Basic Electrical Engineering - Organized
Basic Electrical Engineering - Organized
Basic Electrical Engineering - Organized
Objectives:
1. To understand the basic concepts of electrical circuits & networks and their analysis which is the
foundation for all the subjects in the electrical engineering discipline.
2. To emphasize on the basic elements in electrical circuits and analyze Circuits using Network
Theorems.
3. To analyze Single-Phase AC Circuits.
4. To illustrate Single-Phase Transformers and DC Machines.
5. To get overview of basic electrical installations and calculations for energy consumption.
UNIT –I:
Introduction to Electrical Circuits: Concept of Circuit and Network, Types of elements, R-L-C
Parameters, Independent and Dependent sources, Source transformation and Kirchhoff’s Laws
UNIT –II:
Network Analysis: Network Reduction Techniques- Series and parallel connections of resistive networks,
Star–to-Delta and Delta-to-Star Transformations for Resistive Networks, Mesh Analysis, and Nodal
Analysis,
Network Theorems: Thevenin’s theorem, Norton’s theorem and Superposition theorem and Illustrative
Problems.
UNIT-III:
Single Phase A.C. Circuits: Average value, R.M.S. value, form factor and peak factor for sinusoidal wave
form, Complex and Polar forms of representation. Steady State Analysis of series R-L-C circuits. Concept of
Reactance, Impedance, Susceptance, Admittance, Concept of Power Factor, Real, Reactive and Complex
power, Illustrative Problems.
UNIT –IV:
Electrical Machines (elementary treatment only):
Single phase transformers: principle of operation, constructional features and emf equation.
DC. Generator: principle of operation, constructional features, emf equation. DC Motor: principle of
operation, Back emf, torque equation.
UNIT –V:
Electrical Installations:
Components of LT Switchgear: Switch Fuse Unit (SFU), MCB, ELCB, Types of Wires and Cables,
Earthing. Elementary calculations for energy consumption and battery backup.
TEXT BOOKS:
1. Engineering Circuit Analysis - William Hayt, Jack E. Kemmerly, S M Durbin, Mc Graw Hill
Companies.
2. Electric Circuits - A. Chakraborty, Dhanipat Rai & Sons.
3. Electrical Machines – P.S. Bimbra, Khanna Publishers.
REFERENCE BOOKS:
1. Network analysis by M.E Van Valkenburg, PHI learning publications.
2. Network analysis - N.C Jagan and C. Lakhminarayana, BS publications.
3. Electrical Circuits by A. Sudhakar, Shyammohan and S Palli, Mc Graw Hill Companies.
4. Electrical Machines by I.J. Nagrath & D. P. Kothari, Tata Mc Graw-Hill Publishers.
Outcomes:
At the end of the course students, would be able to
1. Apply the basic RLC circuit elements and its concepts to networks and circuits.
2. Analyze the circuits by applying network theorems to solve them to find various electrical parameters.
3. Illustrate the single-phase AC circuits along with the concept of impedance parameters and power.
4. Understand the Constructional Details and Principle of Operation of DC Machines and Transformers
5. Understand the basic LT Switch gear and calculations for energy consumption.
PREFACE
Engineering institutions have been modernizing and updating their curriculum to keep pace
with the continuously developing technological trends so as to meet the correspondingly changing
educational demands of the industry. As the years passed by, multi-disciplinary education system also has
become more and more relevant in the present global industrial development. Thus, just as Computer
Systems & Applications, Basic Electrical Engineering also has become an integral part of all the industrial
and engineering sectors be it infrastructure, power generation, minor & major Industries, Industrial Safety or
process industries where automation has become an inherent part. Accordingly, several universities have
been bringing in a significant change in their graduate programs of engineering starting from the first year to
meet the needs of these important industrial sectors to enhance the employability of their graduates. Thus, at
college entry level itself Basic Electrical Engineering has become the first Multidisciplinary core
engineering subject for almost all the other core engineering branches like Civil, Mechanical, Production
engineering, Industrial Engineering, Aeronautical, Instrumentation, Control Systems and Computer
Engineering. As a further impetus, since for understanding of this subject a practical knowledge is equally
important, a laboratory course is also added in the curriculum. The chapters are so chosen that the student
comprehends all the important theoretical concepts with good practical insight.
This handbook of Digital notes for Basic Electrical Engineering is brought out in a simple
and lucid manner highlighting the important underlying concepts & objectives along with sequential steps to
understand the subject.
INDEX
SNO. TOPIC PAGE NO.
UNIT –I INTRODUCTION TO ELECTRICAL CIRCUITS
Average value, R.M.S. value, form factor and peak factor 51-57
for sinusoidal wave form.
Steady State Analysis of series R-L-C circuits. 58-64
Earthing 95-98
UNIT-I
INTRODUCTION TO ELECTRICAL CIRCUITS
• Electric Circuit
• Electric Network
• Current
• Voltage
• Power
So, it is imperative that we gather some basic knowledge on these terms before proceeding further. Let’s
start with Electric Circuit.
Electric Circuit
An electric circuit contains a closed path for providing a flow of electrons from a voltage
source or current source. The elements present in an electric circuit will be in series connection, parallel
connection, or in any combination of series and parallel connections.
Electric Network
An electric network need not contain a closed path for providing a flow of electrons from a
voltage source or current source. Hence, we can conclude that "all electric circuits are electric networks"
but the converse need not be true.
Current
The current "I" flowing through a conductor is nothing but the time rate of flow of charge.
Mathematically, it can be written as
Where,
• Q is the charge and its unit is Coloumb.
• t is the time and its unit is second.
As an analogy, electric current can be thought of as the flow of water through a pipe. Current is measured
in terms of Ampere. In general, Electron current flows from negative terminal of source to positive
terminal, whereas, Conventional current flows from positive terminal of source to negative terminal.
MRCET EAMCET CODE: MLRD www.mrcet.ac.in 8
DEPARTMENT OF HUMANITIES AND
SCIENCES BASIC ELECTRICAL ENGINEERING
Electron current is obtained due to the movement of free electrons, whereas, Conventional current is
obtained due to the movement of free positive charges. Both of these are called as electric current.
Voltage
The voltage "V" is nothing but an electromotive force that causes the charge (electrons) to
flow. Mathematically, it can be written as
Where,
• W is the potential energy and its unit is Joule.
• Q is the charge and its unit is Coloumb.
As an analogy, Voltage can be thought of as the pressure of water that causes the water to flow through a
pipe. It is measured in terms of Volt.
Power
The power "P" is nothing but the time rate of flow of electrical energy. Mathematically, it
can be written as
Where,
• W is the electrical energy and it is measured in terms of Joule.
• t is the time and it is measured in seconds.
We can re-write the above equation a
Therefore, power is nothing but the product of voltage V and current I. Its unit is Watt.
Types of Network Elements
We can classify the Network elements into various types based on some parameters.
Following are the types of Network elements −
• Active Elements and Passive Elements
• Linear Elements and Non-linear Elements
• Bilateral Elements and Unilateral Elements
• Lumped Elements and Distributed Elements
In the above figure, the current (I) is flowing from terminals A to B through a passive element having
impedance of Z Ω. It is the ratio of voltage (V) across that element between terminals A & B and current (I).
In the above figure, the current (I) is flowing from terminals B to A through a passive element having
impedance of Z Ω. That means the current (–I) is flowing from terminals A to B. In this case too, we will
get the same impedance value, since both the current and voltage having negative signs with respect to
terminals A & B.
Unilateral Elements are those that allow the current in only one direction. Hence, they offer different
impedances in both directions.
We discussed the types of network elements in the previous chapter. Now, let us identify the nature of
network elements from the V-I characteristics given in the following examples.
Example 1
The V-I characteristics of a network element is shown below.
• Similarly, in the third quadrant, the values of both voltage (V) and current (I) have negative values.
So, the ratios of voltage (V) and current (I) produce positive impedance values.
Since, the given V-I characteristics offer positive impedance values, the network element is a Passive
element.
Step 3 − Verifying the network element as bilateral or unilateral.
For every point (I, V) on the characteristics, there exists a corresponding point (-I, -V) on the given
characteristics. Hence, the network element is a Bilateral element.
Therefore, the given V-I characteristics show that the network element is a Linear, Passive, and Bilateral
element.
Example 2
The V-I characteristics of a network element is shown below.
Distributed elements are those which are not electrically separable for analytical purposes.
For example a transmission line has distributed parameters along its length and may extend for hundreds
of miles.
R-L-C Parameters
Resistor
The main functionality of Resistor is either opposes or restricts the flow of electric current.
Hence, the resistors are used in order to limit the amount of current flow and / or dividing (sharing) voltage.
Let the current flowing through the resistor is I amperes and the voltage across it is V volts. The symbol of
resistor along with current, I and voltage, V are shown in the following figure.
According to Ohm’s law, the voltage across resistor is the product of current flowing through it and the
resistance of that resistor. Mathematically, it can be represented as
So, we can calculate the amount of power dissipated in the resistor by using one of the formulae mentioned
in Equations 3 to 5.
Inductor
In general, inductors will have number of turns. Hence, they produce magnetic flux when
current flows through it. So, the amount of total magnetic flux produced by an inductor depends on the
current, I flowing through it and they have linear relationship.
Mathematically, it can be written as
Where,
• Ψ is the total magnetic flux
• L is the inductance of an inductor
Let the current flowing through the inductor is I amperes and the voltage across it is V volts. The symbol of
inductor along with current I and voltage V are shown in the following figure.
According to Faraday’s law, the voltage across the inductor can be written as
From the above equations, we can conclude that there exists a linear relationship between voltage across
inductor and current flowing through it.
We know that power in an electric circuit element can be represented as
By integrating the above equation, we will get the energy stored in an inductor as
So, the inductor stores the energy in the form of magnetic field.
Capacitor
In general, a capacitor has two conducting plates, separated by a dielectric medium. If
positive voltage is applied across the capacitor, then it stores positive charge. Similarly, if negative voltage
is applied across the capacitor, then it stores negative charge.
So, the amount of charge stored in the capacitor depends on the applied voltage V across it and they have
linear relationship. Mathematically, it can be written as
Where,
• Q is the charge stored in the capacitor.
• C is the capacitance of a capacitor.
Let the current flowing through the capacitor is I amperes and the voltage across it is V volts. The symbol
of capacitor along with current I and voltage V are shown in the following figure.
We know that the current is nothing but the time rate of flow of charge. Mathematically, it can be
represented as
From the above equations, we can conclude that there exists a linear relationship between voltage across
capacitor and current flowing through it.
We know that power in an electric circuit element can be represented as
By integrating the above equation, we will get the energy stored in the capacitor as
So, the capacitor stores the energy in the form of electric field.
Types of Sources
Active Elements are the network elements that deliver power to other elements present in an
electric circuit. So, active elements are also called as sources of voltage or current type. We can classify
these sources into the following two categories −
• Independent Sources
• Dependent Sources
IndependentSources
As the name suggests, independent sources produce fixed values of voltage or current and
these are not dependent on any other parameter. Independent sources can be further divided into the
following two categories −
The V-I characteristics of an independent ideal voltage source is a constant line, which is always equal to
the source voltage (VS) irrespective of the current value (I). So, the internal resistance of an independent
ideal voltage source is zero Ohms.
Hence, the independent ideal voltage sources do not exist practically, because there will be some internal
resistance.
Independent practical voltage source and its V-I characteristics are shown in the following figure.
There is a deviation in the V-I characteristics of an independent practical voltage source from the V-I
characteristics of an independent ideal voltage source. This is due to the voltage drop across the internal
resistance (RS) of an independent practical voltage source.
Independent Current Sources
An independent current source produces a constant current. This current is independent of
the voltage across its two terminals. Independent ideal current source and its V-I characteristics are shown
in the following figure.
The V-I characteristics of an independent ideal current source is a constant line, which is always equal to
the source current (IS) irrespective of the voltage value (V). So, the internal resistance of an independent
ideal current source is infinite ohms.
Hence, the independent ideal current sources do not exist practically, because there will be some internal
resistance.
Independent practical current source and its V-I characteristics are shown in the following figure.
There is a deviation in the V-I characteristics of an independent practical current source from the V-I
characteristics of an independent ideal current source. This is due to the amount of current flows through
the internal shunt resistance (RS) of an independent practical current source.
Dependent Sources
As the name suggests, dependent sources produce the amount of voltage or current that is
dependent on some other voltage or current. Dependent sources are also called as controlled sources.
Dependent sources can be further divided into the following two categories −
Dependent current sources are represented with an arrow inside a diamond shape. The magnitude of the
current source can be represented outside the diamond shape. We can observe these dependent or controlled
sources in equivalent models of transistors.
Source Transformation Technique
We know that there are two practical sources, namely, voltage source and current source. We
can transform (convert) one source into the other based on the requirement, while solving network
problems.
The technique of transforming one source into the other is called as source transformation technique.
Following are the two possible source transformations −
Practical voltage source consists of a voltage source (VS) in series with a resistor (RS). This can be
converted into a practical current source as shown in the figure. It consists of a current source (IS) in
parallel with a resistor (RS).
The value of IS will be equal to the ratio of VS and RS. Mathematically, it can be represented as
Practical current source consists of a current source (IS) in parallel with a resistor (RS). This can be
converted into a practical voltage source as shown in the figure. It consists of a voltage source (VS) in series
with a resistor (RS).
The value of VS will be equal to the product of IS and RS. Mathematically, it can be represented as
In this chapter, we will discuss in detail about the passive elements such as Resistor, Inductor, and
Capacitor. Let us start with Resistors.
Kirchhoff’s Laws
Network elements can be either of active or passive type. Any electrical circuit or network
contains one of these two types of network elements or a combination of both.
Now, let us discuss about the following two laws, which are popularly known as Kirchhoff’s laws.
Where,
• Im is the mth branch current leaving the node.
• In the above figure, the branch currents I1, I2 and I3 areentering at node P. So, consider negative
signs for these three currents.
• In the above figure, the branch currents I4 and I5 areleaving from node P. So, consider positive signs
for these two currents.
The KCL equation at node P will be
In the above equation, the left-hand side represents the sum of entering currents, whereas the right-hand
side represents the sum of leaving currents.
In this tutorial, we will consider positive sign when the current leaves a node and negative sign when it
enters a node. Similarly, you can consider negative sign when the current leaves a node and positive sign
when it enters a node. In both cases, the result will be same.
Note − KCL is independent of the nature of network elements that are connected to a node.
Kirchhoff’s Voltage Law
Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of voltages around a loop or
mesh is equal to zero.
A Loop is a path that terminates at the same node where it started from. In contrast, a Mesh is a loop that
doesn’t contain any other loops inside it.
Where,
• Vn is the nth element’s voltage in a loop (mesh).
• N is the number of network elements in the loop (mesh).
The above statement of KVL can also be expressed as "the algebraic sum of voltage sources is equal to the
algebraic sum of voltage drops that are present in a loop." Let us verify this statement with the help of the
following example.
Example
Write KVL equation around the loop of the following circuit.
The above circuit diagram consists of a voltage source, VS in series with two resistors R1 and R2. The
voltage drops across the resistors R1 and R2 are V1 and V2 respectively.
Apply KVL around the loop.
In the above equation, the left-hand side term represents single voltage source VS. Whereas, the right-hand
side represents the sum of voltage drops. In this example, we considered only one voltage source. That’s
why the left-hand side contains only one term. If we consider multiple voltage sources, then the left side
contains sum of voltage sources.
In this tutorial, we consider the sign of each element’s voltage as the polarity of the second terminal that is
present while travelling around the loop. Similarly, you can consider the sign of each voltage as the polarity
of the first terminal that is present while travelling around the loop. In both cases, the result will be same.
Note − KVL is independent of the nature of network elements that are present in a loop.
In this chapter, let us discuss about the following two division principles of electrical quantities.
The above circuit diagram consists of an input current source IS in parallel with two resistors R1 and R2.
The voltage across each element is VS. The currents flowing through the
resistors R1 andR2 are I1 and I2 respectively.
The KCL equation at node P will be
From equations of I1 and I2, we can generalize that the current flowing through any passive element can be
found by using the following formula.
This is known as current division principle and it is applicable, when two or more passive elements are
connected in parallel and only one current enters the node.
Where,
• IN is the current flowing through the passive element of Nth branch.
• IS is the input current, which enters the node.
• Z1, Z2, …,ZN are the impedances of 1st branch, 2ndbranch, …, Nth branch respectively.
Voltage Division Principle
When two or more passive elements are connected in series, the amount of voltage present
across each element gets divided (shared) among themselves from the voltage that is available across that
entire combination.
Consider the following circuit diagram.
The above circuit diagram consists of a voltage source, VS in series with two resistors R1 and R2. The
current flowing through these elements is IS. The voltage drops across the resistors R1and R2 are V1 and
V2 respectively.
The KVL equation around the loop will be
From equations of V1 and V2, we can generalize that the voltage across any passive element can be found by
using the following formula.
This is known as voltage division principle and it is applicable, when two or more passive elements are
connected in series and only one voltage available across the entire combination.
Where,
• VN is the voltage across Nth passive element.
• VS is the input voltage, which is present across the entire combination of series passive elements.
• Z1,Z2, …,Z3 are the impedances of 1st passive element, 2nd passive element, …, Nth passive element
respectively.
UNIT-II
NETWORK ANALYSIS
It has a single voltage source (VS) and three resistors having resistances of R1, R2 and R3. All these
elements are connected in series. The current IS flows through all these elements.
The above circuit has only one mesh. The KVL equation around this mesh is
The equivalent circuit diagram of the given circuit is shown in the following figure.
That means, if multiple resistors are connected in series, then we can replace them with an equivalent
resistor. The resistance of this equivalent resistor is equal to sum of the resistances of all those multiple
resistors.
Note 1 − If ‘N’ inductors having inductances of L1, L2, ..., LN are connected in series, then the equivalent
inductance will be
Note 2 − If ‘N’ capacitors having capacitances of C1, C2, ..., CNare connected in series, then the equivalent
capacitance will be
It has a single current source (IS) and three resistors having resistances of R1, R2, and R3. All these elements
are connected in parallel. The voltage (VS) is available across all these elements.
The above circuit has only one principal node (P) except the Ground node. The KCL equation at this
principal node (P) is
The equivalent circuit diagram of the given circuit is shown in the following figure.
That means, if multiple resistors are connected in parallel, then we can replace them with an equivalent
resistor. The resistance of this equivalent resistor is equal to the reciprocal of sum of reciprocal of each
resistance of all those multiple resistors.
Note 1 − If ‘N’ inductors having inductances of L1, L2, ..., LN are connected in parallel, then the equivalent
inductance will be
Note 2 − If ‘N’ capacitors having capacitances of C1, C2, ..., CNare connected in parallel, then the equivalent
capacitance will be
Example Problems:
fig(a)
Solution:
Also, the 1 ohm and 5ohms resistors are in series; hence their equivalent resistance is
Thus the circuit in Fig.(b) is reduced to that in Fig. (c). In Fig. (b), we notice that the two 2 ohms resistors
are in series, so the equivalent resistance is
This 4 ohms resistor is now in parallel with the 6 ohms resistor in Fig.(b); their equivalent resistance is
The circuit in Fig.(b) is now replaced with that in Fig.(c). In Fig.(c), the three resistors are in series. Hence,
the equivalent resistance for the circuit is
Solution:
In the given network 4 ohms, 5 ohms and 3 ohms comes in series then equivalent resistance is
4+5 + 3 = 12 ohms
From fig(c), 3 ohms and 3 ohms are in series, equivalent resistance is 6 ohms
From fig(d), 6 ohms and 6 ohms are in parallel, equivalent resistance is 3 ohms
From fig(e), 4 ohms, 3 ohms and 3 ohms are in series .Hence Req = 4+ 3+ 3 =10 ohms
Delta Network
Consider the following delta network as shown in the following figure.
The following equations represent the equivalent resistance between two terminals of delta network, when
the third terminal is kept open.
Star Network
The following figure shows the equivalent star network corresponding to the above delta
network.
The following equations represent the equivalent resistance between two terminals of star network, when the
third terminal is kept open.
By using the above relations, we can find the resistances of star network from the resistances of delta
network. In this way, we can convert a delta network into a star network.
Star to Delta Transformation
In the previous chapter, we discussed about the conversion of delta network into an
equivalent star network. Now, let us discuss about the conversion of star network into an equivalent delta
network. This conversion is called as Star to Delta Conversion.
In the previous chapter, we got the resistances of star network from delta network as
By using the above relations, we can find the resistances of delta network from the resistances of star
network. In this way, we can convert star network into delta network.
Example problems:
Solution: The equivalent delta for the given star is shown in fig(b), where
Solution: Delta connected resistors 25 ohms, 10 ohms and 15 ohms are converted in to star as shown in
given figure.
The given circuit thus reduces to the circuit shown in below fig.
Mesh Analysis:
Mesh analysis provides general procedure for analyzing circuits using mesh currents as the circuit
variables. Mesh Analysis is applicable only for planar networks. It is preferably useful for the circuits that
have many loops .This analysis is done by using KVL and Ohm's law.
In Mesh analysis, we will consider the currents flowing through each mesh. Hence, Mesh analysis is also
called as Mesh-current method.
A branch is a path that joins two nodes and it contains a circuit element. If a branch belongs to only one
mesh, then the branch current will be equal to mesh current.
If a branch is common to two meshes, then the branch current will be equal to the sum (or difference) of
two mesh currents, when they are in same (or opposite) direction.
Procedure ofMesh Analysis
Follow these steps while solving any electrical network or circuit using Mesh analysis.
• Step 1 − Identify the meshes and label the mesh currents in either clockwise or anti-clockwise
direction.
• Step 2 − Observe the amount of current that flows through each element in terms of mesh currents.
• Step 3 − Write mesh equations to all meshes. Mesh equation is obtained by applying KVL first and
then Ohm’s law.
• Step 4 − Solve the mesh equations obtained in Step 3 in order to get the mesh currents.
Now, we can find the current flowing through any element and the voltage across any element that is
present in the given network by using mesh currents.
Example
Find the voltage across 30 Ω resistor using Mesh analysis.
Step 1 − There are two meshes in the above circuit. The mesh currents I1 and I2 are considered in clockwise
direction. These mesh currents are shown in the following figure.
Step 2 − The mesh current I1 flows through 20 V voltage source and 5 Ω resistor. Similarly, the mesh
current I2 flows through 30 Ω resistor and -80 V voltage source. But, the difference of two mesh currents,
I1 and I2, flows through 10 Ω resistor, since it is the common branch of two meshes.
Step 3 − In this case, we will get two mesh equations since there are two meshes in the given circuit. When
we write the mesh equations, assume the mesh current of that particular mesh as greater than all other mesh
currents of the circuit. The mesh equation of first mesh is
Network Theorems:
Introduction:
Any complicated network i.e. several sources, multiple resistors are present if the single element response is
desired then use the network theorems. Network theorems are also can be termed as network reduction
techniques. Each and every theorem got its importance of solving network. Let us see some important
theorems with DC and AC excitation with detailed procedures.
Thevenin’s Theorem and Norton’s theorem (Introduction) :
Thevenin’s Theorem and Norton’s theorem are two important theorems in solving Network problems having
many active and passive elements. Using these theorems the networks can be reduced to simple equivalent
circuits with one active source and one element. In circuit analysis many a times the current through a
branch is required to be found when it’s value is changed with all other element values remaining same. In
such cases finding out every time the branch current using the conventional mesh and node analysis methods
is quite awkward and time consuming. But with the simple equivalent circuits (with one active source and
one element) obtained using these two theorems the calculations become very simple. Thevenin’s and
Norton’s theorems are dual theorems.
Thevenin’s Theorem Statement:
Any linear, bilateral two terminal network consisting of sources and resistors(Impedance),can
be replaced by an equivalent circuit consisting of a voltage source in series with a resistance
(Impedance).The equivalent voltage source VTh is the open circuit voltage looking into the terminals(with
concerned branch element removed) and the equivalent resistance RTh while all sources are replaced by their
internal resistors at ideal condition i.e. voltage source is short circuit and current source is open circuit.
(a) (b)
Figure (a) shows a simple block representation of a network with several active / passive elements with the
load resistance RL connected across the terminals ‘a & b’ and figure (b) shows the Thevenin's equivalent
circuit with VTh connected across RTh & RL .
Fig.(a)
Example: Find VTH, RTH and the load current and load voltage flowing through RL resistor as shown in fig.
by using Thevenin’s Theorem?
Solution:
The resistance RL is removed and the terminals of the resistance RL are marked as A & B as shown in the
fig. (1)
Fig.(1)
Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (VTH). We have already
removed the load resistor from fig.(a), so the circuit became an open circuit as shown in fig (1). Now we
have to calculate the Thevenin’s Voltage. Since 3mA Current flows in both 12kΩ and 4kΩ resistors as this
is a series circuit because current will not flow in the 8kΩ resistor as it is open. So 12V (3mA x 4kΩ) will
appear across the 4kΩ resistor. We also know that current is not flowing through the 8kΩ resistor as it is
open circuit, but the 8kΩ resistor is in parallel with 4k resistor. So the same voltage (i.e. 12V) will appear
across the 8kΩ resistor as 4kΩ resistor. Therefore 12V will appear across the AB terminals.
So, VTH = 12V
Fig (2)
All voltage & current sources replaced by their internal impedances (i.e. ideal voltage sources short circuited
and ideal current sources open circuited) as shown in fig.(3)
Fig(3)
Calculate /measure the Open Circuit Resistance. This is the Thevenin's Resistance (RTH)We have Reduced
the 48V DC source to zero is equivalent to replace it with a short circuit as shown in figure (3) We can see
that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:
8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)
RTH = 8kΩ + [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]
RTH = 8kΩ + 3kΩ
RTH = 11kΩ
Fig(4)
Connect the RTH in series with Voltage Source VTH and re-connect the load resistor across the load
terminals(A&B) as shown in fig (5) i.e. Thevenin's circuit with load resistor. This is the Thevenin’s
equivalent circuit.
VTH
Fig (5)
Now apply Ohm’s law and calculate the load current from fig 5.
IL = VTH/ (RTH + RL)= 12V / (11kΩ + 5kΩ) = 12/16kΩ
IL= 0.75mA
And VL = ILx RL= 0.75mA x 5kΩ
VL= 3.75V
Norton’s Theorem Statement:
Any linear, bilateral two terminal network consisting of sources and resistors(Impedance),can
be replaced by an equivalent circuit consisting of a current source in parallel with a resistance
(Impedance),the current source being the short circuited current across the load terminals and the resistance
being the internal resistance of the source network looking through the open circuited load terminals.
(a) (b)
Figure (a) shows a simple block representation of a network with several active / passive elements with the
load resistance RL connected across the terminals ‘a & b’ and figure (b) shows the Norton equivalent
circuit with IN connected across RN & RL .
Main steps to find out IN and RN:
• The terminals of the branch/element through which the current is to be found out are marked as say a
& b after removing the concerned branch/element.
• Open circuit voltage VOC across these two terminals and ISC through these two terminals are found
out using the conventional network mesh/node analysis methods and they are same as what we
obtained in Thevenin’s equivalent circuit.
• Next Norton resistance RN is found out depending upon whether the network contains dependent
sources or not.
a) With dependent sources: RN = Voc / Isc
b) Without dependent sources : RN = Equivalent resistance looking into the concerned terminals
with all voltage & current sources replaced by their internal impedances (i.e. ideal voltage
sources short circuited and ideal current sources open circuited)
• Replace the network with IN in parallel with RN and the concerned branch resistance across the load
terminals(A&B) as shown in below fig
Example: Find the current through the resistance RL (1.5 Ω) of the circuit shown in the figure (a)
below using Norton’s equivalent circuit.
Fig(a)
Solution: To find out the Norton’s equivalent ckt we have to find out IN = Isc ,RN=Voc/ Isc. Short the 1.5Ω
load resistor as shown in (Fig 2), and Calculate / measure the Short Circuit Current. This is the Norton
Current (IN).
Fig(2)
We have shorted the AB terminals to determine the Norton current, IN. The 6Ω and 3Ω are then in parallel
and this parallel combination of 6Ω and 3Ω are then in series with 2Ω.So the Total Resistance of the circuit
to the Source is:-
2Ω + (6Ω || 3Ω) ….. (|| = in parallel with)
RT = 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)]
RT = 2Ω + 2Ω
RT = 4Ω
IT = V / RT
IT = 12V / 4Ω= 3A..
Now we have to find ISC = IN… Apply CDR… (Current Divider Rule)…
ISC = IN = 3A x [(6Ω / (3Ω + 6Ω)] = 2A.
ISC= IN = 2A.
Fig(3)
All voltage & current sources replaced by their internal impedances (i.e. ideal voltage sources short circuited
and ideal current sources open circuited) and Open Load Resistor. as shown in fig.(4)
Fig(4)
Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (RN) We have Reduced the
12V DC source to zero is equivalent to replace it with a short circuit as shown in fig(4), We can see that 3Ω
resistor is in series with a parallel combination of 6Ω resistor and 2Ω resistor. i.e.:
3Ω + (6Ω || 2Ω) ….. (|| = in parallel with)
RN = 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]
RN = 3Ω + 1.5Ω
RN = 4.5Ω
Fig(5)
Connect the RN in Parallel with Current Source IN and re-connect the load resistor. This is shown in fig (6)
i.e. Norton Equivalent circuit with load resistor.
Fig(6)
Now apply the Ohm’s Law and calculate the load current through Load resistance across the terminals
A&B. Load Current through Load Resistor is
IL = IN x [RN / (RN+ RL)]
IL= 2A x (4.5Ω /4.5Ω +1.5kΩ)
IL = 1.5A IL = 1. 5A
Superposition Theorem:
The principle of superposition helps us to analyze a linear circuit with more than one current
or voltage sources sometimes it is easier to find out the voltage across or current in a branch of the circuit by
considering the effect of one source at a time by replacing the other sources with their ideal internal
resistances.
Superposition Theorem Statement:
Any linear, bilateral two terminal network consisting of more than one sources, The total
current or voltage in any part of a network is equal to the algebraic sum of the currents or voltages in the
required branch with each source acting individually while other sources are replaced by their ideal internal
resistances. (i.e. Voltage sources by a short circuit and current sources by open circuit)
Steps to Apply Super position Principle:
1. Replace all independent sources with their internal resistances except one source. Find the output
(voltage or current) due to that active source using nodal or mesh analysis.
Fig.(a)
Solution: Applying the superposition theorem, the current I2 in the resistance of 3 Ω due to the voltage
source of 20V alone, with current source of 5A open circuited [ as shown in the figure.1 below ] is given by
:
Fig.1
I2 = 20/(5+3) = 2.5A
Similarly the current I5 in the resistance of 3 Ω due to the current source of 5A alone with voltage source of
20V short circuited [ as shown in the figure.2 below ] is given by :
Fig.2
I5= 5 x 5/(3+5) = 3.125 A
The total current passing through the resistance of 3Ω is then = I2 + I5= 2.5 + 3.125 = 5.625 A
Let us verify the solution using the basic nodal analysis referring to the node marked with V in fig.(a).Then
we get :
𝑉 − 20 𝑉
+ =5
5 3
3V-60+5V=15× 5
8V-60=75
8V=135
V=16.875
The current I passing through the resistance of 3Ω =V/3 = 16.875/3 = 5.625 A.
UNIT-III
SINGLE PHASE A.C. CIRCUITS
• Average value, R.M.S. value, form factor and peak factor for sinusoidal wave form.
• Steady State Analysis of series R-L-C circuits.
• Concept of Reactance, Impedance, Susceptance, Admittance.
• Concept of Power Factor, Real, Reactive and Complex power.
• Illustrative Problems.
RMS VALUE:
• The RMS (Root Mean Square) value (also known as effective or virtual value) of of an alternating
current (AC) is the value of direct current (DC) when flowing through a circuit or resistor for the
specific time period and produces same amount of heat which produced by the alternating current (AC)
when flowing through the same circuit or resistor for a specific time.
• The value of an AC which will produce the same amount of heat while passing through in a heating
element (such as resistor) as DC produces through the element is called R.M.S Value.
• In short,
• The RMS Value of an Alternating Current is that when it compares to the Direct Current, then both AC
and DC current produce the same amount of heat when flowing through the same circuit for a specific
time period.
Hence, the RMS value of the current is (while putting I = Im Sin θ):
Now,
If the maximum value of alternating current is “IMAX“, then the value of converted DC current through
rectifier would be “0.637 IM” which is known as average value of the AC Sine wave (IAV).
Average Value of Current = IAV = 0.637 IM
Average Value of Voltage = EAV = 0.637 EM
The Average Value (also known as Mean Value) of an Alternating Current (AC) is expressed by that Direct
Current (DC) which transfers across any circuit the same amount of charge as is transferred by that
Alternating Current (AC) during the same time.
Keep in mind that the average or mean value of a full sinusoidal wave is “Zero” the value of current in first
half (Positive) is equal to the the next half cycle (Negative) in the opposite direction. In other words, There
are same amount of current in the positive and negative half cycles which flows in the opposite direction, so
the average value for a complete sine wave would be “0”. That’s the reason that’s why we don’t use average
value for plating and battery charging. If an AC wave is converted into DC through a rectifier, It can be used
for electrochemical works.
Thus, the average value of a sinusoidal wave over a complete cycle is zero.
(b) Average value of current over a half cycle
Peak Factor:
Peak Factor is also known as Crest Factor or Amplitude Factor.
It is the ratio between maximum value and RMS value of an alternating wave.
Form Factor:
The ratio between RMS value and Average value of an alternating quantity (Current or Voltage) is known as
Form Factor.
Waveform
• The path traced by a quantity (such as voltage or current) plotted as a function of some variable (such
as time, degree, radians, temperature etc.) is called waveform.
Cycle
1. One complete set of positive and negative values of alternating quality (such as voltage and current) is
known as cycle.
2. The portion of a waveform contained in one period of time is called cycle.
3. A distance between two same points related to value and direction is known as cycle.
4. A cycle is a complete alternation.
Period
• The time taken by a alternating quantity (such as current or voltage) to complete one cycle is called
its time period “T”.
• It is inversely proportional to the Frequency “f” and denoted by “T” where the unit of time period is
second.
• Mathematically;
T = 1/f
Frequency
• Frequency is the number if cycles passed through per second. It is denoted by “f” and has the unit
cycle per second i.e. Hz (Herts).
• The number of completed cycles in 1 second is called frequency.
• It is the number of cycles of alternating quantity per second in hertz.
• Frequency is the number of cycles that a sine wave completed in one second or the number of cycles
that occurs in one second.
f = 1/T
Amplitude
• The maximum value, positive or negative, of an alternating quantity such as voltage or current is
known as its amplitude. Its denoted by VP, IP or EMAX and IMAX.
• Alternation
• One half cycle of a sine wave (Negative or Positive) is known as alternation which span is 180°
degree.
As already said, when the current flowing through a pure resistance changes, no back emf is set up,
therefore, applied voltage has to overcome the ohmic drop of i R only:
The instantaneous power delivered to the circuit in question is the product of the instantaneous values of
applied voltage and current.
Where V and I are the rms values of applied voltage and current respectively.
Thus for purely resistive circuits, the expression for power is the same as for dc circuits. From the power
curve for a purely resistive circuit shown in Fig. 4.1 (b) it is evident that power consumed in a pure resistive
circuit is not constant, it is fluctuating.
However, it is always positive. This is so because the instantaneous values of voltage and current are always
either positive or negative and, therefore, the product is always positive. This means that the voltage source
constantly delivers power to the circuit and the circuit consumes it.
An inductive circuit is a coil with or without an iron core having negligible resistance. Practically pure
inductance can never be had as the inductive coil has always small resistance. However, a coil of thick
copper wire wound on a laminated iron core has negligible resistance arid is known as a choke coil.
When an alternating voltage is applied to a purely inductive coil, an emf, known as self-induced emf, is
induced in the coil which opposes the applied voltage. Since coil has no resistance, at every instant applied
voltage has to overcome this self-induced emf only.
From the expressions of instantaneous applied voltage and instantaneous current flowing through a purely
inductive coil it is observed that the current lags behind the applied voltage by π/2 as shown in Fig. 4.2 (b)
by wave diagram and in Fig 4.2 (c) by phasor diagram.
Inductive Reactance:
ωL in the expression Imax = Vmax/ωL is known as inductive reactance and is denoted by XL i.e., XL = ω L
If L is in henry and co is in radians per second then XL will be in ohms.
When a dc voltage is impressed across the plates of a perfect condenser, it will become charged to full
voltage almost instantaneously. The charging current will flow only during the period of “build up” and will
cease to flow as soon as the capacitor has attained the steady voltage of the source. This implies that for a
direct current, a capacitor is a break in the circuit or an infinitely high resistance.
In Fig. 4.4 a sinusoidal voltage is applied to a capacitor. During the first quarter-cycle, the applied voltage
increases to the peak value, and the capacitor is charged to that value. The current is maximum in the
MRCET EAMCET CODE:MLRD www.mrcet.ac.in 61
DEPARTMENT OF ELECTRICAL AND
ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING
beginning of the cycle and becomes zero at the maximum value of the applied voltage, so there is a phase
difference of 90° between the applied voltage and current. During the first quarter-cycle the current flows in
the normal direction through the circuit; hence the current is positive.
In the second quarter-cycle, the voltage applied across the capacitor falls, the capacitor loses its charge, and
current flows through it against the applied voltage because the capacitor discharges into the circuit. Thus,
the current is negative during the second quarter-cycle and attains a maximum value when the applied
voltage is zero.
The third and fourth quarter-cycles repeat the events of the first and second, respectively, with the difference
that the polarity of the applied voltage is reversed, and there are corresponding current changes.
In other words, an alternating current flow in the circuit because of the charging and discharging of the
capacitor. As illustrated in Figs. 4.4 (b) and (c) the current begins its cycle 90 degrees ahead of the voltage,
so the current in a capacitor leads the applied voltage by 90 degrees – the opposite of the inductance current-
voltage relationship.
Let an alternating voltage represented by v = Vmax sin ω t be applied across a capacitor of capacitance C
farads.
Since the capacitor current is equal to the rate of change of charge, the capacitor current may be
obtained by differentiating the above equation:
From the equations of instantaneous applied voltage and instantaneous current flowing through capacitance,
it is observed that the current leads the applied voltage by π/2, as shown in Figs. 4.4 (b) and (c) by wave and
phasor diagrams respectively.
Capacitive Reactance:
1/ω C in the expression Imax = Vmax/1/ω C is known as capacitive reactance and is denoted by XC i.e.,
XC = 1/ω C
If C is in farads and ω is in radians/s, then Xc will be in ohms.
Hence power absorbed in a purely capacitive circuit is zero. The same is shown graphically in Fig. 4.4 (b).
The energy taken from the supply circuit is stored in the capacitor during the first quarter- cycle and returned
during the next.
The energy stored by a capacitor at maximum voltage across its plates is given by the expression:
This can be realized when it is recalled that no heat is produced and no work is done while current is flowing
through a capacitor. As a matter of fact, in commercial capacitors, there is a slight energy loss in the
dielectric in addition to a minute I2 R loss due to flow of current over the plates having definite ohmic
resistance.
The power curve is a sine wave of double the supply frequency. Although it raises the power factor from
zero to 0.002 or even a little more, but for ordinary purposes the power factor is taken to be zero. Obviously
the phase angle due to dielectric and ohmic losses decreases slightly.
The applied voltage, being equal to phasor sum of VR and VC, is given in magnitude by-
Voltage triangle and impedance triangle Fig. 4.19 are shown in Figs. 4.19 (a) and 4.19 (b) respectively.
The above relations can easily be followed by referring to the power diagram shown in Fig. 4.7 (a).
Reactance is essentially inertia against the motion of electrons. It is present anywhere electric or magnetic
fields are developed in proportion to applied voltage or current, respectively; but most notably in capacitors
and inductors. When alternating current goes through a pure reactance, a voltage drop is produced that is
90o out of phase with the current. Reactance is mathematically symbolized by the letter “X” and is
measured in the unit of ohms (Ω).
Impedance is a comprehensive expression of any and all forms of opposition to electron flow, including
both resistance and reactance. It is present in all circuits, and in all components. When alternating current
goes through an impedance, a voltage drop is produced that is somewhere between 0 o and 90o out of
phase with the current. Impedance is mathematically symbolized by the letter “Z” and is measured in the
unit of ohms (Ω), in complex form
Admittance is also a complex number as impedance which is having a real part, Conductance
(G) and imaginary part, Susceptance (B).
(it is negative for capacitive susceptance and positive for inductive susceptance)
Susceptance (symbolized B) is an expression of the ease with which alternating current (AC) passes
through a capacitance or inductance
UNIT-IV
ELECTRICAL MACHINES
Dc Generator
• Principle of Operation
• Constructional Features
• EMF Equation
Dc Motor
• Principle of Operation
• Back EMF
• Torque Equation
Single Phase Transformer
• Principle of Operation
• Constructional Features
• EMF Equation
• Simple Problems
DC GENERATOR
Principle of DC Generator
There are two types of generators, one is ac generator and other is DC generator. Whatever
may be the types of generators, it always converts mechanical power to electrical power. An AC generator
produces alternating power. A DC generator produces direct power. Both of these generators produce
electrical power, based on same fundamental principle of Faraday's law of electromagnetic induction.
According to this law, when a conductor moves in a magnetic field it cuts magnetic lines of force, due to
which an emf is induced in the conductor. The magnitude of this induced emf depends upon the rate of
change of flux (magnetic line force) linkage with the conductor. This emf will cause a current to flow if the
conductor circuit is closed.
Hence the most basic tow essential parts of a generator are
1. a magnetic field
2. conductors which move inside that magnetic field.
Now we will go through working principle of DC generator. As, the working principle of ac generator is not
in scope of our discussion in this section.
In the figure above, a single loop of conductor of rectangular shape is placed between two
opposite poles of magnet.
Let's us consider, the rectangular loop of conductor is ABCD which rotates inside the
magnetic field about its own axis ab. When the loop rotates from its vertical position to its horizontal
position, it cuts the flux lines of the field. As during this movement two sides, i.e. AB and CD of the loop
cut the flux lines there will be an emf induced in these both of the sides (AB and BC) of the loop.
As the loop is closed there will be a current circulating through the loop. The direction of the current can be
determined by Fleming’s right hand Rule. This rule says that if you stretch thumb, index finger and middle
finger of your right hand perpendicular to each other, then thumbs indicates the direction of motion of the
conductor, index finger indicates the direction of magnetic field i.e. N - pole to S - pole, and middle finger
indicates the direction of flow of current through the conductor. Now if we apply this right-hand rule, we
will see at this horizontal position of the loop, current will flow from point A to B and on the other side of
the loop current will flow from point C to D.
Now if we allow the loop to move further, it will come again to its vertical position, but now upper side of
the loop will be CD and lower side will be AB (just opposite of the previous vertical position). At this
position the tangential motion of the sides of the loop is parallel to the flux lines of the field. Hence there
will be no question of flux cutting and consequently there will be no current in the loop.
If the loop rotates further, it comes to again in horizontal position. But now, said AB side of the loop comes
in front of N pole and CD comes in front of S pole, i.e. just opposite to the previous horizontal position as
shown in the figure beside.
Here the tangential motion of the side of the loop is perpendicular to the flux lines, hence rate of flux cutting
is maximum here and according to Fleming’s right hand rule, at this position current flows from B to A and
on other side from D to C.Now if the loop is continued to rotate about its axis, every time the side AB comes
in front of S pole, the current flows from A to B and when it comes in front of N pole, the current flows
from B to A. Similarly, every time the side CD comes in front of S pole the current flows from C to D and
when it comes in front of N pole the current flows from D to C.
If we observe this phenomena in different way, it can be concluded, that each side of the loop comes in front
of N pole, the current will flow through that side in same direction i.e. downward to the reference plane and
similarly each side of the loop comes in front of S pole, current through it flows in same direction i.e.
upwards from reference plane. From this, we will come to the topic of principle of DC generator.
Now the loop is opened and connected it with a split ring as shown in the figure below. Split ring are made
out of a conducting cylinder which cuts into two halves or segments insulated from each other. The external
load terminals are connected with two carbon brushes which are rest on these split slip ring segments.
Working Principle of DC Generator
It is seen that in the first half of the revolution current flows always 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. 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 wave from of the current through the load circuit is as shown in the figure. This current is
unidirectional.
This is basic working principle of DC generator, explained by single loop generator model. The position of
the brushes of DC generator is so arranged 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 right angle to the plane of the lines of force. It is so
become in that position, the induced emf in the coil is zero.
Construction of DC Generator
During explaining working principle of DC Generator, we have used a single loop DC
generator. But now we will discuss about practical construction of DC Generator.
A DC generator has the following parts
1. Yoke
2. Pole of generator
3. Field winding
4. Armature of DC generator
5. Brushes of generator and Commentator
6. Bearing
Yoke of DC Generator
Yoke or the outer frame of DC generator serves two purposes,
1. It holds the magnetic pole cores of the generator and acts as cover of the generator.
2. It carries the magnetic field flux.
In small generator, yoke are made of cast iron. Cast iron is cheaper in cost but heavier than steel. But for
large construction of DC generator, where weight of the machine is concerned, lighter cast steel or rolled
steel is preferable for constructing yoke of DC generator. Normally larger yokes are formed by rounding a
rectangular steel slab and the edges are welded together at the bottom. Then feet, terminal box and hangers
are welded to the outer periphery of the yoke frame.
Pole Cores and Pole Shoes
Let's first discuss about pole core of DC generator. There are mainly two types of construction available.
One: Solid pole core, where it is made of a solid single piece of cast iron or cast steel.
Two: Laminated pole core, where it made of numbers of thin, limitations of annealed steel which are riveted
together. The thickness of the lamination is in the range of 0.04" to 0.01". The pole core is fixed to the inner
periphery of the yoke by means of bolts through the yoke and into the pole body. Since the poles project
inwards they are called salient poles. The pole shoes are so typically shaped, that, they spread out the
magnetic flux in the air gap and reduce the reluctance of the magnetic path. Due to their larger cross-section
they hold the pole coil at its position.
Pole Coils: The field coils or pole coils are wound around the pole core. These are a simple coil of insulated
copper wire or strip, which placed on the pole which placed between yoke and pole shoes as shown.
Armature Core
The purpose of armature core is to hold the armature winding and provide low reluctance
path for the flux through the armature from N pole to S pole. Although a DC generator provides direct
current but induced current in the armature is alternating in nature. That is why, cylindrical or drum shaped
armature core is build up of circular laminated sheet. In every circular lamination, slots are either die - cut or
punched on the outer periphery and the key way is located on the inner periphery as shown. Air ducts are
also punched of cut on each lamination for circulation of air through the core for providing better cooling.
Up to diameter of 40", the circular stampings are cut out in one piece of lamination sheet. But above 40",
diameter, number of suitable sections of a circle is cut. A complete circle of lamination is formed by four or
six or even eight such segment.
Armature Winding
Armature winding are generally formed wound. These are first wound in the form of flat
rectangular coils and are then pulled into their proper shape in a coil puller. Various conductors of the coils
are insulated from each other. The conductors are placed in the armature slots, which are lined with tough
insulating material. This slot insulation is folded over above the armature conductors placed in it and
secured in place by special hard wooden or fiber wedges. Two types of armature windings are used – Lap
winding and Wave winding.
Commutator
The commentator plays a vital role in DC generator. It collects current from armature and
sends it to the load as direct current. It actually takes alternating current from armature and converts it to
direct current and then send it to external load. It is cylindrical structured and is build up of wedge-shaped
segments of high conductivity, hard drawn or drop forged copper. Each segment is insulated from the shaft
by means of insulated commutator segment shown below. Each commentator segment is connected with
corresponding armature conductor through segment riser or lug.
Brushes
The brushes are made of carbon. These are rectangular block shaped. The only function of
these carbon brushes of DC generator is to collect current from commutator segments. The brushes are
housed in the rectangular box shaped brush holder or brush box. As shown in figure, the brush face is placed
on the commutator segment which is attached to the brush holder.
Bearing
For small machine, ball bearing is used and for heavy duty DC generator, roller bearing is used. The bearing
must always be lubricated properly for smooth operation and long life of generator.
Armature winding
Basically armature winding of a DC machine is wound by one of the two methods, lap
winding or wave winding. The difference between these two is merely due to the end connections and
commutator connections of the conductor. To know how armature winding is done, it is essential to know
the following terminologies -
1. Pole pitch: It is defined as number of armature slots per pole. For example, if there are 36 conductors
and 4 poles, then the pole pitch is 36/4=9.
2. Coil span or coil pitch (Ys): It is the distance between the two sides of a coil measured in terms of
armature slots.
3. Front pitch (Yf): It is the distance, in terms of armature conductors, between the second conductor
of one coil and the first conductor of the next coil. OR it is the distance between two coil sides that
are connected to the same commutator segment.
4. Back pitch (Yb): The distance by which a coil advances on the back of the armature is called as back
pitch of the coil. It is measured in terms of armature conductors.
5. Resultant pitch (Yr): The distance, in terms of armature conductor, between the beginning of one coil
and the beginning of the next coil is called as resultant pitch of the coil.
Armature winding can be done as single layer or double layer. It may be simplex, duplex or multiplex, and
this multiplicity increases the number of parallel paths.
In wave winding, a conductor under one pole is connected at the back to a conductor which occupies an
almost corresponding position under the next pole which is of opposite polarity. In other words, all the coils
which carry e.m.f in the same direction are connected in series. The following diagram shows a part of
simplex wave winding.
Now,
▪ Average emf generated per conductor is given by dΦ/dt (Volts) ... eq. 1
▪ Flux cut by one conductor in one revolution = dΦ = PΦ ….(Weber),
▪ Number of revolutions per second (speed in RPS) = N/60
▪ Therefore, time for one revolution = dt = 60/N (Seconds)
▪ From eq. 1, emf generated per conductor = dΦ/dt = PΦN/60 (Volts) …..(eq. 2)
Above equation-2 gives the emf generated in one conductor of the generator. The conductors are connected
in series per parallel path, and the emf across the generator terminals is equal to the generated emf across
any parallel path.
Therefore, Eg = PΦNZ / 60A
For simplex lap winding, number of parallel paths is equal to the number of poles (i.e. A=P),
Therefore, for simplex lap wound dc generator, Eg = PΦNZ / 60P
For simplex wave winding, number of parallel paths is equal to 2 (i.e P=2),
Therefore, for simplex wave wound dc generator, Eg = PΦNZ / 120
DC MOTOR
Working or Operating Principle of DC Motor
A DC motor in simple words is a device that converts electrical energy (direct current
system) into mechanical energy. It is of vital importance for the industry today, and is equally important for
engineers to look into the working principle of DC motor in details that has been discussed in this article. In
order to understand the operating principle of DC motor we need to first look into its constructional feature.
The very basic construction of a DC motor contains a current carrying armature which is connected to the
supply end through commutator segments and brushes. The armature is placed in between north south poles
of a permanent or an electromagnet as shown in the diagram above.
As soon as we supply direct current in the armature, a mechanical force acts on it due to
electromagnetic effect of the magnet. Now to go into the details of the operating principle of DC motor its
important that we have a clear understanding of Fleming’s left hand rule to determine the direction of force
acting on the armature conductors of DC motor.
If a current carrying conductor is placed in a magnetic field perpendicularly, then the conductor
experiences a force in the direction mutually perpendicular to both the direction of field and the current
carrying conductor. Fleming’s left hand rule says that if we extend the index finger, middle finger and
thumb of our left hand perpendicular to each other, in such a way that the middle finger is along the
direction of current in the conductor, and index finger is along the direction of magnetic field i.e. north to
south pole, then thumb indicates the direction of created mechanical force. For clear understanding the
principle of DC motor we have to determine the magnitude of the force, by considering the diagram below.
We know that when an infinitely small charge dq is made to flow at a velocity ‘v’ under the influence of an
electric field E, and a magnetic field B, then the Lorentz Force dF experienced by the charge is given by:-
From the 1st diagram we can see that the construction of a DC motor is such that the
direction of current through the armature conductor at all instance is perpendicular to the field. Hence the
force acts on the armature conductor in the direction perpendicular to the both uniform field and current is
constant.
So if we take the current in the left hand side of the armature conductor to be I, and current at right hand
side of the armature conductor to be -I, because they are flowing in the opposite direction with respect to
each other.
Therefore, we can see that at that position the force on either side is equal in magnitude but opposite in
direction. And since the two conductors are separated by some distance w = width of the armature turn, the
two opposite forces produces a rotational force or a torque that results in the rotation of the armature
conductor.
Now let's examine the expression of torque when the armature turn crate an angle of α (alpha) with its initial
position.
The torque produced is given by,
Where, α (alpha) is the angle between the plane of the armature turn and the plane of reference or the initial
position of the armature which is here along the direction of magnetic field.
The presence of the term cosα in the torque equation very well signifies that unlike force the torque at all
position is not the same. It in fact varies with the variation of the angle α (alpha). To explain the variation of
torque and the principle behind rotation of the motor let us do a step wise analysis.
Step 1:
Initially considering the armature is in its starting point or reference position where the angle α = 0.
Since, α = 0, the term cos α = 1, or the maximum value, hence torque at this position is maximum given by
τ = BILw. This high starting torque helps in overcoming the initial inertia of rest of the armature and sets it
into rotation.
Step 2:
Once the armature is set in motion, the angle α between the actual position of the armature and its reference
initial position goes on increasing in the path of its rotation until it becomes 90o from its initial position.
Consequently the term cosα decreases and also the value of torque.
The torque in this case is given by τ = BILwcosα which is less than BIL w when α is greater than 0o.
Step 3:
In the path of the rotation of the armature a point is reached where the actual position of the rotor is exactly
perpendicular to its initial position, i.e. α = 90o, and as a result the term cosα = 0.
The torque acting on the conductor at this position is given by,
i.e. virtually no rotating torque acts on the armature at this instance. But still the armature does not come to a
standstill, this is because of the fact that the operation of DC motor has been engineered in such a way that
the inertia of motion at this point is just enough to overcome this point of null torque. Once the rotor crosses
over this position the angle between the actual position of the armature and the initial plane again decreases
and torque starts acting on it again.
Torque Equation of DC Motor
When a DC machine is loaded either as a motor or as a generator, the rotor conductors carry
current. These conductors lie in the magnetic field of the air gap. Thus each conductor experiences a force.
The conductors lie near the surface of the rotor at a common radius from its center. Hence torque is
produced at the circumference of the rotor and rotor starts rotating. The term torque as best explained by Dr.
Huge d Young is the quantitative measure of the tendency of a force to cause a rotational motion, or to bring
about a change in rotational motion. It is in fact the moment of a force that produces or changes a rotational
motion.
The equation of torque is given by,
The DC motor as we all know is a rotational machine, and torque of DC motor is a very important
parameter in this concern, and it’s of utmost importance to understand the torque equation of DC motor for
establishing its running characteristics.
To establish the torque equation, let us first consider the basic circuit diagram of a DC motor, and its
voltage equation.
Referring to the diagram beside, we can see, that if E is the supply voltage, Eb is the back emf produced and
Ia, Ra are the armature current and armature resistance respectively then the voltage equation is given by,
But keeping in mind that our purpose is to derive the torque equation of DC motor we multiply both sides of
equation (2) by Ia.
Now Ia2.Ra is the power loss due to heating of the armature coil, and the true effective mechanical power that
is required to produce the desired torque of DC machine is given by,
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Where, P is no of poles,
φ is flux per pole,
Z is no. of conductors,
A is no. of parallel paths,
and N is the speed of the DC motor.
The torque we so obtain, is known as the electromagnetic torque of DC motor, and subtracting the
mechanical and rotational losses from it we get the mechanical torque.
Therefore,
Which is constant for a particular machine and therefore the torque of DC motor varies with only flux φ and
armature current Ia.
The Torque equation of a DC motor can also be explained considering the figure below.
Which is constant for a particular machine and therefore the torque of DC motor varies with only flux φ and
armature current Ia.
TRANSFORMER
Introduction
The transformer is a device that transfers electrical energy from one electrical circuit to
another electrical circuit. The two circuits may be operating at different voltage levels but always work at
the same frequency. Basically transformer is an electro-magnetic energy conversion device. It is commonly
used in electrical power system and distribution systems. It can change the magnitude of alternating voltage
or current from one value to another. This useful property of transformer is mainly responsible for the
widespread use of alternating currents rather than direct currents i.e., electric power is generated,
transmitted and distributed in the form of alternating current. Transformers have no moving parts, rugged
and durable in construction, thus requiring very little attention. They also have a very high efficiency as
high as 99%.
A transformer is a static device of equipment used either for raising or lowering the voltage
of an a.c. supply with a corresponding decrease or increase in current. It essentially consists of two
windings, the primary and secondary, wound on a common laminated magnetic core as shown in Fig 1. The
winding connected to the a.c. source is called primary winding (or primary) and the one connected to load is
called secondary winding (or secondary). The alternating voltage V1 whose magnitude is to be changed is
applied to the primary.
Depending upon the number of turns of the primary (N1) and secondary (N2), an alternating e.m.f. E2 is
induced in the secondary. This induced e.m.f. E2 in the secondary causes a secondary current I2.
Consequently, terminal voltage V2 will appear across the load.
Constructional Details
Depending upon the manner in which the primary and secondary windings are placed on the
core, and the shape of the core, there are two types of transformers, called
Core
The core is built-up of thin steel laminations insulated from each other. This helps in reducing
the eddy current losses in the core, and also helps in construction of the transformer. The steel used for core
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is of high silicon content, sometimes heat treated to produce a high permeability and low hysteresis loss. The
material commonly used for core is CRGO (Cold Rolled Grain Oriented) steel. Conductor material used for
windings is mostly copper. However, for small distribution transformer aluminum is also sometimes used.
The conductors, core and whole windings are insulated using various insulating materials depending upon
the voltage.
Insulating Oil
In oil-immersed transformer, the iron core together with windings is immersed in insulating oil. The
insulating oil provides better insulation, protects insulation from moisture and transfers the heat produced in
core and windings to the atmosphere.
The transformer oil should possess the following qualities:
(a) High dielectric strength,
(b)Low viscosity and high purity,
(c)High flash point, and
(d)Free from sludge.
Transformer oil is generally a mineral oil obtained by fractional distillation of crude oil.
Tank and Conservator
The transformer tank contains core wound with windings and the insulating oil. In large transformers small
expansion tank is also connected with main tank is known as conservator. Conservator provides space when
insulating oil expands due to heating. The transformer tank is provided with tubes on the outside, to permits
circulation of oil, which aides in cooling. Some additional devices like breather and Buchholz relay are
connected with main tank. Buchholz relay is placed between main tank and conservator. It protect the
transformer under extreme heating of transformer winding. Breather protects the insulating oil from
moisture when the cool transformer sucks air inside. The silica gel filled breather absorbs moisture when air
enters the tank. Some other necessary parts are connected with main tank like, Bushings, Cable Boxes,
Temperature gauge, Oil gauge, Tapings, etc.
Principle of Operation
When an alternating voltage V1 is applied to the primary, an alternating flux ϕ is set up in the core. This
alternating flux links both the windings and induces e.m.f.s E1 and E2 in them according to Faraday’s laws
of electromagnetic induction. The e.m.f. E1 is termed as primary e.m.f. and E2 is termed as secondary e.m.f.
Note that magnitudes of E2 and E1 depend upon the number of turns on the secondary and primary
respectively.
If N2 > N1, then E2 > E1 (or V2 > V1) and we get a step-up transformer.
If N2 < N1, then E2 < E1(or V2< V1) and we get a step-down transformer.
If load is connected across the secondary winding, the secondary e.m.f. E2 will cause a current I2 to flow
through the load. Thus, a transformer enables us to transfer a.c. power from one circuit to another with a
change in voltage level.
The following points may be noted carefully
(a)The transformer action is based on the laws of electromagnetic induction.
(b)There is no electrical connection between the primary and secondary.
(c) The a.c. power is transferred from primary to secondary through magnetic flux.
(d) There is no change in frequency i.e., output power has the same frequency as the input power.
(e)The losses that occur in a transformer are:
(a) core losses—eddy current and hysteresis losses
(b) copper losses—in the resistance of the windings
In practice, these losses are very small so that output power is nearly equal to the input primary power. In
other words, a transformer has very high efficiency
E.M.F. Equation of a Transformer
Consider that an alternating voltage V1 of frequency f is applied to the primary as shown in Fig.2.3. The
sinusoidal flux ϕ produced by the primary can be represented as:
ϕ=ϕm sinωt
When the primary winding is excited by an alternating voltage V1, it is circulating alternating current,
producing an alternating flux ϕ.
ϕ - Flux
ϕm - maximum value of flux ,
N1 - Number of primary turns ,
N2 - Number of secondary 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
The instantaneous e.m.f. e1 induced in the primary is –
The flux increases from zero value to maximum value ϕm in 1/4f of the time period that is in 1/4f seconds.
The change of flux that takes place in 1/4f seconds = ϕm - 0 = ϕm webers
Voltage Ratio
Voltage transformation ratio is the ratio of e.m.f induced in the secondary winding to the
e.m.f induced in the primary winding.
This ratio of secondary induced e.m.f to primary induced e.m.f is known as voltage transformation ratio
1. If N2>N1 i.e. K>1 we get E2>E1 then the transformer is called step up transformer.
2. If N2< N1 i.e. K<1 we get E2< E2 then the transformer is called step down transformer.
3. If N2= N1 i.e. K=1 we get E2= E2 then the transformer is called isolation transformer or 1:1 Transformer.
Current Ratio
Current ratio is the ratio of current flow through the primary winding (I1) to the current
flowing through the secondary winding (I2). In an ideal transformer -
Apparent input power = Apparent output power.
V1I1 = V2I2
Volt-Ampere Rating
i) The transformer rating is specified as the products of voltage and current (VA rating).
ii) On both sides, primary and secondary VA rating remains same. This rating is generally expressed in KVA
(Kilo Volts Amperes rating)
UNIT –V
ELECTRICAL INSTALLATIONS
in this case, the cable consists of 19 strands, each strand has a diameter of 1.12mm. The
conductor is made with aluminium, insulation is made with PVC, is covered with PVC
sheathing, and is used for 1100Vsupply system.
Earthing of Grounding:
The process of connecting the metallic frame (i.e., non- current carrying part) of electrical equipment or
some electrical part of the system (e.g., neutral point in a star-connected system, one conductor of the
secondary of a transformer, etc.) to the earth (i.e., soil) is called grounding or Earthing. The potential of the
earth is to be considered zero for all practical purposes. Earthing is to connect any electrical equipment to
earth with a very low resistance wire, making it to attain earth’s potential, This ensures safe discharge of
electrical energy due to failure of the insulation line coming in contact with the casing, etc. Earthing brings
the potential of the body of the equipment to zero i.e., to the earth’s potential, thus ptotecting the operating
personnel against electrical shock.
The earth resistance is affected by the following factors :
(a) Material properties of the earth, wire and the electrode
(b) Temperature and moisture content of the soil
(c) Depth of the pit
(d) Quantity of the charcoal used
Necessity of Earthing:
The requirement for provision of earthing can be listed as follows :
(1) To protect the operating personnel from the danger of shock.
(2) To maintain the line voltage constant, under unbalanced load condition.
(3) To avoid risk of fire due to earth leakage current through unwanted path.
(4) Protection of the equipments.
(5) Protection of large buildings and all machines fed from overhead lines
against lighting.
Methods of Earthing:
The various methods of earthing in common use are
(i) Plate earthing
(ii) Pipe earthing
(iii) Rod earthing
(iv) Strip or wire earthing
(i) Plate earthing:
In this method either a copper plate of 60cm × 60cm ×3.18 or GI plate of 60cm × 60cm × 6.35 is used for
earthing. The plate is buried into the ground not less than 3m from the ground level. The earth plate is
embedded in alternate layers of coal and salt for a thickness of 15cm as shown in figure (12.4). In addition,
water is poured for keeping the earth’s electrode resistance value below a maximum of 5Ω. The earth wire is
securely bolted to the earth plate.
A cement masonry chamber is built with a cast iron cover for easy regular maintenance
When compared to the plate earth system the pipe earth system can carry larger leakage currents due to
larger surface area is in contact with the soil for given electrode size. This system also enables easy
maintenance as the earth wire connection is housed at the ground levels.
(iii) Rod earthing:
It is the same method as pipe earthing, A copper rod of 12.5cm (1/2 inch) diameter or 16mm (0.6in)
diameter of galvanized steel or hollow section 25mm (1 inch) of GI pipe of length above 2.5m (8.2 ft) are
buried upright in the earth manually or with the help of a pneumatic hammer. The length of embedded
electrodes in the soil reduces earth resistance to a desired value.
Earth Resistance:
The earth resistance should be kept as low as possible so that the neutral of any electrical system, which is
earthed, is maintained almost at the earth potential. The earth resistance for copper wire is 1Ω and that of
GI wire less than 3Ω. The typical value of the earth resistance at large power stations is 0.5Ω , major sub-
stations is 1Ω, small sub-stations is 2 Ω and in all other cases 5 Ω.
The resistance of the earth depends on the following factors
Condition of soil.
ii. Moisture content of soil.
iii. Temperature of soil.
iv. Depth of electrode at which it is embedded.
v. Size, material and spacing of earth electrode.
vi. Quality and quantity of coal and salt in the earth pit.
Difference Between Earth Wire and Neutral Wire:
Neutral Wire :
(i) In a 3-phase 4-wire system, the fourth wire is a neutral wire.
(ii) IT acts a return path for 3-phase currents when the load is not balanced.
(iii) IN domestic single phase AC circuit, the neutral wire acts as return path for the line
current.
Earth Wire :
(i) Earth wire is actually connected to the general mass of the earth and metallic body of the
equipment
(ii) It is provided to transfer any leakage current from the metallic body to the earth.
Fuse:
The electrical equipment are designed to carry a particular rated value of current under normal conditions.
Under abnormal conditions such as short circuits, overload, or any fault; the current rises above this value,
damaging the equipment and sometimes resulting in fire hazard. Fuses come into operation under fault
conditions.
A fuse is short piece of metal, inserted in the circuit, which melts when excessive current flows through
it and thus breaks the circuits. Under normal operating conditions it designed to carry the full load current.
If the current increases beyond this designed value due to any of the reasons mentioned above, the fuse
melts, isolating the power supply from the load.
(a) Desirable characteristics of a Fuse Element:
The material used foe fuse wires must have the following characteristics:
i. Low melting point e.g., tin, lead.
ii. High conductivity e.g., copper.
iii. Free from deterioration due oxidation e.g., silver.
iv. Low cost e.g., tin, copper.
(b) Materials:
Material used are tin lead or silver having low melting points. Use of copper or iron is dangerous, though
Circuit Breaker:
Electrical circuits breaker is a switching device which can be operated manually and automatically for the
controlling and protection of electrical power system, respectively. The modern power system deals with a
huge power network and huge numbers of associated electrical equipment. During shirt circuits fault or any
other type of electrical fault, this equipment, as well as the power network, suffer a high stress of fault
current, which in turn damage the equipment and networks permanently. For saving this equipment and the
power networks, the fault current should be cleared from the system as quickly as possible. Again, after the
cleared, the system must come to its normal working condition as soon is possible for supplying reliable
quality power to the receiving ends. The circuits breaker is the special device all the required switching
operations during current carrying condition.
A circuits breaker essentially consists of fixed and moving contacts, called electrodes. Under normal
operating conditions, these contacts remain closed and will not open automatically until and unless the
system becomes faulty. The contacts can be opened manually or by remote control whenever desired. When
a fault occurs in any part of the system, the trip coils of the breaker get energized and the moving contacts
are pulled apart by some mechanism, thus opening the circuits.
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if circuits is overload for a long time, the bi -metallic strip becomes over heated and deformed. This
deformation of bi-metallic strip causes displacement of latch point. The moving contact of the MCB is so
arranged by means of spring, with this latch point, that a little displacement of latch causes releases of
spring and makes the moving contact to move for opening the MCB. The current coil or trip coil placed in
such a manner that during SC faults, the MMF of that coil causes its plunger to hit the same latch point and
force the latch to be displaced. Hence, the MCB will open in the same manner. Again when operating lever
of the MCB is operated by hand, that means when we make the MCB at off position manually, the same
latch point is displaced as a result moving contact separated from fixed contact in same manner. So,
whatever may be the operating mechanism, i.e., may be due to deformation of bi-metallic strip or may be
due to increased MMF of trip coil or may be due to manual operation - actually the same latch point is
displaced and the deformed spring is released, which is ultimately responsible for movement of the moving
contact. When the moving contacts is separated from fixed contact, there may be a high chance of arc. This
are then goes up thorough the arc runner and enters into arc splitters and is finally quenched. When we
switch on the MCB, we actually reset the displaced operating latch to its previous on position and make the
MCB ready for another switch off or trip operation.
These are available in single pole, double pole, triple pole, and four pole versions with neutral poles, if
required. The normal current ratings are available from 0.5-63 A with a symmetrical short circuit rupturing
capacity of 3-10kA, at a voltage level of 230/440v. MCBs are generally designed to trip within 2.5
millisecond when an over current fault arises. In case of temperature rise or over heating it may take 2
seconds to 2 min. For the MCB to trip.
Advantages:
i. MCBs are replacing the re-wireable switch i.e., fuse units for low power domestic and industrial
applications.
ii. The disadvantages of fuses, like low SC interrupting capacity (say 3kA), Etc. Are overcome with high
SC breaking capacity of 10kA.
iii. MCB is combination of all three functions in a wiring system like switching, overload and short circuits
protection. Overload protection can be obtained by using bi-metallic strips whereas shorts circuits
protection can be obtained by using solenoid
Earth Leakage Circuits Breaker (ELCB):
None of the protection devices like MCB, MCCB, etc. Can protect the human life against electric shocks or
avoid fire due to leakage current. The human resistance noticeably drops with an increase in voltage. It also
depends upon the duration of impressed voltage and drops with increase in time. As per IS code, a contact
potential of 65V is within tolerable limit of human body for 10 seconds, whereas 250V can be withstood by
human body for 100 milliseconds. The actual effect of current thorough human body varies from person to
person with reference to magnitude and duration. The body resistance at 10V is assessed to be 19 kΩ for 1
second and 8kΩ for 15 min. At 240V, 3 to 3.6 kΩ for dry skin and 1-1.2 kΩ for wet skin.
An Earth Leakage Circuits Breakers (ELCB) is a device used to directly detect currents leaking to earth
from an installation and cut the power. There are two types of ELCBs:
neutral wire, since the current which flows from the phase will be returned back to the neutral. When a fault
occurs, a small current will flow to the ground also. This makes an unbalanced between line and neutral
currents and creates an unbalanced magnetic field. This induces a current through the secondary winding,
which is connected to the sensing circuits. This will sense the leakage and send a signal to the tripping
system and trips the contact.
rings or metal watchbands when working with electrical equipment as they cause a strong electric shock.|
kilowatts, megawatts, gigawatts, and time which is measured in an hour. Joule is the smallest unit of energy.
But for some bigger calculation, some better unit it required. So, the unit used for electrical energy is watt-
hour.
Electrical energy is the product of electrical power and time, and it measured in joules. It is defined as “1
joule of energy is equal to 1 watt of power is consumed for 1 second”. I.e.,
Energy = Power × Time
1 Joule = 1 watt × 1 second
Watts are the basic unit of power in which electrical power is measured or we can say that rate at which
electrical current is being used at a particular moment.
Watt-hour is the standard unit used for measurement of energy, describing the amount of watts used over a
time. It shows how fast the power is consumed in the period of time.
Energy in watt hours = Power in watts × Time in hours
Kilowatt-hour is simply a bigger unit of energy when large appliance drawn power in kilowatts. It can be
described as one kilowatt hour is the amount of energy drawn by the 1000 watts appliance when used for an
hour.
Where, One kilowatt = 1000 watts
Energy in kilowatt hours = Power in kilowatts × Time in hours
The electrical supply companies take electric energy charges from their consumer per kilowatt hour unit
basis.
This kilowatt hour is board of trade (BOT) unit.
Illustration for Energy Consumption:
A consumer uses a 10 kW geezer, a 6 kW electric furnace and five 100 W bulbs for 15 hours. How many
units (kWh) of electrical energy have been used?
Explanation : Given that
Load – 1 = 10 kW geezer
Load – 2 = 6 kW electric furnace
Load – 3 = 500 watt (five 100 watt bulbs)
Total load = 10kW + 6kW + 0.5kW = 16.5kW
Time taken = 155 hours
Energy consumed = Power in kW × Time in hours
= 16.5 × 15 = 247.5 kWh
For above electrical energy consumption, the tariff can be calculated as follows :
1 unit = 1kWh