When a cage induction motor is fed from a variable voltage for speed control the following
observations may be made.

a. The torque curve beyond the maximum torque point has a negative shape. A stable operating
point in this region is not possible for constant torque load.

b. The voltage controlled must be capable of withstanding high starting currents. The range of
speed control is rather limited.
c. The motor power factor is poor.

25. Classify the type of loads driven by the motor.

The type of load driven by the motor influences the current drawn and losses of the motor as
the slip various. The normally occurring loads are
1. Constant torque loads.
2. Torque varying proportional to speed.
3. Torque varying preoperational to the square of the speed.
4.
26. What are the disadvantages of constant torque loads?
The constant torque loads are not favored due to increase in the losses linearly with slip
and becoming maximum at s= 1.0. This is obvious form the variation of flux as the voltage is
varied for speed control. To maintain constant torque the motor draws heavy current resulting in
poor torque/ampere, poor efficiency ad poor power factor at low speeds.

27. In which cases, torque versus speed method is suitable.

Torque versus speed method is suitable only for the following cases.
a. For short time operations where the duration of speed controls ids defined.

b. For speed control of blowers or pumps having parabolic or cubic variations

of torque with speed. This is not suitable for constant torque loads due to
increases and heating.

28. How is the speed of a squirrel cage induction motor controlled?

The speed of a squirrel cage induction motor can be controlled very effectively by varying
the stator frequency. Further the operation of the motor is economical and efficient, if it operates
at very small slips. The speed of the motor is therefore, varied by varying the supply frequency
and maintaining the rotor frequency at the rated value or a value corresponding to the required
torque on the linear portion of the torque-speed curve.

29. Why the control of a three-phase indication motor is more difficult than D.C. motors.
The control of a three-phase induction motor, particularly when the dynamic performance
involved is more difficult than D.C. motors. This is due to a. Relatively large internal resistance
of the converter causes voltage fluctuations following load fluctuations because the capacitor
cannot be ideally large.
b. In a D.C. motor there is a decoupling between the flux producing magnetizing current and
torque producing armature current. They can be independently controlled. This is not the case
with induction motors.
c. An induction motor is very poorly damped compared to a D.C. motor.

30. Where is the V/f control used?

The V/f control would be sufficient in some applications requiring variable torque, such as
centrifugal pumps, compressors and fans. In these, the torque varies as the square of the speed.
Therefore at small speeds the required torque is also small and V/f control would be sufficient to
drive these leads with no compensation required for resistance drop. This is true also for the case
of the liquid being pumped with minimal solids.

PART-B

1. (i) Discuss briefly separate controlled mode of synchronous motor in detail (ii) Explain self
control of synchronous motor drive in detail
2. (i) Explain margin angle control of synchronous motor drive. (ii)Describe briefly the power
factor angle control of synchronous motors with relevant vector diagram.
3. (i) Explain commutator less Dc motor. (ii) Describe closed loop speed control of load
commutated inverter synchronous motor drive and explain it.
4. (i) Describe the open loop v/f control of VSI fed synchronous motor in detail (ii)Describe the
CSI fed synchronous motor drive in detail.
5. Describe the closed loop operation of permanent magnet synchronous motor drive in details.
6. Discuss the construction and working of permanent magnet synchronous motor with neat
diagram
7. (i) Name the various types of permanent magnet synchronous motor and explain it .
(ii)Describe the vector control of sinusoidal SPM in constant torque region.
8. A 3phase, 400V, 50Hz, 6pole star connected round rotor synchronous motor has Zs=0+j2Ω
Load torque proportional to speed squared is 340Nm at rated synchronous speed. The speed of the
motor is lowered by keeping v/f constant and maintaining unity pf by field control of the motor.
For the motor operation at 600 rpm, calculate a) supply voltage b) armature current c) excitation
angle d) load angle e) pull out torque. Neglect rotational losses.
9. A 7MW, three phase 12 kV star connected 6 pole 50Hz 0.9 leading power factor synchronous
motor has Xs= 10Ω and Rs=0. The rated field current is 40A. The machine is controlled by
variable frequency control at constant V/f ratio up to the base speed and at constant V above base
speed. Evaluate (i) Torque (ii) The field current for the rated armature current 750rpm and 0.8
leading power factor.
10. A 500kW, 3 phase, 3.3 kV, 50 Hz, 0.8 lagging power factor, 4 pole, star connected
synchronous motor has the following parameters Xs=15Ω, Rs=0.Rated field current s 10A.
Calculate armature current and power factor at half the rated torque and field current.

UNIT – IV

SYNCHRONOUS MOTOR DRIVES

PART-A

1. What are the components of the applied voltage to the induction motor?
The applied voltage to the induction motor has two components at low frequencies. They
are
a. Proportional to stator frequency.
 b. To compensate for the resistance drop in the stator.
The second component deepens on the load on the motor and hence on rotor frequency.
2. What is indirect flux control?
The method of maintaining the flux constant by providing a voltage boost proportional to
slip frequency is a kind of indirect flux control. This method of flux control is not desirable if very
good dynamic behaviour is required.

3. What is voltage source inverter?

Voltage source inverter is a kind of D.C. link converter, which is a two stage conversion
device.

4. What is the purpose of inductance and capacitance in the D.C. link circuit?
The inductance in the D.C. link circuit provides smoothing whereas the capacitance
maintains the constancy of link voltage. The link voltage is a controlled quality.

5. What are the disadvantages of square wave inverter in induction motor drive?
Square wave inverters have commutation problems at very low frequencies, as the D.C.
link voltage available at these frequencies cannot charge the commutating capacitors sufficiently
enough to commutate the thrusters. Those puts a limit on the lower frequency of operation. To
extend the frequency towards zero, special charging circuits must be used.

6. What is slip controlled drive?

When the slip is used as a controlled quantity to maintain the flux constant in the motor the
drive is called slip enrolled drive. By making the slip negative (i.e., decreasing the output
frequency of the inverter) The machine may be made to operate as a generator and the energy of
the rotating parts fed back to the mains by an additional line side converter or dissipated in a
resistance for dynamic barking. By keeping the slip frequency constant, braking at constant torque
and current can be achieved. Thus braking is also fast.

7. What are the effects of harmonics in VSI fed induction motor drive?
The motor receives square wave voltages. These voltages have harmonic components. The
harmonics of the stator current cause additional losses and heating. These harmonics are also
responsible for torque pulsations. The reaction of the fifth and seventh harmonics with the
fundamental gives rise to the seventh harmonic pulsations in the torque developed. For a given
induction motor fed from a square wave inverter the harmonic content in the current tends to
remain constant independent of input frequency, with the rang of operating frequencies of the
inverter.

8. What is a current source inverter?

In a D.C. link converter, if the D.C. link current is controlled, the inverter is called a
current source inverter, The current in the D.C. link is kept constant by a high inductance and he
capacitance of the filter is dispensed with . A current source inverter is suitable for loads which
present a low impedance to harmonic currents and have unity p.f.

9. Explain about the commutation of the current source inverter.

The commutation of the inverter is load dependent. The load parameters form a part of the
commutation circuit. A matching is therefore required between the inverter and the motor.
Multimotor operation is not possible. The inverter must necessarily be a force commutated one as
the induction motor cannot provide the reactive power for the inverter. The motor voltage is
almost sinusoidal with superimposed spikes.

10. Give the features from which a slip controlled drive is developed.
The stator current of an induction motor operating on a variable frequency, variable
voltage supply is independent of stator frequency if the air gap flux is maintained constant.
However, it is a function of the rotor frequency. The torque developed is also a function of rotor
frequency. The torque developed is also a function of rotor frequency only. Using these features a
slip controlled drive can be developed employing a current source inverter to feed an induction
motor.

11. How is the braking action produced in plugging?

In plugging, the barking torque is produced by interchange any two supply terminals, so
that the direction of rotation of the rotating magnetic field is reversed with respect to the rotation
of the motor. The electromagnetic torque developed provides the braking action and brings the
rotor to a quick stop.

12. Where is rotor resistance control used?

Where the motors drive loads with intermittent type duty, such as cranes, ore or coal
unloaders, skip hoists, mine hoists, lifts, etc. slip-ring induction motors with speed control by
variation of resistance in the rotor circuit are frequently used. This method of speed control is
employed for a motor generator set with a flywheel (Ilgner set) used as an automatic slip regulator
under shock loading conditions.

13. What are the advantages and disadvantages of rotor resistance control?
Advantage of rotor resistance control is that motor torque capability remains unaltered
even at low speeds. Only other method which has this advantage is variable frequency control.
However, cost of rotor resistance control is very low compared to variable frequency control.

Major disadvantage is low efficiency due to additional losses in resistors connected in the
rotor circuit.

14. Where is rotor resistance control used?

Where the motors drive loads with intermittent type duty, such as cranes, ore or coal
unloaders, skip hoists, mine hoists, lifts, etc. slip-ring induction motors with speed control by
variation of resistance in the rotor circuit are frequently used. This method of speed
control is employed for a motor generator set with a flywheel (Ilgner set) used as an automatic slip
regulator under shock loading conditions.

15. What are the advantages and disadvantages of rotor resistance control?
Advantage of rotor resistance control is that motor torque capability remains unaltered
even at low speeds. Only other method which has this advantage is variable frequency control.
However, cost of rotor resistance control is very low compared to variable frequency control.
Major disadvantage is low efficiency due to additional losses in resistors connected in the rotor
circuit.

16. How is the resistance in the output terminals of a chopper varied?

The resistance connected across the output terminals of a chopper can be varied from O to
R by varying the time ratio of the chopper. When the chopper is always OFF, the supply is always
connected to the resistance R. The time ratio in this case is zero and the effective resistance
connected in R. Similarly when the chopper is always ON, the resistance is short circuited. The
time ratio in the case is unity and the effective resistance connected is 0. Hence by varying the
time ratio from 0 to 1, the value of resistance can be varied from R to O.

17. What is the function of inductance L and resistance R in the chopper resistance circuit?
A smoothing inductance L is used in the circuit to maintain the current at a constant value.
Any short circuit in the chopper does not become effective due to L.
The value of R connected across the chopper is effective for all phases and its value can be related
to the resistance to be connected in each phase if the conventional method has been used. The
speed control range is limited by the resistance.

18. What are the disadvantages and advantages of chopper controlled resistance in the rotor
circuit method?
The method is very inefficient because of losses in the resistance. It is suitable for
intermittent loads such as elevators. At low speeds, in particular the motor has very poor
efficiency. The rotor current is non-sinusoidal. They harmonics of the rotor current produce torque
pulsations. These have a frequency which is six times the slip frequency.
Because of the increased rotor resistance, the power factor is better.

19. How is the range of speed control increased?

The range of speed control can be increased if a combination of stator voltage control and
rotor resistance control is employed. Instead of using a high resistance rotor, a slip ring rotor with
external rotor resistance can be used when stator voltage control is used for controlling the speed.

20. Why the static scherbius drive has a poor power factor?
Drive input power is difference between motor input power and the power fed back.
Reactive input power is the sum of motor and inverter reactive power. Therefore, drive has a poor
power factor throughout the range of its options.

21. How is super synchronous speed achieved?

Super synchronous speed can be achieved if the power is fed to the rotor from A.C. mains.
This can be made possible by replacing the converter cascade by a cycloconverter. A
cyclo converter allows power flow in either direction making the static sherbets drive operate at
both sub and supper synchronous speeds.

22. Give the features of static scherbius drive

The torque pulsations and other reactions are minimal. The performance of the drive
improves with respect to additional losses and torque pulsations. A smooth transition is possible
from sub to super synchronous speeds without any commutation problems. Speed reversal is not
possible. A step up transformer may be interposed between the lines and the converter, to reduce
the voltage rating of the converter.

23. Where is Kramer electrical drive system used?

Some continuous rolling mills, large air blowers, mine ventilators, centrifugal pumps and
any other mechanisms including pumps drives of hydraulic dredgers require speed adjustment in
the range from 15 to 30% below or above normal. If the induction motor is of comparatively big
size (100 to 200 KW) it becomes uneconomical to adjust speed by mean's pf external resistances
due to copper losses as slip power is wasted as heat in the retort circuit resistance. In these case ,
the Kramer electrical drive system is used , where slip power recovery takes places.

24. What is the use of sub synchronous converter cascades?

Sub synchronous converter cascades have been used, till now, in applications requiring
one quadrant operation. These can be employed for drives where at least one electrical barking
is required. A four quadrant operation can also be made possible in these cascades, using suitable
switching.

25. How is the speed control obtained in static Kramer drive?

For speed control below synchronous speed, the slip power is pumped back to the supply,
where as for the case of speed above synchronous speed, additional slip power is injected into the
rotor circuit.

26. What is static Kramer drive?

Instead of wasting the slip power in the rotor circuit resistance, it can be converted to 60
Hz A.C. and pumped back to the line. The slip power controlled drive that permits only a sub
synchronous range of speed control through a converter cascade is know as static Kramer drive.

27. What is the use and functions of step down transformer is static Kramer drive?
For a restricted speed range closer to synchronous speed, the system power factor can be
further improved by using a step -down transformer.
The step-down transformer has essentially two functions: besides improving the line power factor,
it also helps to reduce the converter power ratings.

28. What are the advantages of static Kramer drive?

The static Kramer drive has been very popular in large power pump and fan-type drives,
where the range of speed control is limited near, but below the synchronous speed. The drive
system is very efficient and the converted power rating is low because t has to handle only the slip
power, In fact, the power rating becomes lower with a more restricted range of speed control. The
additional advantages are that the drive system has D.C. machine like characteristics and the
control is very simple.

29. What are the causes of harmonic currents in static Kramer drive?
The rectification of slip power causes harmonic currents in the rotor, and these harmonics
are reflected to the stator by the transformer action of the machine. The harmonic currents are also
injected into the A.C. line by the inverter. As a result, the machine losses are increased and some
amount of harmonic torque is produced. Each harmonic current in the rotor will create a reading
magnetic field and its direction of rotation will depend on the order pf the harmonic.

PART-B

1. (i)Discuss briefly separate controlled mode of synchronous motor in detail (ii)Explain self
control of synchronous motor drive in detail
2. (i)Explain margin angle control of synchronous motor drive. (ii)Describe briefly the power
factor angle control of synchronous motors with relevant vector diagram
3. (i) Explain commutator less Dc motor. (ii)Describe closed loop speed control of load
commutated inverter synchronous motor drive and explain it.
4. (i)Describe the open loop v/f control of VSI fed synchronous motor in detail (ii)Describe the
CSI fed synchronous motor drive in detail.
5. Describe the closed loop operation of permanent magnet synchronous motor drive in details.
6. Discuss the construction and working of permanent magnet synchronous motor with neat
diagram
7. (i)Name the various types of permanent magnet synchronous motor and explain it .
(ii)Describe the vector control of sinusoidal SPM in constant torque region.
8. A 3phase, 400V, 50Hz, 6pole star connected round rotor synchronous motor has Zs=0+j2Ω
Load torque proportional to speed squared is 340Nm at rated synchronous speed. The speed of the
motor is lowered by keeping v/f constant and maintaining unity pf by field control of the motor.
For the motor operation www.Vidyarthiplus.com www.Vidyarthiplus.com at 600 rpm, calculate a)
supply voltage b) armature current c) excitation angle d) load angle e) pull out torque. Neglect
rotational losses
9. A 7MW, three phase 12 kV star connected 6 pole 50Hz 0.9 leading power factor synchronous
motor has Xs= 10Ω and Rs=0. The rated field current is 40A. The machine is controlled by
variable frequency control at constant V/f ratio up to the base speed and at constant V above base
speed. Evaluate(i) Torque (ii) The field current for the rated armature current 750rpm and 0.8
leading power factor.
10. A 500kW, 3 phase, 3.3 kV, 50 Hz, 0.8 lagging power factor, 4 pole, star connected
synchronous motor has the following parameters Xs=15Ω, Rs=0.Rated field current s 10A.
Calculate armature current and power factor at half the rated torque and field current.

UNIT – V
DESIGN OF CONTROLLERS FOR DRIVES
PART-A
1. Give the four modes of operation of a Scherbius drive
The four modes of operation of static Scherbius drive are, Sub synchronous motoring.
Sub synchronous regeneration Super synchronous motoring Super synchronous regeneration.

2. Give the use of synchronous motors.

Synchronous motors were mainly used in constant speed applications. The development of
semiconductor variable frequency sources, such as inverters and cycloconverters, has allowed
their use in draft fane, main line traction, servo drives, etc.
3. How are the stator and rotor of the synchronous motor supplied?
The stator of the synchronous motor is supplied from a thyristor power converter capable
of providing a variable frequency supply. The rotor, depending upon the situation, may be
constructed with slip rings, where it conforms to a conventional rotor. It is supplied with D.C.
through slip rings. Sometimes rotor may also be free from sliding contacts (slip rings), in which
case the rotor is fed from a rectifier rotating with rotor.

4. What is the difference between an induction motor and synchronous motor?

An induction motor operates at lagging power factor and hence the converter supplying the
same must invariable is a force commutated one. A synchronous motor, on the other hand, can be
operated at any power factor by controlling the field current.

5. List out the commonly used synchronous motors. Commonly used synchronous motors are,
a. Wound field synchronous motors.
 b. Permanent magnet synchronous motors
c. Synchronous reluctance synchronous motors.
d. Hysterias motors.

6. Mention the main difference between the wound field and permanent magnet motors.
When a wound filed motor is started as an induction motor, D.C. field is kept off. In case
of a permanent magnet motor, the field cannot be 'turned off’.

7. Give the advantages and applications of PMSM.

a. High efficiency b. High power factor c. Low sensitivity to supply voltage variations.
The application of PMSM is that it is preferred of industrial applications with large duty cycle
such as pumps, fans and compressors.

8. Give the uses of a hysteresis synchronous motor.

Small hysteresis motors are extensively used in tape recorders, office equipment and fans.
Because of the low starting current, it finds application in high inertia application such as
gyrocompasses and small centrifuges.

9. Mention the two modes employed in variable frequency control

Variable frequency control may employ and of the two modes. a. True synchronous mode
b. Self-controlled mode

10. Define load commutation

Commutation of thyristors by induced voltages pf load is known as load commutation.

11. List out the advantages of load commutation over forced commutation.
Load commutation has a number of advantages over forced commutation It does not
require commutation circuits
Frequency of operation can be higher
It can operate at power levels beyond the capability of forced commutation.

12. Give some application of load commutated inverter fed synchronous motor drive.
Some prominent applications of load commutated inverter fed synchronous motor drive
are high speed and high power drives for compressors, blowers, conveyers, steel rolling mills,
main-line traction and aircraft test facilities.
13. How the machine operation is performed in self-controlled mode?
For machine operation in the self-controlled mode, rotating filed speed should be the same
as rotor speed. This condition is relaised by making frequency of voltage induced in the armature.
Firing pulses are therefore generated either by comparison of motor terminal voltages or by rotor
position sensors.

14. What is meant by margin angle of commutation?

The difference between the lead angle of firing and the overlap angle is called the margin
angle of commutation. If this angle of the thyristor, commutation failure occurs. Safe
commutation is assured if this angle has a minimum value equal to the turn off angle f the
thyristor.

15. What are the disadvantages of VSI fed synchronous motor drive?
 VSI synchronous motor drives might impose fewer problems both on machine as well as
on the system design. A normal VSI with 180° conduction of thyristors required forced
commutation and load commutation is not possible.

16. How is PNM inverter supplied in VSI fed synchronous motor?

When a PWM inverter is used, two cases may arise the inverter may be fed from a
constant D.C. source in which case regeneration is straight forward. The D.C. supply to the
inverter may be obtained from a diode rectifier. In this case an additional phase controlled
converter is required on the line side.

17. What is D.C. link converter and cycloconverter?

D.C. link converter is a two stage conversion device which provides a variable voltage,
variable frequency supply.
Cycloconverter is a single stage conversion device which provides a Variable voltage, variable
frequency supply.

18. What are the disadvantages of cycloconverter?

A cycloconverter requires large number of thyristors and ts control circuitry is complex.
Converter grade thyristors are sufficient but the cost of the converter is high.

19. What are the applications of cycloconverter?

A cycloconverter drive is attractive for law speed operation and is frequently employed in
large, low speed reversing mils requiring rapid acceleration and deceleration. Typical applications
are large gearless drives, e.g. drives for reversing mills, mine heists, etc.

20. Give the application of CSI fed synchronous motor.

Application of this type of drive is in gas turbine starting pumped hydroturbine starting,
pump and blower drives, etc.

21. What are the disadvantages of machine commutation?

The disadvantages of machine commutation are,
a. Limitation on the speed range. b. The machine size is large
c. Due to overexciting it is underutilized.
21. What is the use of an auxiliary motor?
Sometimes when the power is small an auxiliary motor can be used to run up the
synchronous motor to the desired speed.
23. What are the advantages of brushless D.C. motor?
The brushless D.C. motor is in fact an inverter-fed self controlled permanent synchronous
motor drive. The advantages of brushless D.C. motor are low cost, simplicity reliability and good
performance.

24. When can the synchronous motor be load commutated?

When the synchronous motor operates at a leading power factor thyristors of load side
converter can be commutated by the motor induced voltages same way as the thyristors of a line
commutated converter are commutated by line voltages.

25. What are the characteristics of self controlled mode operated synchronous motor?

a) It operates at like dc motor also commutator less motor.

 b) These machines have better stability behavior.
c) Do not have oscillatory behavior.

26. What are the characteristics of true synchronous mode operated synchronous motor?
The motor behaves like conventional synchronous motor (i.e) hunting oscillations exists.
The change in frequency is slow enough for rotor to truck the changes.
Multi motor operation is possible here.

27. What is meant by sub synchronous speed operation?

The sub synchronous speed operation means the SRIM speed can be controlled below the
Synchronous speed. i.e) the slip power is fed back to the supply.

28. What is meant by super synchronous speed operation?

The super synchronous speed operation means the SRIM speed can be controlled above
the synchronous speed. i.e) the supply is fed back to the rotor side.

PART-B

1. Derive and explain from basic principles the transfer function for separately excited DC motor
load system with converter fed armature voltage control.
2. Explain the closed loop operation of armature voltage control method and field weakening
mode control for DC drive.

3. Describe the step by step procedure for the design of current controller.

4. Give the design procedure for speed controller of an electrical drive system with necessary
diagrams.

5. Discuss the use of simulation software package for design of controller for drives.

6. List the factors involved in converter selection and equations involved in controller
characteristics.

7. A 50KW, 240V, 1700 rpm separately excited DC motor is controlled by a converter. The field
current is maintained at If=1.4A and the machine back EMF constant is Kv=.91VA rad/sec.The
armature resistance is Rm=0.1Ω and thev is constriction constant is B=0.3Nm/rad/sec. The
amplification of the speed sensor is K1=95mV/rad/sec and the gain of the power controller is
K2=100. Calculate (i)The reference voltage Vr to drive the motor at the rated speed. (ii)If the
reference voltage is kept unchanged, determine the speed at which the motor develops rated
torque.

8. Discuss the current controller design using (i) P controller and (II) PI controller for a separately
excited dc motor drive systems.

9. Design a speed controller Dc motor drive maintaining the field flux constant. The motor
parameters and ratings are as follows. 220V, 8.3A, 1470 rpm, Ra = 4Ω, J = 0.0607 kg-m2, La =
0.072H,Bt = 0.0869 Nm/rad/sec, Kb = 1.26V/rad/sec The converter is supplied from 230V,
3phase AC at 60 Hz. The converter is linear and its maximum control input voltage is ±10 V. The
tacho generator has the transfer function Gw(s) =( 0.065)/(1+0.002s). The speed reference voltage
has a maximum of 10V. The maximum current permitted in the motor is 20A.

10. Using suitable block diagram explain the following controls. (i) Current limit control.
 EE8602 - PROTECTION AND SWITCHGEAR

UNIT I

PROTECTION SCHEMES

PART A

1. How does the over voltage surge affect the power system?
The over voltage of the power system leads to insulation breakdown of the equipments. It
causes the line insulation to flash over and may also damage the nearby transformer, generators
and the other equipment connected to the line.

2. What are symmetrical components?

It is a mathematical tool to resolve unbalanced components into balanced components. The
symmetrical components of three phase system are, i) Positive sequence components. ii)
i)Negative sequence components iii) i)Zero sequence components.

3. Define negative sequence component.

It has three vectors equal in magnitude and displaced from each other by an angle 120
degrees and has the phase sequence in opposite to its original phasors.

4. State the essential qualities of protection.

i) Reliability ii) Selectivity iii)Fastness of operation and iv) Discrimination.

5. Give the consequences of short circuit or What are the effects of short circuit faults in
power system if uncleared? (Nov/Dec 2018)
When a short-circuit occurs, the current in the system increases to an abnormally high
value while the system voltage decreases to a low value. The heavy current due to short-circuit
causes excessive heating which may result in fire or explosion. Sometimes short-circuit takes the
form of an arc and causes considerable damage to the system. If the voltage remains low for even
a few seconds, the consumer’s motors may shut down and generators on the power system may
become unstable.

6. What is the need of relay coordination?

The operation of a relay should be fast and selective, i.e., it should isolate the fault in the shortest
possible time causing minimum disturbance to the system. Also, if a relay fails to operate, there
should be sufficiently quick backup protection so that the rest of the system is protected. By
coordinating relays, faults can always be isolated quickly without serious disturbance to the rest of
the system.

7. Define: energizing quantity.

The electrical quantity i.e., current or voltage either alone or in combination with other
electrical quantities required for the functioning of the relay. The quantity either current or voltage
which is the input to the relay energizes the trip coil of the relay which in turn trips the circuit in
case of faults.

8. What is protected zone? (Apr/May 2015)

 Protected zones are those which are directly protected by a protective system such as relays,
fuses or switchgears. When a fault occurs in a zone, it can be immediately detected and isolated
by a protection scheme which is dedicated to that particular zone. To limit the extent of the fault,
power system protection is arranged in zones. Ideally, the zones of protection should overlap, so
that no part of the power system is left unprotected.

9. What are the various faults that would affect an alternator?

i)Phase to phase faults ii) Phase to earth faults iii) Inter turn faults iv) Earth faults v)Fault
between turns vi) Loss of excitation due to fuel failure vii) Over speed viii) Loss of drive ix)
Vacuum failure resulting in condenser pressure rise, resulting in shattering of the turbine low
pressure casing.

10. State the significance of double line fault.

Double line to ground fault occurs when two lines are short circuited and is in contact with the
ground. This type of fault occurrence ranges from 15 to 25%. It has no zero sequence component
and the positive and negative sequence networks are connected in parallel. Since zero sequence
components are absent there is no circulating current.

11. What is primary protection? (Nov/Dec 2017)

Primary protection is the protection in which the fault occurring in a line will be cleared by its
own relay and circuit breaker. It serves as the first line of defense.

12. What are the different types of earthing ? (Apr/May 2015)

i) Resistive earthing
ii) Reactance earthing
iii) Resonant earthing

13. State the significance of single line to ground fault.

In single line to ground fault all the sequence networks are connected in series. All the
sequence currents are equal and the fault current magnitude is three times its sequence currents.

14. Differentiate between a fuse and a circuit breaker.

Fuse is a low current interrupting device. It is a copper or an aluminum wire. Circuit
breaker is a high current interrupting device and it act as a switch under normal operating
conditions.
15. What is surge absorber? How do they differ from surge diverter? (Nov/Dec 2011)
Surge absorber is a device designed to protect electrical equipment from transient high
voltage to limit the duration and amplitude of the following current. Surge diverter discharges the
impulse surge to the earth and dissipates energy in the form of heat.

16. Define the term “insulation coordination” (Nov/Dec 2011)

 The selection of the insulation strength of equipment in relation to the voltages, which can
appear on the system for which the equipment is intended and taking into account the service
environment and the characteristics of the available protective device.

17. What are the various types of faults occurring in a power system (May/June 2017 )
(Nov/Dec 2017)
Series Fault: a) One open conductor fault b) Two open conductor fault Shunt Fault: (a)
Symmetrical or balanced fault (i) Three phase Fault (LLLG)
(b) Unsymmetrical or unbalanced fault ( i) Line to line fault(LL)(ii) Line to ground fault (LG)(iii)
Double line to ground fault.(LLG).

18. How are arcing grounds avoided? (May/June 2012)

The presence of inductive and capacitive currents in the isolated neutral system leads to
formation of arcs called as arcing grounds. The surge voltage due to arcing ground can be
removed by using the arc suppression coil or Peterson coil. The arc suppression coil has an iron
cored tapped reactor connected in neutral to ground connection. The reactor of the arc suppression
coil extinguishes the arcing ground by neutralizing the capacitive current.

19. What are the effects of power system faults? (Nov/Dec 2012)
Increase in current above rated value, Insulation failure, Equipment damage.

20. What is back up protection? (Nov/Dec 2012)

Back up protection is the second line of defence, which operates if the primary protection
fails to activate within a definite time delay.

21. What is meant by pick-up current? (May/June 2013)(Nov/Dec 2014)

The minimum current at which the relay armature is attracted to close the trip circuit is
called pick-up current. In most of the relays, the pick up current is also indicated with the relay.

22. Write the sources of fault power. (Nov/Dec 2013)

The fault power can be originated from the generation (faults in alternator) or transmission
(short circuit) or from the distribution side (loads). Also the fault power can be from external
sources like lightning.

23. List out the duties of fault limiting reactors. (Nov/Dec 2013)
The duties of fault limiting reactors are to limit the fault current and to eliminate the arcing
ground.

24. What are the functions of protective relays? (May/June 2013) (Apr/May 2015)
To detect the fault and initiate the operation of the circuit breaker and to isolate the
defective element from the rest of the system, thereby protecting the system from damages
occurring due to fault.

25. What is the necessity for earthing? (Nov/Dec 2014) (Nov/Dec 2015)
When earthing is provided it ensures the safety of personnel against electrical shocks and
avoids accidents. The potential of earthed body does not reach to dangerously high value above
earth since it is connected to earth. Also the earth fault current flows through the earthing and may
cause operation of fuse or an earth relay.
 26. What is the difference between short circuit and an
overload?(Nov/Dec2015)(May/June2016)
On the occurrence of short circuit, the voltage at the point of fault falls to zero and the current in
the network increases abnormally to a higher value. But in the case of overload reduction in the
terminal voltage of the equipment occurs but the voltage will never fall to zero. Similarly the
current also increases to a higher value but not as high as in the case of short circuit.

28. What is the difference between primary and back up protection? (May/June 2016)
Primary protection is the protection in which the fault occurring in a line will be cleared by its
own relay and circuit breaker. It serves as the first line of defense. Instantaneous relays are used.
Back up protection is the second line of defense, which operates if the primary protection fails to
activate within a definite time delay. Relays with definite time lag is used.

29. Why earth wire is provided in overhead transmission lines? (May/June 2016)
Earthing wire usually consists of a Low Resistance wire connected to earth or buried into Earth.
It's nothing but a Low Resistance path. Whenever there is a fault or abnormal operation or any
external activities, the current flows through the earth wire and charges are discharged into the
ground. If a fault occurs, current follows through earth wire first and the electrical equipment is
protected.

30. What do you mean by dead spot in zones of protection?

In practice, various protective zones are overlapped. The overlapping of protective zones is
done to ensure complete safety of each and every element of the system. The zone which is
unprotected is called dead spot. The zones are overlapped and hence there is no chance of
existence of a dead spot in a system. If there are no overlaps, then dead spot may exist which
means the circuit breakers lying within the zone may not trip even though the fault occurs. This
may cause damage to the healthy system.

30. State the difference between circuit breaker and switch. (May/June 2017)

Circuit breaker Switch

A mechanical switching device capable of A mechanical switching device capable of
making , carrying and breaking currents under making , carrying and breaking currents under
normal conditions and abnormal conditions like normal conditions but not breaking under
short circuit. abnormal conditions such as short circuit.
It is an automatic device A switch is operated manually.

31. Why protection scheme is required for power system? (April/May 2018)
An electrical power system consists of generators, transformers, transmission lines and
distribution stations etc., Short circuits and other abnormalities often occur in power systems
which cause heavy short circuit currents. The heavy current associated with short circuits will
cause damage to the equipment if suitable protective relays and circuit breakers are not provided.

32. Write down the importance of symmetrical components for fault current calculation.
(April/May 2018)
 The method of symmetrical components is used to simplify fault analysis by converting a
three-phase unbalanced system into two sets of balanced phasors and a set of single-phase
phasors, or symmetrical components. These sets of phasors are called the positive-, negative-, and
zero-sequence components. These components allow for the simple analysis of power systems
under faulted or other unbalanced conditions. Once the system is solved in the symmetrical
component domain, the results can be transformed back to the phase domain.

33. How protective relays are classified based on functions? (Nov/Dec 2018)
The protective relays are classified in the following few categories.
• Directional Over current Relay
• Distance Relay
• Over voltage Relay
• Differential Relay
• Reverse Power Relay

PART B

1. What do you understand by a zone of protection? Discuss various zones of protection.

(Nov/Dec 2015) (April/may 2018) (Nov/Dec 2018)
2. (i) Discuss briefly the role of protective relays in a modern power system.
(ii) Describe the essential qualities of protective relaying system (May/June 2012)
3. Explain the essential qualities of protection and explain them in detail. (May/June 20
4. Briefly explain the various methods of overvoltage protection of overhead transmission line.
5. What is a Peterson coil? Explain the protective function performed by this device with
necessary diagram. (May/June 2016)
6. Discuss the need and compare various methods of neutral earthing? (Nov/Dec 2019)
7. Classify the different types of faults in power system. Which of these are more
frequent? (Nov/Dec 2015) )
8. Explain the overlapping of protective zones with neat sketch. (May/June 2016)
9. (i) Explain in detail about the need and different types of earthing scheme.
(ii) A 132kV, 3 phase, 50 cycles, overhead line, 50km, long has a capacitance to earth
for each line of 0.0157 µF/km. Determine the inductance and kVA rating of the arc
suppression coil. (Nov/Dec 2016)
10. Explain the method of calculating fault current using symmetrical components.
11. i)Explain in detail about different protection schemes. (May/June 2017)
12. Explain arc suppression coil earthing with neat diagram. (May/June 2017)
13. Explain the various methods of neutral grounding. (Nov/ Dec 2017)
14. Briefly discuss the nature of occurrence and types of fault in the power system.
 UNIT II

ELECTROMAGNETIC RELAYS

PART A

1. Name the different kinds of over current relays.

Induction type non-directional over current relay, Induction type directional over current
relay & current differential relay.

2. Define operating time of a relay.

It is the time which elapses from the instant at which actuating quantity exceeds the relay
pick up value to the instant at which the relay closes its contacts.

3. Define resetting time of a relay.

It is the time which elapses from the moment the actuating quantity falls below its reset
value to the instant when the relay comes back to its normal (initial) position.

4. What is ‘Time grading’ of relays. (Nov/Dec 2018)

It is the setting of time of operation of various relays protecting the different sections of a
line. It is set so that the relay which is nearest to the fault location alone will operate first and
clear the fault.

5. What are over and Under current relays?

Over current relays are those that operate when the current in a line exceeds a
predetermined value. (e.g.: Induction type non-directional/directional over current relay,
differential over current relay) whereas Undercurrent relays are those which operate whenever
the current in a circuit/line drops below a predetermined value.(e.g.: differential over-voltage
relay).

6. What is biased differential beam relay?

The biased beam relay is designed to respond to the differential current in terms of its
fractional relation to the current flowing through the protected zone. It is essentially an over-
current balanced beam relay type with an additional restraining coil. The restraining coil
produces a bias force in the opposite direction to the operating force.

7. Give the limitations of Merz Price protection.

Since neutral earthing resistances are often used to protect circuit from earth-fault currents,
it becomes impossible to protect the whole of a star-connected alternator. If an earth-fault
occurs near the neutral point, the voltage may be insufficient to operate the relay. Also it is
extremely difficult to find two identical CT’s. In addition to this, there always an inherent phase
 difference between the primary and the secondary quantities and a possibility of current through
the relay even when there is no fault.

8. Why is an under frequency relay required in a power system?(May/June 2012)

(Nov/Dec 2013) (Nov/Dec2014)
An under frequency relay is one which operates when the frequency of the system (usually
an alternator or transformer) falls below a certain value. Under frequency relays are used to
shed automatically certain portion of load whenever the system frequency falls to such a low
level which threatens the stability of the power system.

9. What are the features of directional relay?

High speed operation; high sensitivity; ability to operate at low voltages; adequate short-
time thermal ratio; burden must not be excessive.

10. What is static relay?

It is a relay in which measurement or comparison of electrical quantities is made in a static
network which is designed to give an output signal when a threshold condition is passed which
operates a tripping device.

11. What is a programmable relay?

A static relay which has one or more programmable units such as microprocessors or
microcomputers embedded in its circuit is called a programmable relay.

12. What are the advantages of static relay over electromagnetic relay? (Nov/Dec 2011)
(May/June 2014) (Nov/Dec 2014)
i) Low power consumption as low as 1mW ii) No moving contacts; hence associated
problems of arcing, contact bounce, erosion, replacement of contacts iii) No gravity effect on
operation of static relays. Hence can be used in vessels ie, ships, aircrafts etc.
iv) A single relay can perform several functions like over current, under voltage, single phasing
protection by incorporating respective functional blocks. This is not possible in electromagnetic
relays v) Static relay is compact.

13. What are the different types of over current relays

i) Definite time
ii)Inverse definite minimum time(IDMT)
iii)Very Inverse iv)Extremely Inverse.

14. What is earth fault protection?

A ground fault (earth fault) is any failure that allows unintended connection of power circuit
conductors with the earth. Such faults can cause objectionable circulating currents, or may
energize the housings of equipment at a dangerous voltage. Under such condition residual current
flowing to the ground is calculated. Such a protective scheme used for the protection of an
element of a power system against earth faults is called as earth fault protection.

15. List out the applications of static relays. (Nov/Dec 2012) ) (May/June 2016)
i) Protection of generators
ii) Protection of transformers
iii) Protection of transmission lines, and
 iv) Protection of motors.

16. What is meant by directional relay? (May/June 2012)

A directional relay detects the whether the point of fault lies in the forward or reverse
direction with respect to relay location. The relay which is able to sense the direction of power
flow and act for a particular direction of power flow is called directional relay.

17. What is meant by differential relay? (May/June 2013) (Apr/May 2015)

A differential relay is one that operates when the phasor difference of two or more
similar electrical quantities exceeds a predetermined value. It has two coils viz., operating coil
which produces operating torque and restraining coil which produces restraining torque.

18. What are the types of fuses? (Nov/Dec 2013)

a) Low voltage fuses i) Semi-enclosed rewireable fuse ii)HRC fuse
b) High voltage fuses i) cartridge type ii)liquid type iii)metal clad type.

19. List out the different types of distance relay.(May/June 2014)

Dependent on the ratio of V and I there are three types of distance relays which are
i) Impedance relay which is based on measurement of impedance Z ii) Reactance relay which is
based on measurement of reactance X iii)Admittance or Mho relay which is based on
measurement of component of admittance Y.

20. In what way distance relay is superior to over current protection? (Nov/Dec 2015)
Distance relays are preferred to overcurrent relays because they are not nearly so much
affected by changes in short-circuit-current magnitude as overcurrent relays are, and, hence, are
much less affected by changes in generating capacity and in system configuration. This is
because distance relays achieve selectivity on the basis of impedance rather than current.

21. Where are negative sequence relays employed?

Negative sequence relays are employed for negative sequence protection of generators
against the unbalanced load condition. The negative phase sequence filter along with the over
current relay provides the necessary protection against the unbalanced loads.
22. Write the effects of arc resistance.
The effect of arc resistance is most significant on short lines where the reach of the relay
setting is small. It can be a problem if the fault occurs near the end of the reach.
High fault-arc resistances tend to occur during midspan flashovers to ground on transmission
lines carried on wood poles without earth wires. These problems can usually be overcome by
using relays having different shaped characteristics.

23. What is the significance of PSM and TSM? (Nov/Dec 2016)

Time setting multiplier TSM: TSM determines the operating time of the relay. Lower the
value of TSM, lower will be the operating time.
Plug setting multiplier PSM: The plug position ensures the current setting value of the relay.
Plug setting multiplier (PSM) indicates the severity of the fault.
 24. A relay is connected to 400/5 ratio current transformer with current setting of 150%.
Calculate the plug setting multiplier when circuit carries a fault current of 4000A.
(Nov/Dec 2016)
Fault Current = 4000A
C.T. ratio = 400/5
Fault current in the relay coil = 4000 * (5/400) = 50A
Plug Setting Multiplier (PSM) = Fault Current in the relay coil / (Rated secondary
C.T. Current * Current setting)
Plug Setting Multiplier (PSM) = 50 / (5*1.5) = 6.667

25. Why shaded ring is provided in induction disc relay? (May/June 2017)
In the induction disc relay, a metal disc is allowed to rotate between two electromagnets.
The shaded pole structure is generally actuated by current flowing in a single coil on a magnetic
structure containing an air gap. The air gap flux produced by this current is split into two out-of-
phase components by a so called “shading ring” generally of copper, that encircles part of the
pole face of each pole at the air gap.

26. Give the principle of negative sequence relay. (Nov/Dec 2017)

A relay which protects the electrical system from negative sequence component is called a
negative sequence relay or unbalance phase relay. The actuating quantity is negative sequence
current. When the negative sequence current exceeds a certain value, the relay operates. This is
used to protect electrical machines against overheating due to unbalanced currents.

27. Write the torque equation of the universal relay. (Nov/Dec 2017)

where K1, K2, K3 are the tap setting or constant of Voltage V and current I. The K4 is the
mechanical restraint due to spring or gravity.

28. Mention the principle of operation of distance relay. (April/may 2018)

There is one voltage element from PT and a current element fed from CT of the system.
The deflecting torque is produced by secondary current of CT and restoring torque is produced
by voltage of potential transformer. In normal operating condition, restoring torque is more than
deflecting torque. Hence relay will not operate. But in faulty condition, the current becomes
quite large whereas voltage becomes less. Consequently, deflecting torque becomes more than
restoring torque and dynamic parts of the relay starts moving which ultimately close the No
contact of relay. Hence clearly operation or working principle of distance relay depends upon
the ratio of system voltage and current.

29. Determine the plug setting multiplier of a 5 ampere, 3 second over current relay
having a current setting of 125% and a time setting multiplier of 0.6 connected to
supply circuit through a 400/5 current transformer when the circuit carries a fault
current of 4000A. (April/may 2018)
Plug Setting Multiplier = Fault current in relay coil/(Rated CT secondary current
* Current Setting)
Fault current in relay coil = 4000*(5/400) = 50A. Therefore, PSM= 50/(5*1.25)=8
30. What are the factors affecting the performance of differential relays? (Nov/Dec 2018)
 Phasor sum of currents
 CT ratio
 Polarity of transformers

PART- B

1. Describe the construction and operation of over current relay with directional Scheme.
(June 2014) (Nov/Dec 2015). (May/June 2016)
2. i) Discuss the operating principle, constructional features area of applications of
directional relay. How do you implement directional feature in the over current relay.
3. (i) What are the different inverse-time characteristics of over current relays and mention
how to characteristics can be achieved in practice for an electromagnetic relay? (Nov/Dec
2018)
4. Explain the principle of percentage biased differential relay with necessary diagrams. Also
discuss its application. (May/June 2012)
5. Describe the Principe, of operation of various differential relays with neat sketches.
(Nov/Dec 2019).
6. i) With neat block diagram, explain the construction and operating principle of
electromagnetic relay. ii) Describe the operation of over current relay with directional
feature.(Nov/Dec 2013) (Nov/Dec 2018)
7. Describe the operating principles and characteristics of impedance and mho relays.
(Nov/Dec 2013)
8. Explain the operation of i) Negative sequence relay ii) Static relay. iii) under frequency
relay (Apr/May 2015) (April/may 2018)
9. Explain in what way distance protection is superior to over current protection for the
protection of transmission lines. (Nov/Dec 2013,2014) (May/June 2014)
10. i) Explain with the help of neat diagram the construction and working of induction type
directional power relay. (Nov/Dec 2015).
11. What is universal torque equation? Using this equation derive the following operating
characteristics. i) Impedance relay ii) Reactance relay iii) Mho relay. (May/June 2013)
(Nov/Dec 2015)(May/June 2016)
12. Draw and explain about differential protection of transmission lines. (Apr/May 2015)
(April/may 2018)
13. Explain the construction and operating principle of impedance type distance relay
with R-X diagram. (May/June 2017) (April/may 2018)
14. With necessary sketches discuss in detail about electromagnetic attraction type relays.
(May/June 2017) (Nov/ Dec 2017)
15. Describe the construction and principle of operation of non-directional induction
type over current relay. (Nov/ Dec 2017) (Nov /Dec 2019)

16. Explain MHO relay characteristic on the R- X diagram. Discuss the range setting of
various distance relays placed on a particular location. (May/June 2016)
 UNIT III

APPARATUS PROTECTION

PART A

1. What are the causes of over speed and how alternators are protected from it? (April/may
2018)
Sudden loss of all or major part of the load causes over-speeding in alternators. Modern
alternators are provided with mechanical centrifugal devices mounted on their driving shafts to trip the
main valve of the prime mover when a dangerous over- speed occurs.

2. What are the uses of Buchholz’s relay?

Buccholz relay is used to give an alarm in case of incipient (slow-developing) faults in the
transformer and to disconnect the transformer from the supply in the event of severe internal faults. It is
usually used in oil immersion transformers with a rating over 750KVA.

3. What are the various faults that would affect an alternator? (May/June 2013) (Apr/May
2015) )(May/June 2016)
(a) Stator faults i) Phase to phase faults ii) Phase to earth faults iii) Stator inter turn faults (b)
Rotor faults i)Rotor earth faults ii)Field over loading iii) Heating of rotor c)Abnormal Running
Conditions i) Over speeding ii) Over loading iii) Unbalanced Loading iv)Over voltage v)Failure of
Prime mover.

4. What are faults associated with a transformer?

a) Overheating b) Winding Faults i)phase to phase fault ii) Earth fault iii) Interturn faults
c)Open circuits d)Through faults e)Over fluxing.

5. What are the main safety devices available with transformer? (May/June 2012)
Oil level gauge, sudden pressure delay, oil temperature indicator, winding temperature
indicator.

6. What are the limitations of Buchholz relay? (May/June 2017)

(a) Only fault below the oil level are detected.
(b) Mercury switch setting should be very accurate, otherwise even for vibration; there can be a
false operation.
(c) The relay is of slow operating type, which is unsatisfactory.

7. What are the problems arising in differential protection in power transformer and how are
they overcome? (May/June 2012) (Nov/Dec 2015)
i) Difference in lengths of pilot wires on either sides of the relay. This is overcome by
connecting adjustable resistors to pilot wires to get equipotential points on the pilot wires. ii)
Difference in CT ratio error difference at high values of short circuit currents that makes the relay to
operate even for external or through faults. This is overcome by introducing bias coil.iii) Tap changing
alters the ratio of voltage and currents between HV and LV sides and the relay will sense this and act.
Bias coil will solve this. iv) Magnetizing inrush current will be identified as short circuit current. A
harmonic restraining unit is added to the relay which will block it when the transformer is energized.

8. What is REF relay?

It is Restricted Earth Fault relay. When the fault occurs very near to the neutral point of the
transformer, the voltage available to drive the earth circuit is very small, which may not be sufficient to
activate the relay, unless the relay is set for a very low current. Hence the zone of protection in the
winding of the transformer is restricted to cover only around 85%. Hence the relay is called REF relay.

3
9. What is over fluxing protection in transformer? (Nov/Dec 2016)
If the turn’s ratio of the transformer is more than 1:1, there will be higher core loss and the
capability of the transformer to withstand this is limited to a few minutes only. This phenomenon is
called over fluxing.

10. Why bus-bar protection is needed? (May/June 2013)

(i) Fault level at bus-bar is high (ii) The stability of the system is affected by the faults in the bus
zone.(iii) A fault in the bus bar causes interruption of supply to a large portion of the system network.

11. What are the causes of bus zone faults?

i) Failure of support insulator resulting in earth fault ii) Flashover across support
insulator during over voltage iii ) Heavily polluted insulator causing flashover iv) Earthquake,
mechanical damage etc.

12. What are the problems in bus zone differential protection?

i)Large number of circuits, different current levels for different circuits for external faults ii)
Saturation of CT cores due to dc component and ac component in short circuit currents. The saturation
introduces ratio error.iii) Sectionalizing of the bus makes circuit complicated. iv) Setting of relays need
a change with large load changes.

13. What are the disadvantages of time graded protection?

i) Time lag is not desirable on short circuits ii) Not suitable for ring main distribution
iii) Difficult to coordinate & needs changes with new connection iv) Not suitable for long distance
relaying.

14. How does the over voltage surge affect the power system?
The over voltage of the power system leads to insulation breakdown of the equipments. It
causes the line insulation to flash over and may also damage the nearby transformer, generators and the
other equipment connected to the line.

15. What is the general connection rule for Current transformers in differential protection?
If the windings of the power transformer are delta connected then the current transformers are
star connected and if the windings of the power transformer are star connected then the current
transformers are delta connected.

16. Write the coordination equation for inverse over-current relay?

TA=TB+CBB+OA+F
Where TA operating time of relay at station A,TB operating time of relay at station B,CBB
operating time of circuit breaker at station B,OA over travel time of relay at station A,F factor of safety.

17. Explain why secondary of current transformer should not be open. (Nov/Dec 2011)(Dec 2014)
(Apr/May 2015) (May/June 2016)
Current transformers generally work at a low flux density. Core is then made of very good
metal to give small magnetizing current. On open-circuit, secondary impedance now becomes infinite
and the core saturates. This induces a very high voltage in the primary upto approximately system volts
and the corresponding volts in the secondary will depend on the number of turns. Since secondary of
CT has more turns compared to the primary, the voltage generated on the open-circuited CT will be
high, leading to flashovers. Hence as a safety precaution, CT secondary should not be open-circuited.

18. What is meant by time graded system protection? (Nov/Dec 2018)

In a time graded system, the operating time of the relay is increased from the far end of
protected circuit towards the generating source. Definite time over current relays are used which after a
preset time will trip the circuit. The difference in time setting of the two adjacent relays are kept at 0.5s.
This difference is to cover the operating time of CB and errors in CT and relay.
3
19. Write the function of earth fault relay. (Nov/Dec 2012)
Earth fault relay is used for the protection of an element of a power system against earth faults.
Earth relay calculates the residual current. If the residual current is zero the relay will not operate.
Restricted earth fault relay is used in differential protection which will not operate for external faults.

20. What is meant by relay operating time? (Nov/Dec 2012)

It is the time which elapses from the instant at which actuating quantity exceeds the relay pick
up value to the instant at which the relay closes its contacts.

21. What are the different types of zones of protection? (Nov/Dec 2013)
i) Primary protection and ii) backup protection and Unit and Non-Unit protection.

22. State the methods of protection of busbars. (Nov/Dec 2014) (Nov/Dec 2016)
i) Frame leakage protection of bus bar ii) Circulating current protection of bus bar
iii) High impedance differential protection of bus bar

23. List the applications of current transformer. (May/June 2014)

i) To the supply the stepped down current to the relay coil in the event of any
overloading or short-circuiting of the equipment lines.
ii) To measure power of a load in conjunction with a wattmeter. The secondary of the CT is
connected to the current coil of the wattmeter.
iii) To measure large currents in conjunction with medium/Small range meters.

24. Give examples of Unit and Non – Unit Protection Schemes (Nov/Dec 2015)
The concept of 'Unit Protection', whereby sections of the power system are protected
individually as a complete unit without reference to other sections.
eg. Differential Protection, Overcurrent Protection. eg. Non – Unit Protection: Distance Protection.

25. What are the difficulties encountered through differential protection? (May/June
2017)
Though the saturation in Current transformer is avoided, there exist difference in the
C.T. characteristics due to ratio error at high values of short circuit currents. This causes an appreciable
difference in the secondary currents which can operate the relay. So the relay operates for external
faults. Due to the difference in lengths of the pilot wires on both sides, the unbalance condition may
result. Due to the magnetizing current inrush current in transformers which may be as great as 10 times
the full load current of the transformer, the differential relay may operate falsely.

26. What is the need of instrumentation transformer? (May/June 2017)

Instrument transformers are high accuracy electrical devices used to isolate or transform voltage
or current levels. The most common usage of instrument transformers is to measure high voltage or
high current (with common meters) by safely isolating secondary control circuitry from the high
voltages or currents.

27. Why secondary of a transformer should not be opened? (Nov/Dec 2017)

The secondary side of a current transformer should never be kept in open condition because,
when kept open, there is a very high voltage found across the secondary side. This high voltage causes
a high magnetizing current to build up on the secondary side which in turn causes high flux and makes
the core to saturate.

28. List the types of bus bar protection. (Nov/Dec 2017)

Frame-Earth Protection
Differential Protection for Sectionalized Bus bars High-Impedance Differential Protection Low-
Impedance Differential
Protection Digital Bus bar Protection

29. What are the protection methods used for transmission lines?(April/may 2018)
Over current protection; Simple Impedance Relay; Mho relay; Reactance relay
3
30. In the event of fault in generator windings, the field excitation should be suppressed as early
as possible. Why? (Nov/Dec 2018)
Failure of excitation that is failure of field system in the generator makes the generator run at a
speed above the synchronous speed. In that situation the generator or alternator becomes an IG which
draws magnetizing current from the system. Although this situation does not create any problem in the
system immediately but over loading of the stator and overheating of the rotor due to continuous
operation of the machine in this mode may create problems in the system in long-run. Therefore special
care should be taken for rectifying the field or excitation system of the generator immediately after
failure of that system. The generator should be isolated from rest of the system till the field system is
properly restored.

31. Which type of protection is used for EHV and UHV lines? (Nov/Dec 2018)
Carrier current protective scheme Pilot wire protective scheme

PART- B

1. Discuss how the generator is protected against an inter turn fault with necessary diagram.
2. Explain what is meant by distance protection and why it is superior to other types of protection
for an overhead transmission line.

3. i) Describe the construction and working of Buchholz relay. (April/may 2018)

(ii) Discuss the time graded over current protection for parallel feeders.

4. i) Explain with the neat diagram the application of Merz-price circulating current principle for
protection of alternator. (Apr/May 2015)
ii) What is the role of instrument transformer in protective schemes? (May 2013)

5. i) Describe the differential protective schemes of transformer (May/June2014)

(ii) Enumerate the protective scheme employed for the bus bar. (Nov/Dec 2013)

6. Briefly discuss the protective devices used for the protection of a large transformer.
7. Explain impedance relay characteristics on the R-X diagram. Also discuss the range setting of
three impedance relays placed at a particular location. (Nov/Dec 2014)
8. Explain about carrier aided protection of transmission lines and various relays associated with it
(Apr/May 2015) (Nov/Dec 2015)
9. Why is harmonic restrained differential relay used for protecting large size transformer?
Describe its working and construction. (Nov/Dec 2015) (Nov/Dec 2018)
10. With neat sketches, explain the different types of protective schemes for transmission lines.
(May/June 2016)
11. Draw and explain protection scheme of an A.C. induction motor. (Nov/Dec 2016)
12. (i) A generator is protected by restricted earth fault protection. The generator ratings are
13.2kV, 10 MVA. The percentage of winding protected against phase to ground is 85%.
The relay setting is such that it trips for 20% out of balance. Calculate the resistance to be
added in the neutral to ground connection.
(i) Explain a protection scheme for protection of transformer against incipient fault.
13. Give a detailed explanation for protection of transformer using differential protection which
includes associate faults. (May/June 2014) (Nov/ Dec 2017)
14. Give a detailed explanation about CT’s and PT’s and its application to power systems.
15. Give a brief account on the protection of generator using differential and biased differential
protection scheme. (Nov/ Dec 2017)
16. With neat sketches, explain the different types of protective schemes for motors.
17. A star connected, 3 phase , 10 MVA, 6.6 kV alternator has a per phase reactance of 10%. It is
protected by Merz-price circulating current principle which is set to operate for fault
currents not less than 175A. Calculate the value of earthing resistance to be provided in
order to ensure that only 10% of the alternator winding remains unprotected. (April/may
2018) 3
 18. An alternator rated at 10kV protected by the balanced circulating current system has its neutral
grounded through a resistance of 10ohms protective relay is set to operate when there is an
out of balance of 1.8A in the pilot wires which are connected to the secondary windings of
1000/5 CT ratio. Determine the percentage of windings which remain unprotected and
minimum value of earthing resistance to protect 80% of the winding. (Nov/Dec 2018)
19. Identify and explain the different protection schemes necessary for the protection of 3 phase
alternators with suitable circuit diagram. (Nov/ Dec 2012) (Apr/May 2015)
20. i) Explain the factors which cause difficulty in applying Merz-Price circulating current
principle to a power transformer.
(ii) A three phase transformer of 220/11000 line volts is connected in star/delta. The
protective transformers on 220 V side have a current ratio of 600/5. What should be the current
transformer ratio on 11000 V side? (Nov/Dec 2011)
21. i) With aid of neat schematic diagram, describe the percentage differential
Protection scheme of a transformer and generator. (Nov/Dec 2014).
(ii) Describe the differential pilot wire method of protection of feeder.

UNIT IV

STATIC RELAYS AND NUMERICAL PROTECTION

PART-A

1. Define static relay? (Nov/Dec 2017)

It is a relay in which measurement or comparison of electrical quantities is made in a
static network which is designed to give an output signal when a threshold condition is passed which
operates a tripping device.

2. What is CPMC?
It is combined protection, monitoring and control system incorporated in the static system. For
example static relays employ microprocessor units which incorporate protection principles such as over
current, inverse time etc., in their operation. Also these units sense the fault current each and every
time. The fault current can also be controlled by changing the code embedded into the processor.

3. What are the advantages of static relay over electromagnetic relay? (Nov/Dec 2018)
 Low power consumption as low as 1mW
 No moving contacts; hence associated problems of arcing, contact bounce, erosion,
replacement of contacts
 No gravity effect on operation of static relays. Hence can be used in vessels ie, ships,
aircrafts etc.
 A single relay can perform several functions like over current, under voltage, single phasing
protection by incorporating respective functional blocks. This is not possible in
electromagnetic relays
 Static relay is compact. Superior operating characteristics and accuracy
 Static relay can think , programmable operation is possible with static relay
 Effect of vibration is nil, hence can be used in earthquake-prone areas
 Simplified testing and servicing. Can convert even non-electrical quantities to electrical in
conjunction with transducers.

4. What is pick up value?

It is the minimum current in the relay coil at which the relay starts to operate. The relay should
not operate when the current does not exceed the pick up value.

3
5. Define target.
It is the indicator used for showing the operation of the relay. This helps the operator to know
the cause of tripping of the circuit breaker.

6. Define blocking.
Blocking is preventing the relay from tripping due to its own characteristics or due to additional
relays. False operation of relay may lead to unnecessary opening of circuit.

7. What are the advantages of numerical relays over conventional relays? (May/June 2014)
(May/June 2015) (Nov/Dec 2016)
No moving parts and therefore no friction Easy to replace and service. Numeric relays are not
affected by gravity Are compact and has modular arrangement Various characteristics can be obtained.

8. What are the drawbacks of analogue and active analogue filters? (May/June 2014)
They are bulky, especially inductors require large space; High precision components are needed
making them expensive; Their characteristics drift with respect to time and temperature; Filters for very
low frequencies need impracticably high component values; They are not programmable and adaptable.

9. Draw the block diagram of FIR and IIR filter

10. Compare FIR and IIR filters

S.No FIR Filter IIR Filter

1. Difficult to control and have no Always make a linear phase.
particular phase

2. FIR is always stable IIR can be unstable

3. FIR has no limited cycles Can have limited cycles
4. FIR has no analog history IIR is derived from analog.
5. FIR can always be made casual IIR filters make polyphase implementation
possible
6. FIR filters are Finite IR filters which IIR is infinite and used for applications
are required for linear- wherelinear characteristics are not of
phase characteristics. concern.

11. What is Fourier analysis?

The analysis of a complex waveform expressed as a series of sinusoidal functions, the frequencies of
which form a harmonic series. If a function is periodic, then it can be written as a discrete sum of
trigonometric or exponential functions with specific frequencies.

12. What is discrete Fourier transform?

As the name implies, the Discrete Fourier Transform (DFT) is purely discrete: discrete-time
data sets are converted into a discrete-frequency representation. This is in contrast to the DTFT that
4
uses discrete time, but converts to continuous frequency. Since the resulting frequency information is
discrete in nature, it is very common for computers to use DFT calculations when frequency
information is needed.

12. What is Aliasing?

Aliasing is a phenomenon where the high frequency components of the sampled signal interfere
with each other because of inadequate sampling. It results in loss of signal and its place will be taken
by a different lower frequency wave.

13. What is sampling?

Sampling is the process of converting a signal (for example, a function of continuous time or
space) into a numeric sequence (a function of discrete time or space).

14. What is sample and hold circuit?

A sample and hold circuit is an analog device that samples (captures, grabs) the voltage of a
continuously varying analog signal and holds (locks, freezes) its value at a constant level for a specified
minimum period of time. Sample and hold circuits and related peak detectors are the elementary analog
memory devices. They are typically used in analog-to-digital converters to eliminate variations in input
signal that can corrupt the conversion process.

15. What is digital filter?

In signal processing, a digital filter is a system that performs mathematical operations on a
sampled, discrete-time signal to reduce or enhance certain aspects of that signal.

16. Draw the block diagram of static differential relay

17. Draw the block diagram of numerical relay

4
18. Draw the block diagram of static directional relay

19. Draw the block diagram of static over current relay

20. Draw the block diagram of static relay

21. What are the building blocks of static relay? or What are the basic circuits in static relay?
(April/may 2018)

Rectifier, Comparator, Amplifier, Transducer are some of the building blocks of static relay.

22. What is least error squared technique? (May/June 2015)

The least error squared technique is directly related to the Fourier technique. If a given function
were to be synthesized by using a dc component, a sine wave of fundamental frequency and harmonics
of this fundamental, then the amplitudes of various components given by the Fourier analysis are the
ones which give the least squared error. We can directly find out the amplitudes of the components by
using the LES technique.

23. List out the applications of static relays. (Nov/Dec 2016) (May/June 2016)
 Ultra high speed protection of EHV AC transmission lines utilizing distance protection.
 In over current and earth fault protection schemes
 As main element in differential relay

24. State Nyquist–Shannon sampling theorem (May/June 2017)

If a function x(t) contains no frequencies higher than B hertz, it is completely determined by
giving its ordinates at a series of points spaced 1/(2B) seconds apart. A band limited signal can be

4
reconstructed exactly if it is sampled at a rate atleast twice the maximum frequency component in it.
fs≥2fm where fs = sampling frequency;
fm = frequency of the signal which is reconstructed.

25. Write about numerical transformer differential protection. (May/June 2017)

It provides fast and selective tripping for two winding transformer. It quickly discriminates
between faults that occur in the protected zone and those occurring outside this zone and thus provides
selective and fast tripping. The faults within protected zone are short circuit between turns, windings
and cables and earth faults inside transformer housing and protected zone. It discriminates between
above internal faults and the operational conditions like inrush, over-fluxing and faults external to
protected zone using numerical algorithms.

26. What is phase comparator? (Nov/Dec 2017)

A phase detector or phase comparator is a frequency mixer, analog multiplier or logic circuit
that generates a voltage signal which represents the difference in phase between two signal inputs. It is
an essential element of the phase-locked loop (PLL).

27. List out the general characteristics of numerical protection. (April/may 2018)
 The numerical relay relies on one system for all approach and use indication
on LCD for relay activation, ensuring less space.
 Since the numerical relay system relies on software, customized modifications can be made for
getting the desired protection features. This saves the cost of replacing hardware. Fewer
interconnections ensure reliability.
 The range of operation of traditional models is narrow while numerical relays are diverse and
evolution adaptable.
 It also has the feature of auto resetting and self-diagnosis.
 The benefit of using microprocessor based relays in the numerical system is that it gives
minimum burden on the instrument transformers. The sensitivity of the system is pretty nifty
and boasts a high pickup ratio.

PART B

1. Explain in detail the numerical over current protection of transmission line. Derive the
necessary equations. (May/June 2015)
2. Explain with a neat diagram the numerical transformer differential protection scheme.
3. (May/June 2015)
4. How will you synthesize a mho relay using static phase comparator? (Nov. 16)
5. Explain the numerical over current protection and numerical transformer differential protection.
(Nov/Dec 2016)
6. With a neat sketch discuss in detail about the synthesis of reactance relay using phase
comparator. (May/June 2017)
7. Explain with neat block diagram the operation static relay and list its advantages and
disadvantages. (Nov/ Dec 2017)
8. Describe the operation of static instantaneous over current relay with neat diagram. (Nov/ Dec
2019)
9. (a) Compare static relays with numerical relays. (b) Explain the advantages of
numerical relays. (April/may 2018)
10. Describe the construction, working principle and operation of static over current relay.
11. Discuss in detail, integrating and instantaneous type static amplitude comparators.
12. Illustrate your answer with appropriate circuits and waveforms. (Nov/Dec 2018)
13. How static over current relays are different from electromechanical relays?
14. Explain how the operation of instantaneous relay is achieved using electronic relays?

4
 UNIT V

CIRCUIT BREAKERS

PART A

1. What is dielectric test of a circuit breaker?

It consists of over voltage withstand test of power frequency lightning and impulse voltages.
Tests are done for both internal and external insulation with switch in both open and closed conditions.

2. Define composite testing of a circuit breaker.

In this method the breaker is first tested for its rated breaking capacity at a reduced voltage and
afterwards for rated voltage at a low current. It is the combination of both field type testing station and
laboratory type testing station. This method does not give a proper estimate of the breaker performance.

3. What is making capacity? (Nov/Dec 2015)

It is the capacity of the circuit breaker to be closed onto a short circuit.
It is expressed as 1.414 X 1.8 X symmetrical breaking capacity = 2.55 X symmetrical breaking
capacity.

4. What are the advantages of synthetic testing methods?

i) The breaker can be tested for desired transient recovery voltage and RRRV.
ii) Both test current and test voltage can be independently varied. This gives flexibility to the
test
iii) The method is simple iv) With this method a breaker capacity (MVA) of five times of that
of the capacity of the test plant can be tested.

5. Write are the types of test conducted on circuit breakers. (May/June 2012) (Apr/May
2015)
• Type test ii) Routine test iii) Reliability test iv) Commissioning test
Type test can be classified into mechanical performance test, thermal test, dielectric test and
short circuit tests.

6. What are the characteristic of SF6 gas?

It has good dielectric strength and excellent arc quenching property. It is inert, non- toxic, non
inflammable and heavy. At atmospheric pressure, its dielectric strength is 2.5 times that of air.
At three times atmospheric pressure, its dielectric strength is equal to that of the transformer oil.

7. Give the advantage of SF6 circuit breaker over air blast circuit breaker (May/June 2013)
(Apr/May 2015) (May/June 2016)
High electro negativity, compactness, reduced switching over voltages, reduced insulation
time, superior arc interruption and increased safety.

8. What is meant by electro negativity of SF6 gas?

SF6 has high affinity for electrons. When a free electron comes and collides with a neutral
gas molecule, the electron is absorbed by the neutral gas molecule and negative ion is formed.
This is called as electro negativity of SF6 gas.

9. What are the demerits of using oil as an arc quenching medium?

i) The air has relatively inferior arc quenching properties
ii) The air blast circuit breakers are very sensitive to variations in the rate of rise of restriking
voltage
iii) Maintenance is required for the compression plant which supplies the air blast
10. What are the advantages of air blast circuit breaker over oil circuit breaker?
4
 i) The risk of fire is diminished.
ii) The arcing time is very small due to rapid buildup of dielectric strength between contacts.
iii) The arcing products are completely removed by the blast whereas oil deteriorates with
successive operations.

11. What are the types of air blast circuit breaker?

i)Arial-blast type ii) Cross blast iii) Radial-blast

12. What are the disadvantages of MOCB over a bulk oil circuit breaker?
i) The degree of carbonization is increased due to smaller quantity of oil
ii) There is difficulty of removing the gases from the contact space in time
iii) The dielectric strength of the oil deteriorates rapidly due to high degree of carbonization.

13. What are the advantages of MOCB over a bulk oil circuit breaker?
i) It requires lesser quantity of oil
ii) It requires smaller space
iii) There is a reduced risk of fire
iv) Maintenance problem are reduced.

14. What are the advantages of oil as arc quenching medium?

 It absorbs the arc energy to decompose the oil into gases, which have excellent cooling
properties
• It acts as an insulator and permits smaller clearance between line conductors and earthed
components
15. What are demerits of MOCB?

i) Short contact life

ii) Frequent maintenance
iii) Possibility of explosion
iv) Larger arcing time for small currents
v) Prone to restricts

16. Mention different types of circuit breakers? (May/June 2012)

i) Air break circuit breaker ii) Oil circuit breaker iii) Minimum oil circuit breaker
iv) Air blast circuit breaker v) SF6 circuit breaker vi)Vacuum circuit breaker
17. What are the different types of oil circuit breakers?
i) Plain break oil circuit breakers
ii) Arc control circuit breakers
iii) Minimum oil circuit breakers
18. What are the advantages of using vacuum as an arc interrupting medium?
Vacuum offers the utmost insulating strength. Interruption occurs in the first current
zero. So it has superior arc quenching properties than any other medium. Also the dielectric
strength of vacuum is superior to those of porcelain, oil, air and SF6.
19. Write any two properties of contact material used in vacuum circuit breaker?
i) Good electrical conductivity to pass normal load currents without overheating.
ii) Good thermal conductivity to dissipate rapidly the large heat generated during arcing.

20. What are the basic requirements of circuit breaker? (Nov/Dec 2011)
i)To make or break a circuit either manually or by remote control under normal conditions ii)
Break a circuit automatically under fault condition iii) Make a circuit automatically either manually
or by remote control after the fault is cleared.

21. Write the difference between fuse and circuit breaker. (Nov/Dec 2012)
Fuse is a low current interrupting device. It is a copper or an aluminum wire. Circuit breaker is
a high current interrupting device and it act as a switch under normal operating conditions.
4
22. Enumerate breaking capacity of circuit breaker. (Nov/Dec 2012) (Nov/Dec 2014)
The capacity of the circuit breaker which can break under specified conditions of recovery
voltage. The breaking capacity of a circuit breaker is expressed in MVA and given as 1.732 X (rated
voltage in kV) X (rated current in kA).

23. Write the ratings of the circuit breaker. (Nov/Dec 2013)

Circuit breaker has three ratings. i) Breaking capacity ii) Making capacity and iii) Short
time capacity.

24. Define the opening time of circuit breaker.(May/June 2014)

The time interval which is passed in between the energization of the trip coil to the instant
of contact separation is caused the opening times. It is dependent on fault current level.

25. What is meant by current chopping? (Nov/Dec 2014)

At the time of interruption of a large fault current, the arc energy is high enough to keep
ionized until the arc is interrupted at natural current zero. On the other hand, while interrupting
small inductive currents such as unloaded currents of transformers and currents of shunt reactor,
there is a possibility of overvoltage depending on the value of the chopping current. This small
inductive current is interrupted just before natural current zero and thus induces high transient
voltages, which is known as current chopping.

26. What is resistance switching? (Nov/Dec 2013)

It is the method of connecting a resistance in parallel with the contact space (arc). The inserted
resistance reduces the re striking voltage frequency and it diverts part of the arc current. It
assists the circuit breaker in interrupting the magnetizing current and capacity current.

27. What is an arc?

Arc is a phenomenon occurring when the two contacts of a circuit breaker separate under
heavy load or fault or short circuit condition.

28. Give the two methods of arc interruption? (May/June 2012) (Apr/May 2015)
i) High resistance interruption:-the arc resistance is increased by elongating, and splitting
the arc so that the arc is fully extinguished. ii)Current zero method:-The arc is interrupted at
current zero position that occurs100 times a second in case of 50Hz power system frequency in
AC.

29. What is restriking voltage? (Nov/Dec 2011) (May/June 2017)

It is the transient voltage appearing across the breaker contacts at the instant of arc being
extinguished.

30. What is meant by recovery voltage? (Nov/Dec 2011) (May/June 2013)

The power frequency RMS voltage appearing across the breaker contacts after the arc is
extinguished and transient oscillations die out is called recovery voltage.

31. Define the term RRRV? (May/June 2012) (Apr/May 2015)

The transient voltage which appears across the circuit breaker contacts at the instant of arc
extinction is called restriking voltage. RRRV is the Rate of Rise of Restriking Voltage,
expressed in volts per microsecond. It is the rate at which the restriking voltage changes per
microsecond. It is closely associated with natural frequency of oscillation.

32. What is the main problem of the circuit breaker?

4
 When the contacts of the breaker are separated, an arc is struck between them. This arc
delays the current interruption process and also generates enormous heat which may cause
damage to the system or to the breaker itself. This is the main problem.

33. What are the factors the arc resistance depends upon?
i) Degree of ionization ii)Length of the arc iii) Cross section area of the arc.

34. Mention the details circuit breaker rating

i) Rated voltage & rated current ii)Rated Frequency iii)Rated breaking capacity,
symmetrical & asymmetrical iv)Rated making capacity v)Rated short time current vi)Rated
operating duty.

35. What are the factors the ARC phenomenon depends upon? (May/June 2013)
i) The nature and pressure of the medium ii) The external ionizing and de-ionizing agent
present iii) Voltage across the electrodes and its variation with time iv) The nature shape &
separation of electrodes v) The nature and shape of vessel and its position in relation to the
electrodes.

36. Define symmetrical breaking capacity. (Nov/Dec 2017)

The symmetrical value of breaking capacity is the value of the symmetrical breaking current
which the circuit breaker is capable of breaking at the stated recovery voltage and restriking voltage
under prescribed condition.

37. What are the two theories explaining current zero interruption?
i)Recovery rate theory or voltage race theory or slepain’s theory. ii) Energy balance
theory or Cassie’s theory.

38. What are the factors the recovery voltage depends upon? (Nov/Dec 2011)
i) Power factor, ii) Armature reaction &iii) Circuit condition.

39. What is the basic requirement of DC circuit breaking?

Lengthening of the arc is basic requirements of D.C circuit breaker. Loss of energy
increases with increasing length of arc and more power will be required to maintain the arc.

40. What are the problems associated with DC circuit breakers?

i) Natural current zero does not occur as in the case of A.C circuit breakers. ii) The
amount of energy to be dissipated during the short interval of breaking is very high as compared to
conventional A.C circuit breakers.

41. What is the purpose of protective spark gap?

A protective spark gap can be used across the circuit breaker to reduce the size of
commutation capacitor. The spark gap acts as an energy dissipating device for high frequency
currents.

42. List out the various methods of arc interruptions. (Nov/Dec 2012) (Apr/May 2015)
i) High resistance interruption ii) Current zero method.

43. How do you classify the circuit breakers? (Nov/Dec 2012) (Nov/Dec 2014)

• Air break circuit breaker, ii)Oil circuit breaker, iii)Air blast circuit breaker iv)SF6
circuit breaker, and v)Vacuum circuit breaker.

44. What is meant by auto reclosing? (Nov/Dec 2013) (May/June 2016)

In electric power distribution, an auto recloser is a circuit breaker equipped with a mechanism
that can automatically close the breaker after it has been opened due to a fault.
4
 45. Write the function of isolating switch. (Nov/Dec 2013) (May/June 2016)
A disconnector, disconnect switch or isolator switch is used to ensure that an electrical circuit
is completely de-energised for service or maintenance. The disconnector is usually not intended for
normal control of the circuit, but only for safety isolation. Disconnector can be operated either
manually or automatically (motorized disconnector).

46. Mention any two advantages of vacuum circuit breakers.(Nov/Dec 2014)

i) They are compact in size and have longer life. ii) There are no fire hazards.iii) No
generation of gas during and after operation. iv) They require less maintenance and quiet in
operation. v) They can successfully withstand lightning surges.

47. Why current chopping is not required in MOCB?

MOCB has superior arc quenching properties when compared to air blast circuit
breakers due to the cooling oil and hence there is no special mechanism required for current chopping.

48. How does a circuit breaker differ from a switch? (Nov/Dec 2015) (May/June 2016)
Switches are not automatic as they need to be manually turned on or off while circuit breakers
just trips off on certain conditions. Switches allow users to cut off power supply to a certain area or
equipment while circuit breakers are more preventive in nature. Circuit breakers are essentially
automatic off switches designed for a very specific purpose, which is to prevent unnecessary
electrical circuit damage.

49. Name the materials used for contacts of vacuum circuit breakers. (Nov/Dec 2015)
Compounds of copper and Chromium are used most widely for making the contacts of circuit
breakers.

50. What is the difference between re-striking voltage and recovery voltage? (Nov/Dec 2016)
Re-striking voltage: It is the transient voltage appearing across the breaker contacts at the
instant of arc being extinguished. Recovery voltage: The power frequency RMS voltage
appearing across the breaker contacts after the arc is extinguished and transient oscillations die
out is called recovery voltage.

51. State the difference between D.C. and A.C. circuit breaking. (Nov/Dec 2016)
DC circuit breaker, like their name suggests, is used for the protection of electrical devices
that operate with direct current. The main difference between direct current and alternating
current is that in DC the voltage output is constant, while in AC it cycles several times per
second.

52. What is rupturing Capacity? (May/June 2017)

Rupturing capacity is the current that a fuse, circuit breaker, or other electrical apparatus is
able to interrupt without being destroyed or causing an electric arc with unacceptable duration. The
prospective short-circuit current which can occur under short circuit conditions should not exceed the
rated breaking capacity of the apparatus. This theory states that the rate at which positive ions and
electrons recombine to form neutral molecules is compared with rate of rise of restriking voltage and
if the restriking voltage rises more rapidly than the dielectric strength, gap space breaks down and arc
strikes again persists.

53. What are the factors responsible for increase in arc resistance? (April/may 2018)

The arc resistance increases

 When ionized particle between contact decreases.
 As the separation between contact increases and length of the arc also increases.
 With decrease in cross section area of the arc.
4
54. A circuit breaker is rated as 1500A, 1000MVA, 3second, 3 phase oil circuit breaker . Find the
rated making current. (April/may 2018)
Given:
Breaking capacity= 1000MVA; Breaking current = 1500A;
Soln:
Making current = 2.55*breaking current=2.55*1500=3825A.

55. Why rate of rise of restriking voltage plays an important role in circuit breaker operation?
(Nov/Dec 2018)
The rate of rise of restriking voltage denotes the rate at which transient voltage increases or
decreases. This factor plays an important role in circuit breaker operation since it decides the
interruption of current by the circuit breaker. Transient recovery voltage depends upon natural
frequency and power factor.

56. Why oil circuit breakers are not suitable for heavy current interruption at low voltages?
(Nov/Dec 2018)
Oil circuit breakers cannot be used for heavy current interruption because high current causes
arc which produces flammability of oil. Thus it requires high maintenance.

PART B

1. Explain with neat sketch, the construction and working of minimum oil circuit breaker. What are
its main advantages and disadvantages? (May/June 2016)
i) Describe the various types of rating of circuit breaker (May/June 2012) (May/June
2013)
2. With a neat block diagram, explain the construction, operating principle and
3. applications of SF6 circuit breaker. What are its advantages over other circuit breaker?
(Apr/May 2015) (May/June 2014) (Dec 2014) (Nov/Dec 2015).
4. With neat sketches, explain the construction and operating principle of air break and minimum oil
circuit breaker. (16) (Nov/Dec 2013) (Nov/Dec 2015)
5. Compare the performance, characteristics and application of different types of circuit breaker.
6. What are the requirements of a contact material for a vacuum circuit breaker? Why current
chopping is not a serious problem with this circuit breaker? (May 2015)
7. With neat sketch explain the principles of axial blast circuit breaker. Enumerate the advantages
and disadvantages of air blast circuit breakers. (Apr/May 2015) (Nov/Dec 2018)
8. What are the various types of operating mechanism which are used for opening and closing of the
contacts of a CB? Discuss their merits and demerits.
9. i)Explain how arc is initiated and sustained when the circuit breaker contacts break ii)What is
current chopping? Explain how can the effect of current chopping be minimized?
10. Derive an expression for the rate of rise of restriking voltage in a circuit breaker ii)Describe the
operating principle of DC circuit breaker.
10. Discuss the recovery rate theory and energy balance theory of arc interruption in a circuit breaker.
(Apr/May 2015)
11. Discuss the problems associated with the interruptions of low inductive current and the fault
occurs nearer to the substation. (Nov/ Dec 2012) (Nov/Dec 2014) (Nov/Dec 2015)
12. Explain the phenomenon of current chopping and capacitive current breaking with diagram and
waveforms. (16) (May/June 2016) (May/June 2017)
13. Derive an expression for restriking voltage and RRRV in terms of system voltage, inductance up
to the fault location and bushing to earth capacitance of the circuit breaker. (Apr/May 2015)
(Nov/Dec 2014) (May/June 2016) (Nov/Dec 2016)
14. Explain the methods of arc interruption. (May/June 2014) (Nov/Dec 2011)
15. What are the comparative merits and demerits of circuit breaker? (Apr/May 2015)
16. Discuss the different types of rating of circuit breakers along with the significance
4
 and features.
17. Explain about current zero interruption theories. (May/June 2016)
(i) With a neat sketch explain the principle of vacuum circuit breaker.
(ii) Explain the phenomenon of interruption of capacitive current in a circuit breaker.
18. In short circuit test on a 3 pole, 132kV, circuit breaker, the following observations are made.
Power factor for fault = 0.4, recovery voltage 0.9 times full line value, the breaking current
symmetrical, frequency of oscillation of restriking voltage 16kHz. Assume neutral is grounded
and fault is not grounded. Determine RRRV. (Nov/Dec 2016)
19. With neat diagram explain the construction and working principle of Air blast Circuit breaker and
Vacuum circuit breaker. (May/June 2017) (April/may 2018)
20. Write short notes on i) Current Chopping ii) Resistance switching. (Nov/ Dec 2017)

5
 EE8691- EMBEDDED SYSTEMS

UNIT-1

INTRODUCTION TO EMBEDDED SYSTEMS

PART-A

1. Define Embedded System. What are the components of embedded system?

An Embedded system is one that has computer hardware with software embedded in it as one
of its most important component.

The three main components of an embedded system are

1. Hardware
2. Main application software
3. RTOS.

2. In what ways CISC and RISC processors differ?

CISC RISC
1. It provides number of addressing It provides very few number of addressing
modes modes
2. It has a micro programmed unit with It has a hard wired unit without a control
a control memory memory
3. An easy compiler design Complex compiler design
4. Provides precise and intensive Provides precise and intensive calculations
calculations slower than a RISC faster than a RISC

3. Define system on chip (SOC) with an example.

Embedded systems are being designed on a single silicon chip called system on chip. SOC is
a new design innovation for embedded system

Ex. Mobile phone.

4. Give any two uses of VLSI designed circuits

A VLSI chip can embed IPs for the specific application besides the ASIP or a GPP core. A
system ona VLSI chip that has all of needed analog as well as digital circuits.

Eg. Mobile phone.

5. List the important considerations when selecting a processor.

1. Instruction set
2. Maximum bits in an operand
3. Clock frequency
4. Processor ability.

6. What are the types of embedded system?

5
  Small scale embedded systems
 Medium scale embedded systems
 Sophisticated embedded systems.

7. Classify the processors in embedded system?

General purpose processor
 Microprocessor
 Microcontroller
 Embedded processor
 Digital signal processor
 Media processor
Application specific system processor
Multiprocessor system using GPP and ASSP GPP core or ASIP core integrated into
either an ASIC or a VLSI circuit or an FPGA core integrated with processor unit in a VLSI chip.

8. What are the important embedded processor chips?

 ARM 7 and ARM 9
 i 960
 AMD 29050

9. Name some DSP used in embedded systems?

 TMS320Cxx
 SHARC
 5600xx

10. Name some of the hardware parts of embedded systems?

 Power source
 Clock oscillator circuit
 Timers
 Memory units
 DAC and ADC
 LCD and LED displays
 Keyboard/Keypad

11. What are the various types of memory in embedded systems?

 RAM (internal External)
 ROM/PROM/EEPROM/Flash
 Cache memory

12. What are the points to be considered while connecting power supply rails with
embedded system?
 A processor may have more than two pins of Vdd and Vss
 Supply should separately power the external I/O driving ports, timers, and clock and
 From the supply there should be separate interconnections for pairs of Vdd and Vss
pins analog ground analog reference and analog input voltage lines.

13. What is watch dog timer?

Watch dog timer is a timing device that resets after a predefined timeout.

14. What are the two essential units of a processor on a embedded system?
 Program Flow control Unit
 Execution Unit

15. What does the execution unit of a processor in an embedded system do?

5
 The EU includes the ALU and also the circuits that execute instructions for a program control
task. The EU has circuits that implement the instructions pertaining to data transfer operations and
data conversion from one form to another.

16. Give examples for general purpose processor.

 Microcontroller
 Microprocessor

17. Define microprocessor.

A microprocessor is a single VLSI chip that has a CPU and may also have some other units
for example floating point processing arithmetic unit pipelining and super scaling units for faster
processing of instruction.

PART B

1. What are the challenges of Embedded System? (Apr/May 2016)

2. Explain various architectural features of 8051 micro processor.
3. Explain ARM Processor, Architecture, Instruction set Programming (Nov/Dec 2015)
4. Discuss the data operations of ARM processor. (Apr/May 2017)
5. (i) What are the various problems that must be taken into account in embedded system design?
(ii) List the hardware units that must be present in the embedded systems.
6. (i) Explain the various data operations involved in ARM.
(ii) Implement an if statement in ARM.
7. With the help of a neat block diagram explain the 8051 processor.
8. (i) List the various types of instructions in 8051.
(ii) Write an 8051 C program to toggle all the bits of P0, P1, and P2continuously with a 250 ms
delay. Use the sfr keyword to declare the port address.
9. Explain the major levels of embedded system design process with an example.

UNIT II

EMBEDDED NETWORKING

PART-A

1. Differentiate synchronous communication and iso-synchronous communication.

Synchronous communication
When a byte or a frame of the data is received or transmitted at constant time
intervals with uniform phase difference, the communication is called synchronous
communication.
Iso-synchronous communication
Iso -synchronous communication is a special case when the maximum time interval can
be varied.

2. What are the two characteristics of synchronous communication?

Bytes maintain a constant phase difference
The clock is not always implicit to the synchronous data receiver.

3. What are the three ways of communication for a device?

 Iso-synchronous communication
5
  synchronous communication
 Asynchronous communication

4. Expand a) SPI b) SCI

 SPI—serial Peripheral Interface
 SCI—Serial Communication Interface

5. Define software timer.

This is software that executes and increases or decreases a count variable on an interrupt
from a timer output or form a real time clock interrupt. A software timer can also generate
interrupt on overflow of count value or on finishing value of the count variable.
6.What is I2C?
I2C is a serial bus for interconnecting ICs .It has a start bit and a stop bit like an UART. It
has seven fields for start,7 bit address, defining a read or a write, defining byte as acknowledging
byte, data byte, NACK and end.
7. What are the bits in I2C corresponding to?
It has seven fields for start,7 bit address, defining a read or a write, defining byte as
acknowledging byte, data byte, NACK and end.
8. What is a CAN bus? Where is it used?
CAN is a serial bus for interconnecting a central Control network. It is mostly used in
automobiles. It has fields for bus arbitration bits, control bits for address and data length data
bits, CRC check bits, acknowledgement bits and ending bits.
9. What is USB? Where is it used?
USB is a serial bus for interconnecting a system. It attaches and detaches a device from the
network. It uses a root hub. Nodes containing the devices can be organized like a tree structure.
It is mostly used in networking the IO devices like scanner in a computer system.
10. What are the features of the USB protocol?
A device can be attached, configured and used, reset, reconfigured and used, share the
bandwidth with other devices, detached and reattached.
11. Explain briefly about PCI and PCI/X buses.
PCI and PCI/X buses are independent from the IBM architecture .PCI/X is an extension
of PCI and support 64/100 MHZ transfers. Lately, new versions have been introduced for the
PCI bus architecture.
12. Why are SPCI parallel buses important?
SPCI serial buses are important for distributed devices. The latest high speed
sophisticated systems use new sophisticated buses.
13. What is meant by UART?
UART stands for universal Asynchronous Receiver/Transmitter.
1. UART is a hardware component for translating the data between parallel and serial
interfaces.
2. UART does convert bytes of data to and from asynchronous start stop bit.
3. UART is normally used in MODEM.

14. What does UART contain?

1. A clock generator.
2. Input and Output start Registers
3. Buffers.
4. Transmitter/Receiver control.
5
15. What is meant by HDLC?
o HDLC stands for “High Level Data Link Control”.
o HDLC is a bit oriented protocol.
o HDLC is a synchronous data Link layer.

16. Name the HDLC’s frame structure?

17.List out the states of timer?

There are eleven states as follows
 Reset state
 Idle state
 Present state
 Over flow state
 Over run state
 Running state
 Reset enabled state / disabled
 Finished state
 Load enabled / disabled
 Auto reload enabled / disabled
 Service routine execution enabled / disabled.

18. Name some control bit of timer?

1. Timer Enable
2. Timer start
3. Up count Enable
4. Timer Interrupt Enable.

19. What is meant by status flag?

Status flag is the hardware signal to be set when the timer reaches zeros.
20. List out some applications of timer devices?
 Real Time clock
 Watchdog timer
 Input pulse counting
 TDM
 Scheduling of various tasks

21.State the special features on I2C?

 Low cost
 Easy implementation
 Moderate speed (upto 100 kbps).

22.What are disadvantages of I2C?

 Slave hardware does not provide much support
 Open collector drivers at the master leads to be confused.

23.What are the two standards of USB?

 USB 1.1
5
  USB 2.0

24.Draw the data frame format of CAN?

25. What is the need of Advanced Serial High Speed Buses?

If the speed in the rate of “Gigabits per second‟ then there is a need of Advanced Serial
HighSpeed Buses.
26.What is meant by ISA?
ISA stands for Industry standard Architecture.
Used for connecting devices following IO addresses and interrupts vectors as per IBM pc
architecture.
27.What is meant by PCI-X?
 PCI X offers more speed over PCI.
 30 times more speed than PCI.

28.Define CPCI?
 CPCI stands for Compact peripheral component Interfaces.
 CPCI is to be connected via a PCI.
 CPCI is used in the areas of Telecommunication Instrumentation and data communication
applications.

29.Define half-duplex communication.

Transmission occurs in both the direction, but not simultaneously.
30. Define full duplex communication.
Transmission occurs in both the direction, simultaneously.
31. Define Real Time Clock (RTC)?
Real time clock is a clock which once the system stats does not stop and cant be reset and
its count value cannot be reloaded.

32. Define Time-out or Time Overflow?

A state in which the number of count inputs exceeded the last acquirable value and on
reaching that state, an interrupt can be generated.

33. Why do we need at least one timer in an ES?

The embedded system needs at least on timer device. It is used as a system clock.
PART B

1. (i) Along with the structure of a cache memory explain it.

(ii) Explain the memory management units and address translation techniques.
2. Explain in detail the programming of input and output devices.
3. (i) List the various input and output devices and explain them in detail.
(ii) Explain the various memory devices along with component interfacing.
4. Explain the various interrupt handling mechanisms.
5. Discuss the memory devices and its interfacing concepts.(Apr/May 2016)
6. Discuss the I/O devices & its interfacing concepts? (Apr/May 2018)
7. Explain the following memory systems:
(i) Two-level cache 5
 (ii) Direct mapped cache
(iii) Set- associative cache. (Nov/Dec 2014)
8. Draw a UML sequence diagram for copying characters from an input to an output device
using interrupt-driven I/O. the diagram should include the two devices and the two I/O
handlers. (Nov/Dec 2017)

UNIT-III

EMBEDDED FIRMWARE DEVELOPMENT ENVIRONMENT

PART-A

1. What are the states of a process?

 Running
 Ready
 Waiting

2. What is the function in steady state?

Processes which are ready to run but are not currently using the processor are in the 'ready' state.
3. Define scheduling.
This is defined as a process of selection which says that a process has the right to use
the processor at given time.
4. What is scheduling policy?
It says the way in which processes are chosen to get promotion from ready state to running state.
5. Define hyper period?
It refers the duration of time considered and also it is the least common multiple of all
the process.
6. What is schedulability?
It indicates any execution schedule is there for a collection of process in the system's
functionality.
7. What are the types of scheduling?
1. Time division multiple access scheduling.
2. Round robin scheduling.

8. What is cyclostatic scheduling?

In this type of scheduling, interval is the length of hyper period 'H'. For this interval, a
cyclostatic schedule is separated into equal sized time slots.
9. Define round robin scheduling?
This type of scheduling also employs the hyper period as an interval. The processes are run in
the given order.

10. What is scheduling overhead?

It is defined as time of execution needed to select the next execution process.

11. What is meant by context switching?

The actual process of changing from one task to another is called a context switch.

12. Define priority scheduling?

A simple scheduler maintains a priority queue of processes that are in the runnable state.
5
 13. What is rate monotonic scheduling?
Rate monotonic scheduling is an approach that is used to assign task priority for a
preemptive system.
14. What is critical instant?
It is the situation in which the process or task posses‟ highest response time.
15. What is critical instant analysis?
It is used to know about the schedule of a system. Its says that based on the periods given,
the priorities to the processes has to be assigned.
16. Define earliest deadline first scheduling?
This type of scheduling is another task priority policy that uses the nearest deadline as
the criterion for assigning the task priority.
17. What is IDC mechanism?
It is necessary for a 'process to get communicate with other process' in order to attain a
specific application in an operating system.
18. What are the two types of communication?
1. Blocking communication
2. Non blocking communication

PART-B

1. With appropriate diagrams explain multiple tasks and multiple processes.

2. Explain in detail the various scheduling policies with example.(Apr/May 2016)
3. Explain the following
a) Inter process communication
b) Context Switching (Apr/May 2014)
4. Explain in detail the various performance issues.
5. Compare RMF and EDS with suitable example.
6. Explain the following :
(i) Blocking interprocess communication
(ii) Non blocking interprocess communication
(iii) Shared memory communication (Nov/Dec 2017)

UNIT-IV

RTOS BASED EMBEDDED SYSTEM DESIGN

PART-A

1. Name the important terms of RTOS?

 Task State Scheduler
 Shared data Reentrancy

2. Define process.
Process is a computational unit that processes on a CPU under the control of a scheduling
kernel of an OS. It has a process structure, called Process control block. A process defines a
sequentiallyexecuting program and its state.
3. What is meant by PCB?
5
 Process Control Block‟ is abbreviated as PCB.PCB is a data structure which contains all the
information and components regarding with the process.
4. Draw the process state transitions?

5. Define task and Task state.

A task is a set of computations or actions that processes on a CPU under the control of a
scheduling kernel. It also has a process control structure called a task control block that saves at
the memory. It has a unique ID. It has states in the system as follows: idle, ready, running, blocked
and finished.
6. Define Task Control Block (TCB)
A memory block that holds information of program counter, memory map, the signal
dispatch table, signal mask, task ID, CPU state and a kernel stack.

7. What is a thread?
Thread is a concept in Java and UNIX and it is a light weight sub process or process in an
application program. It is controlled by the OS kernel. It has a process structure, called thread
stack, at the memory. It has a unique ID .It have states in the system as follows: stating, running,
blocked and finished.
8. Define Inter process communication.

An output from one task passed to another task through the scheduler and use of signals,
exception, semaphore, queues, mailbox, pipes, sockets, and RPC.

9. What is shared data problem?

If a variable is used in two different processes and another task if interrupts before the
operation on that data is completed then the value of the variable may differ from the one expected
if the earlier operation had been completed .This ids known as shared data problem.
10. Define Semaphore.
Semaphore provides a mechanism to let a task wait till another finishes. It is a way of
synchronizing concurrent processing operations. When a semaphore is taken by a task then that
task has access to the necessary resources. When given the resources unlock. Semaphore can be
used as an event flag or as a resource key.
11. Define Mutex.
A phenomenon for solving the shared data problem is known as semaphore. Mutex is a
semaphore that gives at an instance two tasks mutually exclusive access to resources.

12. Differentiate counting semaphore and binary semaphore.

Binary semaphore
5
 When the value of binary semaphore is one it is assumed that no task has taken it and that it
has been released. When the value is 0 it is assumed that it has been taken.

Counting semaphore
Counting semaphore is a semaphore which can be taken and given number of times. Counting
semaphores are unsigned integers.
13. What is Priority inversion?
A problem in which a low priority task inadvertently does not release the process for a higher
priority task.
14. What is Deadlock situation?
A set of processes or threads is deadlocked when each process or thread is waiting for a
resource to be freed which is controlled by another process.
15. Define Message Queue.
A task sending the multiple FIFO or priority messages into a queue for use by another task
using queue messages as an input.
16. Define Mailbox and Pipe.

A message or message pointer from a task that is addressed to another task.

17. Define Socket.
It provides the logical link using a protocol between the tasks in a client server or peer to peer
environment.
18. Define Remote Procedure Call.
A method used for connecting two remotely placed methods by using a protocol. Both
systems work in the peer to peer communication mode and not in the client server mode.
19. What are the goals of RTOS?
 Facilitating easy sharing of resources
 Facilitating easy implantation of the application software
 Maximizing system performance
 Providing management functions for the processes, memory, and I/Os and for other
functions for which it is designed.
 Providing management and organization functions for the devices and files and file like
devices.
 Portability
 Interoperability
 Providing common set of interfaces.

20. What is RTOS?

An RTOS is an OS for response time controlled and event controlled processes. RTOS is an
OS for embedded systems, as these have real time programming issues to solve.
21. List the functions of a kernel.
 Process management
 Process creation to deletion
 Processing resource requests
 Scheduling
 IPC
 Memory management
 I/O management
 Device management

22. What are the two methods by which a running requests resources?
6
  Message
 System call

23.What are the functions of device manager?

 Device detection and addition
 Device deletion
 Device allocation and registration
 Detaching and deregistration
 Device sharing

24. List the set of OS command functions for a device

 Create and open
 Write
 Read
 Close and delete.

PART-B

1. (i) Explain the various software development tools with their applications.
(ii) List the features of
(a) Source code engineering tool
(b) Integrated Development Environment
2. Explain in detail programming the embedded systems in assembly and C.
3. Discuss the concept of programming in assembly language Vs High-level language.
4. Implement the multi state timed system with an example.
5. Implement the multi state Input/Timed system with an example.
6. Explain the time-out mechanisms with examples.
7. Explain
(i) ROM Emulators
(ii) Remote Debuggers (Nov/Dec 2016)
8. Explain the multi-state systems and function sequences. (Nov/Dec 2017)
9. (i) Discuss the concept of programming in assembly language Vs high level language (Apr/May
2018)
(ii) Discuss the tools used to download the embedded software into the target system (Apr/May
2014)
10. Discuss the various debugging techniques and debugging challenges. (Apr/May
2017)

UNIT-V

EMBEDDED SYSTEM APPLICATION AND DEVELOPMENT

PART-A

6
1. What is a PIC?
PIC refers to Programmable Intelligent Computer. PIC is microprocessor lies inside a personal
computer but significantly simpler, smaller and cheaper. It can be used for operating relays,
measuring sensors etc.

2. What are the main elements inside a PIC?

 Processing engine,
 Program memory,
 Data memory and
 Input / Output.

3. What are the types of program memory in a PIC?

 Read-only,
 EPROM and
 EEPROM,
 Flash

4. What is MBasic Compiler Software?

From version 5.3.0.0 onward, Basic Micro offers one version of its MBasic compiler, the
“Professional” version. MBasic runs under Microsoft‟s Windows operating system in any
version from Windows 95 to Windows XP. The computer requires an RS-232 port for
connection to the ISP-PRO programmer board.
5. Define pseudo-code.
Pseudo-code is a useful tool when developing an idea before writing a line of true code or
when explaining how a particular procedure or function or even an entire program

6. What is design technology?

Design technology involves the manner in which we convert our concept of desired system
functionality into an implementation. Design methodologies are used in taking the decisions
at the time of designing the large systems with multiple design team members.

7. What are the goals of design process? (Apr/May 2011)

A design process has several important goals beyond function, performance, and power.
Theyare time to market, design cost and quality

8. What is a design flow?

A design flow is a sequence of steps to be followed during a design.

9. Define successive refinement design methodology.

In successive refinement design methodology, the system is built several times. A first
system is used as a rough prototype, and successive models of the system are further refined.
This methodology makes sense when you are relatively unfamiliar with the application domain
for which you are building the system

10. What are the phases in water fall development model?

The waterfall development model consists of five major phases; they are requirements
analysis, architecture, coding, testing and maintenance.

11. What are the elements of concurrent engineering?

Cross-functional teams, Concurrent product realization, Incremental information sharing and
use, integrated project management & Early and continual supplier involvement.
6
12. What are requirements and specification?
Requirements are informal descriptions of what the customer wants, while specifications are
more detailed, precise, and consistent descriptions of the system that can be used to create the
architecture.

13. What are the several tests met by a good set of requirements?
The several tests that should be met by a good set of requirements are Correctness,
Unambiguousness, Completeness, Verifiability, Consistency, Modifiability and Traceability.
14. What does the acronym CRC stands for?
CRC stands for Classes, Responsibilities and Collaborators.

 Classes - define the logical groupings of data and functionality.

 Responsibilities - describe what the classes do.
 Collaborators -are the other classes with which a given class works.

15. What are the steps to be followed in a CRC card methodology?

 Develop an initial list of classes:, Write an initial list of responsibilities and
collaborators
 Create some usage scenarios, Walk through the scenarios Refine the classes,
responsibilities, and collaborators
16. What is the need for quality assurance (QA)?
The quality assurance (QA) process is vital for the delivery of a satisfactory system.

17. What are the observations about quality management based on ISO 9000?
Process is crucial, Documentation is important & Communication is important

18. What are the five levels of maturity in capability maturity model?
Initial, Repeatable, Defined, Managed and Optimizing are the five levels of maturity in
CMM.

19. What is a design review?

Design review is a simple, low-cost way to catch bugs early in the design process. A design
review is simply a meeting in which team members discuss a design, reviewing how a component
of the system works.

20. Give the members of the design review team.

Designers, Review leader, Review scribe and Review audience are the members of the design
review team.
21. What is the role of a review scribe in a design review?
The review scribe records the minutes of the meeting so that designers and others know
which problems need to be fixed.

22. Give the role of the review leader in a design review team.
The review leader coordinates the pre-meeting activities, the design review itself, and the
post-meeting follow-up. During the meeting, the leader is responsible for ensuring that the meeting
runs smoothly.

23. What are the potential problems to be looked for by the audience of a design review
meeting?
1. Is the design team’s view of the component’s specification consistent with
2. The overall system specification, or has the team misinterpreted somthing? Is the
6
 interface specification correct?
3. Does the component’s internal architecture work well?
4. Are there coding errors in the component?
5. Is the testing strategy adequate?

24. Why is the verification of specification very important?

Verifying the requirements and specification is very important for the simple reason that bugs
in the requirements or specification can be extremely expensive to fix later on. A bug introduced
in the requirements or specification and left until maintenance could force an entire redesign of
the product

25. What is prototype?

Prototype is the model of the system being designed. Prototypes are a very

PART-B

1. Draw and explain basic system of an Automatic chocolate vending system.

2. Discuss with the diagram task synchronization model for a specific application.
3. Explain in detail the design issues and techniques in embedded system development.
4. (i) Explain the case study of an embedded system for a smart card.
(ii) Explain the features of Vx Works.
5. Discuss the complete design of typical embedded system. (Apr/May 2017)
6. Discuss the embedded system design issues and techniques. (Apr/May 2016)
7. Explain the advanced techniques for specification with an example. (Nov/Dec 2014)
8. (i) Explain the case study of Audio players.
(ii) Explain the design example of software modem (Apr/May 2018)

6
 DESIGN OF ELECTRICAL APPARATUS

UNIT-1

DESIGN OF FIELD SYSTEM AND ARMATURE

PART-A

1. What is design? What are the conditions for an optimum design?

Design may be defined as the creative physical realization of theoretical concepts. Engineering
design is the application of science, technology and invention to produce machines to perform the
specified task with optimum economy and efficiency.
A design is said to be optimum when the following conditions are satisfied:
 Minimum loss (or) Maximum efficiency
 Minimum cost
 Minimum volume
 Minimum weight.

2. Mention the different circuits available in a machine.

 Magnetic circuit
 Electric circuit
 Dielectric circuit
 Thermal circuit
 Mechanical parts.

3. What are the major design considerations? (M’13, Dec 2015 & May 2018)
The major considerations to evolve a good design are:
 Cost: Low initial cost and reasonable operating cost.
 Durability: Quality of lasting for a long time
 Compliance: Suffering with performance criteria as laid down in the specifications.
 Weight: Lower

4. What are the limitations in design?

 Saturation
 Temperature rise
 Stress on insulation
 Efficiency
 Mechanical precision of air gap
6
  Commutation
 Power factor
 Specifications
 Stress & strain on rotating parts and bearings.

5. What are the major design factors?

The major design factors of electrical machines are as follows
Magnetic circuit or the flux path:
Should establish required amount of flux using minimum MMF. The core losses should be
less.
Electric circuit or windings:
Should ensure required EMF is induced with no complexity in winding arrangement. The copper
losses should be less.

Insulation:
Should ensure trouble free separation of machine parts operating at different potential
and confine the current in the prescribed paths.
Cooling system or ventilation:

Should ensure that machine operates at the specified temperature.

Machine parts: Should be robust.

6. What are the different conducting materials used in rotating machines?(D’18)

The most common conducting materials used in rotating machines are
Copper – usually used in windings of electric machines
Aluminum – substitute for copper due to low cost - Armature and field winding
Iron and Steel- Pole core and Armature core

Alloys of copper – Bronze – brush holders, commutator segment, cage windings

Copper silver alloy - bearings.

7. Mention the requirements of highly conducting materials.

 Highest possible conductivity (least resistivity)
 Least possible temperature coefficient of resistance
 Adequate mechanical strength, in particular, high tensile strength and degree of
flexibility (absence of brittleness)
 Rollability and drawability
 Good weldability and solderability
 Adequate resistance to corrosion

6
8. What is a soft magnetic material?
Soft magnetic materials are easy to magnetize and demagnetize. These materials are used
for making temporary magnets. Soft magnetic materials with narrow hysteresis loop are called
soft magnetic material.
9. What are the causes of failure of insulation?
a. Weak points in the insulator b. Effect of aging and mechanical fatigue
c. Mica migration d. Tracking.

10. List out the classification of the resistivity materials.

Resistivity materials are classified as Nickel, Sliver, and Iron.
 The first group consists of materials used in precision measuring instruments and in
making standard resistances boxes.
 The second group consists of materials from which resistance elements are made for all
kinds of rheostats and similar control devices.
 The third group consists of materials suitable for making high temperature elements for
electric furnaces, heating devices and loading rheostats.
Should ensure trouble free separation of machine parts operating at different potential and confine
the current in the prescribed paths.
Cooling system or ventilation:
Should ensure that machine operates at the specified temperature.
Machine parts: Should be robust.

11. What are the different conducting materials used in rotating machines?(D’18)
The most common conducting materials used in rotating machines are
Copper – usually used in windings of electric machines
Aluminum – substitute for copper due to low cost - Armature and field winding
Iron and Steel- Pole core and Armature core
Alloys of copper – Bronze – brush holders, commutator segment, cage windings
Copper silver alloy - bearings.
12. Mention the requirements of highly conducting materials.
Highest possible conductivity (least resistivity)
Least possible temperature coefficient of resistance
Adequate mechanical strength, in particular, high tensile strength and degree of flexibility
(absence of brittleness)
Rollability and drawability
Good weldability and solderability
Adequate resistance to corrosion.

13. What is a soft magnetic material?

6
 Soft magnetic materials are easy to magnetize and demagnetize. These materials are used for
making temporary magnets. Soft magnetic materials with narrow hysteresis loop are called soft
magnetic material.
14. What are the causes of failure of insulation?
a. Weak points in the insulator b. Effect of aging and mechanical fatigue
c. Mica migration d. Tracking.
15. List out the classification of the resistivity materials.
Resistivity materials are classified as Nickel, Sliver, and Iron.
The first group consists of materials used in precision measuring instruments and in making
standard resistances boxes.
The second group consists of materials from which resistance elements are made for all kinds
of rheostats and similar control devices.
The third group consists of materials suitable for making high temperature elements for electric
furnaces, heating devices and loading rheostats.

16. What are the different types of magnetic materials according to their degree of magnetism?
(May 2011, Dec 2011 & May 2014)
Magnetic materials are classified in to three groups as:
Ferromagnetic materials
Relative permeability (µr) >1
Values depend upon the magnetizing force
Paramagnetic materials
Relative permeability (µr) is only slightly greater than unity. The value of susceptibility is
positive for these materials
17. What are the constituents of magnetic circuits of DC Machine? (D’18)
 The various elements in the flux path of salient pole machine are poles, pole shoes, air
gap, armature teeth, armature core and yoke.
 The various elements in the flux path of non- salient pole machines are: Stator core,
Stator teeth, Air gap, Rotor teeth and Rotor core.

18. Write any two similarities & difference between magnetic and electric circuits.
Similarities:
 In an electric circuit, the emf circulates current in a closed path, similarity in a
magnetic circuit, the mmf creates flux in a closed path.
 In electric circuit the flow of current is opposed by resistance of the circuit. Similarity
in a magnetic circuit the creation of flux is opposed by reluctance of the circuit.
Differences:
 When the current flows in a circuit the energy is spent continuously, whereas in
magnetic circuit the energy is needed only to create the flux but not to maintain it.
 Current actually flows in a magnetic circuit,6 whereas the flux does not flow in a
 magnetic circuit but it is only assumed to flow.

16. What is meant by magnetic circuit calculation?

Magnetic circuit calculation involves two types of problems as follows:

 It is required to determine the mmf needed to establish a desired flux at a given point in a
magnetic circuit.
 The flux or flux density is unknown and is required to be determined for a given geometry
of the magnetic circuit and specified mmf.
The magnetic circuit is split up into convenient parts which may be connected in series
or parallel. The flux density is calculated in every part and mmf per unit length, ‘at’ is found
by consulting B-at curves. The calculations of mmf in the magnetic parts are simple whereas
in case of air gap and tapered teeth, the calculations are complex.

17. List the various steps involved in the estimation of mmf for a section of magnetic
circuit.
The various steps in the estimation of mmf for a section of magnetic circuit are:
 Determine the flux in the concerned section.
 Calculate the area of cross section of the section.
 Calculate the flux density in the section.
 From B-at curve of the magnetic material, determine the mmf per metre (at) for the
calculate flux density.
The mmf of the section is given by the product of length of the section and mmf per
metre

18. What is carter’s coefficient? Write down the carter’s coefficient of d.c. machine. (N/D’15,
M’19)
 The carter’s coefficient is a parameter that can be used to estimate the contracted or effective
slot pitch in case of armatures with open or semi enclosed slots. It is a function of the ratio of
w0 / lg where w0 is slot opening and lg is air-gap length. The carter’s coefficient is also used
to estimate the effective length of armature when ducts are employed. In this case it will be a
function of wd / lg .-where wd is width of the ducts and lg is air-gap length.
In electrical machine design the carter’s coefficient is used to estimate the air-gap expansion (or
contraction) factor for slots and ducts.
19. Why equalizer connections are not needed in wave winding?
In simplex wave winding there are only two parallel paths. The conductors forming a
parallel path will be distributed equally under all poles. Hence both the parallel paths are equally
affected by the asymmetry in the magnetic circuit and so there is no circulating current.
Therefore, there is no necessity for equalizer connections.
20. What is split coil?
The split coils will have more than two coil sides, when all the top coil sides of a coil are
lying in one slot and their corresponding bottom coil sides are accommodated in two different
slots then the coil is called split coil.
21. What is progressive and retrogressive winding?
 In wave winding, ff after one round of the armature the coil falls in a slot right to its
starting slot the winging is called Progressive wave winding.
 If after one round of the armature the coil falls in a slot left to its starting slot the
winging is called Retrogressive wave winding.

22. What is the purpose of dummy coil?

The wave winding is possible only with particular number of conductors and slots
combinations. It is not always possible to have the standard stampings in the winding shop
6
 consist of the number of slots according to the design requirements. In such cases dummy
coils are employed. This coil is placed in the slots to give the machine the mechanical
balance but they are not electrically connected to the rest of the winding
23. Mention the two types of armature winding used in DC machine and compare.

S. No. Lap winding Wave winding

1 Number of parallel path Number of parallel path

(A)= Poles (P) (A)= 2 Always
2 Number of brush sets required is Number of brush sets required is
equal to number of poles always equal to two

3 Preferable for high current, Preferable for high voltage,

low voltage capacity low current capacity
generators generators
4 Normally used for generators of Preferred for generators of
capacity more than 500A capacity less than 500A

PART-B

1. Calculate the apparent flux density at a section of the teeth of an armature of a D.C machine from the
following data at that section. Slot pitch=24mm, slot width=tooth width =12mm, length of armature
core including five ducts of 10mm each=0.38m,iron stacking factor=0.92. True flux density in the teeth
at that section is 2.2T for which the mmf is 70000AT/m. (Nov.2012),(MAY 2016)
2. Define the terms specific electric loading and specific magnetic loading as applied to electrical
machines. What are the considerations in the choice of these for D.C machines? (NOV 2013),(NOV
2015)
3. A 6-pole D.C. machine has the following design data. Armature diameter=30cm,armature
core length=15cm,length of air gap at pole center=0.25cm,flux per pole=12milliweb. Field form
factor=0.65. Calculate the amp. turns required for the air gap a) if the armature surface is smooth b)
if the armature surface is slotted and the gap expansion factor is 1.2

4 . Explian the real and apparent flux densities. Discuss about the various leakage fluxes. (NOV
2011)

5. Find the apparent flux density at a section of the tooth in the following case when the real tooth
flux density at that section is 2.157. Gross armature length =32cm. Number of ventilating ducts
=4,each 1cm wide, tooth width at the section=1.2cm,slot width with parallel sides=1cm.
Permeability of the tooth corresponding to real tooth density=35.8x10-6 (NOV 2012)

7
6 . Determine the air gap length of a D.C machine from the following data. Gross core
length=0.12m,no.of ducts =one of 10mm width, slot pitchy =25mm, Carters coefficient for slots
and ducts=0.32, gap density at pole center =0.7T. Field mmf per pole = 3900AT, mmf require for
iron parts ofmagnetic circuit =800AT. (MAY 2016)

7. The diameter and length of a 500kw, 500V, 450r.p.m, 6-pole D.C generator are 84cm and 30cm
respectively. If it is lap wound with 660 conductors estimate the specific electric and magnetic
loadings. (APR 2011)

8. A salient pole machine with semi closed slots has a core length (including 4 ducts each of 10mm)
of 0.32m, pole arc of 0.19m, slot pitch of 65.4mm, slot opening of 5mm,air gap length of 4mm and
a flux per pole of 0.052 wb. Assume Carters coefficient as 0.18 and 0.28 for opening/gap ratio of 1
and 2 respectively.

9. A D.C machine has the following dimensions: cross section of pole body =0.08m2,length of pole
=0.25m, cross section8of yoke =0.05m2, mean flux path in. yoke =0.9m (pole to pole). Cross section
of armature core =0.04m2. Length of flux path in core =0.45m (pole to pole), area of pole face =0.12m2,
air gap length=0.005m. Find the mmf per pole to give a flux of 0.1 wb/pole. Take relative permeability
of iron as 1200. Neglect leakage. (APR 2012)
10. Discuss quantitatively the effects of slots and ventilating ducts upon the reluctance of the air gap
of a D.C machine.
11. Discuss quantitatively the effects of slots and ventilating ducts upon the reluctance of the air
gap of a D.C machine.
12. Find the permeability at the root of the tooth of a D.C machine armature with the following
data.(NOV 2008)
13. Slot pitch 2.1cm, tooth width at the root 1.07cm, gross length 32cm, stacking factor 0.9, real flux
density 2.25T, and apparent flux density at the root 2.36T.

14. Draw the magnetic circuit of a D.C machine. Derive an expression for the total mmf per pole.
(NOV 2009)
15. What are the major groups of electrical conducting materials? Describe the properties and
applications of those materials.(MAY 2013),(MAY 2014)

16. Describe the methods of measurement of temperature rise in various parts of an electrical
machine.(MAY 2013)

17. Explain in detail the various cooling methods of electrical machines.(MAY 2014)

18. Discuss the advantages of hydrogen cooling.(MAY 2015),(MAY2016)

19. The exciting coil of an electromagnet has a cross section of 120×50 and a length of mean turn
0.8m. It dissipates 150 W continuously. Its cooling surface is 0.125 and specific heat dissipation
is 30 W/-degree. Calculate the final steady temperature rise of the coil surface. Also calculate the
hot spot temperature rise of the coil if the thermal resistivity of insulating material used is 8 ohm
m. The space vector is 0.56.(MAY 2015)

20. Describe the classification of insulating materials7 used for electrical machines.(MAY 2015).
21. The temperature rise of a transformer is 25 degree C. after one hour and37.5 degree C after two
hours of starting from cold conditions. Calculate its final steady temperature rise and the heating time
constant. If its temperature falls from the final steady

22. Describe any two methods used for determination of motor rating for variable load drives(NOV
2015)

UNIT-II

DESIGN OF TRANSFOMERS

PART-A

1. List the various types of transformers.

The transformer can be classified based on construction, applications, frequency
range, number of winding and type of connection
Based on construction the transformers are classified as:
Core type, Shell type
Based on applications the transformers are classified as:
Distribution transformers, Instrumental transformers Power transformers Electronic
Transformers Special Transformers.

Based on frequency range the transformers are classified as:

Power frequency transformer, Wide band transformer, Audio frequency transformer, Narrow
band transformer, Pulse transformer, UHF transformer
Based on number of windings the transformer are classified as;
 Auto transformer
 Two winding transformer
Based on the type of construction the transformers are classified as:
Single Phase transformer
Three Phase transformer
2. What are the salient features of distribution transformers?
The salient features of distribution transformers are:
The distribution transformers will have low iron loss and higher value of copper loss.
The capacity of transformers will be up to 500KVA.
The transformers will have plain walled tanks or provided with cooling tubes or
radiators.
The leakage reactance and regulation will be low.

3. Write about distribution & power transformers and mention its use?
Distribution Transformer:
A distribution transformer is a transformer that provides the final voltage transformation
in the electric power distribution system, stepping down the voltage used in the distribution
lines to the level used by the customer.
If mounted on a utility pole, they are called as pole-mount transformers. If the
distribution lines are located at ground level or underground, distribution transformers are
mounted on concrete pads and locked in steel cases, thus known as pad-mount transformers.
7
Uses:
The transformers used at load centers to step down distribution voltage to a standard
service voltage required for consumers.
Power Transformer:
The transformers used in sub stations and generating stations for step down or step
4.How does the design of distribution transformer differ from that of a power
transformer?
The distribution transformers are designed to have low iron loss and higher copper loss,
whereas in power transformers, the copper loss will be lesser than iron loss.
The distribution transformers are designed to have the maximum efficiency at a load much lesser
than full load, whereas the power transformers are designed to have maximum efficiency at or
near full load.
In distribution transformer the leakage reactance is kept low to have better regulation,
whereas the power transformer in leakage reactance is kept high to limit the short circuit current.

5. What is transformer bank?

A transformer bank consists of three independent single phase transformers withtheir
primary and secondary windings connected either is star or delta.
6. What is yoke section of a transformer? What is its use?
The sections of the core which connect the limbs are called yoke. The yoke is used to
provide a closed path for flux.

7. Mention the uses of distribution transformer.

The distribution transformers are used at load centres to step down the transmission line
voltage to a standard voltage required for consumers.
8. Why is low voltage winding placed near the core in transformers?
The winding & core are both made of metals and so on insulation has to be placed between
them. The thickness of insulation depends upon voltage rating of the winding. In order to
reduce the insulation requirement, the low voltage winding is placed near the core.

9. Write the advantages of shell type transformers over core type transformers. (Or)
Distinguish between core and shell type transformers. (N/D’13) (M/J’16)
(M/J’12)

S.No Core Type Shell Type

1 The winding encircles the The core encircles the most part of
core. the winding.
2 Cylindrical type coils are Generally, multi layer disc type or
used. sandwich coils are used.
Easy in design and Comparatively complex.
3 construction.
Has low Mechanical
4 High Mechanical Strength
Strength
5 Reduction of leakage Reduction of leakage reactance is
reactance is not easily highly
possible. possible.
6 The assembly can be easily It cannot be easily dismantled for repair
dismantled for repair work. work.

7
7 Better heat dissipation from Heat is not easily dissipated from
winding. (natural cooling) winding since it is surrounded by core.
(no natural cooling)
Has longer mean length of core It is not suitable for EHV requirements.
and shorter length of coil turn.
8 Hence best suited for EHV
(Extra High voltage)
requirements.

7
10. List the advantages of star connected transformers.
The phase voltage is smaller, and hence the major insulation required between
windings and earth is smaller. The number of turns is smaller with the result the amount of
insulation used is smaller. In case of star connection, both phase and line voltages are
available for four wire supply. This is useful for distribution purposes at lower voltages
where both 1-phase and 3-phase loads have to be supplied.

11. Mention the advantage and disadvantages of 3-Phase transformers.

Advantages of 3-Phase transformers:
A three-phase transformer is lighter, occupies lesser space, cheaper and more efficient
than a bank of 1-phase transformers. The reason for lower cost of 3-Phase
transformers: savings in cost of iron core of the tank and oil of the bushings
In case of 3-Phase transformers, there is only one unit to install and operate. Hence
installation and operational costs are smaller for 3-Phase units.
Disadvantages of 3-Phase transformers:
It is more difficult to transport a 3-Phase transformer as the weight per unit is more
In this event of a fault in any phase of a 3-Phase transformer, the fault is transferredto
the other two phases. Therefore, the whole unit needs a replacement.

12. What is the range of current density used in the design of transformer windings?
The choice of current density depends on the allowable temperature rise, copper loss
and method of cooling. The range of current density for various types of transformers is given
below.
δ=1.1 to 2.2 A/mm2- For distribution transformers
δ=1.1 to 2.2 A/mm2- For small power transformers with self-oil cooling
δ=2.2 to 3.2 A/mm2- For large power transformers with self-oil cooling or air-blast
δ=5.4 to 6.2 A/mm2- For large power transformers with forced circulation of oil or
with water cooling coils
13. List the optimum quantities for transformer design.
Transformer may be designed to make one of the following quantities as minimum:
 Total volume
 Total weight
 Total cost
 Total losses

14. What are the advantages of using higher flux density in core?
a. Reduction in core and yoke section
b. Reduction in mean length of LV and HV winding
c. Reduction in mean cost.
15. What are the disadvantages of using higher flux density in core?
a. Increased magnetizing current & Iron loss
b. Saturation of magnetic material
 c. Low efficiency
Increased temperature Rise
16. Why is the area of yoke of a transformer usually kept 15-20% more thanthat of
core?
The area of yoke is taken as 15 to 25 % larger than that of core of transformers
using hot rolled silicon steel. This reduces the value of flux density obtained in the yoke,
thereby resulting in the reduction of iron loss and the magnetizing current. In order to
compensate for reduction in flux, the number of turns has to be increased. Therefore, the core
area is not increased.
The section of the yoke may be taken as rectangular or it may be stepped. In rectangular
section yokes, the depth of the yoke is equal to the depth of the core. In stepped section, the
depth of the yoke is equal to the width of the largest stamping.
17. Why is low voltage winding placed near the core in transformers?
The winding & core are both made of metals and so on insulation has to be placed
between them. The thickness of insulation depends upon voltage rating of the
winding.
In order to reduce the insulation requirement, the low voltage winding is placed near
the core.
The low voltage winding is placed closer to the core because the high voltage
winding requires more space due to its MV insulation. If designers put the LV
winding outside the MV winding, the tank would be larger because the MV winding
size would be about the same size. More insulating oil would be required, and the
leakage reactance would be significantly higher.
Remember, higher leakage reactance means that the voltage regulation (more aptly
voltage drop) under load will be higher, which is not desirable.
18. What are the advantages of stepped cores? (M’15, D’14, 17, M’18, D’18)
For same area of cross section, the stepped cores will have lesser diameter of
circumscribing circle than square cores. This results in reduction in length of mean turn
of the winding with consequent reduction in both copper cost and copper loss.
19. Why are the cores of large transformers built-up of cross section? (M15)
The excessive leakage fluxes produced during short circuit and over loads develop
severe mechanical stress in the coils. On circular coils these forces are radial andthere is
no tendency for the coil to change its shape but on the rectangular coils the forces are
perpendicular to the conductors and tend to deform the coil in circular.
20. Why the efficiency of transformer is so high? (May 2016)
Since, transformer is a Static device (i.e. no moving parts in it) hence it is the most
efficient machines ever made by man. However, in rotating machines there are various losses
like friction and windage, losses in commutator etc. So, yes, Transformers are highly
efficient devices.
21. Why is transformer yoke designed for low flux density? (Dec 2015)
The transformer yoke is designed for low flux density because, higher flux density
gives rise to the following disadvantages:
 Increased magnetizing current and iron losses,
Saturation of magnetic material
Lower efficiency due to higher no-load losses
Higher temperature rise of transformers

22. How the magnetic curves are used for calculating the no-load current of the
transformer? (May2017)
The B-H curve can be used to find the mmf per metre for the flux densities in yoke and
core. The loss curve can be used to estimate the iron loss per kg for the flux densities in yoke
and core.
23. What are the factors affecting the choice of flux density of core in a transformer?
(May2011)
(i) Core and yoke area (ii) Overall size and weight of the transformer (iii)
Magnetizing current (iv) Iron loss (v) saturation and temperature rise.

24. What is meant by stacking factor? (May2012 & Dec2014)

In transformers, the core is made up of laminations and the laminations are insulated
from each other by a thin coating of varnish. Hence when the laminations are stacked
to form the core, the actual iron area will be less than the core area. The ratio of iron
area and total core is called stacking factor. The value is usually 0.9.
25. Give the necessary expression for yoke design of transformers.
For rectangular section of yokes,

Ay  D y  H y
Area of yoke A y =Depth of yoke x height of yoke
Ay-(1.15 to 1.25) Agi for transformers using hot rolled steel
Ay =Agi for transformer using grain-oriented steel

PART-B

1. A 4 pole, 400V, 960r.p.m shunt motor has an armature of 0.3m in diameter and 0.2m in
length .The commutator diameter is o.22m. Give full details of a suitable winding including the
number slots, number of commutator segments, and number of conductors in each slot for an
average flux density of 0.55wb/m2 in the gap. (NOV 2011)
2. Discuss the total design steps of D.C.machines. Briefly describe each step.

3. Estimate the main dimensions of a 4 pole 100kW, 1500 r.p.m D.C generator, assuming
specific electric and specific magnetic loading as 19000 amp. conductors per m and 0.4T
respectively. Length of the armature is equal to pole pitch. (APR 2011)

4. The field coil of a D.C. machine is to dissipate about 5kW at a supply voltage of 200V. The
winding space is 25x20 cm2. Take a winding space factor of 0.6 and specific resistance of
0.02 /m/mm2. Determine the number of turns of the coil, cross section of the conductor and the
total ampere-turns. Take the mean length of turn as 2m. (APR 2011)

5. A field coil has an internal diameter of 0.3m and an external diameter of 0.4m and a length of
0.175m. The outside surface can dissipate 1000W/m2 and the cooling effect of outer surface can be
neglected. The potential across is 100V. Calculate the ampere turns of the coil. Assume space
factor =0.6 and resistivity of copper 0.02 michroohm-m.
6. Design suitable commutator for a 500kW, 500V, 8pole, 375 r.p.m D.C shunt generator.
D=80cm,L=25cm. No. of commutator segments=192.

7. Determine suitable number of conductors, slots and commutator segments for a 4pole, 440V d.c.
motor that runs at 720 r.p.m at no load. The flux per pole is 0.024wb and the armature may be
designed to have wave winding. (APR 2010)

8. Calculate the main dimensions of an armature suitable for 4 pole, 20kw, and 1500r.p.m D.C.
generator. Assume amp. conductor /m as 18000, average gap density =0.387 and
efficiency= 90%.(NOV 2010).

9. A rectangular field coil is to produce an mmf of 7500 amp. turns when dissipating 220kw
at a temperature of 600C. The inner dimensions of the coil are 10cmx24cmxd15cmhigh. The heat
dissipation is 30w/m2/0C. from the outer surface, neglecting top and bottom surface of the
coil.
10. Temperature of ambient air is 200C. Calculate the thickness of the coil and the current
density. Resistivity of the conductor is 0.02 ohm per m and mm2. (APR 2011)

11. A 5kw 250V 4pole 1500r.p.m shunt generator is designed to have a square pole face. The
loadings are: average flux density in the gap=0.42wb/m2, ampere conductors per meter=15000.
Find the main dimensions of the machine. Assume full load efficiency=0.75 and ratio of pole
arc to pole pitch=0.65. (APR 2010), (MAY 2015)

12. Determine the main dimensions, number of poles and length of air gap of a 600kw, 500V,
950rpm generator. Assume average flux density as 0.5wb/m2 and ampere conductor per meter as
35200. The peripheral speed should not exceed 37 meter per second and the armature mmf per pole
should be below 7550Ats. The mmf required for and gap is 50% of armature mmf and gap
contraction factor is 1.15.
13. Calculate the main dimensions of a 20kw, 1000rpm D.C motor. Given, Bav=0.37T and
ac=16000amp.conductors/m. Make the necessary assumptions.
 UNIT-III

DESIGN OF DC MACHINES

PART-A

1. What are the main dimensions of rotating machine?

The main dimensions of a rotating machine are the
 Armature diameter or stator bore, D
 Armature or stator core length, L.

2. What is output equation?

The equation which relates the power output to the main dimensions (D and L),
specific loadings (Bav and ac) and speed (n) of a machine is known as output equation.
3. Write the output equation of DC machine and output co-efficient.
Pa = C0 D2L n , in kW
C0 = π 2 Bav ac x 10-3 in kW/m3 – rps
where
Pa – power developed in armature of dc
machineC0 – output co-efficient
D-Diameter of the machine, L-Length of the machine, n-Speed
in rps Bav-Specific Magnetic Loading, ac – Specific Electric
Loading
4. What is square pole criterion?
The square pole criterion states that for a given flux and cross-section area of pole,
the length of mean turn of field winding is minimum, when the periphery forms a square.
This implies that the length L must be approximately equal to pole arc, ie)L = b = ψ τ
5. State the parameters governing slot utilization factor or slot space factor.
 Voltage rating
 Thickness of insulation
 No. of conductors per slot
 Area of cross-section of the conductor
 Dimensions of conductor.

6. List the different types of slots that are used in rotating machines.
 Parallel sided slots with flat bottom
 Tapered slots with flat bottom
 Parallel sided slots with circular bottom
 Tapered slots with circular bottom
 Circular slots
 7. List the factors governing the length of armature core in dc machines. (D’14)
. Cost
 Voltage Ventilation
 between adjacent commutator segments
 Specific magnetic loading.

8. What are the factors to be considered for the choice of specific magnetic
loading?
 Flux density in teeth
 Frequency of flux reversals
 Size of machine

9. What are the factors to be considered for the choice of specific electricloading?
 Temperature rise
 Size of machine
 Speed of machine
 Voltage
 Armature reaction
 Commutation

10. What is meant by magnetic loading? (May 2012)

Total amount of flux available in air gap of armature periphery is called magnetic
loading.It is denoted by Bav. The value lies between 0.4 to 0.8 Tesla for a DC machine
11. What are the effects of higher specific electric loading in dc machines?
 Reduction in armature size
 Reduction in magnetic material and cost.
 Armature copper losses are increased.

12. Mention any two guiding factors for the choice of number of poles.
Frequency of flux reversal
Weight of iron
Weight of copper
Length of commutator.
Labour charge
Flash Over
Distribution of Distorted field form.

13. Why square pole is preferred? (May-2016, 2015, Dec-2013)

If the cross-section of the pole body is square than the length of the mean turn of field
winding is minimum. Hence to reduce the copper requirement a square cross- section is preferred
for the poles of dc machine.
14. Define specific electric and magnetic loading (Dec 2017 & Dec 2018)
Specific Electric Loading: The total amount of ampere conductors available at the
armature periphery per unit length is called specific electric loading.
 It is given by ac = Izz/πD, A/m, where Iz= current hrough conductor, z – no.of
conductors, D-diameter of the core.
Specific Magnetic loading: The average flux density over the air gap of a machine is
known as specific magnetic loading. Bav = Total flux around the air gap/Area of
flux path at the air gap = PΦ/πDL, where P- no.of poles, Φ- flux per pole,
D-diameter of the core , L- armature core length

15. State the advantages of having larger number of poles in DC Machines.

The following are the advantages of having larger number of poles in Dc machines.
There is a reduction in (i) Weight of armature core and yoke
(ii) Cost of armature and field conductors
(iii) Overall length and diameter of the machine
(iv) Length of commutator
16. Mention the factors that govern the choice of number of armature slots in a dc
machine.
 Slot pitch (mechanical difficulties)
 Cooling of armature conductors
 Flux pulsations
 Commutation
 Cost

17. What are the factors to be considered for deciding the slot dimensions?
 Flux density in tooth
 Flux pulsations
 Eddy current loss in conductors & Reactance voltage.

18. Discuss the parameters governing the length of commutator.

The length of the commutator depends upon the number of brushes and clearance between
the brushes. The surface area required to dissipate the heat generated by the commutator
losses is provided by keeping sufficient length of commutator.
19. What are the factors that influence the choice of commutator
diameter?(M’19)
 Peripheral speed
 Peripheral voltage gradient should be limited to 3 V/mm
 No. of coils in armature.

20. What are the materials used for brushes in DC machines?

 Natural graphite
 Hard carbon
 Electro graphite
 Metal graphite
21. Find out the cross-sectional area of armature conductor of a four pole wave
wound DC machine with armature current 200 A and armature winding
current density 4.5 A/mm2.
Current per conductor = 200/2 = 100 A
Cross sectional area of conductor = 100/4.5 = 22.2 mm2.
22. What is reactance voltage?
The reactance voltage is the induced voltage due to leakage flux. It is not beneficial.
The reactance voltage will affect commutation and produce sparking in DC machines.
The reactance voltage will increase the regulation in case of transformers and rotating
machines.
23. State the parameters governing slot utilization factor or slot space factor.
 Voltage rating
 Thickness of insulation
 No. of conductors per slot
 Area of cross-section of the conductor
 Dimensions of conductor.

24. Name any two methods to reduce armature reaction. (May 2011)
(i) Increasing the length of air gap at pole tips (ii) Increasing reluctance of poletips
(iii) Compensating winding (iv) Interpoles
25. What is meant by peripheral speed? Write the expression for peripheral speed
of a rotating machine.(May 2012 & Dec 2013)
The peripheral speed is a translational speed that may exist at the surface of the
rotor,while it is rotating. It is translational speed equivalent to the angular speed at the
surface of the rotor.
Peripheral speed Va= πDrn in m/sec.
26. What is meant by magnetic loading? (May 2012)
Total amount of flux available in air gap of armature periphery is called magnetic
loading. It is denoted by Bav. The value lies between 0.4 to 0.8 Tesla

PART B

1. Determine the dimensions of core and window for a 5KVA 50Hz single phase, core
type transformer. A rectangular core is used with long side twice as long as short side.
The window height is 3times the width. Voltage per turn=1.8V. Window space factor
=0.2. The current density in the conductor =1.8A/mm2, the flux density in the
core=1.00wb/m2. (NOV 2012),(NOV 2013)
2. A 250KVA, 6600/400V, 3 phase core type transformer has a total loss of 4800 watts on
full load. The transformer tank is 1.25m in height and 1mx0.5m in plan. Design a suitable
scheme for cooling tubes if the average temperature is to be limited to 35 0C. The diameter
of the tube is 50mm and tubes are spaced 75mm from each other. The average height of
the tube is 1.05m. (NOV 2012),(NOV 2013),(MAY 2015)(MAY 2016)

3. What is the object of employing a reduced flux density in the yoke of a transformer?

4. Discuss the relative merits and demerits of using water and oil for forced cooling of
transformer.
5. The tank of a 300KVA transformer is 100cmx45cmx150cmhigh. If the full load loss of
transformer is 8kw,find the suitable arrangement of cooling tubes having a diameter of
5cm and an average height of 100cm in order to keep the average temperature of the
external surface at 350C above the ambient temperature. (NOV 2011)
6. Determine the dimensions of limb and yoke of a 200KVA 50Hz single-phase
transformer. A cruciform core is used with distance between adjacent limbs equal to
1.6time the width of core laminations. Assume voltage per turn of 14V,maximum flux
density 1.1T,window space factor 0.32, current density 3A/mm2 and a stacking factor of
0.9. (NOV 2011)

7. The tank of a 500KVA, 11KV/415V, 3phase 50Hz transformer is 160cm in height and
55x110cm in plan. The full load loss of the transformer amounts to 5kw. Determine a
suitable arrangement of cooling tubes each having a diameter of 5cm and an average
height of 105cm in order to limit the temperature rise of the external surface at 350C
above ambient temperature. (APR 2012).

8. The voltage per turn of a 400KVA, 6.6KV/415V delta/star three phase core type power
transformer is 18.7volt. Calculate the number of turns per phase of the LV and HV
windings and suggest suitable cross section area of the conductor required. Assume a
current density of about 2.85A/mm2. (APR 2012)

9. The tank of a 250KVA single-phase oil filled, self-cooled transformer is 100cm in

height and 40x70cm in plan. Total loss to be dissipated on full load =3kw. Determine the
 arrangement of 5cm diameter cooling tubes spaced about 6cm between centers and
averaging 80cm in length. Take mean temperature rise of the tank as 350C. Sketch the
plan showing the arrangement of tubes. (APR 2011).

10.Estimate the main core dimensions, number of turns in the two windings and the
conductor sections in a 25KVA, 3 phase, 6.6KV/440V, 50H
1 2z delta-star core type
transformer with the following data: Stepped core with space factor=0.56. Space factor
for windings =0.25, voltage per turn=9V,current density=3.26A/mm2, maximum flux
density=1.1T. (APR 2011)
11. Determine the dimensions of core and yoke of a 200KVA 50Hz single-phase core
type transformer. A cruciform core is used with distance between adjacent limbs equal to
1.6 times the width of core lamination. Assume voltage per turn of 12 V, maximum flux
density=1.1T,window space factor of 0.32, current density of 3A/mm2 and a stacking
factor of 0.9. (NOV 2010)
12. The tank of a 150KVA oil immersed natural cooled transformer has dimensions
100cm x 60cm x 120cm height. Design a suitable arrangement of cooling tubes of mean
length 100cm to limit temperature rise to 350C, if the full load losses to be dissipated are
6kw. Make suitable assumptions wherever necessary. (NOV 2010)

UNIT-IV

DESIGN OF INDUCTION MOTORS

PART-A

1. Write down the output equation for 3-phase induction motor (Dec2011, May 2013)
The equation for input kVA is considered as output equation in induction motors.
The input kVA , Q=C0D2Lns in kVA
C0= output coefficient
D,L-main dimensions
ns= speed in rps.
The rating of induction motor is sometimes given in horse power. This rating refers
to the power output at the shaft of the motor.
kVA input , Q=(HP x 0.746)/(η x cos φ)
2. Why an induction is motor also called as a rotating transformer? (Dec 2018)
 The principle of operation of induction motor is electromagnetic induction, which is similar
to that of a transformer. The stator winding is equivalent to primary of a transformer and the rotor
winding is equivalent to short circuited secondary of a transformer. In transformer the secondary
is fixed but in induction motor it is allowed to rotate. The difference is that, the normal
transformer is an alternating flux transformer, while the induction motor is a rotating flux
transformer. Hence the induction motor is also called as a rotating transformer.
3. Write down the equation for output coefficient in an induction motor. (M’19)
The output coefficient of induction motor can be expressed as,
C0= 11kwsBav ac x 10-3 in kVA/m3
Where, C0= output coefficient
Bav – Average flux density
ac- ampere conductors
Kws- winding factor
The value of output coefficient depends on the value of specific electric loading (ac)
and specific magnetic loading (Bav)
4. Name the different types of induction motor. How do they differ from each other?
The two different types of induction motor are:
 Three phase induction motor
 Squirrel cage induction motor and
 Slip ring induction motor
 Single phase induction motor
The stator is identical for both types but they differ in the construction of rotor. The
squirrel cage rotor has copper or aluminium bars mounted on rotor slots and short
circuited at both ends by end rings. The slip ring rotor carries a three-phase winding.
One end of each phase is connected to a slip ring and other ends are star connected.
5. List the advantages of slip ring rotor over cage rotor.
 The main advantage of a slip ring induction motor is that its speed can be controlled
easily.
 "Pull-out torque" can be achieved even from zero rpm.
 It has a high starting torque when compared to squirrel cage induction motor.
Approximately 200 - 250% of its full-load torque.
 A squirrel cage induction motor takes 600% to 700% of the full load current, but a
slip ring induction motor takes a very low starting current approximately 250% to
350% of the full load current
6. Why is fractional slot winding not used in induction motor? (May 2019)
Windings with fractional number of slots per pole per phase create asymmetrical mmf
distribution around the air gap and favour the creation of noise in the motor. Therefore
fractional slot windings are not used in the stator of induction motors. The total number of
slots is also a multiple of number of phases. So, integral slot windings are used. In integral
slot winding, the total number of slots is chosen such that the number of slots per pole per
phase is an integer
 7. How can a 3-phase induction motor be designed for various design features?
The ratio of core length to pole pitch (ration L/τ) for various design features are: Minimum
cost- 1.5 to 2
Good power factor-1.0 to 1.25
Good efficiency -1.5
Good overall length -1
For best power factor the pole pitch τ is chosen such that,
  0.18L
The value of L/τ lies between 0.6 and 2 depending upon the size of machine and the
characteristics desired. For normal design, the diameter should be chosen that the
peripheral speed does not exceed about 30m/s.
8. What is slot space factor?
The slot space factor is the ratio of conductor (or copper) area per slot and slot area. It gives
an indication of the space occupied by the conductors and the space available for insulation. The
slot space factor for induction motor varies for 0.25 to 0.4. High voltage machines have lower
space factors owing to large thickness of insulation.

9. List the factors to be considered for the choice of specific magnetic loading.
The choice of specific magnetic loading depends on:
 Power factor – With large values of Bav results in magnetizing current will be high,
which results in poor power factor
 Iron loss - With large values of Bav increased iron loss and decreased efficiency
 Over load capacity
 With large values of Bav provides, large values of flux per pole, the turns per phase
will be less and so the leakage reactance will be less. Lower value of leakage
reactance results in higher overload capacity.

10. What are the factors to be considered for the choice of specific electric loading?
The choice of specific electric loading depends on:
 Copper loss
 Temperature rise- Large values of ac results in higher copper losses and higher
temperature rise
 Voltage rating –Machine with high voltage rating smaller values of ac should be
preferred. Since for high voltage machines the space required for insulation is large.
 Over load capacity- For high overload capacity lower values of ac should be
selected.
11. Comment on the selection of values of overload capacity of induction motor. What
is its impact if the “ac” value is higher?
A large value of ampere conductors would result in large number of turns per phase.
This would mean that, the leakage reactance of the machine becomes high and the
diameter of the circle diagram is reduced resulting in reduced value of overload
capacity. Therefore higher the values of a.c, lower would be the overload capacity.
 Hence the value of ampere conductors per metre depends upon the size of the motor,
voltage of stator winding, type of ventilation and the overload capacity desired. It varies
between 5000 to 45000 ampere conductors per meter
12. Mention the factors to be considered for the choice of number of slots of an
induction machine. (May 2015)
The factors to be considered for choice of number of slots of an induction machine
are
 Slot loading- slot loading should not exceed 750 amp. conductor
 Slot pitch - The slot pitch should lie between 15mm and 25mm
 Type of winding
 For integral slot winding the stator slots should be a multiple number of
slots per pole per phase.
 For double layer winding, the conductors per slot should be even.
Harmonic torque- Certain combinations of stator and rotor slots give rise to
harmonic torques which results in crawling and cogging. To avoid these
undesirable effects the difference between stator and rotor slots should not be
equal
13. List the factors to be considered for estimating the length of air gap in
induction motor.
The following factors are to be considered for estimating the length of air-gap:
 Power factor
 Overload capacity
 Pulsation loss
 Unbalanced magnetic pull
 Cooling.

14. What happens when the air-gap of an induction motor is doubled?(D’16)

(i) If the air gap of an induction motor is doubled then the mmf and magnetizing current
approximately doubles. Because in induction motor the air gap requires large mmf when
compared to rest of the magnetic circuit.
(ii) Also the increase in air gap length increases the overload capacity, offers better
cooling, reduces noise and reduces unbalanced magnetic pull.
15. Give the different formulas for calculating the length of air gap in induction
motors.
 For small induction motor,
lg  0.2  2 DLmm
 Alternate formula for small induction motor
lg  0.125  0.35D  L  0.015Va mm

 Alternate formula to use

 lg  1.6 D  0.25mm

lg  0.2  Dmm


 For machines with journal bearings

Where, D, L – main dimensions expressed in metre
Va= peripheral speed in metre per second.
16. Give the reasons for the rise of imbalanced magnetic pull in induction motors.
The design of frames has special significance in the case of induction motor of large
dimensions on account of the relatively small air gaps used in these machines. For induction
motors, the frame should be strong and rigid both during the construction and after assembly
of the machine. This is because the length of air gap is very small and if the frame is not rigid,
the rotor will not remain concentric with stator giving rise to imbalanced magnetic pull
17. Write down the rules for selecting rotor slots of an squirrel cage IM.
(May/June16) (Nov/Dec15)
The following general rules should be concerning the choice of rotor slots for squirrel cage
machines
 The number of rotor slots should never be equal to the stator slots, but it must be either
smaller or larger.
 The difference between stator and rotor slots should not be equal to 3p to avoid
synchronous cusps.
 The difference between the number of stator and rotor slots should not be equal to 3p for
3 phase machines in order to avoid magnetic locking.
 The difference between number of stator slots and rotor slots should not be equal to 1.2,
(p+1) or (p+2) to avoid noise and vibrations.

18. Where is mush winding used? (or) Where is basket winding used in
Induction motors? (May15)

The mush windings are used in small induction motors of ratings less than 5 HP having
circular conductors. This is a single layer winding where all the coils have the same coil span
(unlike concentric winding where the coils have different span). For small induction motors of
the slip ring type, it is normal practice to use mush windings for rotor housed in semi closed
slots. The coils are roughly formed outside the machine and dropped one by one, into the
previously insulated slots.
 For large motors, a double layer bar type wave winding is used. This winding has generally
two bars per slot. The bars are pushed through partially closed slots and are bent to shape at the
other end.
19. Show a relation between D and L for best power factor. (May 2016)
For best power fator the relation between D and L is as follows:
τ= 0.18L
where τ- pole pitch, L=Length of stator core.
20. What are the factors to be considered for selecting the number of slots in
induction machine stator? (May 2012)(Dec 2017)
The factors to be considered for selecting the number of slots is tooth pulsation loss,
leakage reactance, magnetizing current, iron loss and cost. Also, the number of slots should be
multiple of slots per pole per phase for integral slot winding.
21. Why rotor slots are skewed?
i.) To reduce noise and vibrations, ii.)To avoid tendency of cogging,
iii.) To reducetorque defects
22. How can crawling be prevented by design in an induction motor? (M’14)
Crawling is a phenomenon in which the induction motor cannot accelerate upon its full
speed but continues to run at a speed lesser than sub synchronous speed. This action is due to
the fact that, flux wave produced by a stator winding is not purely sinusoidal. Instead, it is a
complex wave consisting of a fundamental wave and odd harmonics like 3rd, 5th, 7th etc.
Slotting produces harmonics of the order 6Aq± 1 in a three-phase machine, where A is any
integer. Therefore, to prevent crawling, it is necessary to avoid the values of rotor slots
exceeding stator slots by about 15-30%.
23. What is meant by cogging? (or) Why in an induction motor, the number of
stator slots should never be equal to the number of rotor slots? (Dec 2018)
When the number of rotor slots is equal to the number of stator slots, the speeds of all the
harmonics produced by the stator slotting, coincides with the speed of corresponding rotor
harmonics. Thus, the harmonics of every order would try to exert synchronous torques at their
corresponding synchronous speeds and the machine would refuse to start. This is known as
cogging. Therefore, the number of stator slots should never be equal to the number of rotor slots
in an induction motor.
 PART B

1. Estimate the stator core dimensions, number of stator slots and number of stator conductors per
slot for a 100kw, 3300V, 50Hz, 12 pole star connected slip ring induction motor. Assume average
gap density =0.4wb/m2; Conductors per metre =25000A/m; Efficiency =0.9; Power factor =0.9
and winding factor =0.96. Choose main dimensions to give best power factor.
2. The slot loading should not exceed 500-ampere conductors.(MAY 2013), (NOV 2013).

3. Explain the design of rotor bars and end rings of induction motor.(NOV 2013).
4. Explain the design of induction motor using circle diagram. (Nov.2003)
5. Estimate the main dimensions, air gap length, stator slots, stator turns per phase and cross
sectional area of stator and rotor conductors for a 3 phase, 15HP, 400V, 6 pole, 50Hz, 975 rpm
induction motor. The motor is suitable for star delta starting. Average flux density
=0.45wb/m2, specific electric loading =20000 amp. Conductors/m. The ratio of core length to
pole arc =0.85; full load efficiency =0.9; power factor =0.85. (APR 2009)

6. Discuss the various factors to be considered in the design of induction motor. Discuss the end
ring current briefly.
7. An 11kw, 3phase, 6 pole, 50Hz, delta connected induction motor has 54 stator slots, each
containing 9 conductors. Calculate the value of bar current and end ring currents. The number
rotor bars are 64. The machine has an efficiency of 0.85 and a power factor of 0.86 at full
load. The rotor mmf is assumed as 84% of stator mmf. Also find the bar and end ring current
sections if the current density is 5A/m2. (APR 2009).

8. Derive an expression for the equivalent resistance of cage rotor referred to stator per phase of
three-phase induction motor.
9. Describe the importance of dispersion coefficient and power factor in the design of induction
motor. Explain why the air gap of an induction motor is made as small as possible. (NOV
2009)
10. State and discuss the factors, which influence the ratio of length to diameter of the armature
core of a 3phase induction motor.
11. Bring out the main difference in the design procedure of a three-phase cage induction
motor with three-phase slip ring induction motor.

12. A 3.75kw, 440V, 3phase, 4 pole induction motor with 375 turns per phase in the stator
has the following data: rotor slots=30; rotor bar size =8.5x6mm2; length of each bar
=12.5cm; end ring size =10x15mm2;mean diameter of the end ring =12.5cm. The bars
andend rings are made of copper. Calculate the rotor resistance. (NOV 2006)

9
 UNIT – V
DESIGN OF SYNCHRONOUS MACHINES
PART-A

1. Name the two types of synchronous machines.

Based on the construction the synchronous machines are classified as,
 Salient pole machines
 Cylindrical rotor machines

2. What is run away speed of Synchronous Machine?(M’11, ‘12, D’13’17’18)

Speed at which the prime mover would have if suddenly unloaded when working at
its rated load is known as runaway speed.

3. List the disadvantages of high Bav.

• Higher losses and temperature rise
• Reduced efficiency
• Increased transient short circuit.

4. What is critical speed of alternator?

When the rotor of the alternator has an eccentricity, it may have a deflection while
rotating. This deflection will be maximum at a speed called critical speed. When a rotor
with eccentricity passes through critical speed, severe vibrations are developed
5. Why alternators are rated in kVA? (May-2015)
The kVA rating of ac machine depends on power factor of load. The power factor
in turn depends on the operating conditions. The operating conditions differ from place to
place. Therefore the kVA rating is specified for all machines
6. How is cylindrical pole different from salient pole in a synchronous machine?
(May 2015, Dec 2018)
i.) Cylindrical pole are non-projecting pole whereas the salient pole machines are
projecting pole.
ii.) Cylindrical rotor construction is used for turbo alternators which are driven by high
speed steam or gas turbines whereas the salient pole construction is used for generators driven
by hydraulic turbine since these turbines operate at relatively low speeds.
7. What are the factors to be considered for the choice of specific magnetic
loading in synchronous machines?(M’16, Dec-2015,May 2019)
o Iron loss
o Stability
o Voltage rating
o Parallel operation
o Transient short circuit current

9. What are the factors to be considered for the choice of specific electric
loadingin synchronous machines?
 Copper loss
 Synchronous reactance
 Temperature rise
  Stray load losses
 Voltage rating

10. State the factors for separation of D and L for cylindrical rotor machine.
The separation of D and L in cylindrical rotor machine depends on the
following factors.
Peripheral speed
Number of poles.
Short Circuit Ratio.

11. Define SCR of a Synchronous machine. (Dec 2015, May 2019)

The SCR is defined as the ratio of field current required to produce rated voltage on
open circuit to field current required to calculate rated current at short circuit.
12. What are the effects of SCR on machine performance? (May 2017)
For high stability and low regulation, the value of SCR should be high, which requires
large airgap. When the length of airgap is large, the mmf requirement will be high and so
the field system will be large. Hence the machine will be costlier.
13. What are the advantages of large airgap in synchronous machines?
 Reduction in armature reaction
 Better cooling
 Small value of regulation
 Lower tooth pulsation loss
 Higher value of stability
 Less noise
 Smaller unbalanced magnetic pull.

14. List the influence of airgap on the performance of the synchronous

machine.
The synchronous machines usually have large airgap. With increase in airgap length
the following factors decreases.
Armature reaction
Regulation
Noise
Tooth pulsation loss
Unbalanced magnetic pull
Sensitivity to load variation.

15. What are advantages of low values of SCR?

Cost of control equipments is reduced
Reduction in size of machines
Reduction in overall cost of machines.

16. What are the disadvantages of low value of SCR?

Poor stability limit
Poor voltage regulation
Unsatisfactory parallel operation
17. Give expression for armature ampere turns/ pole.
ATa =1.35 Tph . Iph. Kw / P

Tph = turns/phase; Iph = current/Phase: Kw = winding factor, P= Pair of poles.

18. How the value of SCR affects the design of alternator? (May 2012)
 For high stability and low regulation the value of SCR should be high, which requires
larger airgap. When the length of airgap is large, the mmf requirement will be high and so the
field system will be large. Hence the machine will be costlier.
19. What is the limiting factor for the diameter of synchronous machines?
The limiting value of peripheral speed is 175 m / sec for cylindrical rotor
machines and 80 m/sec for salient pole machines.
20. Write the expression for air gap length in cylindrical rotor machine?
For smooth cylindrical rotor the air gap length,
lg = 0.5 (SCR) ac τ Kf x 10 -6 / Kg Bavg

where, SCR-short circuit ratio, ac- specific electric loading, τ- pole pitch, Kf – form factor,
Kg- gap contraction factor, Bavg- average flux density
21. What is the effect of high value of dispersion coefficient?
The effect of having high value of dispersion coefficient is as follows: Poor power
factor, Reduced over – load capacity, reduced output power.
22. Give the need for damper winding in synchronous machine? (M’18)
The primary function of damper winding in a synchronous machine is to prevent the
phenomenon called hunting. If load on a synchronous machine is changed suddenly, its
power angle has to change accordingly. However, change in power angle doesn't occur
smoothly. The rotor starts swinging around the new power angle.
23. How the dimensions of induction generator differ from that of an induction
motor? (May 2015)
Dimensions of induction generator and induction motor are same. Energy
conversionprocess is reversible. Therefore, induction motor can operate as induction
generator.

PART B
1. Determine suitable stator dimensions for a 500KVA, 50Hz, 3Φ, alternator to run at 375
rpm.Take mean gap density over the pole pitch as 0.55T, the electric loading as 250 amp.
Conductors per cm and assume a peripheral speed not exceeding 35m/s.
2. Deduce an expression for the output equation of a synchronous machine (alternator). Also
derive its output coefficient.
3. Discuss briefly the factors, which influence the air gap length of a 3phase synchronous machine.

4. Explain the factors taken into account in the design of field winding of a salient pole alternator.
5. Determine for a 250KVA, 12 pole, 1100V, 500 rpm, 3-phase alternator i) air gap diameter
ii) core length iii) no. of stator conductors iv) no. of stator slots v) cross section of stator
conductors. Assume average gap density as 0.6wb/m2 and the specific electric loading 30,000 a.c/m.
(APR 2009)

6. Describe the important constructional features of the rotating field systems of slow speed alternators
and turbo alternators. (APR 2012)
7. A 500KVA, 3.3kV, 50Hz, 600 rpm, 3 Phase salient pole alternator ahs 200 turns per phase. Estimate
the length of the air gap if the average flux density is 0.55T, the ratio of pole arc to pole pitch is 0.65,
 short circuit ratio is 1.5, gap expansion factor is 1.15, mmf required for the gap is 80% of no0 load
field mmf and the winding factor is 0.955. (APR 2012)

8. Explain how the open circuit characteristics is to be obtained from the design data for a salient pole
alternator. (NOV 2012)

9. Discuss the factors leading to the choice of length of air gap in alternator design.
10. Explain what steps are taken to ensure that that an alternator shall generate an emf, the waveform
of which shallbe close approximation to a sine wave.
11. Explain the influence of air gap length on the performance of synchronous machines with regard to
i) voltage regulation ii) stability iii) synchronizing power.
12. What is a short circuit ratio as connected with synchronous machines? Mention the usual magnetic
loading factors. (APR 2012)(MAY 2016)
 EE8005- SPECIAL ELECTRICAL MACHINES
UNIT-1
STEPPER MOTORS
PART-A
1. Define: Stepper motor?
Stepper motor is a motor which rotates step by step and not by continuous rotation. When the
stator is excited using a DC supply the rotor poles align with the stator poles in opposition such
that the reluctance is minimum.

2. What are the advantages of Stepper motor?

No feedback is normally required for either position control or speed control. Positional
control is non – cumulative. Stepping motor is compatible with modern digital equipment.

3. Mention the different types of stepper motor. (Dec 2018)

Variable Reluctance stepper motor (Single stack, Multi stack), Permanent magnet stepper motor,
Hybrid stepper motor, Outer rotor stepper motor.
4. Define: Step Angle of stepper motor.( June 2016)
A stepping motor rotates through a fixed angle for every pulse. The rated value of this
angle is called the step angle and expressed in degrees.

5. Define: Holding torque and Detent torque of stepper motor.(Dec 2018)

Holding torque is defined as the maximum static torque that can be applied to the shaft
of an excited motor without causing continuous rotation.
It is defined as the maximum static torque that can be applied to the shaft of an unexcited
motor without causing continuous rotation.

6. Define: Resolution of stepper motor.

It is defined as the accuracy of positioning of the rotor pole at a particular step angle with
respect to stator pole. Most modern drives implement micro stepping to increase resolution
and motion smoothness. Most common stepper resolution is 200 full steps per revolution
but when driven for example with 16-microstep drive, the resulting resolution is 1600 steps
per revolution (1/1600 revolutions).

7. Define: Pull – in torque of stepper motor.

They are alternatively called the starting characteristics and refer to the range of frictional load
torque at which the motor can start and stop without losing steps for various frequencies in a
pulse strain.

8. Define: Pull – out torque of stepper motor.

These are alternatively called the slewing characteristics. After the test, motor is started
by a specified driver in the specified excitation mode in the self-starting range; the pulse
frequency is gradually increased; the motor will eventually run out of synchronism. The relation
between the frictional load torque and the minimum pulse frequency with which the motor can
synchronize is called pull – out characteristics.
9. Define: Slewing frequency of stepper motor.
This is defined as the maximum frequency (stepping rate) at which the loaded
motor can run without losing steps is alternatively called the maximum slewing
frequency. The range of stepping rate which the motor follows without missing a step
indicates that the motor is started and synchronized. This area of operation of the stepper
motor is called slew range. The motor is said to be operating in slewing mode.

10. Define: Stepping frequency of stepper motor.

The speed of rotation of a stepping motor is given in terms of the number of steps
per second and the term stepping rate is often used to indicate speed. Stepper motors have
a natural frequency of operation. When the excitation frequency matches the resonance,
steps may be missed and stalling is more likely to occur.

11. Define: Maximum starting torque of stepper motor.

This is alternatively called as maximum pull – in torque and is defined as the maximum
frictional load with which the motor can start and synchronize with the pulse train of frequency
as low as 10

12. Mention the features of stepper motor.

The various features of stepper motor are small step angle, High positioning accuracy,
High torque inertia ratio, Stepping rate, Pulsefrequenc

13. Why interleaving is done in a stepper motor?

Interleaving is done in . The stepper motor to decrease the step angle and thus increasing
the resolution. Interleaving is half stepping, when coil pairs are energized simultaneously in
order to get double resolution.

14. Write short notes on VR type stepper motor.

It is a basic type of stepping motor in which the motor step by step rotation is achieved
when the rotor teeth and stator teeth are in alignment such that the magnetic reluctance is
minimized and this state provides a rest or equilibrium position.

15. Write short notes on PM type stepper motor.

A stepping motor using permanent magnet in the rotor for step movement is called a
permanent magnet motor. The Permanent Magnet Stepper Motor has a stator construction similar
to that of the single stack variable

16. What are the different modes of excitation? (Nov 17)

Stepper drives control how a stepper motor operates; there are three commonly used excitation
modes for stepper motors, full step, half step and micro stepping. These excitation modes have an
effect on both the running properties and torque the motor delivers.
17. In what way closed loop control is advantageous when compared to open loop control
in stepper motor.
Closed loop control is more accurate, oscillatory motions are avoided for certain speed ranges,
Speed remains constant for high inertial load, follows the input pulses at stepping frequency are
some of the advantagesover open loop control. But it is costly and complex.

18. Calculate the stepping angle for a 3 phase, 24 pole permanent magnet stepper motor. (Dec 12)
Step angle β = 360/ (no. of stator phases * no. of rotor teeth) =5º.

19. Define torque constant of a stepper motor.

The torque constant of the stepper motor is defined as the initial slope of the torque
current curve of the stepper motor. The torque constant, Kt, ofa motor is a very useful parameter
for sizing and controlling motors showing a linear speed / torque relationship.

20. What is the function of driver circuit in stepper motor?

The stepper motor is a digital device that needs binary signals for its operation. The power
driver is essentially a current amplifier, since the sequence generator can supply only logic but
not any power.

21. Give the difference between single and multi-stack stepping motors.

Sl.No Single Stack Stepper Motor Multi Stack Stepper Motor

1 The number of stator poles should The stator and rotor have same
be different that of the rotor poles number of poles and same pole
in order to pitch.
have self-starting capability and
bidirectional rotation.
2 In single stack each and every It is used to obtain small step
stator pole carries afield coil. sizes. It consists of m identical
single stack variable reluctance
motor with the rotor mounts on
the single shaft.
22. Distinguish the half step and full step operation of a stepper motor.(Nov 17)

HALF STEP
SL.NO FULL STEP OPERATION
OPERATION
1 It is defined as the alternate It is the one phase on mode operation
one phase on and two phases .It means at that time only one winding
on mode is energized.
operation.
2 Rotor rotates on each step By energizing one stator winding
angle is half of the full step the rotor rotates at some angle.it s
angle. full step
operating.

23. Define the micro stepping mode of stepper motor.

Micro stepping means, the step angle of the VR stepper motor is very small. It is also called
mini stepping. It can be achieved by two phases simultaneously as in 2 phases on mode but with
two currents deliberatelymade unequal.

24. Name the various driver circuits used in stepper motor. (June 2016)

Driver circuits for stepper motor are broadly classified into Unipolar and
Bipolar driver circuits. Based on the supply voltage given to stator windings they are classified as L/R
driver circuit, Chopper drive circuit, H bridge drivecircuit.

PART-B

1. Explain the construction and principle of operation of Variable Reluctance Stepping motor. Also
explain the working of Single stack type, multi stack type and Micro stepping (May 2019).
2. Explain the construction and operation of hybrid Stepping motor.
3. Discuss the static and dynamic characteristics of stepper motor with neat sketch. ( Dec 2018)
4. Explain in detail the drive system of a stepping motor.(April 17,Nov 17)
5. Explain in detail the multi stack Variable Reluctance Stepping motor. (Dec 2016, June 2016)
6. What is the motor torque Tm required to accelerate an initial load of 3x10-4 kg – m2 from f1 =
1000Hz to f2 = 2000Hz during 100msec. The frictional torque Tf is 0.05 Nm and the step angle is 1.8°.
7. With a neat block diagram explain the closed loop operation of stepping motor.
8. a) Explain the concept of torque production in variable reluctance stepping motor.
b) A single stack 3-phase variable reluctance motor has a step angle of 15º. Find the number of stator
and rotor poles.
9. What is the motor torque Tm required to accelerate an initial load of 2x10-4 kg-m2from 500Hz to
1500Hz in 50 ms. the frictional torque is 0.03 Nm and step angle is 1.18º.
10. a) With a neat block diagram explain the microprocessor control of stepper motor.(April 17,Nov 17)
b What are the advantages of closed loop control of stepper motor.
11. Explain the modes of excitation of a stepper motor with neat diagram. (Dec 2016)
12. A stepper motor has revolution of 180 steps/rev. Find the pulse rate required in order to obtain a
speed of 2400rpm. (Dec 2016)
13. Explain the modes of operation of variable reluctance stepper motor. (Dec 2018, May 2019)
14. A stepper motor driven by a bipolar drive circuit has the following parameters: Winding
inductance= 30 mH, rated current = 3A, DC supply = 45V, total resistance in each phase = 15ohm.
When the transistors are turned off, determine (i) the time taken by the phase current to delay to zero
and (ii) the proportion of the stored inductive energy returned to the supply.(Dec 2018)

UNIT – II
SWITCHED RELUCTANCE MOTORS (SRM)
PART – A
1. What is switched reluctance motor?
The switched reluctance motor is a doubly salient, singly excited motor. This means that it
has salient poles on both the rotor and the stator, but only one member (usually the stator) carries
windings. The rotor has no windings, magnets, or cage windings, but is built up from a stack of
salient-pole laminations.

2. What are the advantages of Switched Reluctance motor? (Dec 2018)

Rotor is simple and it tends to have a low inertia. The stator is simple to wind. In most
applications the bulk of the losses appear on the stator, which is relatively easy to cool. Due to
the absence of magnet the maximum permissible rotor temperature may be higher than in PM
motors. Under fault conditions the open circuit voltage and short circuit current are zero or
varying small. Extreme by high speeds are possible.

3. What are the applications of Switched Reluctance motor? (Dec 2016,April 17)
SRM motors are used in Precision position control system for Robotics and Low power
servo motor. High-speed motors have the advantage of high power density, which is an
important issue of traction motors in electric vehicles (EV).
4. What is the difference between Switched Reluctance motor and variable reluctance stepper
motor?

Switched Reluctance motor Variable reluctance stepper motor

Conduction angle for phase current is Stepper motor is usually fed with a
controlled and synchronized with the square wave of phase current without
rotor position, usually by means of a rotor positionfeedback.
shaft position sensor.
 The SRM is designed for high speed It is usually designed with a limited
speed
Closed loop control is necessary Closed loop control is required for
high frequency operation

5. Give the basic features or characteristics of Switched Reluctance motor.

The switched reluctance motor is a doubly-salient, singly-excited motor. This means that it
has salient poles on both the rotor and the stator but only one member (usually the stator) carries
windings. The rotor has no windings, magnets, or cage windings, but is built up from a stack of
salient-pole laminations. Low inertia and simple manufacturing, Losses appear only on the stator
and easy to cool, No magnets and so permissible rotor temperature is higher than in PM motors,
Torque is independent of the polarity of phase current. Reduction in no of semiconductor devices
in controller Open circuit voltage and Short circuit current are zero or very small under faulty
condition, Immune from shoot through failure, High starting torque, extremely high speeds
possible.
6. What are the disadvantages of a Switched Reluctance motor?
The absence of free PM excitation imposes the burden of excitation on the stator windings and the
controllers and increases the per unit copper losses, is limited, torque/ampere is limited, Non
uniform nature of the torque production which leads to torque ripple and may contribute to
acoustic noise.

7. Mention different modes of operation of SRM. (Nov 17)

The different modes of operation of Switched Reluctance motor are Low speed operating mode
and High Speed operating mode. Monitoring of exciting current during low speed operation is
essential because of long duration of each phase period and needs chopping of energization to
restrict each phase current within the semiconductor ratings. Moreover, the developed torque is
controlled by varying the average phase current. During high speed operation, current control is
not essential because the inductance of the winding and the motional back emf induced restrict
the excitation to single pulses of current. Torque is controlled by optimal positioning of these
pulses rather than the current level.
8. Mention the applications of micro stepping VR stepper motor?
Micro stepping operation mode of VR stepper motor is used in following applications.
Printers, digital cameras, CNC machines, toys, robotics, fine position control systems.

9. Write the relations between the speed and fundamental switchingfrequency.

f=nNr=(r.p.m./60)Nr Hz , Nr=No. of rotor poles, If there are q phases there are q Nr steps per
revolution and the step angle or stroke is given by ε=2π/(qNr) rad. The number of stator poles
usually exceeds the number ofrotor poles.

10. What is co-energy?

 In the ψ –i curve of a motor, the area between the curve and horizontal i axis is the co-energy W’
and the other part is the stored field energy Wf. Co-energy is expressed in the same units as energy
and is especially usefulfor calculation of magnetic forces and torque in rotating machines.
11. Give the expression for torque of a Switched Reluctance motor.
The torque is given by T = [∂W’/∂θ] i=const or by T = [∂Wf / ∂θ] ψ=const.
With magnetic saturation negligible and with ψ–i curve straight line,ψ=Li, W’=Wf=(½)Li2z,
T = (½)i2dL/dθ Nm where T is the torque, L is the inductance, Wf is thestored field energy.

12. Why rotor position sensor is essential for the operation of SwitchedReluctance motor?
(Dec 2016)
It is normally necessary to use a rotor position sensor for communication and speed feedback.
The turning ON and OFF operation of the various devices of power semiconductor switching
circuit are influenced by signalsobtained from rotor position sensor.
13. Define: Chopping mode of operation of Switched Reluctance motor.
In this mode, also called low – speed mode, each phase winding gets excited for a period
which is sufficiently long. As the speed is low in position control, the Switched reluctance motor
operates in PWM mode.
Soft-chopping control is used in motoring mode while hard- chopping control is applied
in braking mode to hold on the load torque and to provide stable operation at zero speed.
14. What are the types of power controllers used for Switched Reluctancemotor? (May
2019)
The power semiconductor switching circuits used are 1. Two power semiconductor switching
devices per phase and two diodes. 2. (n+1) power semiconductor switching devices (n+1) diodes.
3. Phase winding using bifilar wires. 4. Split-link circuit used with even-phase number. 5. C-
dumpcircuit.

15. What are the two types of current control techniques?

The two types of current control techniques of Switched Reluctance motor are Hysteresis type
control and PWM type control. Hysteresis type control is used to compensate contact bounce in
switches or noise in an electrical signal. Pulse width modulation (PWM) is a powerful technique
for controlling analog circuits with a microprocessor's digital outputs. PWM is employed in a
wide variety of applications, ranging from measurement and communications to power control
and conversion.
16. What is the step angle of a 5 phase Switched Reluctance motor and commutation frequency in
each phase for the speed of 6000 rpm. SRM having 10 stator poles and 4 rotor poles.
Solution: Step angle =(2π/qNr)=(360°/5*4)=18°.
Commutation frequency at each phase=(Nr*ω)/2π=(4*6000)/60=400Hz.[ω = 2πN].

17. What are the merits of Dump C – Converter?

 The Dump C-Converter topology uses lower number of switching devices and has only one
switch voltage drop, the converter has full regenerative capability, and there is faster
demagnetization of phases during commutation.

18. What are the merits of split power supply Converter?

The merits of split power supply converter are it requires lower number of switching devices,
there is faster demagnetization of phases during commutation.
19. What are the merits of classic converter or power controller in SRM?
Control of each phase is completely independent of the other phases; the energy from the off
going phase is feedback to the source, which results in useful utilization of the energy.

20. Why SR machines popular in adjustable speed drives?

Rotor is simple and it tends to have a low inertia, The stator is simple to wind, In most
applications the bulk of the losses appear on the stator, which is relatively easy to cool, Due to
the absence of magnet the maximum permissible rotor temperature may be higher than in PM
motors, Under fault conditions the open circuit voltage and short circuit current are zero or
varying small, Extreme by high speeds are possible.
21. Mention some position sensors used in switched reluctance motor. (Dec2018)
Switched reluctance motor is always operated with closed loop control. Normally we
have to use a rotor position sensor for commutation and speed feedback. Here the phase
windings are energized by using power semiconductor circuit. Some of the position sensors are
Optical encoder,resolver, Speed sensors and Hall Effect sensor.

PART-B

1. Explain the construction and working of Switched Reluctance motor. (April 17, Nov 17, Dec 2018)
2. With a block diagram explain the importance of closed loop control of Switched Reluctance motor.
3. Describe the Hysteresis type and PWM type current regulator for one phase of Switched Reluctance
motor.
4. Explain in detail about microprocessor based control of Switched Reluctance motor. (April 17)
5. a) Describe the configuration of various power controller circuits to Switched Reluctance motor and
6. Explain the operation of any twos schemes with suitable circuit diagram. (May 2019)
b) State the advantages of sensorless operation. (June 2016)
7. Describe the various operating modes of Switched Reluctance motor.
8.Derive the voltage and torque equation of Switched Reluctance motor and also explain the torque –
speed characteristics of Switched Reluctance motor.( Dec 2016)
9. Describe the construction and working of rotary and linear switched reluctance motors. (June 2016)
10. Discuss the methods of rotor position sensing and sensor less operation.
11. Compare and contrast the performance of SR motor and VR stepper motor.
12. Derive the expression for static torque in SRM. (Dec 2016)
13. Draw a schematic diagram and explain the operation of a “C” dump converter used for the control of
SRM. (May 2019).
 UNIT – III
PERMANENT MAGNET BRUSHLESS D.C. MOTORS
PART – A

1. Why adjustable speed drives are preferred over a fixed speed motor?
The common reasons for preferring an adjustable speed drives over a fixed speed motor are:
Energy saving e.g. Fan or pump flow process, Velocity and position control e.g. Electric train,
portable tools, washing machine, Amelioration of transients: Starting and stopping of motors
produce sudden transients. It can be smoothened using adjustable speeddrives.

2. What is the structure of an adjustable speed drive system?

An adjustable speed drive consists of an electric motor and controller that is used to adjust
the motor's operating speed. The combination of a constant-speed motor and a continuously
adjustable mechanical speed- changing device might also be called an adjustable speed drive.
The general structure of a motion control system or drive consists of the following elements: The
load, the motor, the power electronic converter; and the control.
3. Write briefly about the construction and types of a Brushless DCmachines.

Brushless PM machines are constructed with the electric winding on the stator and PMs on the
rotor. There are several conventional PM machine configurations and other more novel concepts
conceived in recent years to improve performance.
The configuration of a PM machine and the relationship of the rotor to the stator determine the
geometry and the shape of the rotating magnetic field. PM machines in which the magnetic flux travels
in the radial direction are classified as radial-flux machines.
They are cylindrical in shape, and the rotor is usually located inside the stator but can also be
placed outside the stator. PM machines in which the magnetic flux travels in the axial direction are
classified as axial-gap machines. They can have multiple disk or pancake-shaped rotors and stators. The
stator-rotor-stator configuration is typical.
4. What are the advantages of PM machine? (May 2019)
In general, PM machines have a higher efficiency as a result of the passive, PM-based field
excitation. PM machines have the highest power density compared with other types of electric
machines, which implies that they are lighter and occupy less space for a given power rating.
The amount of magnet material that is required for a given power rating is a key cost
consideration. The cost of magnet material is high compared with the cost of the other materials used
in electric motors, and design attributes that minimize the required amount of magnet material are
important considerations in motor selection.
The stators of PM machines are generally fabricated in the same manner as induction
machine stators; however, modifications are sometimes necessary, such as the design of a stator
lamination to accommodate high flux density.
5. What are the types of PM machines?
The types of Permanent magnet machines are
1.Interior PM Machine 2. Surface mounted PM machine.
A type of motor that has a rotor embedded with permanent magnets is called the IPM (interior
permanent magnet) type. Compared with the SPM (surface permanent magnet), this type of
 motor can reduce the risk of a magnet being peeled off by centrifugal force, and take advantage
of reluctance torque. The IPM type allows various structures for embedding permanent magnets.
6. What are the differences between mechanical and electroniccommutator? (Dec 2016)

MECHANICAL ELECTRONIC
COMMUTATOR COMMUTATOR
1. Commutator arrangement is 1. Commutator arrangement is
located in the rotor. located in the stator.
2. Shaft position sensing is 2. It requires a separate rotor position
inherent in the arrangement sensor.
3. Sparking takes place. 3. There is no sparking.
4. It requires regular 4. It requires less maintenance.
maintenance.
5. Sliding contacts between 5. No sliding contacts.
commutator and brushes.

7. What is meant by Permeance coefficient?

The line drawn from the origin through the operating point is called load line and absolute value
of its slope normalized to µ0 is called permeance coefficient.
Permeance coefficient= µrec((1+PrlRg)/(PmoRg))
where µrec=relative recoilpermeability,
Rg=air gap reluctance,
Pmo=internal leakage permeance,
Prl=normalized rotor leakage permeance.
8. Discuss briefly about the types of Permanent Magnets used in electricalmachines. ( May
2019)

PM strength and other key properties: The various types of PMs include the following:
Alnico—a family of magnets made from aluminum, nickel, and cobalt characterized by excellent
temperature stability, high residual induction, and enough energy for a number of industrial and
commercialapplications. Ceramic—a hard, low-cost ferrite made of barium and strontium ferrite
with excellent stability. Ceramic magnets tend to be brittle, hard, and resistant to corrosion.

9. What is commutation?
Commutation is the process in which generated alternating current in the armature winding
of a dc machine is converted into direct current after going through the commutator and the
stationary brushes. it will takes place, the coil undergoing commutation is short circuited by the
brush.
10. Draw the magnetic equivalent circuit of PMBLDC motor.

11. Compare brushless DC motor with P.M. commutator motor.(Dec2018)

Brushless DC motor P.M. Commutator motor

1. No Brushes. Maintenance problems (RFI, 1. Commutator based DCmachines
sparking, ignition and fire accidents) need carbon brushes, so sparking and
eliminated. wearand tear is un avoidable.

2. More cross sectional available for armature 2. Armature winding is insideand the
windings. Conduction of heat through the magnet is on the stator outside.
frame is improved.

3. Increase in electric loading ispossible, 3. Efficiency less.

providing a greater specific torque. Higher
efficiency.
4. Space saving, higher speedpossible, 4. Commutator restrictsspeed.
with reduced inertia.

5. Maximum speed limited byretention of 5. Magnet is on the stator. Noproblem.

magnet against centrifugal force.

6. Shaft position sensor is a must. 6. Not mandatory.

7. Complex electronics forcontroller. 7. Simple

12. Give the emf and torque equations of the square wave BLDCmotor. (Nov 17)
The emf equation is given by E = kφω and the torque equation is given byT = kφI
where k is the armature constant depending on the number of turns in series per phase in the
armature winding,
ω is rotor speed in rad / sec and
φ is the flux ( mainly contributed by the Permanent Magnet onthe rotor). I is the load current.
13. What is meant by demagnetization in PM-BLDC motor?
In the absence of externally applies ampere turn, the magnets operating point is at the
intersection of demagnetization curve and the load line.

Demagnetization curve
13. Write the principle of operation of PM-BLDC motor. (Nov 17)

When a D.C supply is switched on to the motor the armature winding draws a current. The
current distributes with the stator armature winding depends upon rotor position and the devices
turned on. As per faradays law of electromagnetic induction a emf is dynamically induced in the
armature conductors. This back emf as per lenz law opposes the cause. As a result developed torque
reduces. Finally the rotors attain the steady speed.

14. Compare 120 degree and 180 degree operation of BLDC motor.

The 180 degree magnetic arc motor uses 120 degree mode of inverter operation. The motor with
120 degree magnetic arc uses 180 degree modeof inverter operation.
In 180 degree mode of inverter has 1.5 times copper losses but produce same torque with only
2/3 of magnetic material. Motor operation is lessefficient.
15. Give the expression for self and mutual inductances of a BLDC motor.

Self inductance is given by Lg=(ψ/i)=(πμ0 N2 lr1)/(2g”)

where g”= g’+lm/μrec,g’= Kcg,
N=Number of conductors in the slot, I = current,
lm =magnet length in radial direction , g’ = air gap,
g” = air gap including radial thickness of the magnet,
μrec = relative recoil permeability,
Mutualinductance is given by Mg=-(1/3)Lg.
16. What are the types of sensors used with PMBLDC motors?
Hall Effect sensors are most commonly used for speed, position sensing with PMBLDC
motors. Optical Disc based sensors are also used.
Presently rotor position sensors are avoided by using alternative methods called as Sensorless
control methods, which uses terminal emf measurement, third harmonic voltage measurement,
flux estimation and neuro – fuzzy techniques etc.
17. What are the relative merits and demerits of brush less DC motordrives? (Dec 2016)
Merits: Commutator less motor, Specified electrical loading is better, Heat can be easily
dissipated, No sparking takes place due to brush, Source of EMI is avoided.
Demerits: Above 10 kW, the cost of magnet is increase, Due to centrifugalforce the magnet may
come out.
18. What are the difference between conventional DC motor andPMBLDC motor? (April 17)

DC PMBLDC
Brushes are present. Brushes are not present.
Sparking may occur due to brush. Sparking will not occur as brush is
not present.
Brushes tend to produce RF1. RF1 problem does not occur.

There is a need for brushmaintenance. No need of brush maintenance.

19. What are the various kinds of permanent magnets?

There are basically three different types of permanent magnets which are used in small DC
motors; Alnico magnets, Ferrite or ceramic magnet, and Rare - earth magnet (samarium – cobalt
magnet).

20. What is meant by multiphase brushless motors?

A multi-phase brushless motor including a stator having a plurality of drive coils each
corresponding to a specific phase and a rotor having a plurality of field magnet poles of
successively alternating polarity. The stator further has a plurality of Hall generators for
detecting the positions of the rotor and a speed sensor for detecting the rotational speed of the
rotor.

21. Give the uses of sensors in motors.

It is used to identify the position of the rotor and it is also used to excite the coils in proper
manner. For instance, a motor is an electrical machine used to create motion. The device converts
electricity (electrical energy) into motion (mechanical energy). Typically performed by rotating an
object. Rotational sensor: A sensor that measures the turning movement of a wheel for purposes of
calculating distance traveled.
22. List some applications of PMBLDC motor. (June 2016)

Fans, Pump drive, Traction and Hydraulic power steering, precision high speed spindle
drivers for hard disk drivers etc.,

PART B

1. Derive the torque and EMF equations of permanent magnet brushless DC motor. (May 2019)

2. What are the differences between mechanical and electronic commutator?

3. Sketch the structure of controller for permanent magnet brushless DC motor and explain the
functions of various blocks. Also explain the different types of power controllers. (May 2019)

4. Explain in detail the various rotor position sensors used in permanent magnet brushless DC motor.

5. Sketch the torque – speed characteristics of a permanent magnet brushless DC motor. (June 2016)

6. a) Explain in detail the magnetic circuit analysis of brushless DC motor on open circuit.(Nov 17)
b) What are the advantages of BLPM DC motor over conventional DC motor?

7. Explain square wave permanent magnet brushless dc motor drives.

8. Analyze the operation of electronic commutator of PMBLDC motor with neat diagram.

9. A PMBLDC motor has toque constant 0.12 Nm/A referred to DC supply. Find the motors no load
speed when connected to 48V DC supply. Find the stall current and stall torque if armature resistance is
0.15Ω/phase and drop in controller transistor is 2V. (June 2016)

10. A permanent magnet DC commutator motor has a no load speed of 6000 rpm when connected to a
120V dc supply. The armature resistance is 2.5Ω and rotational and iron losses may be neglected.
Determine the speed when the supply voltage is 60V and the torque is 0.5 Nm. (Dec 2016)

11. Illustrate in detail, the operation of PMBLDC motor with 180 magnet arcs and 120 square wave
phase currents.

12. Discuss the hysteresis type current regulation of PMBLDC motor with neat diagram.

13. Discuss the construction and principle of operation of a Permanent magnet dc motor. (May 2016,
Dec 2016, April 17)

14. A brushless PM sine wave motor has an open circuit voltage of 173V at its corner point speed of
3000rpm. It is supplied from a PWM converter whose maximum voltage is 200V rms. Neglecting
resistance and all other losses, estimate the maximum speed at which maximum current can be supplied
to the motor.(May 2019)

15. Derive the relationship between magnetic field intensity and flux density by performing the
magnetic circuit analysis of a brushless dc motor on open circuit. (May 2019)
 UNIT-IV

PERMANENT MAGNET SYNCHRONOUS MOTORS(PMSM)

PART – A

1. Distinguish PM synchronous motor from BLPM DC motors.

PMSM PM Brushless DC motor

Sinusoidal or quasi–sinusoidal distribution Rectangular distribution ofmagnetic flux in the
of magnetic flux inthe air gap air gap
Sinusoidal or quasi-sinusoidal current Rectangular current waveforms
waveforms
Quasi-sinusoidal distribution of stator It has concentrated statorwindings
conductors. (short pitched and distributed
or concentric stator windings)

2. Write in detail the vector control of permanent magnet synchronousmotor.

BLPM SNW motor is usually employs for variable speed applications.
For this we keep V/f constant and vary V and f to get the desired speed and torque. From the
theory of BLPM SNW motor it is known that as thespeed is varied from a very low value up to the
corner frequency, the desired operating point of current is such that Id =0 and I is along q-axis.
3. What are the performances of PMSM when compared with BLDCmotor?
With equal r.m.s. phase currents the torque of the square wave motor exceeds that of
sine wave motor by a factor 1.47. With equal peak currentsthe factor is 1.27. For the same flux-
density flux per pole of a square wave motor exceeds that of a sine wave motor by a factor π/2.
Square wave motor has a slightly better utilization of the peak current capability of the
converter switches. In PMSM three devices conduct at a time (180 degree mode of inverter),
where as in BLDC only two devices conduct at a time in120 degree mode.

4. What is meant by field oriented control of PMSM?

In general for field oriented control the stator currents are transformed into a frame of
reference moving with the rotor flux. In the PMSM the rotor flux is stationary relative to the
rotor. The rotor flux is therefore defined by the mechanical angle of rotation α,this is obtained
from a rotor position sensor. Thus, the control is much easier to implement than in the case of
induction motor.

5. What is meant by slot less motor?

Slot less brushless DC motors represent a unique and compelling subset of motors within the
larger category of brushless DC motors. They are called “slot less” because typical slotted brushless
DC motors contain a stator core of laminated steel composed of slots separated by teeth around
which copper wire is wound. In slot less motor, the stator teeth are removed and resulting space is
partially filled with addition of copper.
6. What are the applications of PMBLDC and PMSM motors?

PMBLDC: (Low rating application) turn table drives for record players, Hard disc drives, Low
cost instruments, Small fans for cooling electronic equipment, (High rating application) Air craft,
Satellite system, Traction system.
PMSM are used in low to medium power (up to several hundred HP) Applications, Fiber
spinning mills, Rolling mills, Cement mills, Ship propeller, Electric Vehicles, Servo and robotic
drives and Starters / generator for air craft engine.

7. What are the features of permanent magnet synchronous motor? (Dec 2018)

Robust, compact and less weight, No field current or rotor current in PMSM, unlike in
induction motor, Copper loss due to current flow whichis largest loss in motors is about half that
of induction motor and High efficiency.
8. Explain the difference between synchronous motor and PMSM.

Synchronous Motor PMSM

3 phase AC or six step voltage or current 3 phase sine wave ac or PWMac is used
source inverter isused as supply. as supply.

This type of motor is used in very large Here it is used in low integral HP
compressor and fandrives. industries drives, fiber spinning mills.

9. What are disadvantages of PMSM relative to the commutator motor?

The disadvantages of PMSM relative to commutator motor are that it requires shaft position
sensing and increased complexion in the electronic controller. Also speed dependence on the
external load torque and the reduced dynamic performances are the other disadvantages.
10. What are the assumptions made in derivation of emf equation forPMSM?
Flux density distribution in the air gap is sinusoidal, Rotor rotates with anuniform
angular velocity of ω m (rad/sec), Armature winding consists of full pitched, concentrated
similarly located coils of equal number of turns.
11. Why PMSM operating in self – controlled mode is known as commutator less DC Motor?
Load side controller performs somewhat similar function as commutator in a DC machine.
The load side converter and synchronous motor combination functions similar to a DC machine.
First it is fed from a DC supply and secondly like a DC machine. The stator and rotor field remain
stationary with respect to each other at all speeds. Consequently, the drive consisting of load side
converter and synchronous motor is known ascommutator less DC motor.
12. Explain the distribution factor for PMSM.
Distribution factor (Kd): Distribution factor is given by the ratio of the MMF performed in the
concentrated windings compared to the distributed windings. Coils in the stator are displaced from each
other bya certain electrical angle, each coil produce sinusoidal MMF with a shift angle β.
 where ‘m’ is no. of coils/pole/phase, β = angle of displacement
13. Write down the expressions for torque of a PMSM.
T= 3 E I sin β / ωm N-m
where
ωm is the angular velocity,
T is the torque produced,
β is the torqueangle or power angle,
I is the current and
E is the induced emf.
14. What the features are of closed- loop speed control of load commutated inverter fed
synchronous motor drive?
The features of closed loop speed control of load commutated inverter fed synchronous motor
drive is that it has Higher efficiency, Four quadrant operation with regeneration braking, Higher
power ratings and run at high speeds ( 6000 rpm).
15. Write down the emf expressions of PMSM.

Eph = 4.44 f Фm KwTph volts.

This is the rms value of induced emf per phase,
where f = Frequency inHertz,
Фm = flux per pole,
Kw = Winding factor,
Tph = Turns per phase.

16. What is meant by self-control in PMSM drive? (April 17)

As the rotor speed changes the armature supply frequency is also changes
proportionally so that the armature field always moves (rotates) at the same speed as the motor.
The armature and rotor field move in synchronism for all operating points. Here accurate tracking of
speed by frequency is realized with the help of rotor position sensor.

17. What are advantages and disadvantages of PMSM?

Advantages: Runs at constant speed, No field winding, no field copper loss, better efficiency,
High power density, Lower rotor inertia, robust construction of rotor, No sliding contact hence
requires less maintenance.
Disadvantages: Loss of flexibilities of field flux control, Demagnetizationeffect and High cost.
18. Define the term load angle.

The phase angle introduced between the induced emf phasor, E and terminal voltage phasor, V
during the load condition of an Alternator iscalled load angle.
In PMSM the load angle is the angle between stator field and rotor field when the machine is
rotated at synchronous speed. It is represented as δ.
19. State the power controllers for PMSM. (Nov 17)
Permanent magnet synchronous motors (PMSM) are typically used for high-performance and
high-efficiency motor drives. High-performance motor control is characterized by smooth
rotation over the entire speed range of the motor, full torque control at zero speed, and fast
acceleration and deceleration. The various power controllers are PWM inverter using power
MOSFETS with microprocessor control. PWM inverter usingBJT’s with microprocessor control
(up to 100 KW).
20. Write the advantages of optical sensors.
Quite suitable for sinusoidal type motor as it is a high resolution sensor. The signal from the
photodiode rises and falls quite abruptly and the sensor outputs are switched high or low so the
switching points are welldefined.
PART-B
1. Derive the EMF and torque equation of BLPM Sine wave motor. (April 17, Nov 17)
2. Write short notes on Volt ampere requirements in PMSM motor.(June 2016)
3. With neat sketch, discuss the torque – speed characteristics of PMSM with necessary phasor diagram
and circle diagram.(Dec 2016, June 2016, May 2019)
4. Draw and explain the phasor diagram of permanent magnet synchronous motor.
5. What are the differences in the constructional features of PMBLDC and PMSM?
6. a) Derive the expression for synchronous reactance of PMSM.
b) Write short notes on armature reaction in PMSM.
7. With a neat diagram, explain the operation of power of controller for permanent magnet synchronous
motor.
8. Explain circle diagram of permanent magnet synchronous motor.
9. Discuss the different rotor configurations of PM synchronous machines.
10. Write a detailed technical note on the following (i) vector control of PMSM (ii) microprocessor
based control.
11. Explain the construction and performance of a permanent magnet synchronous motor with neat
diagrams. (Nov 17)
12. Derive the EMF equation of PMSM. (Dec 2016, June 2016)
13. Derive the torque equation of PMSM with the phasor diagram.(Dec 2016, June 2016)
14. Explain about self-controlled PMSM drive by employing load commutated thyristor inverter.(Dec
18)
15. Explain the microprocessor based control of PMSM with a neat block diagram.(Dec 2018)
 UNIT – V
OTHER SPECIAL MACHINES
PART – A
1. What are SYNREL motors?
Synchronous reluctance motor is similar to three–phase Synchronous motor except the rotor
are demagnetized and made with saliency to increase the reluctance power. It is a motor which
develops torque due to the difference in reluctance of the two axes, namely quadrature and
directaxis.

2. What is the principle of operation of reluctance machine?

1) In reluctance machines, torque is produced by the tendency of the rotor to move to a
position where the inductance of the excited stator winding is maximized (i.e., rotor tooth
aligns with active stator phase to minimize reluctance).
2) The rotor is typically constructed of soft magnetic iron shaped so as to maximize the
variation of inductance with rotor position.
3) Opposite poles form a phase and the phases are magnetically independent of one another. The
machines tend to be noisy; a characteristic that has limited their applications in the past and has
also limited their use currently in vehicles. Research has been on-going for years in an attempt to
address the noise issue, but little has been accomplished in actual noise mitigation. Reluctance
machines are relatively low-cost machines, and they generally do not contain PMs.
4. What are the properties of Reluctance motor?
Combined reluctance and magnet alignment torque, Field weakening capability, under
excited operation for most loaded condition, High inductance, High speed capability and High
temperature capability.
5. What are the various stator current modes used in synchronousreluctance motor?
The various stator current modes used in synchronous reluctance motorare

1. Unipolar current modes

2. Bipolar current modes.
6. Mention the applications of distributed anisotropy cage rotor of synchronous
reluctance motor. (May 2019)
These rotors are used for line – start (constant voltage and frequency) applications. Also
used in processing of continuous sheet or film material, used in regulators and turntables, applied
in wrapping and folding machines and it can be used in synchronized conveyors.

7. What is meant by reluctance torque? (Dec 2016, Dec 2018)

The torque which is exhibited on the rotor due to the difference in Reluctance in the air gap (or) a
function of angular position of rotor withrespect to the stator coil is known as reluctance torque.

8. What are the advantages and disadvantages of Synchronous reluctancemotor?

Advantages :Rotor is simple in construction i.e. very low inertia, Robust, Low torque, ripple,
Can be operated from standard PWM AC Inverters, It can be also built with a standard induction
motor, stator and windings. Disadvantages: It has poor power factor performance and therefore the
efficiency is not as high as permanent magnet motor, The converter kVA requirement is high, The
pull – in and pull – out torque of the motor are weak.

9. What are the types of Synchronous reluctance motor?

Synchronous reluctance motor is classified into three types depending upon the construction of
rotor. They are Salient type or Radial type rotor, Flat type or axial type rotor, Flux Barrier type or
Laminated type rotor.
10. Write the torque equation of Synchronous reluctance motor.

T = (U2/ 2ωs) (1/Xq - 1/Xd) sin 2δ,

U = Supply Voltage,
Is be the supplycurrent which has two components Id and Iq,
Id = Direct axis current,
11. Skewing is required for Synchronous reluctance motor. Justify?
At the time of starting, reluctance motor are subjected to logging due to the saliency of
motor. This can be minimized by the skewing of the rotor parts. Reluctance motors are subject to
cogging, since the locked rotor torque varies with the rotor position, but the effect may be minimized by
skewing the rotor bars and by not having the number of poles.
12. What are the advantages of increasing Ld / Lq ratio in Synchronousreluctance motor?

The advantages of increasing Ld / Lq ratio in synchronous reluctancemotor are:

1. Motor power factor increases

2. I2R losses reduced
3. Reduced volt – ampere ratings of the inverter driving the machine.

13. Compare Synchronous reluctance motor and Induction motor. (Nov17, May 2019)

S.No Synchronous reluctance motor Induction motor

1. Torque generation due to Torque generation due to Lorentz
reluctance principle force
2. Runs at synchronous speed Runs at asynchronousspeed
3. Better efficiency. Efficiency is low.
 4. Low cost. High cost.
5. High power factor. Low power factor.
6. Used for low and medium power Used for high power
application. application.

14. Define: Magnetic flux.

Magnetic flux is the number of magnetic field lines passing through a surface (such as
a loop of wire). The magnetic flux through a closed surface (such as a ball) is always zero.
Magnetic flux is sometimes used by electrical engineers designing systems with
electromagnets or designingdynamos.
15. Define: Reluctance.
The opposition offered to the magnetic flux by a magnetic circuit is called its
reluctance. Reluctance is analogous to resistance. As resistance opposes electric current in
an electric circuit that same work a reluctance does in a magnetic circuit.

16. Define: Permeance.

In electromagnetism, permeance is the inverse of reluctance. In a magnetic circuit,
permeance is a measure of the quantity of magnetic flux for a number of current-turns. It is
a measure of the ease with which flux can be setup in a material.

17. List out any four features of synchronous reluctance motors.

The four features of synchronous reluctance motors are
1. Better efficiency
2. High cost
3. Low power factor
4. Used for low and medium power application.
18. Give some potential application of synchronous reluctance machine.(Dec 16)
It is used for constant speed applications i.e. timing devices, signaling devices,
recording instruments and phonograph, it is used in automatic processors such as in food
processing and packaging industries. Used in high speed applications, Synthetic fiber
manufacturing equipment, Wrapping and folding machines, synchronized conveyors.
19. Write the various design parameters of a synchronous reluctancemotor.
The various design parameters of a synchronous reluctance motor are Power
factor, Copper loss, core loss, Cost and Efficiency. The reluctance synchronous motor
with conventional rotor is simple, robust, requires simple controls but is characterized by
low power factor kW/kVA ratio, low torque density and high torque pulsations.
20. Give the difference between synchronous reluctance motor and switched
reluctance motor.

S.No. Synchronous Switched reluctance motor

reluctance motor
1. Single salient electricmotor Doubly salient electric motor
2. Continuous rotation Designed for continuous rotation
 3. Controller is not necessary. The step pulse are given by external
Hence it is cheap. controller which uses rotor position
sensors
21. What is a vernier motor?

A Vernier motor is an unexcited (or reluctance Type) inductor synchronous

motor. It is also named because it operates on the principle of a vernier. The peculiar
feature of this kind of motor is that a small displacement of the rotor produces a large
displacement of the axes of maximum and minimum permeance. When a rotating
magnetic field is introduced in the air gap of the machine, rotor will rotate slowly and at a
definite fraction of the speed of the rotating field.

22. Write the salient features of Vernier motor.

The nature of a PM vernier motor is analytically surveyed, and substantial
information such as main geometric factors affecting torque and the maximally
obtainable torque for a given current are provided, leading to a new relation for torque per
air gap volume. The peculiar feature of this motor is that a small displacement of rotor
produces a large displacement of the axis of maximum and minimum permeance.

23. State the application of vernier motor.

A vernier motor works as an electric gearing. This kind of motor is attractive in
applications which require low speed and high torque and where mechanical gearing is
undesirable. The stator of a vernier motor has slots and a distributed winding just like the
stator of an ordinary poly phase induction motor.
PART-B

1. Explain the principle of operation and constructional features of Synchronous reluctance

motor. (Dec 2016, June 2016,Nov 17, Dec 2018)
2. Explain the torque – speed and torque – angle characteristics of Synchronous reluctance
motor. (April 17, Dec 2018)
3. Draw and explain the steady state phasor diagram of Synchronous reluctance motor. (Dec
2016, June 2016,Nov 17, Dec 2018)

4. Derive the expression for torque equation of a Synchronous reluctance motor. (Dec 2016,
June 2016)

5. Explain the various types of Synchronous reluctance motor based on rotor construction with
neat sketch. (Axial and Radial type)(Dec 2016, April 17)

6. A 10 HP, 4 pole, 240V, 60Hz, reluctance motor operating under rated load condition has a
torque angle of 30°. Determine (a) Load torque on shaft (b) Torque angle if the voltage drops to
224V (c)For the above torque angle, will the rotor pullout of synchronism.

7. Explain Axial and Radial type synchronous motor.

8. Explain the advantages and disadvantages of synchronous reluctance motor.

9. Explain the applications and properties of synchronous reluctance motor.

10. Compare a reluctance motor with an equivalent induction motor and list out the merits and
demerits of reluctance motor over induction motor.

11. Derive the expression for d- axis synchronous reactance of a permanent magnet synchronous
reluctance motor. (April 17)

12. Explain sequential circuit reluctance motor and Hysteresis motor with neat sketches the
principle of operation and the application.

13. Explain the working, phasor diagram and performance of repulsion type motor. (May 2019)

15. Discuss the applications areas of different special electrical machines. (Dec 2018)

16. Explain the construction and principle of working of a universal motor and mention its
applications. (May 2019)