1. (19ES1EE07) ELEMENTS OF ELECTRICAL AND ELECTRONICS ENGINEERING
COURSE OBJECTIVES: Student shall be able to
• Study and understand the performance of basic electric circuits
• Understand the performance of electrical machines
• Know the utilization of electrical energy in day to day to affairs
• Understand the operation of diode and transistor
COURSE OUTCOMES: After the completion of the course, students will be able to
CO-1: Analyse the performance of electrical circuits
CO-2: Test, analyse and find the applications of different electric machines
CO-3: Know the use of electric power for domestic and industrial purposes
CO-4: Understand the principles of semiconductor devices and their applications
UNIT – I:
Fundamentals of Electrical Circuits: Basic R-L-C parameters, Ohm’s Law, kirchhoff’s Laws,
Series-parallel connections, Star/Delta Transformation, Generation of A.C, Average, RMS
values and Form Factor of Sinusoidal Voltages, AC through RL,RC and RLC, concept of
impedance, power, power factor, simple problems
UNIT – II:
D.C Machines: D.C Generator, Basic Construction, Operation, emf Equation, types, Open
Circuit Characteristics, simple problems. D.C Motor-principle-back emf -Torque equation,
Speed control, Swinburne’s test.
UNIT – III:
Introduction to 3-phase circuits, relation between line and phase quantities in balanced star
and delta networks.
A.C Machines: Single phase transformer: principle-emf equation-types-OC and SC tests-
Voltage Regulation -Efficiency-Simple problems.
2. UNIT – IV:
Three phase induction motor: Working principle – slip- torque equation- Torque slip
characteristics, Principle of Alternator.
Electrical Power Systems and Utilization: Hydro Power Plant: Lay out -Efficiency Calculation,
Illumination: Definitions-Laws of Illumination- working of Incandescent Fluorescent lamps.
UNIT – V:
Electronics Devices: Semiconductor materials, Review of P-N junction, Diode Characteristics,
Basic Operation of Half-wave and Full wave Rectifiers, Zener Diode as Voltage Regulator, BJT,
biasing, Characteristics, applications.
UNIT – VI:
Digital Circuits and Transducers: Logic gates, Combinational Logic circuits, Basic operation of
SR-JK-T and D Flip-Flops, Transducers -Overview - Passive Sensors - Working 24
of Strain Gauge, Pressure Gauge, Dial Gauge - Piezoelectric Accelerometer Model-
1. Principles of Electrical and Electronics Engineering by V. K. Mehta, S. Chand & Co.
2. Fundamentals of Electrical Engineering by Ashafaq Hussain, 2nd Edition, Dhanpat Rai & Co.
3. Electronic Devices and Circuits by S. Salivahanan, N. Suresh Kumar, 3rd Edition, McGraw-
1. Electrical Technology by Edward Hughes, ELBS Longman Publisher
2. Basic Electrical Engineering by D. P. Kothari & I. J. Nagrath, TMH Publications, 2nd Edition.
3. Utilization of Electric Power and Electric Traction by G. C. Garg, Khanna Publishers.
4. Electric Current (1)
• Electric current i = dq/dt. The unit of ampere can be derived as 1 A = 1C/s.
A direct current (dc) is a current that remains constant with time. An
alternating current (ac) is a current that varies sinusoidal with time.
The direction of current flow:
5. Electric Voltage
• Voltage (or potential difference) is the energy required to move a unit charge
through an element, measured in volts (V). Mathematically, (volt) – w is energy
in joules (J) and q is charge in coulomb (C). Electric voltage, vab, is always across
the circuit element or between two points in a circuit.
vab > 0 means the potential of a is higher than potential of b.
vab < 0 means the potential of a is lower than potential of b.
Power and Energy
• The law of conservation of energy ∑ p = 0
• Energy is the capacity to do work, measured in joules (J).
• Mathematical expression
7. Definition of Electrical Resistance
• Resistance (also known as ohmic resistance or electrical resistance) is a
measure of the opposition to current flow in an electrical circuit. Resistance is
measured in ohms, symbolized by the Greek letter omega (Ω).
• When a voltage is applied across a substance there will be an electric current
through it. The applied voltage across the substance is directly proportional to
the current through it. The constant of proportionality is resistance. Hence
resistance is defined as the ratio of the applied voltage to the current through
the substance. Where v is supply voltage I is current
13. Materials used in the field of Electrical Engineering are called the Electrical
Engineering materials. Based on properties and area of applications, Electrical
Engineering materials can be classified as Conductors, Semiconductors, Insulators
Conductors are the materials which have very high conductivity. The number of
free electrons are very high in a conductor at room temperature, which is the basic
reason of high conductivity of conductors.
Examples: Silver, Copper, Gold, Aluminium etc.
Semiconductors are materials which have the conductivity between conductors and
insulators. Semiconductors are the elements of group-III, group-IV and group-IV
elements. Semiconducting materials have covalent bond. At normal temperature the
conductivity of semiconductors is very low. With increase in temperature the
conductivity of semiconductors increases exponentially.
Example: Germanium, Silicon, Gallium Arsenic etc.
14. 3.Insulating Materials
The conductivity of insulating materials is very low. These material are having a
very high resistivity which makes them very suitable to insulate the current
carrying parts from earthed metallic structure. In insulating materials the electrons
are tightly bounded with nucleus. Due to which they cannot be freed for movement
in materials. Due to which the resistivity of insulating materials is very high.
Example:- Plastics, Ceramics, PVC etc.
4. Magnetic Materials
These materials play an important role for existence of various electrical machines.
The magnetic materials having high permeability are used for building the core to
from the low reluctance path for magnetic flux. Magnetic materials can be further
divided in following categories
Ferromagnetic materials: Example: Iron, Cobalt, Nickel.
Paramagnetic material : Example: Aluminum, Platinum, oxygen, Air etc.
Diamagnetic materials : Mostly all metals i.e. silver, copper, gold, hydrogen etc.
Anti ferromagnetic materials : Example: Cr, MNO, FeO, CoO, NiO, Mn etc.
Ferrites : Example: Fe3O4, BaO.6Fe2O3 etc.
15. Classification of circuit element:-
(i) Unilateral and Bilateral element:- If the element property and
characteristic does not change with direction of current, then the element
is called bilateral element; otherwise unilateral element.
(ii) Linear and non linear element:- If the element satisfy homogeneity and
additivity property then element is called linear element, otherwise non
(iii) Active Elements: Active Elements: When the element is capable of
delivering the energy, it is called active element.
Example: Voltage source, Current source, Transistor, Diode, Op-amp etc
(iv) Passive Elements: When the element is not capable of delivering the
energy, it is called passive element.
Example: Resistance, capacitor, inductor etc.
Identify weather the element is: i. Linear or non linear ii. Active or passive iii.
Bilateral or unilateral
17. Inductor:- Inductance is the property of the inductor which opposes the sudden change in
current. Concept:- When a time varying current is flowing through the coil, then magnetic
flux is induced and it is given by
CAPACITANCE: Capacitance is the property of capacitor which opposes the sudden
change in voltage.
Resistance: Resistance is a value that measures how much the component “resist”
the passage of electrical current, the value is measured in ohms (Ω). One way to
R= ρL/A where ρ is the resistivity, a material’s property.
Another form to calculate the resistance is applying Ohm’s law.
V = iR, This constant of proportionality is
27. Convert the following Delta Resistive Network into an equivalent Star Network.
Convert the following Star Resistive Network into an equivalent Delta Network.
28. Generation of AC waveforms
The sine wave is a common type of alternating current (AC) and alternating voltage
The voltage which changes polarity at regular interval of time is known as the
alternating voltage. The one complete cycle of an alternating quantity consists two
half cycles. And the direction of a half cycle changes after every particular interval
of time. The machine which generates the alternating voltage is known as the
Consider the stationary coil places inside the uniform magnetic field. The load is
connected across the coil with the help of brushes and the slip rings. When the coil
rotates in the anticlockwise direction at constant angular velocity ω the
electromotive force induces in the coil. The cross-sectional view of the coil at the
different position is shown in the figure below.
29. The magnitude of the emf induced in the coil depends on the rate of the flux cut by the
conductor. The figure below shows that the no current induces in the coil when they are
parallel to the magnetic line of forces. i.e., at the position (1), (3) and (5). And the total
flux cut by the conductor becomes zero.
The magnitude of the induces emf becomes maximum when the conductor becomes
perpendicular to the magnetic line of force. The conductor cuts the maximum flux at
The direction of the emf induces in the conductor is determined by Fleming’s right-hand
rule. When the coil is at position (2) the emf induces in the outward direction of the
conductor whereas at position (4) the direction of the inducing emf becomes inward.
In other words, the direction of emf induces in the conductor at position (2) and (4)
becomes opposite to each other.
The maximum positive or negative value attained by an alternating quantity in one
complete cycle is called Amplitude or peak value or maximum value. The maximum value of
voltage and current is represented by Em or Vm and Im respectively.
One-half cycle is termed as alternation. An alternation span is of 180 degrees electrical.
When one set of positive and negative values completes by an alternating quantity or it
goes through 360 degrees electrical, it is said to have one complete Cycle.
The value of voltage or current at any instant of time is called an instantaneous value. It is
denoted by (i or e).
The shape obtained by plotting the instantaneous values of an alternating quantity such as
voltage and current along the y-axis and the time (t) or angle (θ=wt) along the x-axis is
called a waveform.
31. Frequency ( f ) is the number of cycles that a sine wave completes in one second .
The more cycles completed in one second, the higher the frequency .
Frequency is measured in hertz (Hz) • Relationship between frequency ( f ) and period (T)
is: f = 1/T
Peak :The peak value of a sine wave is the value of voltage
or current at the positive or negative maximum with
respect to zero • Peak values are represented as: Vp and I p
Peak-to-peak :The peak-to-peak value of a sine wave is the
voltage or current from the positive peak to the negative
peak • The peak-to-peak value is twice the actual voltage
value • Not Often Used • The peak-to-peak values are
represented as: Vpp and Ipp where: Vpp = 2Vp and Ipp = 2Ip
Time period: The time required for a sine wave to complete one full cycle is called the
period (T) • A cycle consists of one complete positive, and one complete negative
alternation T= 1/ f
32. The angular measure measurement of Sine Waves can be done in Degrees or Radians .
A degree is an angular measurement corresponding to 1/360 of a circle or a complete
33. Average Value
Definition: The average of all the instantaneous values of an alternating voltage and currents
over one complete cycle is called Average Value.
If we consider symmetrical waves like sinusoidal current or voltage waveform, the positive half
cycle will be exactly equal to the negative half cycle. Therefore, the average value over a
complete cycle will be zero.
The work is done by both, positive and negative cycle and hence the average value is
determined without considering the signs.
So, the only positive half cycle is considered to determine the average value of alternating
quantities of sinusoidal waves. Let us take an example to understand it.
Divide the positive half cycle into (n) number of
equal parts as shown in the above figure
Let i1, i2, i3…….. in be the mid ordinates
The Average value of current Iav = mean of the mid
34. Average Voltage Equation of sinusoidal wave form
Where: 0 and π are the limits of integration
since we are determining the average value
of voltage over one half a cycle.
Then the area below the curve is finally given
as Area = 2VP. Since we now know the area
under the positive (or negative) half cycle, we
can easily determine the average value of the
positive (or negative) region of a sinusoidal
waveform by integrating the sinusoidal
quantity over half a cycle and dividing by half
For example, if the instantaneous voltage of a
sinusoid is given as: v = Vp.sinθ and the
period of a sinusoid is given as: 2π, then:
35. R.M.S Value
Definition: That steady current which, when flows through a resistor of known resistance
for a given period of time than as a result the same quantity of heat is produced by the
alternating current when flows through the same resistor for the same period of time is
called R.M.S or effective value of the alternating current. In other words, the R.M.S value is
defined as the square root of means of squares of instantaneous values.
Let I be the alternating current flowing through a resistor R for time t seconds, which
produces the same amount of heat as produced by the direct current (Ieff). The base of one
alteration is divided into n equal parts so that each interval is of t/n seconds as shown in the
Let i1, i2, i3,………..in be the mid ordinates
Then the heat produced in
Then, heat produced in first interval = i1
Second interval = i2
Third interval = i3
nth interval = in
Total heat produced in time t
= Rt[(i12 + i2
2 + i3
2 + ……+ in
Since Ieff is considered as the effective value of this current.
Then heat produced by this current in time t = Ieff
36. By definition, equations (1) and (2) are equal. Therefore,
2Rt = Rt[(i1
2 + i2
2 + i3
2 + ……+ in
2 = [(i1
2 + i2
2 + i3
2 + ……+ in
Or Ieff = [(i1
2 + i2
2 + i3
2 + ……+ in
Or Ieff = IRMS = square root of the mean of the squares of the instantaneous
Now, by using the integral calculus the RMS or effective value of the
alternating current over a time period can be calculated as follows:
Total heat produced in time t= Rt[(i12 + i2
2 + i3
2 + ……+ in
t = Ieff
37. RMS Current Equation of sinusoidal wave form
Let i = Im sinωt be the alternating current flowing through a resistance of R ohms for
time t seconds and produces the same heat as produced by Ieff (a direct current).
38. Form Factor
The ratio of the root mean square value to the average value of an alternating quantity
(current or voltage) is called Form Factor.
Mathematically, it is expressed as:
Ir.m.s and Er.m.s are the roots mean square values of the current and the voltage
respectively, and Iav and Eav are the average values of the alternating current and the
For the current varying sinusoidally, the Form Factor is given as:
The value of Form Factor is 1.11
Definition: Peak Factor is defined as the ratio of maximum value to the R.M.S value of an
alternating quantity. Mathematically it is expressed as:
Where, Im and Em are the maximum value of the current and the voltage respectively, and
Ir.m.s and Er.m.s are the roots mean square value of the alternating current and the voltage
respectively. For the current varying sinusoidally, the peak factor is given as:
Pure Resistive AC Circuit:
The circuit containing only a pure resistance of R ohms in the AC circuit is known as Pure
Resistive AC Circuit. The presence of inductance and capacitance does not exist in a purely
resistive circuit. The alternating current and voltage both move forward as well as
backwards in both the direction of the circuit.
In an AC circuit, the ratio of voltage to current depends upon the supply frequency, phase
angle, and phase difference. In an AC resistive circuit, the value of resistance of the resistor
will be same irrespective of the supply frequency.
Let the alternating voltage applied across the circuit be given by the equation Then the
instantaneous value of current flowing through the resistor shown in the figure below will
40. From equation (1) and (3), it is clear that there
is no phase difference between the applied
voltage and the current flowing through a
purely resistive circuit, i.e. phase angle
between voltage and current is zero. Hence, in
an AC circuit containing pure resistance, the
current is in phase with the voltage as shown
in the waveform figure below.
Instantaneous power, p= vi
•P – average power
•Vr.m.s – root mean square value of supply voltage
•Ir.m.s – root mean square value of the current
Hence, the power in a purely resistive circuit is given by:p =VI
41. Pure inductive Circuit
The circuit which contains only inductance (L) and not any other quantities like resistance
and capacitance in the circuit is called a Pure inductive circuit. In this type of circuit, the
current lags behind the voltage by an angle of 90 degrees.
Let the alternating voltage applied to the circuit is given by
As a result, an alternating current i flows through the
inductance which induces an emf in it. The equation
is shown below:
43. Pure Capacitor Circuit
The circuit containing only a pure capacitor of capacitance C farads is known as a Pure
Capacitor Circuit. The capacitors stores electrical power in the electric field, their effect is
known as the capacitance. It is also called the condenser.
44. Instantaneous power is given by p = vi
Hence, from the above equation, it is clear that the average power in the capacitive
circuit is zero.
46. RL,RC, and RLC Circuits
A circuit that contains a pure resistance R ohms connected in series with a coil having a pure
inductance of L (Henry) is known as RL Series Circuit. When an AC supply voltage V is
applied, the current, I flows in the circuit.So, IR and IL will be the current flowing in the
resistor and inductor respectively, but the amount of current flowing through both the elements
will be same as they are connected in series with each other. The circuit diagram of RL Series
Circuit is shown below:
VR – voltage across the resistor R
VL – voltage across the inductor L
V – Total voltage of the circuit
Phasor Diagram of the RL Series Circuit :Current I is taken as a reference. The Voltage drop
across the resistance VR = IR is drawn in phase with the current I. The voltage drop across the
inductive reactance VL =IXL is drawn ahead of the current I. As the current lags voltage by an
angle of 90 degrees in the pure Inductive circuit. The vector sum of the two voltages drops
VR and VL is equal to the applied voltage V.
Now,In right-angle triangle OAB
VR = IR and VL = IXL where XL = 2πfL
Z is the total opposition offered to the flow of alternating current by an RL Series
circuit and is called impedance of the circuit. It is measured in ohms (Ω).
Phase Angle: In RL Series circuit the current lags the voltage by 90 degrees angle
known as phase angle. It is given by the equation:
48. The various points on the power curve are obtained by the product of voltage and current.
If you analyze the curve carefully, it is seen that the power is negative between angle 0 and ϕ
and between 180 degrees and (180 + ϕ) and during the rest of the cycle the power is positive.
The current lags the voltage and thus they are not in phase with each other.
Waveform and Power Curve of the RL Series Circuit
49. RC Series Circuit
A circuit that contains pure resistance R ohms connected in series with a pure capacitor of
capacitance C farads is known as RC Series Circuit. A sinusoidal voltage is applied and
current I flows through the resistance (R) and the capacitance (C) of the circuit.
VR – voltage across the resistance R
VC – voltage across capacitor C
V – total voltage across the RC Series circuit
Phasor Diagram of RC Series Circuit
The phasor diagram of the RC series circuit is
VR = IR and VC = IXC Where XC = I/2πfC
In right triangle OAB,
Z is the total opposition offered to the flow of alternating current by an RC series circuit
and is called impedance of the circuit. It is measured in ohms (Ω).
From the phasor diagram shown above, it is
clear that the current in the circuit leads the
applied voltage by an angle ϕ and this angle is
52. RLC Series Circuit
When a pure resistance of R ohms, a pure inductance of L Henry and a pure capacitance of
C farads are connected together in series combination with each other then RLC Series
Circuit is formed. As all the three elements are connected in series so, the current flowing
through each element of the circuit will be the same as the total current I flowing in the
In the RLC Series circuit
XL = 2πfL and XC = 1/2πfC
When the AC voltage is applied through the RLC
Series circuit the resulting current I flows through
the circuit, and thus the voltage across each
element will be:
VR = IR that is the voltage across the resistance R
and is in phase with the current I.
VL = IXL that is the voltage across the inductance L
and it leads the current I by an angle of 90
VC = IXC that is the voltage across capacitor C and it
lags the current I by an angle of 90 degrees.
The phasor diagram of the RLC series circuit when
the circuit is acting as an inductive circuit that
means (VL>VC) is shown below and if (VL< VC) the
circuit will behave as a capacitive circuit.
53. It is the total opposition offered to the flow of current by an RLC Circuit
and is known as Impedance of the circuit.
Phase Angle From the phasor diagram, the value of phase angle will be
54. Impedance Triangle of RLC Series Circuit
When the quantities of the phasor diagram are
divided by the common factor I then the right angle
triangle is obtained known as impedance triangle.
The impedance triangle of the RL series circuit, when
(XL > XC) is shown below:
If the inductive reactance is greater than the
capacitive reactance than the circuit reactance is
inductive giving a lagging phase angle.
55. Impedance triangle is shown below when the circuit acts as an RC series circuit (XL< XC)
When the capacitive reactance is greater
than the inductive reactance the overall
circuit reactance acts as capacitive and the
phase angle will be leading.
Applications of RLC Series Circuit
•It acts as a variable tuned circuit
•It acts as a low pass, high pass, bandpass,
bandstop filters depending upon the type of
•The circuit also works as an oscillator
•Voltage multiplier and pulse discharge
56. EX1: 200 V, 50 Hz, inductive circuit takes a current of 10A, lagging 30 degree. Find (i) the
resistance (ii) reactance (iii) inductance of the coil.
EX2: A Capacitor of capacitance 79.5μF is connected in series with a non inductive
resistance of 30 across a 100V, 50Hz supply. Find (i) impedance (ii) current (iii) phase angle
57. EX3: A 230 V, 50 Hz ac supply is applied to a coil of 0.06 H inductance and 2.5 resistance
connected in series with a 6.8 μF capacitor. Calculate (i) Impedance (ii) Current (iii)
Phase angle between current and voltage (iv) power factor
EX4: A resistance R, an inductance L=0.01 H and a capacitance C are connected in series.
When an alternating voltage v=400sin( 3000t-20º)is applied to the series combination,
the current flowing is 10 2 sin(3000t-65º). Find the values of R and C.
59. A wound coil that has an inductance of 180mH and a resistance of 35Ω is connected to a
100V 50Hz supply. Calculate: a) the impedance of the coil, b) the current, c) the power
factor, and d) the apparent power consumed. Also draw the resulting power triangle for the
Data given: R = 35Ω, L = 180mH, V = 100V and ƒ = 50Hz.
(a) Impedance (Z) of the coil:
60. UNIT – II:
D.C Machines: D.C Generator, Basic Construction, Operation, emf Equation, types, Open
Circuit Characteristics, simple problems. D.C Motor-principle-back emf-Torque equation,
Speed control, Swinburne’s test.
These can be divided into:
generators – which convert mechanical energy into electrical energy
motors – which convert electrical energy into mechanical energy
Both types operate through the interaction between a magnetic field and a set of windings
A Simple AC Generator
• We noted earlier that Faraday’s law dictates that if a coil of N turns experiences a
change in magnetic flux, then the induced voltage V is given by
• If a coil of area A rotates with respect to a field B, and if at a particular time it is at
an angle to the field, then the flux linking the coil is BAcos, and the rate of
change of flux is given by
63. A Simple DC Generator
The alternating signal from the earlier AC generator could be converted to DC using a
rectifier A more efficient approach is to replace the two slip rings with a single split slip
ring called a commutator.
this is arranged so that connections to the coil are reversed as the voltage from the coil
changes polarity hence the voltage across the brushes is of a single polarity adding
additional coils produces a more constant output.
A dc generator is an electrical machine which converts mechanical energy into direct
current electricity. This energy conversion is based on the principle of production of
dynamically induced emf.
Use of a commutator
64. A simple generator with two coils
Working Principle of DC Generator
We can see that in the first half of
the revolution current always
flows along ABLMCD, i.e., brush no
1 in contact with segment a. In the
next half revolution, in the figure,
the direction of the induced
current in the coil is reversed.
65. But at the same time the position of the segments a and b are also reversed which results
that brush no 1 comes in touch with the segment b. Hence, the current in the load
resistance again flows from L to M. The waveform of the current through the load circuit is
as shown in the figure. This current is unidirectional.
The above content is the basic working principle of DC generator, explained by single loop
generator model. The positions of the brushes of DC generator are so that the change over of
the segments a and b from one brush to other takes place when the plane of rotating coil is at
a right angle to the plane of the lines of force. It is to become in that position, the induced
EMF in the coil is zero.
67. Basic Construction, Operation
Theoretically, a DC generator
can be used as a DC motor
without any constructional
changes and vice versa is also
possible. Thus, a DC generator
or a DC motor can be broadly
termed as a DC machine.
These basic constructional
details are also valid for
the construction of a DC
motor. Hence, let's call this
point as construction of a DC
machine instead of just
'construction of a dc
The above figure shows constructional details of a simple 4-pole DC machine. A DC
machine consists of two basic parts; stator and rotor. Basic constructional parts of a DC
machine are described below.
68. 1.Yoke: The outer frame of a dc machine is called as yoke. It is made up of cast iron or steel. It not only
provides mechanical strength to the whole assembly but also carries the magnetic flux produced by the
2.Poles and pole shoes: Poles are joined to the yoke with the help of bolts or welding. They carry field
winding and pole shoes are fastened to them. Pole shoes serve two purposes; (i) they support field coils
and (ii) spread out the flux in air gap uniformly.
3.Field winding: They are usually made of copper. Field coils are former wound and placed on each pole
and are connected in series. They are wound in such a way that, when energized, they form alternate
North and South poles.
4. Armature core: Armature core is the rotor of a dc machine. It is cylindrical in shape with slots to carry
armature winding. The armature is built up of thin laminated circular steel disks for reducing eddy current
losses. It may be provided with air ducts for the axial air flow for cooling purposes. Armature is keyed to
5.Armature winding: It is usually a former wound copper coil which rests in armature slots. The
armature conductors are insulated from each other and also from the armature core. Armature winding can
be wound by one of the two methods; lap winding or wave winding. Double layer lap or wave windings
are generally used. A double layer winding means that each armature slot will carry two different coils.
6.Commutator and brushes: The function of a commutator, in a dc generator, is to collect the current
generated in armature conductors. Whereas, in case of a dc motor, commutator helps in providing current
to the armature conductors. A commutator consists of a set of copper segments which are insulated from
each other. The number of segments is equal to the number of armature coils. Each segment is connected
to an armature coil and the commutator is keyed to the shaft.
7.Brushes:are usually made from carbon or graphite. They rest on commutator segments and slide on the
segments when the commutator rotates keeping the physical contact to collect or supply the current.
69. E.M.F Equation of DC Generator
We know that the working principle of dc generator, that when conductors begin to cut the
magnetic lines of force and therefore, the e.m.f. induces in the conductors according to
'Faraday's Law of Electromagnetic Induction'. The value of induced e.m.f. depends upon
the lengths of the conductor, the magnetic field strength, and the speed at which the coil
rotates. Let us see the equation for induced e.m.f.
Number of revolutions per minute,= N / 60
70. Time taken for one revolution,dt = 60 / N sec
E.m.f generated per conductor,
Therefore, the total emf E generated between the terminals if given as,
E = Average e.m.f generated per conductor * Number of conductor in each parallel path
Where,A = P for lap winding.
A = 2 for wave winding.
EX-1 A four pole generator, having wave wound armature winding has 51 slots, each slot
containing 20 conductor. What will be the voltage generated in the machine when driven at
1500 rpm assuming the flux per pole to be 7mWb?
71. EX-2 : An 8 pole Dc generator has 500 armature conductors, and a useful flux of 0.05Wb per
pole. what will be the emf generated if it is lap-connected and runs at 1200 rpm? What must
be the speed at which it is to be driven produce the same emf if it is wave-wound?
72. Types of DC Machines
There are 2 main types of DC machines first one is a generator and the second one is a dc
motor, dc motor uses dc current and provides mechanical power and dc generator generates
dc voltage. As we discuss when current-carrying conductor when placed in field emf induced
force apply on it. There are two main categories by which the dc machine gets excited first one
is self-excited and the second one is separately excited. In self-excited machines field current
produced by the machines itself while in separately excited machines the field current is
provided by the separated source. Due to these two excitations methods, the dc machines are
further divided into different types that are described here.
1 .Separately excited DC machine
Application: laboratories for testing as they have a wide range of voltage output.
Used as a supply source of DC motors.
I. Shunt DC machine
II. Series DC machine
III. Compound DC machine
73. Separately Excited DC Machine
If there is no internal voltage induced
at the field windings and there is no
field current so external power source
is connected at the field winding to
produce current and voltage these
winding is known as separate windings.
In the given figure, you see the
detailed circuit of separately excited
74. Shunt DC Machine
In this type of dc machine the field winding (it is located at the stator of a machine) linked
with the armature winding (wound rotor) in parallel.
Due to a parallel connection, the voltage across the field windings is equalled to the supply
voltage in the case of a motor and equals generated voltage in the case of a generator.
So these windings have large no of turns. In a given diagram, you can see the circuit of
shunt dc machines.
Used to charge the battery.
Providing excitation to the
Rsh = Shunt winding resistance
Ish = Current flowing through the shunt field
Ra = Armature resistance
Ia = Armature current
IL = Load current
V = Terminal voltage
Eg = Generated EMF
75. Series Wound DC Machine
In these dc machines, the field windings are connected with the armature windings in
series connection schemes.
As this winding is in series so the armature current also passes through it that has high
value for less power losses the turns of the field windings are less in this machine.
In the given figure, you can see the circuit diagram of these machines.
The traction system,
Here:Rsc = Series winding resistance
Isc = Current flowing through the series field
Ra = Armature resistance
Ia = Armature current
IL = Load current
V = Terminal voltage
Eg = Generated EMF
76. Compound DC Machine
This generator/motor consists of series and shunts windings in its circuitry. 2 windings are
placed at every pole of the machine. The number of turns in the series is less due to the
large value of armature current flowing through it and shunt windings have large no of
There are 2 methods by which these two windings are connected in these machines if the
field windings are in parallel with the armature windings only then the machines are known
as the short shunt compound. In the given figure, you can see these arrangements.
If the field windings is in parallel with the armature and series windings the machines are
known as the long shunt compound machines. In the given figure, you can see resultant
lighting and heavy power
supply, arc welding purpose,
networks, offices, hotels,
homes, schools, etc.
77. If the brush contact drop is included, the
terminal voltage equation is written as:
Terminal voltage is given as:
The shunt field current is given as:
Series field current is given as:
Short Shunt Compound Wound Long Shunt Compound Wound
If the brush contact drop is included, the
terminal voltage equation is written as:
Terminal voltage is given as:
Series field current is given as:
The shunt field current is given as:
78. EX-1: A long shunt compound DC generator delivers a load current of 50A at 500V and has
armature, series field and shunt field resistances of 0.05Ω, 0.03Ω and 250Ω respectively.
Calculate the generated voltage and the armature current. Allow 1V per brush for contact
EX-2: A 230V, 50kW short-shunt compound generator has an armature resistance of Ra =
0.06Ω, series winding resistance of 0.04Ω and shunt field winding of resistance 120Ω.
Calculate the induced armature voltage,E (in Volts) at rated load and terminal voltage. The
total brush contact drop is 2V?
79. EX-3: A compound DC generator has an armature resistance of 0.04Ω, shunt field resistance of
48Ω and series field resistance of 0.02Ω. The field excitation is adjusted to get a rated terminal
voltage of 220V. Find out the internal generated voltage, E(in Volts), when the motor is
supplying 4.4kW at rated voltage for long shunt connection?
EX-4 :A short shunt compound DC generator delivers a load current of 30A at 220V and has
armature, series field and shunt field resistances of 0.05Ω, 0.3Ω and 200Ω respectively.
Calculate the induced emf and the armature current. Allow 1V per brush for contact drop.
80. Open Circuit Characteristic (O.C.C.) (E0/If)
The above figure shows a typical no-load saturation curve or open circuit characteristics for
all types of DC generators.
Open circuit characteristic is also known as
magnetic characteristic or no-load saturation
characteristic. This characteristic shows the relation
between generated emf at no load (E0) and the
field current (If) at a given fixed speed.
The O.C.C. curve is just the magnetization curve and
it is practically similar for all type of generators. The
data for O.C.C. curve is obtained by operating the
generator at no load and keeping a constant speed.
Field current is gradually increased and the
corresponding terminal voltage is recorded.
The connection arrangement to obtain O.C.C. curve
is as shown in the figure below. For shunt or series
excited generators, the field winding is
disconnected from the machine and connected
across an external supply
81. Now, from the emf equation of dc generator, we know that Eg = kɸ. Hence, the generated
emf should be directly proportional to field flux (and hence, also directly proportional to
the field current).
However, even when the field current is zero, some amount of emf is generated
(represented by OA in the figure below). This initially induced emf is due to the fact that
there exists some residual magnetism in the field poles. Due to the residual magnetism, a
small initial emf is induced in the armature. This initially induced emf aids the existing
residual flux, and hence, increasing the overall field flux. This consequently increases the
Thus, O.C.C. follows a straight line. However, as the flux density increases, the poles get
saturated and the ɸ becomes practically constant. Thus, even we increase the If further, ɸ
remains constant and hence, Eg also remains constant. Hence, the O.C.C. curve looks like
the B-H characteristic.
82. D.C Motor-principle-back emf
WORKING PRINCIPLE OF A DC MOTOR
When a current-carrying conductor is placed in a magnetic field, it experiences a torque
and has a tendency to move. or
In other words, when a magnetic field and an electric field interact, a mechanical force is
produced. The DC motor or direct current motor works on that principle. This is known as
The DC motor is the device which converts the direct current into the mechanical work. It
works on the principle of Lorentz Law, which states that “the current carrying conductor
placed in a magnetic and electric field experience a force”. And that force is called the
Lorentz force. The Fleming left-hand rule gives the direction of the force.
Fleming Left Hand Rule
If the thumb, middle finger and the index finger of the left hand are displaced from each
other by an angle of 90°, the middle finger represents the direction of the magnetic field.
The index finger represents the direction of the current, and the thumb shows the
direction of forces acting on the conductor.
The formula calculates the magnitude of the force,
83. where, P=number of poles of dc motor ,Φ= flux per pole ,
Z=total number of armature conductors
N=armature speed ,A=number of parallel paths in armature winding
As all other parameters are constant, therefore, Eb ∝ N
As per Lenz's law, "the induced emf always opposes the cause of its production”. Here, the
cause of generation of back emf is the rotation of armature. Rotation of armature is due to
armature torque. Torque is due to armature current and armature current is due to supply
dc voltage V. Therefore, the ultimate cause of production of Eb is the supply voltage V.
Therefore, back emf is always directed opposite to supply voltage V.
When a dc voltage V is applied across the motor terminals, the armature starts rotating due
to the torque developed in it. As the armature rotates, armature conductors cut the pole
magnetic field, therefore, as per law of electromagnetic induction, an emf called back emf is
induced in them. The back emf (also called counter emf) is given by
84. Torque equation
The turning or twisting force about an axis is called torque.
Consider a wheel of radius R meters acted upon by a circumferential force F new tons as
shown in the figure.
86. Types of DC Motor
A Direct Current Motor, DC is named according to the connection of the field winding with
the armature. Mainly there are two type of DC Motors. One is Separately Excited DC Motor
and other is Self-excited DC Motor.
The self-excited motors are further classified as Shunt wound or shunt motor, Series
wound or series motor and Compound wound or compound motor.
Shunt Wound Motor
This is the most common types of DC
Motor. Here the field winding is
connected in parallel with the armature as
shown in the figure below:
87. Series Wound Motor
In the series motor, the field winding is connected in series with the armature winding.
The connection diagram is shown below:
88. SPEED CONTROL OF DC MOTR
According to the speed equation of a dc motor N ∞ Eb/φ, N = V- Ia Ra/ φ
Thus speed can be controlled by-1.Flux control method: By Changing the flux by
controlling the current through the field winding.
2.Armature control method: By Changing the armature resistance which in turn
changes the voltage applied across the armature, The speed is directly proportional to
the voltage applied across the armature .
The speed of a d.c. motor is controlled
(i) By varying the flux per pole (f) known as flux control method.
(ii) By varying the Ra and is known as armature control method.
(iii) By varying the applied voltage V and is known as voltage control method.
89. i) Field control method:
1. In this field control method the variable is flux (f)
2. The rheostat is placed in series to the field winding, as the field resistance increases
the field current decreases and this weakens the flux
3. The weakening of the flux increases the speed since speed is inversely proportional to
4. Thus using the field control, above base speeds can be controlled.
5. This method is also known as constant power method or variable torque method.
1. This is an easy and convenient method.
2. It is an inexpensive method since very little power is wasted in the shunt field rheostat
due to relatively small value of If
3. The speed control exercised by this method is independent of load on the machine.
1. Only speeds higher than the normal speed can be obtained.
2. There is a limit to the maximum speed obtainable by this method. It is because if the
90. ii) Armature control method
1. In this armature resistance control method the variable is Ra
2. The rheostat is placed in series to the armature winding, as the Ra increases the IaRa drop
increases and this decreases the speed.
3. The decreasing of the back emf decreases the speed since speed is directly
proportional to Eb.
4. Thus using the Ra control method, below base speeds can be controlled.
5. This method is also known as constant torque method or variable power method.
1. A large amount of power is wasted in the controller resistance since it carries full
armature current Ia.
2. The speed varies widely with load since the speed depends upon the voltage drop in
the controller resistance and hence on the armature current demanded by the load.
3. The output and efficiency of the motor are reduced.
4. This method results in poor speed regulation.
91. iii) Voltage control method by Ward-Leonard system
1. This method is used to get the wide range of speed control 10:1.
2. As the speed of the motor is directly proportional to the applied voltage to the
armature, thus by applying the variable voltage the speed is controlled.
3. The armature of the shunt motor M (whose speed is to be controlled) is connected
directly to a d.c. generator G driven by a constant-speed a.c. motor A.
4. The field of the shunt motor is supplied from a constant-voltage exciter E.
5. The field of the generator G is also supplied from the exciter E.
6. The voltage of the generator G can be varied by means of its field regulator.
7. By reversing the field current of generator G by controller FC, the voltage applied to
the motor may be reversed.
1. The speed of the motor can be adjusted through a wide range without resistance losses
which results in high efficiency.
2. The motor can be brought to a standstill quickly, simply by rapidly reducing the
voltage of generator G.
3. The disadvantage of the method is that a special motor-generator set is required for
each motor and the losses in this set are high if the motor is operating under light
loads for long periods.
92. SWINBURNE’S TEST.
This test is a no load test and hence cannot be performed on
The circuit connection is shown in Figure
The machine is run on no load at rated speed which is
adjusted by the shunt field resistance.
1. Economical, because no load input power is sufficient to
perform the test
2. Efficiency can be pre-determined
3. As it is a no load test, it cannot be done on a dc series
1. Change in iron loss from no load to full load is not taken into account. (Because of
Armature reaction, flux is distorted which increases iron losses).
2. Stray load loss cannot be determined by this test and hence efficiency is over
3. Temperature rise of the machine cannot be determined.
4. The test does not indicate whether commutation would be satisfactory when the
machine is loaded.
93. It is a simple indirect test which is applicable where flux' is constant like shunt machine.
The machine is run as motor at no load at its rated speed with the help of shunt field
resistance. The supply voltage, no load input current and field current are measured by
Then the no load armature current Ia0 = Io-Ish amps.
Where Io = No load input current
Ish = Shunt field current
No load input power = V * Io watts
Constant losses (WC) = input - Ar. cu. losses
= V Io – (Iao)2Ra
Where Ra = armature resistance
94. Efficiency of Machine as a motor:
Let motor input Current = I amps
Terminal voltage = V volts
Input Power = VI watts
Output power =Input - Losses
=V I - (WC + (Ia)2Ra) watts
Where Ia = I - Ish Amps
% Efficiency = (output/input) 100
Efficiency of machine as a generator:
Let output current supplied by the generator = I Amps
Terminal voltage = V volts
Output of the generator = V I watts
Input = Output + Losses = V I + (Wc + [Ia]2Ra) watts,
where Ia = I + Ish
% Efficiency of generator = (output/input) 100
95. Efficiency of Machine as a motor:
Let ‘I’ be the line current at which efficiency is to be
Input Power (Pi) = VI watts
Output power (Po) =Input – Total Losses
Total losses = PL = Wc + Ia
Where Ia = I - Ish Amps
Po =V I - (WC + (Ia)2Ra) watts
% Efficiency = (output/input) 100 = (Po / Pi) x 100
Efficiency of machine as a generator:
Let ‘I’ be the line current at which efficiency is to be
Output of the generator (Po) = V I watts
Total losses (PL) = Wc + Ia
Where Ia = I + Ish Amps
Input (Pi)= Output + Losses(PL) = V I + (Wc + [Ia]2Ra) watts,
where Ia = I + Ish
% Efficiency of generator = (output/input) 100 = (Po / Pi) x 100
96. UNIT – III:
Introduction to 3-phase circuits, relation between line and phase
quantities in balanced star and delta networks.
A.C Machines: Single phase transformer: principle-emf equation-
types-OC and SC tests- Voltage Regulation -Efficiency-Simple problems.
99. WHY WE STUDY 3 PHASE SYSTEM ?
•ALL electric power system in the world used 3-phase system to GENERATE, TRANSMIT and
•One phase, two phase, or three phase i can be taken from three phase system rather than
• Instantaneous power is constant (not pulsating).– thus smoother rotation of electrical
machines , High power motors prefer a steady torque
• More economical than single phase – less wire for the same power transfer ,The amount
of wire required for a three phase system is less than required for an equivalent single
104. relation between line and phase quantities in balanced star and delta networks.
Star Connection in a 3 Phase System
•In the Star Connection, the similar ends (either start or finish) of the three windings are
connected to a common point called star or neutral point. The three-line conductors run
from the remaining three free terminals called line conductors.
•The wires are carried to the external circuit, giving three-phase, three-wire star
Connected systems. However, sometimes a fourth wire is carried from the star point to
the external circuit, called neutral wire, forming three-phase, four-wire star connected
•Considering the above figure, the finish terminals a2, b2, and c2 of the three windings are
connected to form a star or neutral point. The three conductors named as R, Y and B run
from the remaining three free terminals as shown in the above figure.
•The current flowing through each phase is called Phase current Iph, and the current flowing
through each line conductor is called Line Current IL. Similarly, the voltage across each
phase is called Phase Voltage Eph, and the voltage across two line conductors is known as
the Line Voltage EL.
105. As the system is balanced, a balanced system
means that in all the three phases, i.e., R, Y and B,
the equal amount of current flows through them.
Therefore, the three voltages ENR, ENY and ENB are
equal in magnitude but displaced from one
another by 120° electrical.
The Phasor Diagram of Star Connection is shown
106. Hence, in a 3 Phase system of star connections, the line current is equal to phase current.
107. Example: A balanced star connected load of (8+j6)Ω per phase is connected to a balanced
3-phase 400 V supply. Find the line current, power factor, power and total volt-amperes.
108. Delta Connection In a 3 Phase System
In Delta (Δ) or Mesh connection, the finished terminal of one winding is connected to start
terminal of the other phase and so on which gives a closed circuit.
The three-line conductors are run from the three junctions of the mesh called Line
•To obtain the delta connections, a2 is connected with b1, b2 is connected with c1 and c2 is
connected with a1 as shown in the above figure.
•The three conductors R, Y and B are running from the three junctions known as Line
•The current flowing through each phase is called Phase Current (Iph), and the current
flowing through each line conductor is called Line Current (IL).
•The voltage across each phase is called Phase Voltage (Eph), and the voltage across two
line conductors is called Line Voltage (EL).
110. The phasor diagram is shown below:
As in the balanced system the three-
phase current I12, I23 and I31 are
equal in magnitude but are
displaced from one another by 120°
112. Definition of Transformer
Electrical transformer is a static device which transforms electrical power from one circuit to
another without any direct electrical connection and without changing frequency of power
but maybe in different voltage levels with the help of mutual induction.
Working Principle of Transformer
The working principle of transformer is very simple. Mutual induction between two or
more windings is responsible for transformation action in an electrical transformer
Basic Theory of Transformer
The alternating current through the winding produces a continually changing and alternating
flux that surrounds the winding. If any other winding is brought nearer to the previous one,
obviously some portion of this flux will link with the second.
As this flux is continually changing in its amplitude and direction, there must be a changing
flux linkage in the second winding or coil.
According to Faraday's law of electromagnetic induction, there must be an EMF induced in
the second. If the circuit of the later winding is closed, there must be a current flowing
through it. This is the most basic thing on which the working principle of transformer stands.
The winding which takes electrical power from the source, is known as the primary
winding. The winding which gives the desired output voltage due to mutual induction is
commonly known as the secondary winding.
113. The rate of change of flux linkage
depends upon the amount of linked flux
with the second winding. So, almost all
flux of primary winding should link to the
secondary winding. This is effectively and
efficiently done by placing one low
reluctance path common to both of the
The form mentioned above of a
transformer is theoretically possible but not
practically, because in open air very tiny
portion of the flux of the first winding will
link with second; so the current that flows
through the closed circuit of later, will be so
small in amount that it will be difficult to
VP – is the Primary Voltage
VS – is the Secondary Voltage
NP – is the Number of Primary Windings
NS – is the Number of Secondary Windings
Φ (phi) – is the Flux Linkage
Principle of Single Phase Transformer
The single-phase transformer works on the principle of Faraday’s Law of Electromagnetic
Induction. Typically, mutual induction between primary and secondary windings is
responsible for the transformer operation in an electrical transformer.
115. Working of Single Phase Transformer
A transformer is a static device that transfers electric power in one circuit to another circuit
of the same frequency. It consists of primary and secondary windings. This transformer
operates on the principle of mutual inductance.
When the primary of a transformer is connected to an AC supply, the current flows in the
coil and the magnetic field build-up. This condition is known as mutual inductance and the
flow of current is as per the Faraday’s Law of electromagnetic induction. As the current
increases from zero to its maximum value, the magnetic field strengthens and is given by
This electromagnet forms the magnetic lines of force and expands outward from the coil
forming a path of magnetic flux. The turns of both windings get linked by this magnetic flux.
The strength of a magnetic field generated in the core depends on the number of turns in
the winding and the amount of current. The magnetic flux and current are directly
proportional to each other.
As the magnetic lines of flux
flow around the core, it
passes through the
secondary winding, inducing
voltage across it. The
Faraday’s Law is used to
determine the voltage
induced across the
secondary coil and it is given
by: N. dɸ/dt
116. Core-type Transformer
In this type of construction, only half
of the windings are wound
cylindrically around each leg of a
transformer to enhance magnetic
coupling as shown in the figure below.
This type of construction ensures that
magnetic lines of force flow across
both the windings simultaneously.
The main disadvantage of the core-
type transformer is the leakage flux
that occurs due to the flow of a small
proportion of magnetic lines of force
outside the core.
Based on Voltage Levels
Commonly used transformer type, depending upon voltage they are classified as:
Step-up Transformer: They are used between the power generator and the power grid. The
secondary output voltage is higher than the input voltage.
Step down Transformer: These transformers are used to convert high voltage primary
supply to low voltage secondary output.
117. Shell-type Transformer
In this type of transformer construction, the primary and secondary windings are
positioned cylindrically on the center limb resulting in twice the cross-sectional area than
the outer limbs.
There are two closed magnetic paths in this type of construction and the outer limb has
the magnetic flux ɸ/2 flowing. Shell type transformer overcomes leakage flux, reduces
core losses and increases efficiency.
To step-down long-distance signals to support both residential and light-commercial
In television sets for voltage regulation
To step-up power in home inverters
To supply power to non-urban areas
To isolate two circuits electrically as primary and secondary are placed far from each
118. As the magnetic flux varies sinusoidally, Φ = Φmax sinωt, then the basic relationship
between induced emf, ( E ) in a coil winding of N turns is given by:
emf = turns x rate of change
ƒ – is the flux frequency in Hertz, = ω/2π
Ν – is the number of coil windings.
Φ – is the amount of flux in webers
This is known as the Transformer EMF Equation. For the primary winding emf, N will be
the number of primary turns, ( NP ) and for the secondary winding emf, N will be the
number of secondary turns, ( NS ).
Transformer EMF Equation
119. EMF Equation of a Transformer
When a sinusoidal voltage is applied to the primary winding of a transformer, alternating
flux ϕm sets up in the iron core of the transformer. This sinusoidal flux links with both
primary and secondary winding. The function of flux is a sine function.
The rate of change of flux with respect to time is
Let ϕm be the maximum value of flux in Weber
f be the supply frequency in Hz
N1 is the number of turns in the primary winding
N2 is the number of turns in the secondary winding
Φ is the flux per turn in Weber
As shown in the above figure that the flux changes from + ϕm to – ϕm in half a cycle of
1/2f seconds. By Faraday’s Law,Let E1 be the emf induced in the primary winding
Where Ψ = N1ϕ
Since ϕ is due to AC supply ϕ = ϕm Sinwt
120. So the induced emf lags flux by 90 degrees.
Maximum valve of emf
But w = 2πf
Root mean square RMS value is
Putting the value of E1max in equation (6) we get
Putting the value of π = 3.14 in the equation (7) we will get the value of E1 as
Now, equating the equation (8) and (9) we get
The above equation is called the turn ratio where K is
known as the transformation ratio.
121. 1. A transformer has 600 turns of the primary winding and 20 turns of the secondary
winding. Determine the secondary voltage if the secondary circuit is open and the primary
voltage is 140 V.
2. A transformer has a primary coil with 1600 loops and a secondary coil with 1000 loops. If
the current in the primary coil is 6 Ampere, then what is the current in the secondary coil.
128. Copper losses : This loss occur in both the primary and secondary windings due to their ohmic
resistance. If I1, I2 are the primary and secondary currents and R1, R2 are the primary and
secondary resistances, respectively.
129. Core or iron losses: When AC supply is given to the primary winding of a transformer an
alternating flux is set up in the core, therefore, hysteresis and eddy current losses occur in
the magnetic core.
Hysteresis and : When the magnetic material is subjected to reversal of magnetic flux, it
causes a continuous reversal of molecular magnets.
This loss can be minimized by using silicon steel material for the construction of core.
Eddy current losses :The emf in the core and circulates eddy currents. This power is
dissipated in the form of heat .
This loss can be minimized by making the core of thin laminations.
130. Voltage regulation: When a transformer is loaded, with a constant supply voltage, the
terminal voltage changes due to voltage drop in the internal parameters of the transformer
i.e., primary and secondary resistances and inductive reactance. The voltage drop at the
terminals also depends upon the load and its power factor. The change in terminal voltage
from no-load to full-load at constant supply voltage with respect to no-load voltage is
known as voltage regulation of the transformer.
Let, E2 = Secondary terminal voltage at no-load.
V2 = Secondary terminal voltage at full-load.
Then, voltage regulation =𝐸2− 𝑉2 /𝐸2 (𝑝𝑒𝑟 𝑢𝑛𝑖𝑡) In the form of percentage,
% Reg = 𝐸2− 𝑉2 /𝐸2 × 100
131. Explain OC and SC tests on a single phase transformer
Purpose of conducting OC and SC tests is to find
i) Equivalent circuit parameters ii) Efficiency iii) Regulation
Open Circuit Test:
1. The OC test is performed on LV side at rated voltage and HV side is kept opened.
2. As the test is conducted on LV side the meters selected will be at low range values like
smaller voltmeter, smaller ammeter and low pf wattmeter
3.As the no-load current is quite small about 2 to 5% of the rated current, the ammeter
required here will be smaller range even after on LV side which are designed
for higher current values.
4. The voltmeter, ammeter and the wattmeter readings V0, I0and W0 respectively are
noted by applying rated voltage on LV side.
5.The wattmeter will record the core loss because of no load input power.
133. Short Circuit Test
This test is carried out to determine the
(i) Copper losses at full load (or at any desired load). These losses are required for the
calculations of efficiency of the transformer.
(ii) Equivalent impedance (Zes or Zep), resistance (Res or Rep) and leakage reactance (Xes or
Xep) of the transformer.
This test is usually carried out on the high-voltage side of the transformer i.e., a wattmeter
W, voltmeter V and an ammeter A are connected in high-voltage winding . The other
winding is then short circuited by a thick strip. A low voltage at normal frequency is applied
to the high voltage winding with the help of on autotransformer so that full-load current
flows in both the windings.
134. Short Circuit Test:
1. The SC test is performed on HV side at rated current and LV side is kept Shorted.
2. As the test is conducted on HV side the meters selected will be at low range values like
smaller voltmeter, smaller ammeter and unity pf wattmeter
3. As the voltage required to circulate the short circuit rated current is very small about 10
to 15% of the rated HV voltage, so the voltmeter required here will be smaller range even
the test is conducted on HV side.
4. The voltmeter, ammeter and the wattmeter readings Vsc, Isc and Wsc respectively are
noted by passing rated current on HV side.
5.The wattmeter will record the copper loss corresponding to the Isc.
137. Variation of voltage regulation and efficiency with respect to load and load power factors
Why Transformers Are Rated In KVA?
From the above transformer tests, it can be seen that Cu loss of a transformer depends on
current, and iron loss depends on voltage. Thus, total transformer loss depends on volt-
ampere (VA). It does not depend on the phase angle between voltage and current, i.e.
transformer loss is independent of load power factor. This is the reason that transformers
are rated in kVA.
138. Ex1:The core loss of the single phase 2200/220‐V, 50‐Hz transformer is measured using the
low‐ voltage side as the primary. The following data are obtained: Voltmeter reading = 240 V
Exciting current = 1.0 A Wattmeter reading = 24 W.
Calculate: a) The core loss, in watts. b) The power factor and phase angle. c) the Core loss
current. d) The lagging (magnetizing) component of the current.