Applied Electricity PDF Lecture Notes (UMaT)

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hjklzxcvbnmrtyuiopasdfghjklzxcvbn appliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYapplied SOLOMON NUNOO ELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECT Department of Electrical and Electronic Engineering RICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITY University of Mines and Technology mqwertyuiopasdfghjklzxcvbnmqwert appliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYapplied Tarkwa ELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECT RICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITY yuiopasdfghjklzxcvbnmqwertyuiopas appliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYapplied ELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECTRICITYappliedELECT 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Electricity Table of Contents TABLE OF CONTENTS CONTENT PAGE CHAPTER 1 CIRCUIT LAWS 1 1.1 Introduction 1 1.2 Ohm’s Law 1 1.3 Kirchhoff’s Voltage Law 1 1.4 Kirchhoff’s Current Law 2 1.5 Circuit Elements in Series 2 1.6 Circuit Elements in Parallel 4 1.7 Voltage Division 5 1.8 Current Division 6 1.9 Network Reduction 7 1.10 Electrical Power and Energy 7 1.10.1 Electrical Power 7 1.10.2 Electrical Energy 8 1.11 Problems 9 1.11.1 Ohm’s Law 9 1.11.2 Kirchhoff’s Laws 10 1.11.3 Power and Energy 11 CHAPTER 2 CIRCUIT THEOREMS (ANALYSIS METHODS) 13 2.1 The Branch Current Method 13 2.2 The Mesh Current Method 13 2.3 The Node Voltage Method 15 2.4 Superposition Theorem 16 2.5 Thévenin’s and Norton’s Theorems 18 2.6 ∆-Y and Y-∆ Conversions 19 2.7 Problems 23 2.7.1 Superposition Theorem 23 2.7.2 Thévenin’s Theorem 24 2.7.3 Norton’s Theorem 24 CHAPTER 3 CAPACITORS AND CAPACITANCE 25 3.1 Electrostatic Field 25 3.2 Electric field strength 26 3.3 Capacitance 26 3.4 Capacitors 27 3.5 Electric Flux Density 27 3.6 Permittivity 28 3.7 The Parallel Plate Capacitor 28 3.8 Capacitors Connected in Parallel and Series 29 3.8.1 Capacitors Connected in Parallel 29 3.8.2 Capacitors Connected in Series 30 University of Mines and Technology, Tarkwa i Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Table of Contents 3.9 Dielectric Strength 31 3.10 Energy Stored 31 3.11 Practical Types of Capacitor 31 3.11.1 Variable Air Capacitors 31 3.11.2 Mica Capacitors 32 3.11.3 Paper Capacitors 32 3.11.4 Ceramic Capacitors 32 3.11.5 Plastic Capacitors 33 3.11.6 Titanium Oxide Capacitors 34 3.11.7 Electrolytic Capacitors 34 3.12 Discharging Capacitors 34 3.13 Problems 34 3.13.1 Charge and Capacitance 34 3.13.2 Electric Field Strength, Electric Flux Density and Permittivity 35 3.13.3 Parallel Plate Capacitor 35 3.13.4 Capacitors in Parallel and Series 36 3.13.5 Energy Stored 36 CHAPTER 4 MAGNETIC CIRCUITS 38 4.1 Magnetic Fields 38 4.2 Magnetic Flux and Flux Density 39 4.3 Magnetomotive Force and Magnetic Field Strength 39 4.4 Permeability and B-H curves 40 4.5 Reluctance 41 4.6 Composite Series Magnetic Circuits 41 4.7 Comparison between Electrical and Magnetic Quantities 41 4.8 Hysteresis and Hysteresis Loss 42 4.9 Problems 42 4.9.1 Magnetic Circuit Quantities 42 4.9.2 Composite Series Magnetic Circuits 43 CHAPTER 5 ELECTROMAGNETISM 45 5.1 Magnetic Field due to an Electric Current 45 5.2 Electromagnets 47 5.2.1 Electric Bell 47 5.2.2 Relay 48 5.2.3 Lifting Magnet 48 5.2.4 Telephone Receiver 49 5.3 Force on a Current-Carrying Conductor 49 5.4 Force on a Charge 51 5.5 Problems 51 5.5.1 Force on a Current-Carrying Conductor 51 5.5.2 Force on a Charge 52 University of Mines and Technology, Tarkwa ii Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Table of Contents CHAPTER 6 ELECTROMAGNETIC INDUCTION 53 6.1 Introduction 53 6.2 Laws of Electromagnetic Induction 53 6.3 Inductance 55 6.4 Inductors 56 6.5 Energy Stored 57 6.6 Inductance of a Coil 58 6.7 Mutual Inductance 58 6.8 Problems 59 6.8.1 Induced emf 59 6.8.2 Inductance 59 6.8.3 Energy Stored 60 6.8.4 Inductance of a Coil 60 6.8.5 Mutual Inductance 60 CHAPTER 7 ALTERNATING VOLTAGE AND CURRENT 62 7.1 Introduction 62 7.2 The AC Generator 62 7.3 Waveforms 63 7.4 AC Values 64 7.5 Equation of a Sinusoidal Waveform 65 7.6 Combination of Waveforms 66 7.7 Rectification 66 7.8 Problems 68 7.8.1 Frequency and Periodic Time 68 7.8.2 AC Values of Non-Sinusoidal Waveforms 68 7.8.3 AC Values of Sinusoidal Waveforms 68 7.8.4 Combination of Periodic Functions 69 CHAPTER 8 FUNDAMENTALS OF ALTERNATING CURRENT CIRCUITS 70 8.1 AC Through Resistance, Inductance and Capacitance 70 8.1.1 Purely Resistive AC Circuits 70 8.1.2 Purely Inductive AC Circuits 70 8.1.3 Purely Capacitive AC Circuits 70 8.2 Series AC Circuits 71 8.2.1 R-L Series Circuits 71 8.2.2 R-C Series Circuits 72 8.2.3 R-L Series Circuits 72 8.2.4 Series Resonance 73 8.2.5 Power in AC Circuits 73 8.2.6 Power Triangle and Power Factor 74 8.3 Parallel AC Circuit 75 8.3.1 R-L Parallel AC Circuits 75 8.3.2 R-C Parallel AC Circuits 76 8.3.3 L-C Parallel AC Circuits 76 University of Mines and Technology, Tarkwa iii Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Table of Contents 8.4 Power Factor Improvement 77 8.5 Problems 77 CHAPTER 9 SIGNAL WAVEFORMS 78 9.1 Introduction 78 9.2 Step Function 78 9.3 The Impulse 79 9.4 Ramp Function 79 9.5 Sinusoidal Function 80 9.6 Decaying Exponential 80 9.7 Time Constant 81 9.8 DC Signal 81 CHAPTER 9 INTRODUCTION TO ELECTRICAL MACHINES 82 10.1 Introduction 82 10.2 Transformers 82 10.2.1 Constructional Features 82 10.2.2 Principle of Operation 84 10.2.3 Three-phase Transformer Connections 86 10.3 DC Machines 86 10.3.1 Constructional Features 86 10.3.2 Principle of Operation of DC Generators 90 10.3.3 Emf Equation 91 10.3.4 Characteristics of DC Generators 93 10.3.5 Types of DC Generators 93 10.3.6 Principle of Operation of DC Motors 96 10.3.7 Torque Equation 98 10.3.8 Characteristics of a DC Motor 99 10.4 Induction Motors 101 10.4.1 Constructional Features of 3-phase Induction Motors 101 10.4.2 Principle of Operation 103 10.4.3 Torque Development 104 10.4.4 Single-phase Induction Motors 106 10.5 Synchronous Machines 107 10.5.1 Constructional Features 108 10.5.2 Principle of Generator Operation 109 10.5.3 EMF Equation 109 10.5.4 Load Characteristics 111 10.5.5 Principle of Motor Operation 111 REFERENCES 114 University of Mines and Technology, Tarkwa iv Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws CHAPTER ONE CIRCUIT LAWS 1.1 Introduction An electric circuit or network consists of a number of interconnected single circuit elements. The circuit will generally contain at least one voltage or current source. The arrangement of elements results in a new set of constraints between the currents and voltages. These new constraints and their corresponding equations, added to the current-voltage relationships of the individual elements, provide the solution of the network. The purpose of defining the individual elements, connecting them in a network, and solving the equations is to analyze the performance of such electrical devices as motors, generators, transformers, electrical transducers, and a host of electronic devices. The solution generally answers necessary questions about the operation of the device under conditions applied by a source of energy. 1.2 Ohm’s Law Ohm’s law states that, “the current, I, flowing in a circuit is directly proportional to the applied voltage, V, provided the temperature remains constant. Thus, I∝V Therefore, V V I= or V = IR or R= R I where R is the constant of proportionality called resistance. 1.3 Kirchhoff’s Voltage Law For any closed path in a network, Kirchhoff’s voltage law (KVL) states that, “the algebraic sum of the voltages is zero.” Some of the voltages will be sources, while others will result from current in passive elements creating a voltage, which is sometimes referred to as a voltage drop. The law applies equally well to circuits driven by constant sources, DC, time variable sources, v(i) and i(t), etc. The mesh current method of circuit analysis introduced in Section 2.2 is based on Kirchhoff’s voltage law. EXAMPLE 1.1 Write the KVL equation for the circuit shown in Figure 1.1. SOLUTION 1.1 Starting at the lower left corner of the circuit, for the current direction as shown, we have − va + v1 + vb + v2 + v3 = 0 − va + iR1 + vb + iR2 + iR3 = 0 vb − va = i(R1 + R2 + R3 ) University of Mines and Technology, Tarkwa 1 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws FIGURE 1.1 Circuit for Example 1.1 1.4 Kirchhoff’s Current Law The connection of two or more circuit elements creates a junction called a node. The junction between two elements is called a simple node and no division of current results. The junction of three or more elements is called a principal node, and here current division does take place. Kirchhoff’s current law (KCL) states that, “the algebraic sum of the currents at a node is zero.” It may be stated alternatively that, “the sum of the currents entering a node is equal to the sum of the currents leaving that node.” The node voltage method of circuit analysis introduced in Section 2.3 is based on equations written at the principal nodes of a network by applying Kirchhoff’s current law. The basis for the law is the conservation of electric charge. EXAMPLE 1.2 Write the KCL equation for the principal node shown in Figure 1.2. FIGURE 1.2 Circuit for Example 1.2 SOLUTION 1.2 i1 − i2 + i3 − i4 − i5 = 0 i1 + i3 = i2 + i4 + i5 1.5 Circuit Elements in Series Three passive circuit elements in series connection as shown in Figure 1.3 have the same current, i. The voltages across the elements are v 1 , v 2 , and v 3. The total voltage, v, is the sum of the individual voltages; v = v 1 + v 2 + v 3. If the elements are resistors, University of Mines and Technology, Tarkwa 2 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws v = iR1 + iR2 + iR3 = i(R1 + R2 + R3 ) = iReq where a single equivalent resistance R eq replaces the three series resistors. The same relationship between i and v will pertain. FIGURE 1.3 Three passive circuit elements connected in series For any number of resistors in series, we have R eq = R 1 + R 2 +… If the three passive elements are inductances, di di di v = L1 + L2 + L3 dt dt dt di = (L1 + L2 + L3 ) dt di = Leq dt Extending this to any number of inductances in series, we have, L eq = L 1 + L 2 +… If the three circuit elements are capacitances, assuming zero initial charges so that the constants of integration are zero, 1 1 1 v= C1 ∫ idt + C2 ∫ idt + C3 ∫ idt  1 1 1  =  + +  ∫ idt C  1 C 2 C3  1 Ceq ∫ = idt 1 1 1 The equivalent capacitance of several capacitances in series is = + +... Ceq C1 C2 University of Mines and Technology, Tarkwa 3 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws EXAMPLE 1.3 The equivalent resistance of three resistors in series is 750.0 Ω. Two of the resistors are 40.0 and 410.0 Ω. What must be the ohmic resistance of the third resistor? SOLUTION 1.3 Req = R1 + R2 + R3 750.0 = 40.0 + 410.0 + R3 R3 = 300.0 Ω EXAMPLE 1.4 Two capacitors, C 1 = 2.0 μF and C 2 = 10.0 μF, are connected in series. Find the equivalent capacitance. SOLUTION 1.4 Ceq = C1C2 = ( )( 2.0 × 10 −6 10.0 × 10 −6 ) = 1.67 µF C1 + C2 2.0 × 10 − 6 + 10.0 × 10 − 6 Note: When two capacitors in series differ by a large amount, the equivalent capacitance is essentially equal to the value of the smaller of the two. 1.6 Circuit Elements in Parallel For three circuit elements connected in parallel as shown in Figure 1.4, KCL states that, “the current, i, entering the principal node is the sum of the three currents leaving the node through the branches.” FIGURE 1.4 Three circuit elements connected in parallel From Figure 1.4, i = i1 + i2 + i3 If the three passive circuit elements are resistances, v v v  1 1 1  1 i= + + =  + + v = v R1 R2 R3  R1 R2 R3  Req For several resistors in parallel, 1 1 1 = + +... Req R1 R2 University of Mines and Technology, Tarkwa 4 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws The case of two resistors in parallel occurs frequently and deserves special mention. The equivalent resistance of two resistors in parallel is given by the product of the two resistors divided by the sum of the two resistors. R1 R2 Req = R1 + R2 EXAMPLE 1.5 Obtain the equivalent resistance of (a) two 60.0-Ω resistors in parallel and (b) three 60.0-Ω resistors in parallel. SOLUTION 1.5 a. Req = (60.0 )2 = 30.0Ω 120.0 1 1 1 1 b. = + + Req = 20.0Ω Req 60.0 60.0 60.0 Note: For n identical resistors in parallel the equivalent resistance is given by R. n Combinations of inductances in parallel have similar expressions to those of resistors in parallel: 1 1 1 L1 L2 = + +... and, for two inductances, Leq = Leq L1 L2 L1 + L2 EXAMPLE 1.6 Two inductances L 1 = 3.0 mH and L 2 = 6.0 mH are connected in parallel. Find L eq. SOLUTION 1.6 1 1 1 = + and L eq = 2.0 mH Leq 3.0 6.0 With three capacitances in parallel, dv dv dv dv dv i = C1 + C2 + C3 = (C1 + C2 + C3 ) = Ceq dt dt dt dt dt For several parallel capacitors, C eq = C 1 + C 2 + …, which is of the same form as resistors in series. 1.7 Voltage Division A set of series-connected resistors as shown in Figure 1.5 is referred to as a voltage divider. The concept extends beyond the set of resistors illustrated here and applies equally to impedances in series, as will be shown in Chapter 7. University of Mines and Technology, Tarkwa 5 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws Since v1 = iR1 and v = i(R1 + R2 + R3 ) , then  R1  v1 = v   R1 + R2 + R3  FIGURE 1.5 A set of series-connected resistors EXAMPLE 1.7 A voltage divider circuit of two resistors is designed with a total resistance of the two resistors equal to 50.0 Ω. If the output voltage is 10 percent of the input voltage, obtain the values of the two resistors in the circuit. SOLUTION 1.7 v1 R1 = 0.1 and 0.1 = v 50.0 from which R 1 = 5.0 Ω and R 2 = 45.0 Ω. 1.8 Current Division A parallel arrangement of resistors as shown in Figure 1.6 results in a current divider. The ratio of the branch current i 1 to the total current i illustrates the operation of the divider. v v v v Since i = + + and i1 = then R1 R2 R3 R1 1 i1 R1 R2 R3 = = i 1 + 1 + 1 R1 R2 + R1 R3 + R2 R3 R1 R2 R3 For a two-branch current divider we have i1 R2 = i R1 + R2 University of Mines and Technology, Tarkwa 6 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws FIGURE 1.6 A parallel arrangement of resistors This may be expressed as follows: The ratio of the current in one branch of a two-branch parallel circuit to the total current is equal to the ratio of the resistance of the other branch resistance to the sum of the two resistances. EXAMPLE 1.8 A current of 30.0 mA is to be divided into two branch currents of 20.0 mA and 10.0 mA by a network with an equivalent resistance equal to or greater than 10.0 Ω. Obtain the branch resistances. SOLUTION 1.8 i1 R2 20 R2 10 R1 R1 R2 Since = then = , = and ≥ 10.0 i R1 + R2 30 R1 + R2 30 R1 + R2 R1 + R2 Solving these equations yields R1 ≥ 15.0 Ω and R2 ≥ 30.0 Ω. 1.9 Network Reduction The mesh current and node voltage methods (Sections 2.2 and 2.3) are the principal techniques of circuit analysis. However, the equivalent resistance of series and parallel branches (Sections 1.4 and 1.5), combined with the voltage and current division rules, provide another method of analyzing a network. This method is tedious and usually requires the drawing of several additional circuits. Even so, the process of reducing the network provides a very clear picture of the overall functioning of the network in terms of voltages, currents, and power. The reduction begins with a scan of the network to pick out series and parallel combinations of resistors. 1.10 Electrical Power and Energy 1.10.1 Electrical Power Power P in an electrical circuit is given by the product of potential difference V and current I. The unit of power is the watt, W. Hence, P = V × I watts From Ohm’s law, V = IR University of Mines and Technology, Tarkwa 7 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws Thus, V2 P = I2R watts or P = watts R There are thus three possible formulae which may be used for calculating power. 1.10.2 Electrical Energy Electrical energy = power × time If the power is measured in watts and the time in seconds then the unit of energy is watt- seconds or joules. If the power is measured in kilowatts and the time in hours then the unit of energy is kilowatt-hours, often called the “unit of electricity”. The “electricity meter” in the home records the number of kilowatt-hours used and is thus an energy meter. EXAMPLE 1.9 Obtain the total power supplied by the 60-V source and the power absorbed in each resistor in the network of Figure 1.7. SOLUTION 1.9 First solve for the equivalent resistance between nodes a-b and c-d giving Rab = 7 + 5 = 12 Ω 12 × 6 Rcd = =4Ω 12 + 6 FIGURE 1.7 Electrical network for Example 1.9 These two equivalents are in parallel (Figure 1.8), giving 4 × 12 Ref = = 3Ω 4 + 12 FIGURE 1.8 Reduced form of Figure 1.7 University of Mines and Technology, Tarkwa 8 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws Then this 3-Ω equivalent is in series with the 7-Ω resistor (Figure 1.9), so that for the entire circuit, Req = 7 + 3 = 10 Ω FIGURE 1.9 Reduced form of Figure 1.8 The total power absorbed, which equals the total power supplied by the source, can now be calculated as V 2 (60 ) 2 PT = = = 360 W Req 10 This power is divided between R ge and R ef as follows: 7 3 Pge = P7 Ω = (360 ) = 252 W Pef = (360) = 108 W 7+3 7+3 Power P ef is further divided between R cd and R ab as follows: 12 4 Pcd = (108 ) = 81 W Pab = (108 ) = 27 W 4 + 12 4 + 12 Finally, these powers are divided between the individual resistances as follows: 6 7 P12Ω = (81) = 27 W P7 Ω = (27 ) = 15.75 W 6 + 12 7+5 12 5 P6 Ω = (81) = 54 W P5 Ω = (27 ) = 11.25 W 6 + 12 7+5 1.11 Problems 1.11.1 Ohm’s Law 1. The current flowing through a heating element is 5 A when a pd of 35 V is applied across it. Find the resistance of the element. [7 Ω] University of Mines and Technology, Tarkwa 9 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws 2. A 60 W electric light bulb is connected to a 240 V supply. Determine (a) the current flowing in the bulb and (b) the resistance of the bulb. [(a) 0.25 A (b) 960 Ω] 3. Graphs of current against voltage for two resistors P and Q are shown in Figure 2.6. Determine the value of each resistor. [2 mΩ, 5 mΩ] FIGURE 1.10 Graph for Problem 3 4. Determine the pd. which must be applied to a 5 kΩ resistor such that a current of 6 mA may flow. [30 V] 1.11.2 Kirchhoff’s Laws 5. Find currents I 3 , I 4 and I 6 in Figure 1.11. [13 2 A; 14 —1 A; 16 3 A] FIGURE 1.11 Electric circuit for Problem 5 6. For the networks shown in Figure 1.12, find the values of the currents marked. [(a) I 1 =4 A, I 2 =-1 A, I 3 =13 A; (b) I 1 =40 A, I 2 =60 A, I 3 =120 A, I 4 =100 A, I 5 =-80 A] FIGURE 1.12 Electric circuit for Problem 6 University of Mines and Technology, Tarkwa 10 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws 7. Use Kirchhoff’s laws to find the current flowing in the 6 Ω resistor of Figure 1.13 and the power dissipated in the 4 Ω resistor. [2.162 A, 42.07 W] FIGURE 1.13 Electric circuit for Problem 7 8. Find the current flowing in the 3 Ω resistor for the network shown in Figure 1.14. Find also the pd. across the 10 Ω and 2 Ω resistors. [2.715 A, 7.410 V, 3.948 V] FIGURE 1.14 Electric circuit for Problem 8 9. For the networks shown in Figure 1.15 find: (a) the current in the battery, (b) the current in the 300 Ω resistor, (c) the current in the 90 Ω resistor, and (d) the power dissipated in the 150 Ω resistor. [(a) 60.38 mA; (b) 15.10 mA; (c) 45.28 mA; (d) 34.20 mW] FIGURE 1.15 Electric circuit for Problem 9 10. For the bridge network shown in Figure 1.16, find the currents I 1 to I 5. [I 1 = 1.26 A, I 2 = 0.74 A, I 3 = 0.16 A, I 4 = 1.42 A, I 5 = 0.59 A] FIGURE 1.16 Electric circuit for Problem 10 1.11.3 Power and Energy 11. The hot resistance of a 250 V filament lamp is 625 Ω. Determine the current taken by the lamp and its power rating. [0.4 A, 100 W] University of Mines and Technology, Tarkwa 11 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 1 – Circuit Laws 12. Determine the resistance of a coil connected to a 150 V supply when a current of (a) 75 mA (b) 300 μA flows through it. [(a) 2 kΩ (b) 0.5 MΩ] 13. Determine the resistance of an electric fire which takes a current of 12 A from a 240 V supply. Find also the power rating of the fire and the energy used in 20 h. [20 Ω, 2.88 kW, 57.6 kWh] 14. Determine the power dissipated when a current of 10 mA flows through an appliance having a resistance of 8 kΩ. [0.8 W] 15. 85.5 J of energy are converted into heat in nine seconds. What power is dissipated? [9.5 W] 16. A current of 4 A flows through a conductor and 10 W is dissipated. What pd exists across the ends of the conductor? [2.5 V] 17. Find the power dissipated when: (a) a current of 5 mA flows through a resistance of 20 kΩ (b) a voltage of 400 V is applied across a 120 kΩ resistor (c) a voltage applied to a resistor is 10 kV and the current flow is 4 mA. [(a) 0.5 W (b) 1 W (c) 40 W] 18. A battery of emf 15 V supplies a current of 2 A for 5 minutes. How much energy is supplied in this time? [9 kJ] 19. In a household during a particular week three 2 kW fires are used on average 25 h each and eight 100 W light bulbs are used on average 35 hours each. Determine the cost of electricity for the week if 1 unit of electricity costs 7 GHp. [GH¢12.46] 20. Calculate the power dissipated by the element of an electric fire of resistance 30 Ω when a current of 10 A flows in it. If the fire is on for 30 hours in a week, determine the energy used. Determine also the weekly cost of energy if electricity costs 7.2 GHp per unit. [3 kW, 90 kWh, GH¢6.48] University of Mines and Technology, Tarkwa 12 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems CHAPTER TWO CIRCUIT THEOREMS (ANALYSIS METHODS) 2.1 The Branch Current Method In the branch current method, a current is assigned to each branch in an active network. Then Kirchhoff’s current law is applied at the principal nodes and the voltages between the nodes employed to relate the currents. This produces a set of simultaneous equations, which can be solved to obtain the currents. EXAMPLE 2.1 Obtain the current in each branch of the network shown in Figure 2.1 using the branch current method. FIGURE 2.1 Electrical network for Example 2.1 SOLUTION 2.1 Currents I 1 , I 2 , and I 3 are assigned to the branches as shown. Applying KCL at node a, I1 = I 2 + I 3 The voltage, V ab , can be written in terms of the elements in each of the branches; Vab = 20 − I1 (5 ) = I 3 (10 ) and Vab = I 2 (2) + 8. Then the following equations can be written 20 − I1 (5 ) = I 3 (10 ) 20 − I1 (5 ) = I 2 (2) + 8 Solving the three equations simultaneously gives I 1 = 2 A, I 2 = 1 A, and I 3 = 1 A. Other directions may be chosen for the branch currents and the answers will simply include the appropriate sign. In a more complex network, the branch current method is difficult to apply because it does not suggest either a starting point or a logical progression through the network to produce the necessary equations. It also results in more independent equations than either the mesh current or node voltage method requires. 2.2 The Mesh Current Method In the mesh current method a current is assigned to each window of the network such that the currents complete a closed loop. They are sometimes referred to as loop currents. Each University of Mines and Technology, Tarkwa 13 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems element and branch therefore will have an independent current. When a branch has two of the mesh currents, the actual current is given by their algebraic sum. The assigned mesh currents may have either clockwise or counterclockwise directions, although at the outset it is wise to assign to all of the mesh currents a clockwise direction. Once the currents are assigned, Kirchhoff’s voltage law is written for each loop to obtain the necessary simultaneous equations. EXAMPLE 2.2 Obtain the current in each branch of the network shown in Figure 2.2 (same as Figure 2.1) using the mesh current method. FIGURE 2.2 Electrical network for Example 2.2 SOLUTION 2.2 The currents I 1 and I 2 are chosen as shown on the circuit diagram. Applying KVL around the left loop, starting at point α, − 20 + 5 I1 + 10(I1 − I 2 ) = 0 and around the right loop, starting at point β, 8 + 10(I 2 − I1 ) + 2I 2 = 0 Rearranging terms, 15 I1 − 10 I 2 = 20 − 10 I1 + 12I 2 = −8 Solving the equations simultaneously results in I 1 = 2 A and I 2 = 1 A. The current in the centre branch, shown dotted, is I 1 – I 2 = 1 A. In Example 2.1 this was branch current I 3. The currents do not have to be restricted to the windows in order to result in a valid set of simultaneous equations, although that is the usual case with the mesh current method. For example, see Problem 4.6, where each of the currents passes through the source. In that problem, they are called loop currents. The applicable rule is that each element in the network must have a current or a combination of currents and no two elements in different branches can be assigned the same current or the same combination of currents. NOTE: The n simultaneous equations of an n-mesh network can be written in matrix form. The matrix equation arising from the mesh current method may be solved by various techniques. One of these, the method of determinants (Cramer’s rule), will be presented University of Mines and Technology, Tarkwa 14 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems here. It should be stated, however, that other techniques are far more efficient for large networks. 2.3 The Node Voltage Method The network shown in Figure 2.3(a) contains five nodes, where 4 and 5 are simple nodes and 1, 2, and 3 are principal nodes. In the node voltage method, one of the principal nodes is selected as the reference and equations based on KCL are written at the other principal nodes. At each of these other principal nodes, a voltage is assigned, where it is understood that this is a voltage with respect to the reference node. These voltages are the unknowns and, when determined by a suitable method, result in the network solution. FIGURE 2.3 Electrical network showing the positions of simple nodes and principal nodes The network is redrawn in Figure 2.3(b) and node 3 selected as the reference for voltages V 1 and V 2. KCL requires that the total current out of node 1 be zero: V1 − Va V1 V1 − V2 + + =0 RA RB RC Similarly, the total current out of node 2 must be zero: V2 − V1 V2 V2 − Vb + + =0 RC RD RE (Applying KCL in this form does not imply that the actual branch currents all are directed out of either node. Indeed, the current in branch 1–2 is necessarily directed out of one node and into the other.) Putting the two equations for V 1 and V 2 in matrix form,  1 1 1 1  V  R + R + R − RC   V1   a R   A B C    A      =    1 1 1 1    Vb   − + +  V  RC RC RD RE   2   RE  Note the symmetry of the coefficient matrix. The 1,1-element contains the sum of the reciprocals of all resistances connected to node 1; the 2,2-element contains the sum of the University of Mines and Technology, Tarkwa 15 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems reciprocals of all resistances connected to node 2. The 1,2- and 2,1-elements are each equal to the negative of the sum of the reciprocals of the resistances of all branches joining nodes 1 and 2. (There is just one such branch in the present circuit.) On the right-hand side, the current matrix contains Va R A and Vb RE , the driving currents. Both these terms are taken positive because they both drive a current into a node. EXAMPLE 2.3 Solve the circuit of Example 2.2 (Figure 2.2) using the node voltage method. SOLUTION 2.3 The circuit is redrawn in Figure 2.4. With two principal nodes, only one equation is required. FIGURE 2.4 Assuming the currents are all directed out of the upper node and the bottom node is the reference, V1 − 20 V1 V1 − 8 + + =0 5 10 2 from which V1 = 10 V. Then, I1 = (10 − 20 ) 5 = −2 A (the negative sign indicates that current I 1 flows into node 1); I 2 = (10 − 8 ) 2 = 1 A ; I 3 = 10 10 = 1 A. Current I 3 in Example 2.2 is shown dotted (i.e. current flowing through the 10 Ω resistor). 2.4 Superposition Theorem The theorem states that, “A linear network which contains two or more independent sources can be analyzed to obtain the various voltages and branch currents by allowing the sources to act one at a time, then superposing the results.” This principle applies because of the linear relationship between current and voltage. With dependent sources, superposition theorem can be used only when the control functions are external to the network containing the sources, so that the controls are unchanged as the sources act one at a time. Voltage sources to be suppressed while a single source acts are replaced by short circuits; open circuits replace current sources. Superposition theorem cannot be directly applied to the computation of power, because power in an element is proportional to the square of the current or the square of the voltage, which is nonlinear. University of Mines and Technology, Tarkwa 16 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems EXAMPLE 2.4 Compute the current in the 23-Ω resistor of Figure 2.5(a) by applying the superposition principle. FIGURE 2.5 Circuits for Example 2.4 SOLUTION 2.4 With the 200-V source acting alone, the 20-A current source is replaced by an open circuit, Fig. 4-11(b). Req = 47 + (27 )(4 + 23) = 60.5 Ω 54 200 IT = = 3.31 A 60.5  27  ' I 23 Ω =  (3.31) = 1.65 A  54  When the 20-A source acts alone, the 200-V source is replaced by a short circuit, Figure 2.5(c). The equivalent resistance to the left of the source is Req = 4 + (27)(47) = 21.15 Ω 74 Then  21.15  (20 ) = 9.58 A " I 23 Ω =   21.15 + 23  The total current in the 23-Ω resistor is ' " I 23Ω = I 23 Ω + I 23 Ω = 11.23 A University of Mines and Technology, Tarkwa 17 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems 2.5 Thévenin’s and Norton’s Theorems Two-terminal networks are called terminally equivalent if the same current flows into both networks when their terminal voltages are equal, and/or if the same voltage appears across both pairs of terminals when identical currents are forced into both networks. Equivalent resistances of resistors in series or parallel are simple examples of such terminal equivalent circuits. Thévenin and Norton expanded the concept of terminal equivalency to include circuits that also contain sources. They showed that any two-terminal network, such as the one in Figure 2.6a, that contains linear resistors and sources (current, voltage, independent or dependent) has a terminal equivalent circuit of the form of either Figure 2.6b or c. FIGURE 2.6 Thévenin and Norton equivalent circuits Thévenin’s theorem states that, “A linear, active, resistive network which contains one or more voltage or current sources can be replaced by a single voltage source and a series resistance.” The voltage is called the Thévenin equivalent voltage and the resistance is called Thévenin equivalent resistance. Norton’s theorem states that, “A linear, active, resistive network which contains one or more voltage or current sources can be replaced by a single current source and a parallel resistance.” The current the Norton equivalent current and the resistance is called the Norton equivalent resistance. The Norton equivalent resistance is the same as the Thévenin equivalent resistance. When terminals ab in Figure 2.6(a) are open-circuited, a voltage will appear between them. From Figure 2.6(b), it is evident that this must be the voltage, V ′ , of the Thévenin equivalent circuit. If a short circuit is applied to the terminals, as suggested by the dashed line in Figure 2.6(a), a current will result. From Figure 2.6(c) it is evident that this current must be I ′ of the Norton equivalent circuit. Now, if the circuits in (b) and (c) are equivalents of the same active network, they are equivalent to each other. It follows that, I ′ = V ′ R′. If both V ′ and I ′ have been determined from the active network, then R′ = V ′ I ′. The important things to remember about Thévenin and Norton equivalent circuits are: 1. The voltage source in the Thévenin equivalent circuit is the open-circuit voltage. University of Mines and Technology, Tarkwa 18 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems 2. The current source in the Norton equivalent circuit is the short-circuit current. 3. The series resistor in the Thévenin circuit is identical to the parallel resistor in the Norton circuit. Thus, the name output resistance is equivalent to either R Th and/or RN. 4. The open-circuit voltage, the short-circuit current, and the resistance R Th = R N are interrelated by Ohm’s law: V Th =I N R Th 2.6 ∆-Y and Y-∆ Conversions In many circuit applications, we encounter components connected together in one of two ways to form a three-terminal network: the “Delta,” or ∆ (also known as the “Pi,” or π) configuration, and the “Y” (also known as the “T”) configuration as shown in Figure 2.7. FIGURE 2.7 Delta and Wye networks It is possible to calculate the proper values of resistors necessary to form one kind of network (∆ or Y) that behaves identically to the other kind, as analyzed from the terminal connections alone. That is, if we had two separate resistor networks, one ∆ and one Y, each with its resistors hidden from view, with nothing but the three terminals (A, B, and C) exposed for testing, the resistors could be sized for the two networks so that there would be no way to electrically determine one network apart from the other. In other words, equivalent ∆ and Y networks behave identically. There are several equations used to convert one network to the other: University of Mines and Technology, Tarkwa 19 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems To convert a Delta (∆) to a Wye (Y) To convert a Wye (Y) to a Delta (∆) R AB R AC R R + R A RC + RB RC RA = R AB = A B R AB + R AC + RBC RC R AB RBC R A RB + R A RC + RB RC RB = RBC = R AB + R AC + RBC RA R AC RBC R A RB + R A RC + RB RC RC = R AC = R AB + R AC + RBC RB ∆ and Y networks are seen frequently in 3-phase AC power systems, but even then they’re usually balanced networks (all resistors equal in value) and conversion from one to the other need not involve such complex calculations. A prime application for ∆-Y conversion is in the solution of unbalanced bridge circuits, such as the one below: Solution of this circuit with Branch Current or Mesh Current analysis is fairly involved, and the Superposition Theorem is of no help, since it has only one source of power. We could use Thévenin’s or Norton’s Theorem, treating R 3 as our load, but what fun would that be? If we were to treat resistors R 1 , R 2 , and R 3 as being connected in a ∆ configuration (R ab , R ac , and R bc , respectively) and generate an equivalent Y network to replace them, we could turn this bridge circuit into a (simpler) series/parallel combination circuit: Selecting Delta (∆) network to convert: University of Mines and Technology, Tarkwa 20 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems After the ∆-Y conversion … If we perform our calculations correctly, the voltages between points A, B, and C will be the same in the converted circuit as in the original circuit, and we can transfer those values back to the original bridge configuration. RA = (12Ω )(18Ω ) = 216 = 6Ω (12Ω ) + (18Ω ) + (6Ω ) 36 RB = (12Ω )(6Ω ) = 72 = 2Ω (12Ω ) + (18Ω ) + (6Ω ) 36 RC = (18Ω )(6Ω ) = 108 = 3Ω (12Ω ) + (18Ω ) + (6Ω ) 36 Resistors R 4 and R 5 , of course, remain the same at 18 Ω and 12 Ω, respectively. Analyzing the circuit now as a series/parallel combination, we arrive at the following figures: University of Mines and Technology, Tarkwa 21 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems RA RB RC R4 R5 E/Volts 4.118 588.24m 1.176 5.294 4.706 I/Amps 686.27m 294.12m 392.16m 294.12m 392.16m R/Ohms 6 2 3 18 12 RB + R4 RC + R5 (R B +R 4 )//(R C +R 5 ) Total E/Volts 588.2 5.882 5.882 10 I/Amps 294.12 392.16m 686.27M 686.27m R/Ohms 20 15 8.571 14.571 We must use the voltage drops figures from the table above to determine the voltages between points A, B, and C, seeing how the add up (or subtract, as is the case with voltage between points B and C): E A − B = 4.706V E A − C = 5.294 V EB − C = 588.24mV Now that we know these voltages, we can transfer them to the same points A, B, and C in the original bridge circuit: University of Mines and Technology, Tarkwa 22 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems Voltage drops across R 4 and R 5 , of course, are exactly the same as they were in the converted circuit. At this point, we could take these voltages and determine resistor currents through the repeated use of Ohm’s Law (I=E/R): 4.706V 5.294V I R1 = = 392.16mA IR2 = = 294.12mA 12Ω 18Ω 588.24mV 5.294V I R3 = = 98.04mA I R4 = = 294.12mA 6Ω 18Ω 4.706V I R5 = = 392.16mA 12Ω 2.7 Problems 2.7.1 Superposition Theorem 1. Use the superposition theorem to find currents I 1 , I 2 and I 3 of Figure 2.8(a). [I 1 = 2 A, I 2 = 3 A, I 3 = 5 A] 2. Use the superposition theorem to find the current in the 8 Ω resistor of Figure 2.7(b). [0.385 A] 3. Use the superposition theorem to find the current in each branch of the network shown in Figure 2.8(c). [10 V battery discharges at 1.429 A 4 V battery charges at 0.857 A Current through 10 Ω resistor is 0.572 A] 4. Use the superposition theorem to determine the current in each branch of the arrangement shown in Figure 2.8(d). [24 V battery charges at 1.664 A 52 V battery discharges at 3.280 A Current in 20 Ω resistor is 1.616 A] FIGURE 2.8 Electric circuits for Problems 1 – 4 University of Mines and Technology, Tarkwa 23 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 2 – Circuit Theorems 2.7.2 Thévenin’s Theorem 5. Use Thévenin’s theorem to find the current flowing in the 14 Ω resistor of the network shown in Figure 2.9. Find also the power dissipated in the 14 Ω resistor. [0.434 A, 2.64 W] FIGURE 2.9 Electric circuit for Problem 5 6. Use Thévenin’s theorem to find the current flowing in the 6 Ω resistor shown in Figure 2.10 and the power dissipated in the 4 Ω resistor. [2.162 A, 42.07 W] FIGURE 2.10 Electric circuit for Problem 6 7. Repeat problems 1 – 4 using Thévenin’s theorem. 8. In the network shown in Figure 2.11, the battery has negligible internal resistance. Find, using Thévenin’s theorem, the current flowing in the 4 Ω resistor. [0.918 A] FIGURE 2.11 Electric circuit for Problem 8 9. For the bridge network shown in Figure 2.12, find the current in the 5 Ω resistor, and its direction, by using Thévenin’s theorem. [0.153 A from B to A] FIGURE 2.12 Electric circuit for Problem 9 2.7.3 Norton’s Theorem 10. Repeat problems 1 – 6, 8 and 9 using Norton’s theorem. University of Mines and Technology, Tarkwa 24 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance CHAPTER THREE CAPACITORS AND CAPACITANCE 3.1 Electrostatic Field Figure 3.1 represents two parallel metal plates, A and B, charged to different potentials. If an electron that has a negative charge is placed between the plates, a force will act on the electron tending to push it away from the negative plate B towards the positive plate, A. Similarly, a positive charge would be acted on by a force tending to move it toward the negative plate. Any region such as that shown between the plates in Figure 3.1, in which an electric charge experiences a force, is called an electrostatic field. The direction of the field is defined as that of the force acting on a positive charge placed in the field. In Figure 3.1, the direction of the force is from the positive plate to the negative plate. Figure 3.1 Electrostatic field Such a field may be represented in magnitude and direction by lines of electric force drawn between the charged surfaces. The closeness of the lines is an indication of the field strength. Whenever a pd is established between two points, an electric field will always exist. Figure 3.2(a) shows a typical field pattern for an isolated point charge, and Figure 3.2(b) shows the field pattern for adjacent charges of opposite polarity. Electric lines of force (often called electric flux lines) are continuous and start and finish on point charges. Also, the lines cannot cross each other. When a charged body is placed close to an uncharged body, an induced charge of opposite sign appears on the surface of the uncharged body. This is because lines of force from the charged body terminate on its surface. FIGURE 3.2 (a) Isolated point charge; (b) adjacent charges of opposite polarity The concept of field lines or lines of force is used to illustrate the properties of an electric field. However, it should be remembered that they are only aids to the imagination. The force of attraction or repulsion between two electrically charged bodies is proportional to the magnitude of their charges and inversely proportional to the square of the distance separating them, University of Mines and Technology, Tarkwa 25 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance q1q2 i.e. force ∝ 2 or where constant k ≈ 9 × 10 9 in air d This is known as Coulomb’s law. Hence the force between two charged spheres in air with their centres 16 mm apart and each carrying a charge of +1.6 μC is given by: )( )2 1.6 × 10 − 6 qq ( force = k 1 2 2 ≈ 9 × 10 9 = 90 newtons d ( 16 × 10 − 3 )2 3.2 Electric field strength Figure 3.3 shows two parallel conducting plates separated from each other by air. They are connected to opposite terminals of a battery of voltage V volts. FIGURE 3.3 Circuit shows parallel conducting plates separated from each other by air There is therefore an electric field in the space between the plates. If the plates are close together, the electric lines of force will be straight and parallel and equally spaced, except near the edge where fringing will occur (see Figure 3.1). Over the area in which there is negligible fringing, V Electric field strength, E = volt/metre d where d is the distance between the plates. Electric field strength is also called potential gradient. 3.3 Capacitance Static electric fields arise from electric charges, electric field lines beginning and ending on electric charges. Thus the presence of the field indicates the presence of equal positive and negative electric charges on the two plates of Figure 3.3. Let the charge be +Q coulombs on one plate and –Q coulombs on the other. The property of this pair of plates which determines how much charge corresponds to a given pd between the plates is called their capacitance: Q capacitance, C = V University of Mines and Technology, Tarkwa 26 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance The unit of capacitance is the farad F (or more usually μF = 10-6 F or pF = 10-12 F), which is defined as the capacitance when a pd of one volt appears across the plates when charged with one coulomb. 3.4 Capacitors Every system of electrical conductors possesses capacitance. For example, there is capacitance between the conductors of overhead transmission lines and also between the wires of a telephone cable. In these examples the capacitance is undesirable but has to be accepted, minimized or compensated for. There are other situations where capacitance is a desirable property. Devices specially constructed to possess capacitance are called capacitors (or condensers, as they used to be called). In its simplest form a capacitor consists of two plates which are separated by an insulating material known as a dielectric. A capacitor has the ability to store a quantity of static electricity. The symbols for a fixed capacitor and a variable capacitor used in electrical circuit diagrams are shown in Figure 3.4. FIGURE 3.4 Symbols for capacitor The charge, Q, stored in a capacitor is given by: Q = I × t coulombs where I is the current in amperes and t the time in seconds. 3.5 Electric Flux Density Unit flux is deified as emanating from a positive charge of 1 coulomb. Thus electric flux ψ is measured in coulombs, and for a charge of Q coulombs, the flux ψ = Q coulombs. Electric flux density D is the amount of flux passing through a defined area A that is perpendicular to the direction of the flux: Q electric flux density, D = coulombs/metre 2 A Electric flux density is also called charge density, σ. University of Mines and Technology, Tarkwa 27 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance 3.6 Permittivity At any point in an electric field, the electric field strength E maintains the electric flux and produces a particular value of electric flux density D at that point. For a field established in vacuum (or for practical purposes in air), the ratio D/E is a constant so, i.e. D = ε0 E where o is called the permittivity of free space or the free space constant. The value of ε 0 is 8.85 × 10 −12 F/m. When an insulating medium, such as rnica, paper, plastic or ceramic, is introduced into the region of an electric field the ratio of D/E is modified: D = ε 0ε r E where ε r , the relative permittivity of the insulating material, indicates its insulating power compared with that of vacuum: flux density in material relative permittivity, ε r = flux density in vacuum ε r has no unit. Typical values of ε r include air, 1.00; polythene, 2.3; rnica, 3 – 7; glass, 5 – 10; water, 80; ceramics, 6 – 1000. The product is called the absolute permittivity, ε , i.e., ε = ε 0ε r The insulating medium separating charged surfaces is called a dielectric. Compared with conductors, dielectric materials have very high resistivities. They are therefore used to separate conductors at different potentials, such as capacitor plates or electric power lines. 3.7 The Parallel Plate Capacitor For a parallel-plate capacitor, as shown in Figure 3.5(a), experiments show that capacitance C is proportional to the area A of a plate, inversely proportional to the plate spacing d (i.e., the dielectric thickness) and depends on the nature of the dielectric: ε 0ε r A Capacitance, C = farads d where ε 0 = 8.85 × 10 −12 F/m (constant) ε r = relative permittivity A = area of one of the plates, in rn2, and University of Mines and Technology, Tarkwa 28 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance d = thickness of dielectric in m FIGURE 3.5 Parallel plate capacitor Another method used to increase the capacitance is to interleave several plates as shown in Figure 6.5(b). Ten plates are shown, forming nine capacitors with a capacitance nine times that of one pair of plates. If such an arrangement has n plates then capacitance C ∝ (n − 1). ε 0ε r A(n − 1) Thus capacitance, C = farads d 3.8 Capacitors Connected in Parallel and Series 3.8.1 Capacitors Connected in Parallel Figure 3.6 shows three capacitors, C 1 , C 2 and C 3 , connected in parallel with a supply voltage V applied across the arrangement. FIGURE 3.6 Circuit showing capacitors connected in parallel When the charging current I reaches point A it divides, some flowing into C 1 , some flowing into C 2 and some into C 3. Hence the total charge Q T (= I × t) is divided between the three capacitors. The capacitors each store a charge and these are shown as Q 1 , Q 2 and Q 3 respectively. Hence QT = Q1 + Q2 + Q3 But Q T = CV, Q 1 = C 1 V, Q 2 = C 2 V and Q 3 = C 3 V. Therefore CV = C 1 V +C 2 V + C 3 V where C is the total equivalent circuit capacitance, University of Mines and Technology, Tarkwa 29 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance That is C = C 1 +C 2 + C 3 It follows that for n parallel-connected capacitors, C = C 1 +C 2 + … + C n i.e. the equivalent capacitance of a group of parallel-connected capacitors is the sum of the capacitances of the individual capacitors. (Note that this formula is similar to that used for resistors connected in series). 3.8.2 Capacitors Connected in Series Figure 3.7 shows three capacitors, C 1 , C 2 and C 3 , connected in series across a supply voltage V. Let the pd across the individual capacitors be V 1 , V 2 and V 3 respectively as shown. FIGURE 3.7 Circuit showing three capacitors arranged in series Let the charge on plate ‘a’ of capacitor Ci be +Q coulombs. This induces an equal but opposite charge of – Q coulombs on plate ‘b’. The conductor between plates ‘b’ and ‘c’ is electrically isolated from the rest of the circuit so that an equal but opposite charge of +Q coulombs must appear on plate ‘c’, which, in turn, induces an equal and opposite charge of – Q coulombs on plate ‘d’, and so on. Hence when capacitors are connected in series the charge on each is the same. In a series circuit: V = V1 + V2 + V3 Q Q Q Q Q Since V = then = + + C C C1 C2 C3 where C is the total equivalent circuit capacitance, 1 1 1 1 i.e., = + + C C1 C2 C3 It follows that for n series-connected capacitors: University of Mines and Technology, Tarkwa 30 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance 1 1 1 1 1 = + + +... + C C1 C2 C3 Cn i.e. for series-connected capacitors, the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. (Note that this formula is similar to that used for resistors connected in parallel) For the special case of two capacitors in series: 1 1 1 C + C2 = + = 1 C C1 C2 C1C2 C1C2 product Hence, C = , i.e. C1 + C2 sum 3.9 Dielectric Strength The maximum amount of field strength that a dielectric can withstand is called the dielectric strength of the material. Vm Dielectric strength, Em = d 3.10 Energy Stored The energy, W, stored by a capacitor is given by 1 W= CV 2 joules 2 Energy Note: Power = time 3.11 Practical Types of Capacitor Practical types of capacitor are characterized by the material used for their dielectric. The main types include: variable air, mica, paper, ceramic, plastic, titanium oxide and electrolytic. 3.11.1 Variable Air Capacitors These usually consist of two sets of metal plates (such as aluminium) one fixed, the other variable. The set of moving plates rotate on a spindle as shown by the end view of Figure 3.9. University of Mines and Technology, Tarkwa 31 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance FIGURE 3.9 Variable air capacitor As the moving plates are rotated through half a revolution, the meshing, and therefore the capacitance, varies from a minimum to a maximum value. Variable air capacitors are used in radio and electronic circuits where very low losses are required, or where a variable capacitance is needed. The maximum value of such capacitors is between 500 pF and 1000 pF. 3.11.2 Mica Capacitors A typical older type construction is shown in Figure 3.10. FIGURE 3.10 Mica capacitor Usually the whole capacitor is impregnated with wax and placed in a bakelite case. Mica is easily obtained in thin sheets and is a good insulator. However, mica is expensive and is not used in capacitors above about 0.2 μF. A modified form of mica capacitor is the silvered mica type. The mica is coated on both sides with a thin layer of silver which forms the plates. Capacitance is stable and less likely to change with age. Such capacitors have a constant capacitance with change of temperature, a high working voltage rating and a long service life and are used in high frequency circuits with fixed values of capacitance up to about 1000 pF. 3.11.3 Paper Capacitors A typical paper capacitor is shown in Figure 3.11 where the length of the roll corresponds to the capacitance required. The whole is usually impregnated with oil or wax to exclude moisture, and then placed in a plastic or aluminium container for protection. Paper capacitors are made in various working voltages up to about 150 kV and are used where loss is not very important. The maximum value of this type of capacitor is between 500 pF and 10 μF. Disadvantages of paper capacitors include variation in capacitance with temperature change and a shorter service life than most other types of capacitor. 3.11.4 Ceramic Capacitors These are made in various forms, each type of construction depending on the value of capacitance required. For high values, a tube of ceramic material is used as shown in the cross section of Figure 3.12. For smaller values the cup construction is used as shown in Figure 3.13, and for still smaller values the disc construction shown in Figure 3.14 is used. Certain ceramic materials have a very high permittivity and this enables capacitors of high University of Mines and Technology, Tarkwa 32 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance capacitance to be made which are of small physical size with a high working voltage rating. Ceramic capacitors are available in the range 1 pF to 0.1 μF may be used in high frequency electronic circuits subject to a wide range of temperatures. FIGURE 3.11 Paper capacitor FIGURE 3.12 Ceramic capacitor FIGURE 3.13 Cup construction of ceramic capacitors FIGURE 3.14 Disc construction of ceramic capacitors 3.11.5 Plastic Capacitors Some plastic materials such as polystyrene and Teflon can be used as dielectrics. Construction is similar to the paper capacitor but using a plastic film instead of paper. Plastic capacitors operate well under conditions of high temperature, provide a precise value of capacitance, a very long service life and high reliability. University of Mines and Technology, Tarkwa 33 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance 3.11.6 Titanium Oxide Capacitors These capacitors have a very high capacitance with a small physical size when used at a low temperature. 3.11.7 Electrolytic Capacitors Construction is similar to the paper capacitor with aluminium foil used for the plates and with a thick absorbent material, such as paper, impregnated with an electrolyte (ammonium borate), separating the plates. The finished capacitor is usually assembled in an aluminium container and hermetically sealed. Its operation depends on the formation of a thin aluminium oxide layer on the positive plate by electrolytic action when a suitable direct potential is maintained between the plates. This oxide layer is very thin and forms the dielectric. (The absorbent paper between the plates is a conductor and does not act as a dielectric.) Such capacitors must always be used on dc and must be connected with the correct polarity; if this is not done the capacitor will be destroyed since the oxide layer will be destroyed. Electrolytic capacitors are manufactured with working voltage from 6 V to 600 V, although accuracy is generally not very high. These capacitors possess a much larger capacitance than other types of capacitors of similar dimensions due to the oxide film being only a few microns thick. The fact that they can be used only on dc supplies limit their usefulness. 3.12 Discharging Capacitors When a capacitor has been disconnected from the supply it may still be charged and it may retain this charge for some considerable time. Thus precautions must be taken to ensure that the capacitor is automatically discharged after the supply is switched off. This is done by connecting a high value resistor across the capacitor terminals. 3.13 Problems Where appropriate, take ε 0 as 8.85 × 10 −12 F/m. 3.13.1 Charge and Capacitance 1. Find the charge on a 10 μF capacitor when the applied voltage is 250 V. [2.5 mC] 2. Determine the voltage across a 1000 pF capacitor to charge it with 2 μC. [2 kV] 3. The charge on the plates of a capacitor is 6 mC when the potential between them is 2.4 kV. Determine the capacitance of the capacitor. [2.5 μF] 4. For how long must a charging current of 2 A be fed to a 5 μF capacitor to raise the pd between its plates by 500 V. [1.25 ms] 5. A steady current of 10 A flows into a previously uncharged capacitor for 1.5 ms when the pd between the plates is 2 kV. Find the capacitance of the capacitor. [7.5 μF] University of Mines and Technology, Tarkwa 34 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance 3.13.2 Electric Field Strength, Electric Flux Density and Permittivity 6. A capacitor uses a dielectric 0.04 mm thick and operates at 30 V. What is the electric field strength across the dielectric at this voltage? [750 kV/m] 7. A two-plate capacitor has a charge of 25 C. If the effective area of each plate is 5 cm2 find the electric flux density of the electric field. [50 kC/m2] 8. A charge of 1.5 μC is carried on two parallel rectangular plates each measuring 60 mm by 80 mm. Calculate the electric flux density. If the plates are spaced 10 mm apart and the voltage between them is 0.5 kV determine the electric field strength. [312.5 μC/m2, 50 kV/m] 9. The electric flux density between two plates separated by polystyrene of relative permittivity 2.5 is 5 μC/m2. Find the voltage gradient between the plates. [226 kV/m] 10. Two parallel plates having a pd of 250 V between them are spaced 1 mm apart. Determine the electric field strength. Find also the electric flux density when the dielectric between the plates is (a) air and (b) mica of relative permittivity 5. [250 kV/m (a) 2.213 μC/m2 (b) 11.063 μC/m2] 3.13.3 Parallel Plate Capacitor 11. A capacitor consists of two parallel plates each of area 0.01 m2, spaced 0.1 mm in air. Calculate the capacitance in picofarads. [885 pF] 12. A waxed paper capacitor has two parallel plates, each of effective area 0.2 m2. If the capacitance is 4000 pF determine the effective thickness of the paper if its relative permittivity is 2. [0.885 mm] 13. Calculate the capacitance of a parallel plate capacitor having 5 plates, each 30 mm by 20 mm and separated by a dielectric 0.75 mm thick having a relative permittivity of 2.3. [65.14 pF] 14. How many plates has a parallel plate capacitor having a capacitance of 5 nF, if each plate is 40 mm by 40 mm and each dielectric is 0.102 mm thick with a relative permittivity of 6. 15. A parallel plate capacitor is made from 25 plates, each 70mm by 120 mm interleaved with mica of relative permittivity 5. If the capacitance of the capacitor is 3000 pF determine the thickness of the mica sheet. [2.97 mm] 16. The capacitance of a parallel plate capacitor is 1000 pF. It has 19 plates, each 50 mm by 30 mm separated by a dielectric of thickness 0.40 mm. Determine the relative permittivity of the dielectric. [1.67] 17. A capacitor is to be constructed so that its capacitance is 4250 pF and to operate at a pd of 100 V across its terminals. The dielectric is to be polythene ( ε r = 2.3 ) which, after allowing a safety factor, has a dielectric strength of 20 MV/m. Find (a) the thickness of the polythene needed, and (b) the area of a plate. [(a) 0.005 mm (b) 10.44 cm2] University of Mines and Technology, Tarkwa 35 Solomon Nunoo MPhil, BSc, MIET, MASEE, MIAENG Applied Electricity Chapter 3 – Capacitors and Capacitance 3.13.4 Capacitors in Parallel and Series 18. Capacitors of 2 μF and 6 μF are connected (a) in parallel and (b) in series. Determine the equivalent capacitance in each case. [(a) 8 iF (b) 1.5 iF] 19. Find the capacitance to be connected in series with a 10 iF capacitor for the equivalent capacitance to be 6 F. [15 μF] 20. Two 6 μF capacitors are connected in series with one having a capacitance of 12 μF. Find the total equivalent circuit capacitance. What capacitance must be added in series to obtain a capacitance of 1.2 μF? [2.4 μF, 2.4 μF] 21. Determine the equivalent capacitance when the following capacitors are connected (a) in parallel and (b) in series: i. 2 μF, 4 μF and 8μF ii. 0.02 μF, 0.05 μF and 0.10 μF iii. 50 pF and 450 pF iv. 0.01 μF and 200 pF [(a) (i) 14 μF (ii) 0.17 μF (iii) 500 pF (iv) 0.0102 μF] [(b) (i) 1 71 μF (ii) 0.0125 μF (iii) 45 pF (iv) 196.1 pF] 22. For the arrangement shown in Figure 3.15 find (a) the equivalent circuit capacitance and (b) the voltage across a 4.5 μF capacitor. [(a) 1.2 μF (b) 100 V] FIGURE 3.15 Circuit diagram for Problem 22 23. Three 12 μF capacitors are connected in series across a 750 V supply. Calculate (a) the equivalent capacitance, (b) the charge on each capacitor and (c) the pd across each capacitor. [(a) 4 μF (b) 3 mC (c) 250 V] 24. If two capacitors having capacitances of 3 μF and 5 μF respectively are connected in series across a 240 V supply, determine (a) the pd across each capacitor and (b) the charge on each capacitor. [(a) 150 V, 90 V (b) 0.45 mC on each] 25. In Figure 3.16 capacitors P, Q and R are identical and the total equivalent capacitance of the circuit is 3 μF. Determine the values of P, Q and R. [4.2 F each] 3.13.5 Energy Stored 26. When a capacitor is connected across a 200 V su

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