Basic Electrical Engineering R-20 PDF

DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING BASIC ELECTRICAL ENGINEERING DIGITAL NOTES MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Instituti...

DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING BASIC ELECTRICAL ENGINEERING DIGITAL NOTES MALLA REDDY COLLEGE OF ENGINEERING & TECHNOLOGY (Autonomous Institution – UGC, Govt. of India) Recognized under 2(f) and 12 (B) of UGC ACT 1956 (Affiliated to JNTUH, Hyderabad, Approved by AICTE-Accredited by NBA & NACC-‘A’ Grade – ISO 9001:2015 Certified) Maisammaguda, Dhulapally (Post Via. Hakimpet), Secunderabad -500100, Telangana State, India. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 1  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING MALLA REDDY COLLEGE OF ENGINEERING AND TECHNOLOGY I Year B. Tech I Sem – CSE/AIML/IOT/CS/DS/IT L T P C 3 0 0 3 (R20A0201) BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 2  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 3  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 4 DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING INDEX SNO. TOPIC PAGE NO. UNIT –I INTRODUCTION TO ELECTRICAL CIRCUITS Concept of Circuit and Network 7-8 Types of elements 9-12 R-L-C Parameters 13-16 Independent and Dependent sources 17-19 Source transformation Technique 20-21 Kirchhoff’s Laws 21-26 UNIT –II NETWORK ANALYSIS Network Reduction Techniques 27-28 Series and Parallel connection of Resistive Networks 28-32 Star–to-Delta and Delta-to-Star Transformations for 32-38 Resistive Networks Mesh Analysis 39-41 Network Theorems: Thevenin’s Theorem 42-45 Norton’s Theorem 45-48 Superposition Theorem 48-50 UNIT-III SINGLE PHASE A.C. 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 Concept of Power Factor, Real, Reactive and Complex 65-66 power Concept of Reactance, Impedance, Susceptance, 66-67 Admittance. UNIT –IV ELECTRICAL MACHINES Dc Generator: Principle of operation 68-72 MRCET EAMCET CODE: MLRD www.mrcet.ac.in 5 DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING Constructional features 72-75 EMF equation 76-77 DC Motor: Principle of operation 77-82 Torque equation 82-83 Back emf 83-84 Single phase transformer: Constructional features 85-86 Principle of operation 86-88 EMF equation 89-91 UNIT –V ELECTRICAL INSTALLATIONS Types of Wires and Cables 92-95 Earthing 95-98 Components of LT Switchgear: Switch Fuse Unit (SFU) 99-101 MCB, ELCB 102-107 Elementary calculations for energy consumption and 107-110 battery backup MRCET EAMCET CODE: MLRD www.mrcet.ac.in 6  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING UNIT-I INTRODUCTION TO ELECTRICAL CIRCUITS  Concept of Circuit and Network  Types of elements  R-L-C Parameters  Independent and Dependent sources  Source transformation Technique  Kirchhoff’s Laws  Simple Problems MRCET EAMCET CODE: MLRD www.mrcet.ac.in 7  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING INTRODUCTION TO ELECTRICAL CIRCUITS Network theory is the study of solving the problems of electric circuits or electric networks. In this introductory chapter, let us first discuss the basic terminology of electric circuits and the types of network elements. Basic Terminology In Network Theory, we will frequently come across the following terms −  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 MRCET EAMCET CODE: MLRD www.mrcet.ac.in 9  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING Active Elements and Passive Elements We can classify the Network elements into either active or passive based on the ability of delivering power.  Active Elements deliver power to other elements, which are present in an electric circuit. Sometimes, they may absorb the power like passive elements. That means active elements have the capability of both delivering and absorbing power. Examples: Voltage sources and current sources.  Passive Elements can’t deliver power (energy) to other elements, however they can absorb power. That means these elements either dissipate power in the form of heat or store energy in the form of either magnetic field or electric field. Examples: Resistors, Inductors, and capacitors. Linear Elements and Non-Linear Elements We can classify the network elements as linear or non-linear based on their characteristic to obey the property of linearity.  Linear Elements are the elements that show a linear relationship between voltage and current. Examples: Resistors, Inductors, and capacitors.  Non-Linear Elements are those that do not show a linear relation between voltage and current. Examples: Voltage sources and current sources. Bilateral Elements and Unilateral Elements Network elements can also be classified as either bilateral or unilateral based on the direction of current flows through the network elements. Bilateral Elements are the elements that allow the current in both directions and offer the same impedance in either direction of current flow. Examples: Resistors, Inductors and capacitors. The concept of Bilateral elements is illustrated in the following figures. 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). MRCET EAMCET CODE: MLRD www.mrcet.ac.in 10  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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. Step 1 − Verifying the network element as linear or non-linear. From the above figure, the V-I characteristics of a network element is a straight line passing through the origin. Hence, it is linear element. Step 2 − Verifying the network element as active or passive. The given V-I characteristics of a network element lies in the first and third quadrants.  In the first quadrant, the values of both voltage (V) and current (I) are positive. So, the ratios of voltage (V) and current (I) gives positive impedance values. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 11  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING  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. Step 1 − Verifying the network element as linear or non-linear. From the above figure, the V-I characteristics of a network element is a straight line only between the points (-3A, -3V) and (5A, 5V). Beyond these points, the V-I characteristics are not following the linear relation. Hence, it is a Non-linear element. Step 2 − Verifying the network element as active or passive. The given V-I characteristics of a network element lies in the first and third quadrants. In these two quadrants, the ratios of voltage (V) and current (I) produce positive impedance values. Hence, the network element is a Passive element. Step 3 − Verifying the network element as bilateral or unilateral. Consider the point (5A, 5V) on the characteristics. The corresponding point (-5A, -3V) exists on the given characteristics instead of (-5A, -5V). Hence, the network element is a Unilateral element. Therefore, the given V-I characteristics show that the network element is a Non-linear, Passive, and Unilateral element. The circuits containing them are called unilateral circuits. Lumped and Distributed Elements Lumped elements are those elements which are very small in size & in which simultaneous actions takes place. Typical lumped elements are capacitors, resistors, inductors. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 12  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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 Where, R is the resistance of a resistor. From Equation 2, we can conclude that the current flowing through the resistor is directly proportional to the applied voltage across resistor and inversely proportional to the resistance of resistor. Power in an electric circuit element can be represented as Substitute, Equation 1 in Equation 3. Substitute, Equation 2 in Equation 3. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 13  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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 Substitute Ψ = LI in the above equation. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 14  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 15  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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 MRCET EAMCET CODE: MLRD www.mrcet.ac.in 16  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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 Independent Sources 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 −  Independent Voltage Sources  Independent Current Sources Independent Voltage Sources An independent voltage source produces a constant voltage across its two terminals. This voltage is independent of the amount of current that is flowing through the two terminals of voltage source. Independent ideal voltage source and its V-I characteristics are shown in the following figure. 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 17  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 18  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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 Voltage Sources  Dependent Current Sources Dependent Voltage Sources A dependent voltage source produces a voltage across its two terminals. The amount of this voltage is dependent on some other voltage or current. Hence, dependent voltage sources can be further classified into the following two categories −  Voltage Dependent Voltage Source (VDVS)  Current Dependent Voltage Source (CDVS) Dependent voltage sources are represented with the signs ‘+’ and ‘-’ inside a diamond shape. The magnitude of the voltage source can be represented outside the diamond shape. Dependent Current Sources A dependent current source produces a current. The amount of this current is dependent on some other voltage or current. Hence, dependent current sources can be further classified into the following two categories −  Voltage Dependent Current Source (VDCS)  Current Dependent Current Source (CDCS) MRCET EAMCET CODE: MLRD www.mrcet.ac.in 19  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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 into a practical current source  Practical current source into a practical voltage source Practical voltage source into a practical current source The transformation of practical voltage source into a practical current source is shown in the following figure 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 into a practical voltage source The transformation of practical current source into a practical voltage source is shown in the following figure. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 20  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING Practical current source consists of a current source (I S) 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.  Kirchhoff’s Current Law  Kirchhoff’s Voltage Law Kirchhoff’s Current Law Kirchhoff’s Current Law (KCL) states that the algebraic sum of currents leaving (or entering) a node is equal to zero. A Node is a point where two or more circuit elements are connected to it. If only two circuit elements are connected to a node, then it is said to be simple node. If three or more circuit elements are connected to a node, then it is said to be Principal Node. Mathematically, KCL can be represented as Where,  Im is the mth branch current leaving the node. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 21  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING  M is the number of branches that are connected to a node. The above statement of KCL can also be expressed as "the algebraic sum of currents entering a node is equal to the algebraic sum of currents leaving a node". Let us verify this statement through the following example. Example Write KCL equation at node P of the following figure.  In the above figure, the branch currents I 1, I2 and I3 areentering at node P. So, consider negative signs for these three currents.  In the above figure, the branch currents I 4 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 22  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING Mathematically, KVL can be represented as 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 23  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING  Current Division Principle  Voltage Division Principle Current Division Principle When two or more passive elements are connected in parallel, the amount of current that flows through each element gets divided(shared) among themselves from the current that is entering the node. Consider the following circuit diagram. 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 MRCET EAMCET CODE: MLRD www.mrcet.ac.in 24  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING 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 N th 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, V S in series with two resistors R1 and R2. The current flowing through these elements is I S. The voltage drops across the resistors R1and R2 are V1 and V2 respectively. The KVL equation around the loop will be  Substitute V1 = IS R1 and V2 = IS R2 in the above equation MRCET EAMCET CODE: MLRD www.mrcet.ac.in 25  DEPARTMENT OF HUMANITIES AND SCIENCES BASIC ELECTRICAL ENGINEERING  Substitute the value of IS in V1 = IS R1. 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 26 DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING UNIT-II NETWORK ANALYSIS  Network Reduction Techniques  Series and Parallel connection of Resistive Networks  Star–to-Delta and Delta-to-Star Transformations for Resistive Networks  Mesh Analysis  Network Theorems: Thevenin’s Theorem  Norton’s Theorem  Superposition Theorem  Problems MRCET EAMCET CODE: MLRD www.mrcet.ac.in 27  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING Network Reduction Techniques: There are two basic methods that are used for solving any electrical network: Nodal analysis and Mesh analysis. In this chapter, let us discuss about the Mesh analysis method. Series and parallel connections of resistive networks: If a circuit consists of two or more similar passive elements and are connected in exclusively of series type or parallel type, then we can replace them with a single equivalent passive element. Hence, this circuit is called as an equivalent circuit. In this chapter, let us discuss about the following two equivalent circuits.  Series Equivalent Circuit  Parallel Equivalent Circuit Series Equivalent Circuit If similar passive elements are connected in series, then the same current will flow through all these elements. But, the voltage gets divided across each element. Consider the following circuit diagram. 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 28  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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 Parallel Equivalent Circuit If similar passive elements are connected in parallel, then the same voltage will be maintained across each element. But, the current flowing through each element gets divided. Consider the following circuit diagram. 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 MRCET EAMCET CODE: MLRD www.mrcet.ac.in 29  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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 L 1, 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 MRCET EAMCET CODE: MLRD www.mrcet.ac.in 30  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING Example Problems: 1) Find the Req for the circuit shown in below figure. fig(a) Solution: To get Req we combine resistors in series and in parallel. The 6 ohms and 3 ohms resistors are in parallel, so their equivalent resistance is  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 2) Find the Req for the circuit shown in below figure. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 31  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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(b), 4 ohms and 12 ohms are in parallel, equivalent is 3 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 Star–to-Delta and Delta-to-Star Transformations for Resistive Networks: Delta to Star Transformation In the previous chapter, we discussed an example problem related equivalent resistance. There, we calculated the equivalent resistance between the terminals A & B of the given electrical network easily. Because, in every step, we got the combination of resistors that are connected in either series form or parallel form. However, in some situations, it is difficult to simplify the network by following the previous approach. For example, the resistors connected in either delta (δ) form or star form. In such situations, we have to convert the network of one form to the other in order to simplify it further by using series combination or parallel combination. In this chapter, let us discuss about the Delta to Star Conversion. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 32  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 33  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING The following equations represent the equivalent resistance between two terminals of star network, when the third terminal is kept open. Star Network Resistances in terms of Delta Network Resistances We will get the following equations by equating the right-hand side terms of the above equations for which the left-hand side terms are same. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 34  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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 Delta Network Resistances in terms of Star Network Resistances Let us manipulate the above equations in order to get the resistances of delta network in terms of resistances of star network.  Multiply each set of two equations and then add. MRCET EAMCET CODE: MLRD www.mrcet.ac.in 35  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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: 1) Convert the Delta network in Fig.(a) to an equivalent star network Solution: MRCET EAMCET CODE: MLRD www.mrcet.ac.in 36  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 2) Convert the star network in fig(a) to delta network Solution: The equivalent delta for the given star is shown in fig(b), where MRCET EAMCET CODE: MLRD www.mrcet.ac.in 37  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 3) Determine the total current I in the given circuit. Solution: Delta connected resistors 25 ohms, 10 ohms and 15 ohms are converted in to star as shown in given figure. R1 = R12 R31 / R12 + R23 + R31 = 10 x 25 / 10 + 15 + 25 = 5 ohms R2 = R23 R12 / R12 + R23 + R31 = 15 x 10 / 10 + 15 + 25 = 3 ohms R3 = R31 R23 / R12 + R23 + R31 = 25 x 15 / 10 + 15 + 25 = 7.5 ohms The given circuit thus reduces to the circuit shown in below fig. The equivalent resistance of (20 + 5) ohms || (10 + 7.5) ohms = 25 x 17.5 / 25 + 17.5 = 10.29 ohms Total resistance = 10.29 + 3 + 2.5 = 15.79 ohms Hence the total current through the battery, I = 15 / 15.79 = 0.95 A MRCET EAMCET CODE:MLRD www.mrcet.ac.in 38  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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 of Mesh 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 39  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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 Step 4 − Finding mesh currents I1 and I2 by solving Equation 1 and Equation 2. The left-hand side terms of Equation 1 and Equation 2 are the same. Hence, equate the right-hand side terms of Equation 1 and Equation 2 in order find the value of I1. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 40  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING Therefore, the voltage across 30 Ω resistor of the given circuit is84 V. Note 1 − From the above example, we can conclude that we have to solve ‘m’ mesh equations, if the electric circuit is having ‘m’ meshes. That’s why we can choose Mesh analysis when the number of meshes is less than the number of principal nodes (except the reference node) of any electrical circuit. Note 2 − We can choose either Nodal analysis or Mesh analysis, when the number of meshes is equal to the number of principal nodes (except the reference node) in any electric circuit. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 41  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 42  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING Main steps to find out VTh and RTh : 1. 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 2. Open circuit voltage VOC across these two terminals is found out using the conventional network mesh/node analysis methods and this would be VTh. 3. Thevenin's resistance RTh is found out by the method depending upon whether the network contains dependent sources or not. a. With dependent sources: RTh = Voc / Isc b. Without dependent sources : RTh = 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) 4. Replace the network with VTh in series with RTh and the concerned branch resistance (or) load resistance across the load terminals (A&B) as shown in below fig. 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) MRCET EAMCET CODE:MLRD www.mrcet.ac.in 43  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (V TH). 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 (R TH)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) MRCET EAMCET CODE:MLRD www.mrcet.ac.in 44  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 45  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING  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) MRCET EAMCET CODE:MLRD www.mrcet.ac.in 46  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING We have shorted the AB terminals to determine the Norton current, I N. 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 MRCET EAMCET CODE:MLRD www.mrcet.ac.in 47  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 48  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 2. Repeat step 1 for each of the other independent sources. 3. Find the total contribution by adding algebraically all the contributions due to the independent sources. Example: By Using the superposition theorem find I in the circuit shown in figure? 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 : MRCET EAMCET CODE:MLRD www.mrcet.ac.in 49  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 50  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 51  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. For a sinusoidal wave , or IRMS = 0.707 x IM , ERMS = 0.707 EM  Actually, the RMS value of a sine wave is the measurement of heating effect of sine wave. For example, when a resistor is connected to across an AC voltage source, it produces specific amount of heat (Fig 2 – a). When the same resistor is connected across the DC voltage source as shown in (fig 2 – b). By adjusting the value of DC voltage to get the same amount of heat generated before in AC voltage source in fig a. It means the RMS value of a sine wave is equal to the DC Voltage source producing the same amount of heat generated by AC Voltage source.  In more clear words, the domestic voltage level in US is 110V, while 220V AC in UK. This voltage level shows the effective value of (110V or 220V R.M.S) and it shows that the home wall socket is capable to provide the same amount of average positive power as 110V or 220V DC Voltage.  Keep in mind that the ampere meters and volt meters connected in AC circuits always showing the RMS values (of current and voltage).  For AC sine wave, RMS values of current and voltage are: IRMS = 0.707 x IM , VRMS = 0.707 VM  Let’s see how to find the R.M.S values of a sine wave.  We know that the value of sinusoidal alternating current (AC) = Im Sin ω θ = Im Sin θ  While the mean of square of instantaneous values of current in in half or complete cycle is: MRCET EAMCET CODE:MLRD www.mrcet.ac.in 52  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING The Square root of this value is: Hence, the RMS value of the current is (while putting I = Im Sin θ): Now, Therefore, We may find that for a symmetrical sinusoidal current: IRMS = Max Value of Current x 0.707 Average Value: If we convert the alternating current (AC) sine wave into direct current (DC) sine wave through rectifiers, then the converted value to the DC is known as the average value of that alternating current sine wave. Fig 4 – Average Value of Voltage MRCET EAMCET CODE:MLRD www.mrcet.ac.in 53  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. Fig 5- Average Value of Current In short, the average value of a sine wave taken over a complete cycle is always zero, because the positive values (above the zero crossing) offset or neutralize the negative values (below the zero crossing.) We know that the standard equation of alternating current is i = Sin ω θ = Im Sin θ  Maximum value of current on sine wave = Im  Average value of current on sine wave = IAV  Instantaneous value of current on sine wave = i  The angle specified fir “i” after zero position of current = θ  Angle of half cycle = π radians  Angle of full cycle = 2π radians MRCET EAMCET CODE:MLRD www.mrcet.ac.in 54  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING (a) Average value of complete cycle: Let i = Sin ω θ = Im Sin θ Thus, the average value of a sinusoidal wave over a complete cycle is zero. (b) Average value of current over a half cycle Average Value of Current (Half Cycle) IAV = 0.637 VM Similarly, the average value of voltage over a half cycle VAV = 0.637 VM What is Peak Voltage or Maximum Voltage Value ? Peak value is also known as Maximum Value, Crest Value or Amplitude. It is the maximum value of alternating current or voltage from the “0” position no matter positive or negative half cycle in a sinusoidal wave as shown in fig 8. Its expressed as IM and EM or VP and IM. Equations of Peak Voltage Value is: VP = √2 x VRMS = 1.414 VRMS MRCET EAMCET CODE:MLRD www.mrcet.ac.in 55  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING VP = VP-P/2 = 0.5 VP-P VP = π/2 x VAV = 1.571 x VAV In other words, It is the value of voltage or current at the positive or the negative maximum (peaks) with respect to zero. In simple words, it is the instantaneous value with maximum intensity. Fig 8 – Peak or Maximum Values of Voltages Peak to Peak Value: The sum of positive and negative peak values is known as peak to peak value. Its expressed as IPP or VPP. Equations and formulas for Peak to Peak Voltage are as follow: VP-P = 2√2 x VRMS = 2.828 x VRMS VP-P =2 x VP VP-P = π x VAV = 3.141 x VAV In other words, the peak to peak value of a sine wave, is the voltage or current from positive peak to the negative peak and its value is double as compared to peak value or maximum value as shown in fig 8 above. 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. For a sinusoidal alternating voltage: MRCET EAMCET CODE:MLRD www.mrcet.ac.in 56  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING For a sinusoidal alternating current: Form Factor: The ratio between RMS value and Average value of an alternating quantity (Current or Voltage) is known as Form Factor. Other Terms Related To AC Circuits 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 MRCET EAMCET CODE:MLRD www.mrcet.ac.in 57  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. Fig 9 – Different Terms used in AC Circuits and Sine Wave Introduction to Single Phase AC Circuit:  In a dc circuit the relationship between the applied voltage V and current flowing through the circuit I is a simple one and is given by the expression I = V/R but in an a c circuit this simple relationship does not hold good. Variations in current and applied voltage set up magnetic and electrostatic effects respectively and these must be taken into account with the resistance of the circuit while determining the quantitative relations between current and applied voltage.  With comparatively low-voltage, heavy- current circuits magnetic effects may be very large, but electrostatic effects are usually negligible. On the other hand with high-voltage circuits electrostatic effects may be of appreciable magnitude, and magnetic effects are also present.  Here it has been discussed how the magnetic effects due to variations in current do and electrostatic effects due to variations in the applied voltage affect the relationship between the applied voltage and current. Purely Resistive Circuit:  A purely resistive or a non-inductive circuit is a circuit which has inductance so small that at normal frequency its reactance is negligible as compared to its resistance. Ordinary filament lamps, water resistances etc., are the examples of non-inductive resistances. If the circuit is purely non-inductive, no reactance emf (i.e., self- induced or back emf) is set up and whole of the applied voltage is utilized in overcoming the ohmic resistance of the circuit.  Consider an ac circuit containing a non-inductive resistance of R ohms connected across a sinusoidal voltage represented by v = V sin wt, as shown in Fig. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 58  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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: And instantaneous current may be expressed as: i = Imax sin ωt From the expressions of instantaneous applied voltage and instantaneous current, it is evident that in a pure resistive circuit, the applied voltage and current are in phase with each other, as shown by wave and phasor diagrams in Figs. 4.1 (b) and (c) respectively. Power in Purely Resistive Circuit: The instantaneous power delivered to the circuit in question is the product of the instantaneous values of applied voltage and current. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 59  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. Purely Inductive Circuit: 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 60  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. Power in Purely Inductive Circuit: Instantaneous power, p = v × i = Vmax sin ω t Imax sin (ωt – π/2) Or p = – Vmax Imax sin ω t cos ω t = Vmax Imax/2 sin 2 ωt The power measured by wattmeter is the average value of p which is zero since average of a sinusoidal quantity of double frequency over a complete cycle is zero. Hence in a purely inductive circuit power absorbed is zero. Physically the above fact can be explained as below: During the second quarter of a cycle the current and the magnetic flux of the coil increases and the coil draws power from the supply source to build up the magnetic field (the power drawn is positive and the energy drawn by the coil from the supply source is represented by the area between the curve p and the time axis). The energy stored in the magnetic field during build up is given as W max = 1/2 L I2max. In the next quarter the current decreases. The emf of self-induction will, however, tends to oppose its decrease. The coil acts as a generator of electrical energy, returning the stored energy in the magnetic field to the supply source (now the power drawn by the coil is negative and the curve p lies below the time axis). The chain of events repeats itself during the next half cycles. Thus, a proportion of power is continually exchanged between the field and the inductive circuit and the power consumed by a purely inductive coil is zero. Purely Capacitive Circuit: 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. The expression for instantaneous charge is given as: q = C Vmax sin ωt Since the capacitor current is equal to the rate of change of charge, the capacitor current may be obtained by differentiating the above equation: MRCET EAMCET CODE:MLRD www.mrcet.ac.in 62  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. Power in Purely Capacitive Circuit: 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 63  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING Resistance — Capacitance (R-C) Series Circuit: Consider an ac circuit consisting of resistance of R ohms and capacitance of C farads connected in series, as shown in Fig. 4.18 (a). Let the supply frequency be of fHz and current flowing through the circuit be of I amperes (rms value). Voltage drop across resistance, VR = I R in phase with the current. Voltage drop across capacitance, VC = I XC lagging behind I by π/2 radians or 90°, as shown in Fig. 4.18 (b). The applied voltage, being equal to phasor sum of VR and VC, is given in magnitude by- The applied voltage lags behind the current by an angle ɸ: If instantaneous voltage is represented by: v = Vmax sin ω t Then instantaneous current will be expressed as: i = Imax sin (ω t + ɸ) And power consumed by the circuit is given by: P = VI cos ɸ MRCET EAMCET CODE:MLRD www.mrcet.ac.in 64  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING Voltage triangle and impedance triangle Fig. 4.19 are shown in Figs. 4.19 (a) and 4.19 (b) respectively. 6. Apparent Power, True Power, Reactive Power and Power Factor: The product of rms values of current and voltage, VI is called the apparent power and is measured in volt- amperes or kilo-volt amperes (kVA). The true power in an ac circuit is obtained by multiplying the apparent power by the power factor and is expressed in watts or kilo-watts (kW). The product of apparent power, VI and the sine of the angle between voltage and current, sin ɸ is called the reactive power. This is also known as wattless power and is expressed in reactive volt-amperes or kilo-volt amperes reactive (kVA R). The above relations can easily be followed by referring to the power diagram shown in Fig. 4.7 (a). MRCET EAMCET CODE:MLRD www.mrcet.ac.in 65  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING Power factor may be defined as: (i) Cosine of the phase angle between voltage and current, (ii) The ratio of the resistance to impedance, or (iii) The ratio of true power to apparent power. The power factor can never be greater than unity. The power factor is expressed either as fraction or as a percentage. It is usual practice to attach the word ‘lagging’ or ‘leading’ with the numerical value of power factor to signify whether the current lags behind or leads the voltage. CONCEPT OF REACTANCE, IMPEDANCE, SUSCEPTANCE AND ADMITTANCE: 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). MRCET EAMCET CODE:MLRD www.mrcet.ac.in 66  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING (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 MRCET EAMCET CODE:MLRD www.mrcet.ac.in 67  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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 MRCET EAMCET CODE:MLRD www.mrcet.ac.in 68  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. Single Loop DC Generator 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in 69  DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING BASIC ELECTRICAL ENGINEERING 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. MRCET EAMCET CODE:MLRD www.mrcet.ac.in