Introduction to Electrical Engineering 22ESC142

Basic Electrical Engineering-22ESC142

2022-23 (MODULE-4)

Mrs . Rekha PS/ Mrs. Vimala C S/

Mr. Kubera U /Mr. Hemanth Kumar

Asst. Professor

Dept. Of EEE, SJBIT

Introduction to Electrical Engineering 22ESC142
SYLLABUS
Introduction to Electrical Engineering
Course Code: 22ESC142/242 CIE Marks 50
Course Type (Theory/Practical Theory SEE Marks 50
/Integrated ) Total Marks 100
Teaching Hours/Week (L:T:P: S) 3:0:0:0 Exam Hours 03
Total Hours of Pedagogy 40 hours Credits 03
Course objectives
 To explain the laws used in the analysis of DC and AC circuits.
 To explain the behavior of circuit elements in single-phase circuits.
 To explain the construction and operation of transformers, DC generators and motors and induction
motors.
 To introduce concepts of circuit protecting devices and earthing.
 To explain electric power generation, transmission and distribution, electricity billing, equipment and
personal safety measures.

Teaching-Learning Process
These are sample Strategies, which teacher can use to accelerate the attainment of the various course outcomes
and make Teaching –Learning more effective
1. Chalk and talk 2. Animated/NPTEL videos 3. Cut sections 4. PPTs

Module-1 (08 Hrs)

Introduction: Conventional and non-conventional energy resources; General structure of electrical
power systems using single line diagram approach.
Power Generation: Hydel, Nuclear, Solar & wind power generation (Block Diagram approach).
DC Circuits: Ohm’s Law and its limitations. KCL & KVL, series, parallel, series-parallel circuits.
Simple Numerical.
Module-2 (08 Hrs)
A.C. Fundamentals:
Equation of AC Voltage and current, waveform, time period, frequency, amplitude, phase, phase
difference, average value, RMS value, form factor, peak factor. (only definitions)
Voltage and current relationship with phasor diagrams in R, L, and C circuits. Concept of Impedance.
Analysis of R-L, R-C, R-L-C Series circuits. Active power, reactive power and apparent power.
Concept of power factor. (Simple Numerical).
Module-3(08 Hrs)
DC Machines:
DC Generator: Principle of operation, constructional details, induced emf expression, types of
generators. Relation between induced emf and terminal voltage. Simple numerical.
DC Motor: Principle of operation, back emf and its significance. Torque equation, types of motors,
characteristics and speed control (armature & field) of DC motors (series & shunt only). Applications
of DC motors. Simple numerical.

Module-4(08 Hrs)
Introduction to Electrical Engineering 22ESC142
Transformers: Necessity of transformer, principle of operation, Types and construction of single-
phase transformers, EMF equation, losses, variation of losses with respect to load. Efficiency and
simple numerical.
Three-phase induction Motors: Concept of rotating magnetic field, Principle of operation,
constructional features of motor, types – squirrel cage and wound rotor. Slip and its significance
simple numerical.
Module-5 (08 Hrs)
Domestic Wiring: Requirements, Types of wiring: casing, capping. Two way and three way control
of load.
Electricity Bill: Power rating of household appliances including air conditioners, PCs, laptops,
printers, etc. Definition of “unit” used for consumption of electrical energy, two-part electricity tariff,
calculation of electricity bill for domestic consumers.
Equipment Safety measures: Working principle of Fuse and Miniature circuit breaker (MCB),
merits and demerits.
Personal safety measures: Electric Shock, Earthing and its types, Safety Precautions to avoid shock.

Course outcome (Course Skill Set)

At the end of the course the student will be able to:
CO1 Understand the concepts of various energy sources and Electric circuits.
CO2 Apply the basic Electrical laws to solve circuits.
CO3 Discuss the construction and operation of various Electrical Machines.
CO4 Identify suitable Electrical machine for practical implementation.
CO5 Explain the concepts of electric power transmission and distribution, electricity billing,
circuit protective devices and personal safety measures.

Suggested Learning Resources:

Books (Title of the Book/Name of the author/Name of the publisher/Edition and Year)
Text Books:
1. Basic Electrical Engineering by D C Kulshreshtha, Tata McGraw Hill, First Edition 2019.
2. A text book of Electrical Technology by B.L. Theraja, S Chand and Company, reprint edition 2014.

Reference Books:
1. Basic Electrical Engineering, D. P. Kothari and I. J. Nagrath, Tata McGraw Hill 4th edition, 2019.
2. Principles of Electrical Engineering & Electronics by V. K. Mehta, Rohit Mehta, S. Chand and
Company Publications, 2nd edition, 2015.
3. Fundamentals of Electrical Engineering by Rajendra Prasad, PHI, 3rd edition, 2014.
 Introduction to Electrical Engineering 22ESC142

Module – 4
Single Phase Transformers

4.1 Transformer is a static device which transfers electrical energy from one electric circuit to another
circuit at desired change in voltage and current, without any change in frequency.

4.2 Necessity of Transformer

The main advantage of alternating current over direct current is that it can be easily increased or reduced
as per requirement during the generation, transmission, distribution and utilization of electric power. This
is made possible by means of transformers. High voltages may be generated and stepped up by means of
transformers to still higher voltages for the transmission lines. Other transformers are employed at suitable
points to step the voltage down to values suitable for motors, lamps, heaters and other loads, The full-load
efficiency of a medium sized transformer is of the order of 97-98 percent, so that the loss at each point of
transmission or distribution is small. As the transformer is a staticapparatus, there are no moving parts
and so the maintenance of a transformer is easy and the amount of supervision is negligible.

4.3 Principle of operation of a single-phase transformer and EMF equation

A transformer operates on the principle of electromagnetic induction as explained by faraday‟s law.
Mutual induction states that when two coils are inductively coupled and if current in one coil is changed
uniformly then an emf gets induced in the other coil.

Fig.1.2 Transformer

The above shows the general arrangement of a transformer. Core is the iron core made of laminated sheets
of about 0.35mm thick insulated from one another by varnish or thin paper. The purpose of laminating
the core is to reduce the power loss due to eddy currents induced by the alternating magnetic flux. The
Introduction to Electrical Engineering 22ESC142
vertical portions of the core are called limbs and the top and bottom portions are called the yokes. Coils P
Introduction to Electrical Engineering 22ESC142
and S are wound on the limbs. Coil P is connected to the supply and therefore called as the primary, coil
S is connected to the load and is called as the secondary.

An alternating voltage applied to Primary drives an alternating current though Primary and this current
produces an alternating flux in the iron core, the mean path of the flux is represented by the dotted line.
This flux links with the coil Secondary and thereby induces an emf in Secondary coil.

EMF Equation
When the primary winding is excited by an alternating voltage V1, it circulates alternating current,
producing an alternating Flux ф. The primary winding has N1 number of turns. The alternating flux ф
linking with the primary winding itself induces an emf in it denoted as E1. The flux links with the
secondary winding through the common magnetic core. It induces emf E2 in the secondary winding and
this is mutually induced emf.

The primary winding is excited by purely sinusoidal alternating voltage. Hence the flux produced is also
sinusoidal in nature having maximum value of ф m as shown in the Fig below.

The various quantities which affect the magnitude of the induced Emf are:
Ф = Flux

Фm =Maximum value of Flux

N1 = Number of primary Winding turns

N2 =Number of secondary Winding turns
f = Frequency of the supply voltage
E1 = R.M.S value of the primary induced E.M.F
E2 = R.M.S value of the secondary induced E.M.F
Introduction to Electrical Engineering 22ESC142
From Faraday‟s law of electromagnetic induction, the average e.m.f induced in each turn is
proportionalto the average rate of change of flux.

Average e.m.f per turn= Average rate of change of flux

Average e.m.f per turn= dф/dt

Now ф = Change in FluX

dt Time required for change in fluX

Consider the 1/4th cycle of the flux as shown in the fig. Complete cycle gets completed in 1/f seconds.
In 1/4th time period, the change in flux is from 0 to фm.

𝑑ф фm−0
= = 4fф
dt 1/4f m

The time for 1/4th time period is 1/4f seconds

⸫ Average e.m.f per turn= 4 f фm volts

As ф is sinusoidal, the induced e.m.f in each turn of both the windings is also sinusoidal in nature. For
sinusoidal quantity,

Form factor =R.M.S value/Average Value= 1.11

R.M.S value =1.11 * Average Value

R.M.S value for induced e.m.f per turn = 1.11 * 4 f ф m

= 4.44 f фm

There are N1 number of primary turns hence the R.M.S value of induced e.m.f of primary denoted as E1
and is given by

E1 = N1 * 4.44 f фm volts

While as there are N2 number of secondary turns the R.M.S value of induced e.m.f of secondary
denoted as E2 is,

E2 = N2 * 4.44 f фm Volts

The expressions of E1 and E2 are called e.m.f equations of a transformer.

 Introduction to Electrical Engineering 22ESC142
4.4 Types and Construction of Transformers
Construction of Single-phase transformer
The core is made up of laminations of silicon steel in such a way as to ensure a continuous magnetic
path, with a minimum of air gap. Silicon content sometimes heat-treated to ensure high permeability
and low hysteresis loss at the usual operating flux densities.

Lamination of the core minimizes eddy current loss. These laminations are insulated from each other by
a thin coating of a suitable varnish. The thickness of laminations ranges from 0.35 mm for a frequency
of 25 Hz to 0.5 mm for a frequency of 50 Hz.

The two main type of transformers are i) Core-type ii) Shell-type

3.13.2 Core type

In this type of transformer, a large part of the core is surrounded by the Windings.Fig(a)
shows the simplified representation of a core-type transformer, where the primary and secondary
windings have been shown wound on the opposite limbs. However, in actual practice, half the primary
and half the secondary windings are situated side by side on each limb, so as to reduce leakage flux,
asshown in Fig(b).

In this type, it has single magnetic circuit and both the coils are placed on both limbs. The low
voltage coil is placed inside near the core while high voltage coil surrounds low voltage coil.

(a) Shell Type

In this type, the windings occupy a smaller portion of the core as shown schematically in
Fig(b). Fig(a) shows the representation of shell type transformer. The primary and secondary
windings are shown located on the central limb.
Introduction to Electrical Engineering 22ESC142

In this type, it has double magnetic circuit and both the windings are placed on central

limb.

4.5 Losses in Transformer

There are two losses in transformer,
3.14.2 Core/Iron Loss or Constant loss
3.14.3 Copper Loss or variable loss

Core Losses: These losses consist of hysteresis and eddy current losses and occur due to the
alternating flux in the transformer core.

Hysteresis Loss: Due to alternating flux set up in the magnetic core of the transformer, itundergoes a
cycle of magnetization and demagnetization. Due to hysteresis effect, there is loss of energy in this
process which is called as Hysteresis loss.

Hysteresis loss, Ph = Kh Bmax 1.6 f v watt

Were, Bmax… Maximum Flux Density

(Wb/m2)f… Frequency (Hz)
Kh …hysteresis
constantv… Volume of
the core
Eddy - Current Loss: Due to the alternating flux in the core, eddy currents flow in the core. Power is
required to maintain these eddy currents. This power is dissipated in the form of heat andis called
eddy current loss.
 Introduction to Electrical Engineering 22ESC142

Eddy current loss, Pe = Ke Bmax 2 f 2 t2 watt

Where, Ke is eddy current constant and t is thickness of the core

(It is apparent that core losses (hysteresis and eddy current losses) depend upon flux density in the core
and supply frequency. As flux density in the core remains practically constant from no-load to full load,
and also supply frequency is constant, it follows that core losses too are constant for a given transformer.
These losses are independent of load which is why these are generally termed constant losses)

Therefore, total iron losses, Pi=Ph+Pe

Core losses can be minimized by using steel of high silicon content for the core and by using
very thin laminations.

Copper losses or I2R losses

These losses occur due to the ohmic resistance in both the primary and secondary windings. If R1 and R2
are the primary and secondary resistances. I1 and 12 are the primary and secondary currents respectively.

Total Cu loss, P cu = 𝑑2𝑑1 + 𝑑2𝑑2 = 𝑑2𝑑01 = 𝑑2𝑑02

1 2 1 2

It is obvious that copper loss is proportional to (current)2. Therefore, Pcu is depends on current drawn by
the load, thus copper losses are called variable losses. This loss can be kept minimum bydesigning the
windings with low resistance values.

Thus, for a transformer total loss is given by,

Total losses = Core losses + Copper losses = Pi +Pcu

4.6 Efficiency of Transformer

The efficiency of a transformer is defined as the ratio of the output power to the input power.

The total losses=Pi+Pcu and Power output= V2 I2

COSФVA rating of the transformer= V2 I2

𝑑2𝑑2 𝑑𝑑𝑑Ф (𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑)𝑑𝑑𝑑Ф

Therefore, ղ = =
𝑑2𝑑2 𝑑𝑑𝑑Ф +𝑑𝑑+𝑑𝑑𝑑 𝑑𝑑 (𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑)𝑑Ф+𝑑𝑑+𝑑𝑑𝑑 𝑑𝑑
Introduction to Electrical Engineering 22ESC142
The above expression is for full load value of current. For fractional load, the load current changes
let „n‟ be the fraction of load, n=actual load/full load

Then new load current is „n‟ times the full load current. The VA rating, Cu loss also changes for
fractional load and are given by,

VA rating= V2 (nI2)=n(V2 I2)=n(VA rating)

𝑑𝑑𝑑 = (𝑑𝑑1)2 𝑑 01 = 𝑑21𝑑2𝑑01 = 𝑑2𝑑𝑑𝑑 𝑑𝑑

Therefore, for a fractional load the efficiency of transformer is given by the above equation.

4.7 Condition for maximum Efficiency

The efficiency is a function of load current I 2 . Assuming COSф is constant and the secondary terminalvoltage
V2 is also assumed constant.

For Maximum Efficiency,

Differentiating the above equation w.r.t I2

(𝑑2𝑑2 𝑑𝑑𝑑Ф + 𝑑𝑑 + 𝑑2𝑑02)(𝑑2𝑑𝑑𝑑Ф )– (𝑑2𝑑2 𝑑Ф )(𝑑2𝑑𝑑𝑑Ф + 2𝑑2𝑑02) = 0

2
Cancelling 𝑑2𝑑𝑑𝑑Ф from both the sides we get,
𝑑2𝑑2 𝑑𝑑𝑑Ф + 𝑑𝑑 + 𝑑2 𝑑02 − 𝑑2 𝑑2 𝑑𝑑𝑑Ф −2𝑑2𝑑02 = 0
2 2

i,e 𝑑𝑑 − 𝑑22 𝑑02 = 0

𝑑𝑑 = 𝑑2𝑑02 = 𝑑𝑑𝑑
2

Condition to achieve maximum efficiency is that,

Copper losses = iron losses i,e Pi = Pcu

 Introduction to Electrical Engineering 22ESC142

1) Problems

2) A 250 kVA, 11000/415 V, 50 Hz single phase transformer has 80 turns on the secondary. Calculate:
(i) The rated primary and secondary currents.
(ii) The number of primary turns.
(iii) The maximum value of flux.
(iv) Voltage induced per turn.

3) The primary winding of a transformer is connected to a 240 V, 50 Hz, supply. The secondary winding has
1500 turns. If the maximum value of the core flux is 0.00207 Wb, determine (i)The secondary induced emf.
(ii)Number of turns in the primary.
(ii) Core area of cross section if the flux density has a maximum value of 0.465 Tesla.

4) A kVA transformer has N1: N2 = 300: 20. The primary winding is connected to a 2200 V, 50 Hz supply.
Calculate:
(i) Secondary voltage on no load.
(ii) Approx. values of primary and secondary currents on full load.
(iii)The maximum value of the flux.

5) A transformer has 1000 turns on its primary and 400 turns on the secondary side. An a.c. voltage of 1250
V, 50Hz is applied to its primary side, with the secondary open circuited. Calculate: i) The secondary emf
(ii) Maximum value of flux density, given that the effective cross-sectional area of core is 60 cm^2.

6) A 40 kVA transformer has a core loss of 450 W and a full load copper loss of 850W. If the load power factor
is 0.8, calculate (i) Efficiency at full load (ii) Load at which copper loss is equal to iron loss (iii) Efficiency
at this Load.

7) The maximum efficiency at full load and unity power factor of a single-phase 25 kVA, 500/1000 V, 50 Hz,
transformer is 98%. Determine its efficiency at, (i)75% load, 0.9 p.f. and (ii) 50% load, 0.8 p.f.

8) A 600 kVA transformer has an efficiency of 92% at full load, unity p.f. and at half load, 0.9 p.f. Determine
its efficiency at 75% of full load and 0.9% p.f.

9) A 40 kVA single phase transformer has core loss of 450 W and full load copper loss of 850 W. If the power
factor of the load is 0.8 calculate (i) Full load efficiency (ii)Maximum efficiency at u.p.f. (iii) load for
maximum efficiency

10) A single phase 25 kVA 1000/2000 V, 50 Hz transformer has maximum efficiency of 98% at full
load u.p.f. Determine its efficiency at: (i)3/4 full load u.p.f. (ii)1/2 full load 0.8 p.f. (iii) 1.25 full load 0.9
p.f.

11) A 600 kVA, 1 transformer has an efficiency of 92% both at full load and half full load u.p.f. Determine its
efficiency at 75% full load, 0.9 p.f.
12) A 25 kVA transformer has an efficiency of 94% of full load unity p.f. and at half full load 0.9 p.f. Determine
the iron loss and full load copper loss.
 Three Phase Induction Motors
4.9 INTRODUCTION

The three phase induction motor is the most widely used a.c. motor. It differs from other type of motors in
that there is no connection from the rotor winding to any source of supply. The necessary voltage and
current in the rotor circuit are produced by induction from the stator winding which is why it is called
induction motor.

Advantages
It is very simple, very robust and rugged, practically unbreakable construction.
Its cost is low.
It is very reliable.
It is highly efficient
It has a fairly good power factor.
Its maintenance requires minimum of attention.
It does not need to be synchronized. It has a simple starting arrangement.
Disadvantage
It is essentially a constant speed motor and the speed cannot be varied easily.
Its speed reduces to some extent with increase in load as in case of D.C. shunt motor.
It has somewhat lesser starting torque as compared to D.C. shunt motor.

4.10 CONSTRUCTION
Three phase induction motor consists of two parts
Stator
Rotor

Stator
It is the stationary part of the motor supporting the entire motor assembly.
This outer frame is made up of a single piece of cast iron in case of small machines.
In case of larger machines they are fabricated in sections of steel and bolted together.

The core is made of thin laminations of silicon steel and flash enamelled to reduce eddy current and
hysteresis losses.
Slots are evenly spaced on the inner periphery of the laminations.
Conductors insulated from each other are placed in these slots and are connected to form a balanced 3 -
phase star or delta connected stator circuit.
Depending on the desired speed the stator winding is wound for the required number of poles. Greater the
speed lesser is the number of poles.

Rotor
They are basically classified into two types based on the rotor construction
Squirrel cage motor
Slip ring motor or phase wound motor
4.10.1 Squirrel cage rotor
Squirrel cage rotors are widely used because of their ruggedness.
The rotor consists of hollow laminated core with parallel slots provided on the outer periphery.
The rotor conductors are solid bars of copper, aluminium or their alloys.
The bars are inserted from the ends into the semi-enclosed slots and are brazed to the thick short circuited
end rings.
This sort of construction resembles a squirrel cage hence the name “squirrel cage induction motor”.
The rotor conductors being permanently short circuited prevent the addition of any external resistance to
the rotor circuit to improve the inherent low starting torque.
The rotor bars are not placed parallel to each other but are slightly skewed which reduces the magnetic hum
and prevents cogging of the rotor and the stator teeth.

4.10.2 Phase wound/slip ring rotor

The rotor in case of a phase wound/ slip ring motor has a 3-phase double layer distributed winding made up
of coils, similar to that of an alternator.
The rotor winding is usually star connected and is wound to the number of stator poles.
The terminals are brought out and connected to three slip rings mounted on the rotor shaft with the brushes
resting on the slip rings.
The brushes are externally connected to the star connected rheostat in case a higher starting torque and
modification in the speed torque characteristics are required.
Under normal running conditions all the slip rings are automatically short circuited by a metal collar
provided on the shaft and the condition is similar to that of a cage rotor. Provision is made to lift the
brushes to reduce the frictional losses. The slip ring and the enclosures are made of phosphor bronze.
4.11 Comparison of the squirrel cage and slip ring rotors
S.No Slip ring rotor Squirrel cage rotor
1. Rotor consist of a 3-phase Rotor consist of bars which are shorted at the
winding similar to stator winding ends with the help of end rings
2. Construction is complicated Construction is very simple
3. Resistance can be added As permanently shorted external resistance
externally cannot be added
4. Slip rings and brushes are present Slip rings and brushes are absent
5. Rotors are very costly Due to simple constructions rotors are cheap
6. High starting torque can be Moderate starting torque which cannot be
obtained controlled
7. Rotor resistance starter can be Rotor resistance starter cannot be used
used
8. Rotor must be wound for same The rotor automatically adjusts itself for the
number of poles as stator same number of poles as that of stator
9. Rotor copper loss is high hence Rotor copper loss is less and hence efficiency is
efficiency is less more
Production of rotating magnetic field
Consider a 3- phase induction motor whose
stator windings mutually displaced from
each other by 120° are connected in delta
and energized by a 3- phase supply.

The currents flowing in each phase will set

up a flux in the respective phases as shown

The corresponding phase fluxes can be

represented by the following equations
 B m sin t 240
R msin t m sin
B m sin 240
Y m
sin t 120
Y m sin 120

The resultant flux at any instant is given by the vector sum of the flux in each of the phases.

(i) When 0 , from the flux waveform diagram ,we have

0 3
R 120 )
m
2
Y km sin( 3
240 )
m
2
B m sin(

The resultant flux is given by,

3 cos(30 ) 1..5 m
r 2* m
2
 3
Y m
2
3
B m
2
r 1.5 m

(ii) When 60o

3
R m
2
3
Y m
2
B 0

(iii)
When 120

Y 0

3
B m
2

(i) When 180

R 0;
3
Y m
2
 3
B 2
From the above discussion it is very clear that when the stator of a 3-phase induction motor is
energized, a magnetic field of constant magnitude (1.5 φ m) rotating at synchronous speed

(Ns) with respect to stator winding is produced.

(b) Rotation of the rotor

Doing the same construction, we get the same result as 𝝋𝑻 = 1.5𝝋𝒎

So the magnitude of 𝝋𝑻 is 1.5𝝋𝒎 , but it has rotated through 60° in space, in clockwise
direction from its previous position.

120 f

NS =

Where f = supply frequency

P = Number of stator poles

ANIMATION INSTRUCTION

Consider a portion of 3- phase induction motor as shown in the above figure which is
representative in nature. The rotating field crosses the air gap and cuts the initially stationary
 rotor conductors. Due to the relative speed between the rotating magnetic field and the initially
stationary rotor,(change of flux linking with the conductor) an e.m.f. is induced in the rotor
conductors, in accordance with the Faraday‟s laws of electromagnetic induction. Current flows in
the rotor conductors as the rotor circuit is short circuited. Now the situation is similar to that of a
current carrying conductor placed in a magnetic field. Hence, the rotor conductors experience a
mechanical force which eventually leads to production of torque. This torque tends to move the
rotor in the same direction as that of the rotating magnetic field.

CONCEPT OF SLIP (S)

According to Lenz‟s law, the direction of rotor current will be such that they tend to oppose the
cause producing it. The cause producing the rotor current is the relative speed between the
rotating field and the stationary rotor. Hence, to reduce this relative speed, the rotor starts
running in the same direction as that of stator field and tries to catch it. In practice the rotor can
never reach the speed of the rotating magnetic field produced by the stator. This is because if
rotor speed equals the synchronous speed, then there is no relative speed between the rotating
magnetic field and the rotor. This makes the rotor current zero and hence no torque is produced
and the rotor will tend to remain stationary. In practice, windage and friction losses cause the
rotor to slow down. Hence, the rotor speed (N) is always less than the stator field speed (NS).
Thus the induction motor cannot run with ZERO SLIP. The frequency of the rotor current

fr = sf. The difference between the synchronous speed (NS) of the rotating stator field and the
actual rotor speed (N) is called the slip speed.

Slip speed = NS – N depends upon the load on the motor

NS - N

% Slip (s) = --------------- * 100

NS

Note: In an induction motor the slip value ranges from 2% to 4%

APPLICATIONS OF INDUCTION MOTORS

Squirrel cage induction motor

Squirrel cage induction motors are simple and rugged in construction, are relatively cheap and
 require little maintenance. Hence, squirrel cage induction motors are preferred in most of the
industrial applications such as in

i) Lathes
ii) Drilling machines
iii) Agricultural and industrial pumps
iv) Industrial drives.

Slip ring induction motors

Slip ring induction motors when compared to squirrel cage motors have high starting torque,
smooth acceleration under heavy loads, adjustable speed and good running characteristics.

They are used in

i) Lifts
ii) Cranes
iii) Conveyors , etc.,

WORKED EXAMPLES

1. A 12 pole, 3 phase alternator is coupled to an engine running at 500 rpm. It supplies an

Induction Motor which ahs a full load speed of 1440 rpm. Find the percentage slop and the
number of poles of the motor.

Solution: NA = synchronous speed of the alternator

PNA 12 X 500

F =------- =---------------- = 50 Hz (from alternator data)

120 120

When the supply frequency is 50 Hz, the synchronous speed can be 750 rpm, 1500 rpm,
3000rpm etc., since the actual speed is 1440 rpm and the slip is always less than 5% the
synchronous speed of the Induction motor is 1500 rpm.
 NS – N 1500 - 1440

s= = = 0.04 OR 4%

NS 1500

120f 120 x 50

NS = = = 1500

P P

P=4

2. A 6 pole induction motor is supplied by a 10 pole alternator, which is driven at 600 rpm. If
the induction motor is running at 970 rpm, determine its percentage slip.

P NA 10 X 600

From alternator date: f =------- = --------------- = 50 Hz

120 120

Synchronous speed of the induction motor

120 f 120 50
NS 1000rpm
P 6
From I.M. data: NS N
100 1000 970
%slip 3%
NS 1000
Basic Electrical Engineering-22ESC142 2022-23