Basic Engineering Circuit Analysis - Irwin Solutions

Basic Engineering Circuit Analysis - Irwin Solutions -...

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Unformatted text preview: Chapter One: Basic Concepts 2 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter One: Basic Concepts 3 SOLUTION: 4 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 5 SOLUTION: 6 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 7 SOLUTION: 8 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 9 SOLUTION: 10 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 11 SOLUTION: 12 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 13 SOLUTION: 14 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 15 SOLUTION: 16 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 17 SOLUTION: 18 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 19 SOLUTION: 20 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 21 SOLUTION: 22 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 23 SOLUTION: 24 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 25 SOLUTION: Continued on next page. 26 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter One: Basic Concepts 27 SOLUTION: Continued on next page. 28 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter One: Basic Concepts 29 SOLUTION: 30 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 31 SOLUTION: 32 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 33 SOLUTION: 34 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter One: Basic Concepts 35 SOLUTION: 36 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter One: Basic Concepts 37 SOLUTION: Chapter Two: Resistive Circuits 46 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter Two: Resistive Circuits 39 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 41 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 43 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 45 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 47 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 49 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 51 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 53 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 55 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 57 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 59 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 61 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 63 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 65 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 67 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 69 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Continued on the next page. Chapter Two: Resistive Circuits 71 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 73 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 75 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 77 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 79 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 81 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 83 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 85 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 87 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 89 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 91 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 93 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 95 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 97 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 99 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 101 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 103 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 105 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 107 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 109 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 111 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 113 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 115 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 117 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 119 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 121 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 123 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 125 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 127 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 129 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 131 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 133 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 135 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 137 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 139 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 141 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 143 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 145 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 147 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 149 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 151 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 153 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 155 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 157 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 159 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Two: Resistive Circuits 161 SOLUTION: 46 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques Chapter Three: Nodal and Loop Analysis Techniques 163 164 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 165 SOLUTION: 166 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 167 SOLUTION: Continued on the next page. 168 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter Three: Nodal and Loop Analysis Techniques 169 SOLUTION: 170 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 171 SOLUTION: 172 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 173 SOLUTION: 174 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 175 SOLUTION: Continued on the next page. 176 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter Three: Nodal and Loop Analysis Techniques 177 SOLUTION: 178 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 179 SOLUTION: 180 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 181 SOLUTION: 182 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 183 SOLUTION: 184 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 185 SOLUTION: 186 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 187 SOLUTION: 188 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 189 SOLUTION: 190 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 191 SOLUTION: 192 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 193 SOLUTION: 194 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 195 SOLUTION: 196 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 197 SOLUTION: 198 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 199 SOLUTION: 200 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 201 SOLUTION: Continued on the next page. 202 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter Three: Nodal and Loop Analysis Techniques 203 SOLUTION: 204 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 205 SOLUTION: 206 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 207 SOLUTION: 208 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 209 SOLUTION: 210 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 211 SOLUTION: 212 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 213 SOLUTION: 214 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 215 SOLUTION: 216 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 217 SOLUTION: 218 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 219 SOLUTION: 220 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 221 SOLUTION: 222 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 223 SOLUTION: 224 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 225 SOLUTION: 226 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Continued on the next page. Chapter Three: Nodal and Loop Analysis Techniques 227 228 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 229 SOLUTION: 230 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 231 SOLUTION: 232 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter Three: Nodal and Loop Analysis Techniques 233 SOLUTION: 234 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 235 SOLUTION: 236 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 237 SOLUTION: 238 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 239 SOLUTION: Continued on the next page. 240 Irwin, Basic Engineering Circuit Analysis, 8/E Chapter Three: Nodal and Loop Analysis Techniques 241 SOLUTION: 242 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 243 SOLUTION: 244 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 245 SOLUTION: 246 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 247 SOLUTION: 248 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 249 SOLUTION: 250 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Continued on the next page. Chapter Three: Nodal and Loop Analysis Techniques 251 252 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 253 SOLUTION: 254 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 255 SOLUTION: 256 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 257 SOLUTION: 258 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 259 SOLUTION: 260 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 261 SOLUTION: 262 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 263 SOLUTION: 264 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 265 SOLUTION: 266 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION: Chapter Three: Nodal and Loop Analysis Techniques 267 SOLUTION: 268 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION Chapter Three: Nodal and Loop Analysis Techniques 269 SOLUTION 270 Irwin, Basic Engineering Circuit Analysis, 8/E SOLUTION Chapter Three: Nodal and Loop Analysis Techniques 271 ______________________________________________________________________ SOLUTION Problem 10.1 Sketch a phasor representation of a balanced three-phase system containing both phase voltages and line voltages if Van = 10045 V rms . Label all magnitudes and assume an abc-phase sequence. Suggested Solution Phase voltages: Van = 10045 V rms Vbn = 100 - 75 V rms Vcn = 100 - 195 V rms Line voltages: Magnitude = 3 phase voltage magnitude Angle = phase voltage angle +30 Vab = 173.275 V rms Vbc = 173.2 - 45 V rms Vca = 173.2 - 165 V rms Imaginary Vab = 173.275 V rms Van = 10045 V rms Vcn = 100-195 V rms Real Vca = 173.2-165 V rms Vbn = 100- 75 V rms Vbc = 173.2- 45 V rms Problem 10.2 A positive-sequence three-phase balanced wye voltage source has a phase voltage of Van = 10020 V . Determine the line voltages of the source. Suggested Solution Vab leads Van by 30 . Therefore, Vab = 100 350 = 173.250 V rms , Vbc = 173.2 - 70 V rms , and Vca = 173.2 - 190 V rms . Problem 10.3 Sketch a phasor representation of an abc-sequence balanced three-phase -connected source, including Vab , Vbc , and Vca if Vab = 2080 V rms . Suggested Solution Vab = 2080 V rms Vbc = 208 - 120 V rms Vca = 208120 V rms Imaginary Vca = 208120 V rms Vab = 2080 V rms Real Vbc = 208-120 V rms Problem 10.4 Sketch a phasor representation of a balanced three-phase system containing both phase voltages and line voltages if Van = 12060 V rms . Label all magnitudes and assume an abc-phase sequence. Suggested Solution Van = 12060 V rms Vbn = 120 - 60 V rms Vcn = 120180 V rms Vab = 20890 V rms Vbc = 208 - 30 V rms Vca = 208 - 150 V rms Imaginary Vab = 20890 V rms Van = 12060 V rms Vcn = 120180 V rms Real Vbc = 208- 30 V rms Vca = 208-150 V rms Vbn = 120- 60 V rms Problem 10.5 A positive-sequence three-phase balanced wye voltage source Van = 24090 V rms . Determine the line voltages of the source. has a phase voltage of Suggested Solution Vab = 3 Van Van + 30 = 415.7120 V rms Vbc = 415.70 V rms Vca = 415.7 - 120 V rms ( ) Problem 10.6 Sketch a phasor representation of a balanced three-phase system containing both phase voltages and line voltages if Vab = 20845 V rms . Label all phasors and assume an abc-phase sequence. Suggested Solution Vab = 20845 V rms Vbc = 208 - 75 V rms Vca = 208165 V rms Van = Vab 3 Vab - 30 = 12015 V rms ( ) Vbn = 120 - 105 V rms Vcn = 120135 V rms Imaginary Vab Vcn Vca Van Real Vbn Vbc Problem 10.7 A positive-sequence balanced three-phase wye-connected source with a phase voltage of 100 V supplies power to a balanced wye-connected load. The per phase load impedance is 40 + j10 . Determine the line currents in the circuit if Van = 0 . Suggested Solution I an = Van 100 + j 0 = = 2.43 - 14 A rms Z 40 + j10 I bn = 2.43 - 134 A rms I cn = 2.43 - 254 A rms Problem 10.8 A positive-sequence balanced three-phase wye-connected source supplies power to a balanced wyeconnected load. The magnitude of the line voltages is 150 V. If the load impedance per phase is 36 + j12 , determine the line currents if Van = 0 . Suggested Solution I an 36 Van j12 Van = 150 3 0 = 86.60 V rms I an = 86.6 + j 0 = 2.28 - 18.43 A rms 36 + j12 I bn = 2.28 - 138.43 A rms I cn = 2.28 - 258.43 A rms Problem 10.9 An abc-sequence balanced three-phase wye-connected source supplies power to a balanced wye-connected load. The line impedance per phase is 1 + j 0 , and the load impedance per phase is 20 + j 20 . If the source line voltage Vab is 1000 V rms , find the line currents. Suggested Solution Van = 100 3 ( 0 - 30 ) = 57.74 - 30 V rms I aA = 57.74 - 30 57.74 - 30 = = 1.99 - 73.6 A rms 21 + j 20 ( 20 + j 20 ) + (1 + j 0 ) I bB = 1.99 - 193.6 A rms I cC = 1.99 - 313.6 A rms Problem 10.10 In a three-phase balanced wye-wye system, the source is an abc-sequence set of voltages with Van = 12060 V rms . The per phase impedance of the load is 12 + j16 . If the line impedance per phase is 0.8 + j1.4 , find the line currents and the load voltages. Suggested Solution Line currents: I AN = Van 12060 12060 = = = 5.566.3 A rms Z + Z L (12 + j16 ) + ( 0.8 + j1.4 ) 21.653.7 I BN = 5.56 - 113.7 A rms I CN = 5.56126.3 A rms Load voltages: VAN = Z Z + Z L Van = 12 + j16 (12060 ) = 111.159.4 V rms 12.8 + j17.4 VBN = 111.1 - 60.6 V rms VCN = 111.1179.4 V rms Problem 10.11 An abc-sequence set of voltages feeds a balanced three-phase wye-wye system. The line and load impedances are 0.6 + j1 and 18 + j14 , respectively. If the load voltage on the a phase is VAN = 114.4718.99 V rms , determine the voltages at the line input. Suggested Solution Ia = VAN = 5.02 - 18.88 A rms 18 + j14 Van = I a ( Z line + Zload ) = I a Z line + VAN = ( 5.02 - 18.88 )( 0.6 + j1) + 114.4718.99 = 12020 V rms Vbn = 120 - 100 V rms Vcn = 120 - 220 V rms Problem 10.12 In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages. The load voltage on the a phase is VAN = 108.5879.81 V rms , Z line = 1 + j1.4 , and Z load = 10 + j13 . Determine the input sequence of voltages. Suggested Solution Ia = VAN = 6.6227.38 A rms Zload Van = I a Z line + VAN = I a ( Zline + Zload ) = 12080 V rms Vbn = 120 - 40 V rms Vcn = 120 - 160 V rms Problem 10.13 A balanced abc-sequence of voltages feeds a balanced three-phase wye-wye system. The line and load impedances are 0.6 + j 0.9 and 8 + j12 , respectively. The load voltage on the a phase is VAN = 116.6310 V rms . Find the line voltage Vab . Suggested Solution Van = VAN Zline + Zload 8.6 + j12.9 = (116.6310 ) = 125.510 V rms 8 + j12 Zload Vab = 3 Van Van + 30 = 217.440 V rms ( ) Problem 10.14 In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages. Zline = 1 + j1.8 , Zload = 14 + j12 , and the load voltage on the a phase is VAN = 398.117.99 V rms . Find the line voltage Vab . Suggested Solution Ia = VAN 398.117.99 = = 21.59 - 22.61 A rms 14 + j12 Zload Van = I a ( Z line + Zload ) = ( 21.59 - 22.61 )(15 + j13.8 ) = 44020 V rms Vab = 440 350 = 762.150 V rms Problem 10.15 An abc-phase sequence balanced three-phase source feeds a balanced load. The system is connected wyewye and Van = 0 . The line impedance is 0.5 + j 0.2 , the load impedance is 16 + j10 , and the total power absorbed by the load is 1836.54 W. Determine the magnitude of the source voltage Van . Suggested Solution P ,load = 1836.54 = 612.18 W 3 Z L = RL + jX L = 16 + j10 2 I aA = 612.18 612.18 = = 38.26 W 16 RL I aA = 6.19 A rms The total impedance is 16.5 + j10.2 = 19.431.72 Therefore, since Van = 0 , I aA = 6.19 - 31.72 A rms and Van = 6.19 19.4 = 120 V rms Problem 10.16 In a balanced three-phase wye-wye system, the total power loss in the lines is 272.57 W. Van = 105.2831.65 V rms and the power factor of the load is 0.77 lagging. If the line impedance is 2 + j1 , determine the load impedance. Suggested Solution If the total line loss is 272.57 W, then the loss per phase is 272.57 = 90.86 W 3 a I aA 2+j ZL A Van n N 2 2 P ,loss = I aA Re {Zline } = 2 I aA = 90.86 W I aA = 6.74 A rms Then, ZL = 105.28 = 15.62 6.74 Z = cos -1 ( 0.77 ) = 39.8 L (lagging pf) So, Z L = 15.6239.8 = 12 + j10 Problem 10.17 In a balanced three-phase wye-wye system the load impedance is 8 + j 4 . The source has phase sequence abc and Van = 1200 V rms . If the load voltage is VAN = 111.62 - 1.33 V rms , determine the line impedance. Suggested Solution I aA = 111.62 - 1.33 = 12.48 - 27.9 A rms 8 + j4 Vline = 1200 - 111.62 - 1.33 = 8.817.12 V rms Zline = 8.817.12 = 0.7145.02 = 0.5 + j 0.5 12.48 - 27.9 Problem 10.18 In a balanced three-phase wye-wye system the load impedance is 10 + j1 . The source has phase sequence abc and the line voltage Vab = 22030 V rms. If the load voltage VAN = 1200 V rms, determine the line impedance. Suggested Solution Z Y = 10 + j1 Van = Vab 3 Vab - 30 = ( ) 220 3 ( 30 - 30 ) = 127.020 V rms Per phase Y circuit: a Rline jXline 10 Van j1 n N A ZY VAN = Van Z Y + Zline V 127.020 - 1 = 0.595.71 Zline = Z Y an - 1 = (10 + j1) VAN 1200 Problem 10.19 In a balanced three-phase wye-wye system, the load impedance is 20 + j12 . The source has an abcphase sequence and Van = 1200 V rms. If the load voltage is VAN = 111.49 - 0.2 V rms, determine the magnitude of the line current if the load is suddenly short circuited. Suggested Solution I aA = 111.49 - 0.2 = 4.78 - 31.16 A rms 20 + j12 Vline = 1200 - 111.49 - 0.2 = 8.522.62 V rms Zline = Vline 8.522.62 = = 1.7833.78 4.78 - 31.16 I aA I aASC = 120 = 67.42 A rms 1.78 Problem 10.20 In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages and Van = 12050 V rms. The load voltage on the a phase is 110.6529.03 V rms and the load impedance is 16 + j 20 . Find the line impedance. Suggested Solution a I aA Zline ZL A Van n N Van = 12050 V rms Z L = 16 + j 20 = 25.651.34 I aA = VAN = 110.6529.03 V rms VAN 110.6529.03 = = 4.32 - 22.31 A rms 25.651.34 ZL Then, Zline = Vline 12050 - 110.6529.03 = = 9.94139.4 4.32 - 22.31 I aA Problem 10.21 In a balanced three-phase wye-wye system, the source is an abc-sequence set of voltages and Van = 12040 V rms. If the a-phase line current and line impedance are known to be 7.10 - 10.28 A rms and 0.8 + j1 , respectively, find the load impedance. Suggested Solution ZT = 12040 = 16.950.28 = 10.8 + j13 7.10 - 10.28 Z L = ZT - Zline = (10.8 + j13) - ( 0.8 + j1) = 10 + j12 Problem 10.22 An abc-sequence set of voltages feeds a balanced three-phase wye-wye system. If Van = 44030 V rms, VAN = 413.2829.78 V rms, and Z line = 1.2 + j1.5 , find the load impedance. Suggested Solution Ia = Van - VAN 44030 - 413.2829.78 = = 13.94 - 17.93 A rms 1.2 + j1.5 Zline VAN 413.2829.78 = = 19.95 + j 21.93 13.94 - 17.93 Ia Zload = Problem 10.23 In a three-phase balanced positive-sequence system a delta-connected source supplies power to a wyeconnected load. If the line impedance is 0.2 + j 0.4 , the load impedance 6 + j 4 , and the source phase voltage Vab = 20840 V rms, find the magnitude of the line voltage at the load. Suggested Solution Vab = 20840 V rms Van = 12010 V rms Then, VAN = 6 + j4 (12010 ) = 113.88.33 V rms 6.2 + j 4.4 VAB = 3 VAN = 3 (113.8 ) = 197.28 V rms Problem 10.24 Given the network shown, compute the line currents and the magnitude of the phase voltage at the load. a 0.4 j0.8 A 8 208-220 V rms 20820 V rms 8 0.4 0.4 j0.8 j0.8 j6 B 208-100 V rms j6 N 8 j6 C c b Suggested Solution Vab = 20820 V rms Zline = 0.4 + j 0.8 ZY = 8 + j6 Per phase Y circuit: a Zline I aA A Van n ZY N Van = Vab 3 Vab - 30 = 120 - 10 V rms ( ) I aA = Van 120 - 10 = = 11.10 - 49.00 A rms 8.4 + j 6.8 Zline + Z Y I bB = 11.10 - 169 A rms I cC = 11.1071 A rms VAN = I aA Z Y = 11.10 ( 8 + j 6 ) = 11.10 10 = 111 V rms VBN = VCN = VAN = 111 V rms Problem 10.25 In a balanced three-phase delta-wye system the source had an abc-phase sequence. The line and load impedances are 0.6 + j 0.3 and 12 + j 7 , respectively. If the line current I aA = 9.6 - 20 A rms, determine the phase voltages of the source. Suggested Solution Van = ( 9.6 - 20 )(12.6 + j 7.3) = 139.7810.09 V rms Vab = 139.78 3 (10.09 + 30 ) = 242.1140.09 V rms Vbc = 242.11 ( 40.09 - 120 ) = 242.11 - 79.91 V rms Vca = 242.11 ( 40.09 + 120 ) = 242.11160.09 V rms Problem 10.26 An abc-phase-sequence three-phase balanced wye-connected 60-Hz source supplies a balanced deltaconnected load. The phase impedance in the load consists of a 20 - resistor in series with a 50-mH inductor, and the phase voltage at the source is Van = 12020 V rms. If the line impedance is zero, find the line currents in the system. Suggested Solution Z = 20 + j 377 ( 0.05 ) Zline = 0 Van = 12020 V rms Y: Van ZY ZY = VAN VAB 3 Z = = = 6.67 + j 6.21 I aA 3 I AB 3 I aA = 13.10 - 23.28 A rms I aA = I bB = 13.10 - 143.28 A rms I cC = 13.1096.72 A rms Problem 10.27 An abc-phase-sequence three-phase balanced wye-connected source supplies a balanced delta-connected load. The impedance per phase in the delta load is 12 + j 9 . The line voltage at the source is Vab = 120 340 V rms. If the line impedance is zero, find the line currents in the balanced wye-delta system. Suggested Solution Zline = 0 Vab = VAB I AB = 120 340 = 13.863.13 A rms 12 + j 9 I aA = 13.86 3 ( 3.13 - 30 ) = 24.01 - 26.87 A rms , I bB = 24.01 - 146.87 A rms , and I cC = 24.0193.13 A rms Problem 10.28 An abc-phase-sequence three-phase balanced wye-connected source supplies power to a balanced deltaconnected load. The impedance per phase in the load is 14 + j12 . If the source voltage for the a phase is Van = 12080 V rms, and the line impedance is zero, find the phase currents in the wye-connected source. Suggested Solution Van = 12080 V rms Vab = 120 3110 V rms Zline = 0 Vab = VAB I AB = VAB 120 3110 = = 11.2769.4 A rms , Zload 14 + j12 I bB = 19.52 - 80.6 A rms and I cC = 19.52159.4 A rms Problem 10.29 An abc-phase-sequence three-phase balanced wye-connected source supplies a balanced delta-connected load. The impedance per phase of the delta load is 10 + j8 . If the line impedance is zero and the line current in the a phase is known to be I aA = 28.10 - 28.66 A rms, find the load voltage VAB . Suggested Solution If I aA = 28.1 - 28.66 A rms , then I AB = 28.1 3 ( -28.66 + 30 ) = 16.221.34 A rms VAB = I AB Z = (16.221.34 )(10 + j8 ) = 207.840 V rms Problem 10.30 An abc-phase-sequence three-phase balanced wye-connected source supplies power to a balanced deltaconnected load. The impedance per phase of the delta load is 14 + j11 . If the line impedance is zero and the line current in the a phase is I aA = 20.2231.84 A rms, find the voltages of the balanced source. Suggested Solution ZY = Z 14 11 = + j = 5.9338.2 3 3 3 Van = I aA Z Y = ( 20.2231.84 )( 5.9338.2 ) = 12070 V rms Vbn = 120 - 50 V rms Vcn = 120190 V rms Problem 10.31 In a balanced three-phase wye-delta system, the source has an abc-phase sequence and Van = 1200 V rms. If the line current is I aA = 4.820 A rms, find the load impedance per phase in the delta. Suggested Solution If Van = 1200 , then VAB = 120 330 V rms and if I aA = 4.820 , then I AB = 4.8 3 50 A rms Zload = VAB = 70.48 - j 25.65 I AB Problem 10.32 In a balanced three-phase wye-delta system, the source has an abc-phase sequence and Van = 12040 V rms. The line and load impedance are 0.5 + j 0.4 and 24 + j18 , respectively. Find the delta currents in the load. Suggested Solution 1 Using Y conversion, Z Y = Z = 8 + j 6 3 I aA = 12040 = 11.283.02 A rms 8.5 + j 6.4 Then, I AB = 11.28 3 ( 3.02 + 30 ) = 6.5133.02 A rms , I BC = 6.51 - 86.98 A rms , and I CA = 6.51153.02 A rms Problem 10.33 In a three-phase balanced delta-delta system, the source has an abc-phase sequence. The line and load impedances are 0.3 + j 0.2 and 9 + j 6 , respectively. If the load current in the delta is I AB = 1540 A rms, find the phase voltages of the source. Suggested Solution Converting to wye sources, Z Y = Z 3 I aA = 15 3 ( 40 - 30 ) = 2610 A rms Van = ( I an )( Zline + Z Y ) = ( 2610 )( 3.3 + j 2.2 ) = 10343.7 V rms Then, Vab = 103 373.7 = 178.573.7 V rms, Vbc = 178.5 - 46.3 V rms, and Vca = 178.5193.7 V rms Problem 10.34 In a balanced three-phase delta-delta system, the source has an abc-phase sequence. The phase angle for the source voltage is Vab = 40 and I ab = 415 A rms. If the total power absorbed by the load is 1400 W, find the load impedance. Suggested Solution PLtotal = 1400 W P = 1400 = 466.67 W 3 Since Vab = 40 and I ab = 415 A rms P = VAB I AB cos ( 40 - 15 ) No line impedance VAB = 466.67 = 128.73 V rms 4 cos 25 ZL = VAB 128.7340 = = 32.1825 I AB 415 Problem 10.35 A three-phase load impedance consists of a balanced wye in parallel with a balanced delta. What is the equivalent wye load and what is the equivalent delta load if the phase impedance of the wye and delta are 6 + j 3 and 15 + j12 , respectively? Suggested Solution Equivalent Y: Z 'Y = Z ' 3 Ztotal = ( 6 + j3)( 5 + j 4 ) 18 + j39 = = 2.77 + j1.78 ( 6 + j3) + ( 5 + j 4 ) 11 + j 7 Equivalent : Z = 3Z Y = 3 ( 2.77 + j1.78 ) = 8.31 + j 5.34 Problem 10.36 In a balanced three-phase system, the abc-phase-sequence source is wye connected and Van = 12020 V rms. The load consists of two balanced wyes with phase impedances of 8 + j 6 and 12 + j8 . If the line impedance is zero, find the line currents and the phase currents in each load. Suggested Solution a I aA A 8 Van j6 12 I1 j8 I2 n N I1 = 12020 = 12 - 16.87 A rms 8 + j6 12020 = 8.32 - 13.69 A rms 12 + j8 I2 = I aA = I1 + I 2 = 20.3 - 15.57 A rms The other currents are shifted by -120 and -240 Problem 10.37 In a balanced three-phase system, the source is a balanced wye with an abc-phase sequence and Vab = 20860 V rms. The load consists of a balanced wye with a phase impedance of 8 + j 5 in parallel with a balanced delta with a phase impedance of 21 + j12 . If the line impedance is 1.2 + j1 , find the phase currents in the balanced wye load. Suggested Solution Converting Z to Z : Y Z = 7 + j 4 Y Then, Z L = ( 8 + j 5 ) ( 7 + j 4 ) = 4.3530.8 Vab = 20860 V rms Van = 208 3 ( 60 - 30 ) = 12030 V rms I aA = Van 12030 = = 20.3 - 3.2 A rms Z L + Zline 5.933.2 VAN = I aA Z L = ( 20.33 - 3.2 )( 4.3530.8 ) = 88.427.6 V rms I AN = VAN 88.427.6 = = 9.37 - 4.4 A rms ZY 8 + j5 I BN = 9.37115.6 A rms I CN = 9.37 - 124.4 A rms Problem 10.38 In a balanced three-phase system, the source is a balanced wye with an abc-phase sequence and Vab = 20850 V rms. The load is a balanced wye in parallel with a balanced delta. The phase impedance of the wye is 5 + j 3 and the phase impedance of the delta is 18 + j12 . If the line impedance is 1 + j 0.8 , find the line currents and the phase currents in the loads. Suggested Solution Z = Y Z = 6 + j4 3 a 1 I aA j0.8 5 Van j3 j4 6 A n N Van = 208 3 ( 50 - 30 ) = 12020 V rms ZL = ( 5 + j3)( 6 + j 4 ) = 3.2232.18 ( 5 + j 3) + ( 6 + j 4 ) 12020 12020 = = 26.67 - 13.94 A rms Zline + Z L 4.533.94 I aA = VAN = I aA Zload = 85.8818.24 V rms Yload : I AN = VAN = 14.73 - 12.72 A rms 5 + j3 load : VAB = 85.88 348.24 V rms I AB = VAB = 6.8814.55 A rms 18 + j12 The other currents are shifted by -120 and -240 Problem 10.39 In a balanced three-phase system the source, which has an abc-phase sequence, is connected in delta and Vab = 20855 V rms. There are two loads connected in parallel. Load 1 is connected in wye and the phase impedance is 4 + j 3 . Load 2 is connected in wye and the phase impedance is 8 + j 6 . Compute the delta currents in the source if the line impedance connecting the source to the loads is 0.2 + j 0.1 . Suggested Solution a 0.8 I aA j0.1 4 Van j3 j6 8 A n N Van = 208 3 ( 55 - 30 ) = 12025 V rms ZL = ( 4 + j3)(8 + j 6 ) = 3.3336.87 = 2.66 + j 2.0 ( 4 + j 3) + ( 8 + j 6 ) 12025 12025 = = 33.8 - 11.29 A rms Zline + Z L 2.86 + j 2.1 33.8 3 ( -11.29 + 30 ) = 19.5118.71 A rms I aA = I ba = I cb = 19.51 (18.71 - 120 ) = 19.51 - 101.29 A rms I ac = 19.51 (18.71 - 240 ) = 19.51 - 221.29 A rms Problem 10.40 In a balanced three-phase system, the source has an abc-phase sequence and is connected in delta. There are two parallel wye-connected loads. The phase impedance of load 1 and load 2 is 4 + j 4 and 10 + j 4 , respectively. The line impedance connecting the source to the loads is 0.3 + j 0.2 . If the current in the a phase of load 1 is I AN1 = 1020 A rms , find the delta currents in the source. Suggested Solution By current division: I AN1 = I aA ( 4 + j 4 ) + (10 + j 4 ) 10 + j 4 I aA = 1527.9 A rms Then, since a source: I ab = - I aA 3 ( + 30 ) = 15 3 - 122.1 = 8.64 - 122.1 A rms I bc = 8.64 - 242.1 A rms I ca = 8.64 - 2.1 A rms a I aA I ab I ca c I bc I cC b I bB Problem 10.41 In a balanced three-phase system, the source has an abc-phase sequence and is connected in delta. There are two loads connected in parallel. The line connecting the source to the loads has an impedance of 0.2 + j 0.1 . Load 1 is connected in wye and the phase impedance is 4 + j 2 . Load 2 is connected in delta, and the phase impedance is 12 + j 9 . The current I AB in the delta load is 1645 A rms . Find the phase voltages of the source. Suggested Solution The Y equivalent circuit is: a 0.2 I aA j0.1 4 Van j2 j3 4 A 16 315 A rms n N Using current division: I aA ( 4 + j 2 ) 8 + j5 = 16 315 I aA = 58.4320.44 A rms Van = I aA ( 0.2 + j 0.1) + 16 315 ( 4 + j 3) = 151.3651.46 V rms ( ) Vab = 262.1681.46 V rms , Vbc = 262.16 - 38.54 V rms , and Vca = 262.16 - 158.54 V rms Problem 10.42 In a balanced three-phase system, the source has an abc-phase sequence and is connected in delta. There are two loads connected in parallel. Load 1 is connected in wye and has a phase impedance of 6 + j 2 . Load 2 is connected in delta and has a phase impedance of 9 + j 3 . The line impedance is 0.4 + j 0.3 . Determine the phase voltages of the source if the current in the a phase of load 1 is I AN1 = 1230 A rms . Suggested Solution The Y equivalent circuit is: a 0.4 I aA j0.3 3 Van j1 j2 6 A I AN1 n N Using current division: I aA ( 3 + j1) 9 + j3 = 1230 A rms I aA = 3630 A rms Van = ( I aA )( 0.4 + j 0.3) + (1230 )( 6 + j 2 ) = 93.1451.93 V rms Vab = 93.14 3 ( 51.93 + 30 ) = 161.3281.93 V rms Vbc = 161.32 ( 81.93 - 120 ) = 161.32 - 38.07 V rms Vca = 161.32 ( 81.93 - 240 ) = 161.32 - 158.07 V rms Problem 10.43 A balanced three-phase delta-connected source supplies power to a load consisting of a balanced delta in parallel with a balanced wye. The phase impedance of the delta is 24 + j12 , and the phase impedance of the wye is 12 + j8 . The abc-phase-sequence source voltages are Vab = 44060 V rms , Vbc = 440 - 60 V rms and Vca = 440 - 180 V rms , and the line impedance per phase is 1 + j 0.8 . Find the line currents and the power absorbed by the wye-connected load. Suggested Solution Vab = 44060 V rms Van = 440 3 ( 60 - 30 ) = 25430 V rms Equivalent single-phase (a-phase) diagram: a 1 I aA j0.8 12 Van j8 j4 8 A n N ZL = (12 + j8)(8 + j 4 ) 20 + j12 = 5.5429.3 ZT = Zline + Z L = 6.830.9 I aA = 25430 = 37.35 - 1 A rms 6.830.9 I bB = 37.35 - 121 A rms and I cC = 37.35119 A rms VAN = I aA Z L = 206.9328.3 V rms I AN - Y = VAN 206.9328.3 = = 14.37 - 5.4 A rms ZY 12 + j8 2 PY,load = 3 (14.37 ) (12 ) = 7.434 kW Problem 10.44 An abc-sequence wye-connected source having a phase-a voltage of 1200 V rms is attached to a wyeconnected load having an impedance of 8070 . If the line impedance is 420 , determine the total complex power produced by the voltage source and the real and reactive power dissipated by the load. Suggested Solution I aA = 1200 1200 = = 1.45 - 67.88 A rms 8070 + 420 31.12 + j 76.55 VL = I aA Z Y = (1.45 - 67.88 )( 8070 ) = 116.182.12 V rms S , S = VI = (1200 )(1.4567.88 ) = 17467.88 VA ST , S = 3S , S = 52267.88 VA S , L = (116.182.12 )(1.4562.88 ) = 168.4670 = 57.62 + j158.30 VA S L = 3S , L = 505.3870 = 172.86 + j 474.9 VA PL = 172.86 W and QL = 474.9 VAR Problem 10.45 The magnitude of the complex power (apparent power) supplied by a three-phase balanced wye-wye system is 3600 VA. The line voltage is 208 V rms. If the line impedance is negligible and the power factor angle of the load is 25 , determine the load impedance. Suggested Solution S = 3 VL I L IL = 3600 208 3 = 9.99 A rms ZY = 208 3 120 25 = 25 = 12.0125 = 10.88 + j 5.08 9.99 9.99 Problem 10.46 A balanced three-phase wye-wye system has two parallel loads. Load 1 is rated at 3000 VA, 0.7 pf lagging, and load 2 is rated at 2000 VA, 0.75 pf leading. If the line voltage is 208 V rms, find the magnitude of the line current. Suggested Solution S1 = 3000 cos -1 ( 0.7 ) = 300045.57 = 2100 + j 2142.43 VA S 2 = 2000 - cos -1 ( 0.75 ) = 2000 - 41.41 = 1500 - j1322.88 VA ST = S1 + S 2 = 3600 + j819.55 = 3692.1112.82 VA IL = 3692.11 208 3 = 10.25 A rms Problem 10.47 Two industrial plants represent balanced three-phase loads. The plants receive their power from a balanced three-phase source with a line voltage of 4.6 kV rms. Plant 1 is rated at 300 kVA, 0.8 pf lagging and plant 2 is rated at 350 kVA, 0.84 pf lagging. Determine the power line current. Suggested Solution S1 = 300 cos -1 ( 0.8 ) = 30036.87 = 240 + j180 kVA S 2 = 350 cos -1 ( 0.84 ) = 35032.86 = 294 + j189.9 kVA ST = S1 + S 2 = 534 + j 369.9 = 649.634.71 kVA IL = 649.6 4.6 3 = 81.53 A rms Problem 10.48 A cluster of loads is served by a balanced three-phase source with a line voltage of 4160 V rms. Load 1 is 240 kVA at 0.8 pf lagging and load 2 is 160 kVA at 0.92 pf lagging. A third load is unknown except that it has a power factor of unity. If the line current is measured and found to be 62 A rms, find the complex power of the unknown load. Suggested Solution ST = 3 VL I L = 3 ( 4160 )( 62 ) = 446730.54 VA S1 = 240 cos -1 ( 0.8 ) = 192000 + j144000 VA S 2 = 160 cos -1 ( 0.92 ) = 147200 + j 62707 VA (192000 + 147200 + P3 ) + j (144000 + 62707 + 0 ) = ST ( 339200 + P3 ) + ( 206707 ) 2 2 = 446730.54 P3 = 56831 W S3 = P3 + j 0 = 568310 VA Problem 10.49 A balanced three-phase source serves two loads: Load 1: 36 kVA at 0.8 pf lagging Load 2: 18 kVA at 0.6 pf lagging The line voltage at the load is 208 V rms at 60 Hz. Find the line current and the combined power factor at the load. Suggested Solution Vab = 208 V rms S1 = 36 cos -1 ( 0.8 ) = 3636.87 = 28.8 + j 21.6 kVA S 2 = 18 cos -1 ( 0.6 ) = 1853.13 = 10.8 + j14.4 kVA S L = S1 + S 2 = 39.6 + j 36.0 = 53.5242.27 kVA S L = 3 Vab I aA I aA = 53.52 103 208 3 = 148.56 A rms pf L = cos = cos ( 42.27 ) = 0.74 lagging Problem 10.50 A balanced three-phase source serves the following loads: Load 1: 48 kVA at 0.9 pf lagging Load 2: 24 kVA at 0.75 pf lagging The line voltage at the load is 208 V rms at 60 Hz. Determine the line current and the combined power factor at the load. Suggested Solution Vab = 208 V rms S1 = 48 cos -1 ( 0.9 ) = 4825.84 = 43.2 + j 20.92 kVA S 2 = 24 cos -1 ( 0.75) = 2441.41 = 18 + j15.87 kVA S L = S1 + S 2 = 61.2 + j 36.79 = 71.4131.02 kVA S L = 3 Vab I aA I aA = 71.41 103 208 3 = 198.21 A rms pf L = cos = cos ( 31.02 ) = 0.86 lagging Problem 10.51 A small shopping center contains three stores that represent three balanced three-phase loads. The power lines to the shopping center represent a three-phase source with a line voltage of 13.8 kV rms. The three loads are: Load 1: 500 kVA at 0.8 pf lagging Load 2: 400 kVA at 0.85 pf lagging Load 3: 300 kVA at 0.90 pf lagging Find the magnitude of the power line current. Suggested Solution Vab = 13.8 kV rms S1 = 500 cos -1 ( 0.8 ) = 50036.87 = 400 + j 300 kVA S 2 = 400 cos -1 ( 0.85 ) = 40031.79 = 340 + j 210.72 kVA S3 = 300 cos -1 ( 0.9 ) = 30025.84 = 270 + j130.76 kVA ST = S1 + S 2 + S3 = 1010 + j 641.48 = 1196.532.42 kVA ST = 3 Vab I aA I aA = 1196.5 13.8 3 = 50.1 A rms Problem 10.52 The following loads are served by a balanced three-phase source: Load 1: 18 kVA at 0.8 pf lagging Load 2: 8 kVA at 0.8 pf leading Load 3: 12 kVA at 0.75 pf lagging The load voltage is 208 V rms at 60 Hz. If the line impedance is negligible, find the power factor of the source. Suggested Solution S1 = 18 cos -1 ( 0.8 ) = 1836.87 = 14.40 + j10.80 kVA S 2 = 8 - cos -1 ( 0.8 ) = 8 - 36.87 = 6.40 - j 4.80 kVA S3 = 12 cos -1 ( 0.75 ) = 1241.41 = 9.00 + j 7.94 kVA ST = S1 + S 2 + S3 = 29.80 + j13.94 = 32.9025.07 kVA pf = cos = cos ( 25.07 ) = 0.91 lagging Problem 10.53 A balanced three-phase source supplies power to three loads. The loads are: Load 1: 30 kVA at 0.8 pf lagging Load 2: 24 kW at 0.6 pf leading Load 3: unknown If the line voltage and total complex power at the load are 208 V rms and 600 kVA , respectively, find the unknown load. Suggested Solution VAB = 208 V rms pf at source = 1.0 ST = 3 VAB I aA cos -1 (1.0 ) = 60.00 kVA I aA = 60 103 208 3 = 166.8 A rms S1 = 30 cos -1 ( 0.8 ) = 3036.87 = 24 + j18 kVA S2 = 24 - cos -1 ( 0.6 ) = 40 - 53.13 = 24 - j 32 kVA 0.6 P3 = 60.0 - 24 - 24 = 12.0 kW Q3 = 0 - 18 + 32 = 14 kVAR PT = P + P2 + P3 1 QT = Q1 + Q2 + Q3 S3 = 12.0 + j14 kVA = 18.44 kVA at 0.65 pf lagging Problem 10.54 A balanced three-phase source serves the following loads: Load 1: 18 kVA at 0.8 pf lagging Load 2: 10 kVA at 0.7 pf leading Load 3: 12 kW at unity pf Load 4: 16 kVA at 0.6 pf lagging The magnitude of the line voltage at the load is 208 V rms at 60 Hz, and the line impedance is 0.02 + j 0.04 . Find the magnitude of the line voltage and power factor at the source. Suggested Solution VAB = 208 V rms Zline = 0.02 + j 0.04 S1 = 18 cos -1 ( 0.8 ) = 1836.87 = 14.4 + j10.8 kVA S 2 = 10 - cos -1 ( 0.7 ) = 10 - 45.57 = 7.0 - j 7.14 kVA S3 = 12 0 = 120 = 12 + j 0 kVA 1.0 S 4 = 16 cos -1 ( 0.6 ) = 1653.13 = 9.6 + j12.8 kVA S L = S1 + S 2 + S3 + S 4 = 43 + j16.46 = 46.0420.95 kVA V2 S L = 3 AN Z L V2 Z L = 3 AN SL VAN = VAB 3 = 208 3 = 120 V rms ; Let VAN = 0 (VAN 0 )2 V 2 1202 = 3 AN = 3 Then, Z L = 3 = 0.9420.95 = 0.88 + j 0.34 SL S L 46.04 - 20.95 Per-phase Y circuit: a I aA A Zline I aA = VAN 1200 = = 127.66 - 20.95 A rms ZL 0.9420.95 Van n ZL Van = I aA [ Z line + Z L ] = 124.721.94 V rms Vab = 3 Van = 216.02 V rms N pf = cos Van - IaA = cos ( 22.89 ) = 0.92 lagging ( ) Problem 10.55 A balanced three-phase source supplies power to three loads. The loads are: Load 1: 18 kW at 0.8 pf lagging Load 2: 10 kVA at 0.6 pf leading Load 3: unknown If the line voltage at the load is 208 V rms, the magnitude of the total complex power is 41.93 kVA and the combined power factor at the load is 0.86 lagging, find the unknown load. Suggested Solution VAB = 208 V rms S L = 3 VAB I aA = 41.93 kVA pf L = 0.86 lagging I aA = 41.93 103 208 3 = 116.39 A rms S = cos -1 ( 0.86 ) = 30.68 L S L = 41.9330.68 = 36.06 + j 21.40 kVA = S1 + S 2 + S3 S1 = 18 cos -1 ( 0.8 ) = 22.536.87 = 18 + j13.5 kVA 0.8 S 2 = 10 - cos -1 ( 0.6 ) = 10 - 53.13 = 6 - j8 kVA S3 = S L - S1 - S 2 = ( 36.06 + j 21.40 ) - (18 + j13.5 ) - ( 6 - j8 ) = 12.06 + j15.9 kVA = 19.96 kVA at 0.60 pf lagging Problem 10.56 A balanced three-phase source supplies power to three loads. The loads are: Load 1: 20 kVA at 0.6 pf lagging Load 2: 12 kW at 0.75 pf lagging Load 3: unknown If the line voltage at the load is 208 V rms, the magnitude of the total complex power is 35.52 kVA and the combined power factor at the load is 0.88 lagging, find the unknown load. Suggested Solution VAB = 208 V rms S L = 3 VAB I aA = 35.52 kVA pf L = 0.88 lagging I aA = 35.52 103 208 3 = 98.6 A rms S = cos -1 ( 0.88) = 28.36 L S L = 35.5228.36 = 31.26 + j16.87 kVA = S1 + S 2 + S3 S1 = 20 cos -1 ( 0.6 ) = 2053.13 = 12 + j16 kVA S2 = 12 cos -1 ( 0.75) = 1641.41 = 12 + j10.58 kVA 0.75 S3 = S L - S1 - S 2 = ( 31.26 + j16.87 ) - (12 + j16 ) - (12 + j10.58 ) = 7.26 - j 9.71 kVA = 12.13 kVA at 0.60 pf leading Problem 10.57 A standard practice for utility companies is to divide its customers into single-phase users and three-phase users. The utility must provide three-phase users, typically industries, with all three phases. However, single-phase users, residential and light commercial, are connected to only one phase. To reduce cable costs, all single-phase users in a neighborhood are connected together. This means that even if the threephase users present perfectly balanced loads to the power grid, the single-phase loads will never be in balance, resulting in current flow in the neutral connection. Consider the 60-Hz, abc-sequence network shown. With a line voltage of 41630 V rms , phase a supplies the single-phase users on A Street, phase b supplies B Street and phase c supplies C Street. Furthermore, the three-phase industrial load, which is connected in delta, is balanced. Find the neutral current. a b c I AN I BN B Street 30 kW pf = 1 A B C I CN C Street 60 kW pf = 1 Three-phase 36 kW pf = 0.5 lagging 2400 V rms n 240 - 120 V rms 240120 V rms I nN A Street 48 kW pf = 1 N Suggested Solution PA = 48, 000 = Van I AN I AN = PA = 2000 A rms Van PB = 125 - 120 A rms VBN PC = 250 - 240 A rms VCN I BN = I CN = I Nn = I AN + I BN + I CN = 108.9783.41 A rms I nN = 108.97 - 96.59 A rms Problem 10.58 A three-phase abc-sequence wye-connected source with Van = 2200 V rms supplies power to a wyeconnected load that consumes 150 kW of power at a pf of 0.8 lagging. Three capacitors are found that each have an impedance of - j 2.0 , and they are connected in parallel with the previous load in a wye configuration. Determine the power factor of the combined load as seen by the source. Suggested Solution Van = 2200 V rms P = 50 kW 1 pf = 0.8 lagging ZC = - j 2 Original situation per phase: S old = 50 cos -1 ( 0.8 ) 0.8 = 62.536.87 = 50 + j 37.5 kVA Corrected situation: Pnew = Pold = 50 kW Qnew = Qold + QC = ( 37.5 103 ) + QC QC = - Van ZC 2 = -24.2 kVAR Qnew = 13.3 kVAR S new = 51.7414.90 kVA pf new = cos new = cos (14.90 ) = 0.97 lagging Problem 10.59 A three-phase abc-sequence wye-connected source with Van = 2200 V rms supplies power to a wyeconnected load that consumes 150 kW of power at a pf of 0.8 lagging. Three capacitors are found that each have an impedance of - j 2.0 , and they are connected in parallel with the previous load in a delta configuration. Determine the power factor of the combined load as seen by the source. Suggested Solution Van = 2200 V rms P = 50 kW 1 pf = 0.8 lagging ZC = - j 2 Z CY = ZC 2 =-j 3 3 Original situation per phase: S old = 50 cos -1 ( 0.8 ) 0.8 = 62.536.87 = 50 + j 37.5 kVA Corrected situation: Pnew = Pold = 50 kW Qnew = Qold + QC = ( 37.5 103 ) + QC QC = - Van 2 Z CY = -72.6 kVAR Qnew = -35.1 kVAR S new = 61.09 - 35.07 kVA pf new = cos new = cos ( -35.07 ) = 0.82 leading Problem 10.60 Find C in the network shown such that the total load has a power factor of 0.9 lagging. + 4.6 kV rms Balanced three-phase source 60 Hz _ C C C Balanced three-phase load 6 MVA 0.8 pf lagging Suggested Solution Vab = 4.6 kV rms pf new = 0.9 lagging S L1 = S L 3 3 S L 3 = 6 MVA f = 60 Hz pf old = 0.8 lagging = 2 cos -1 0.8 = 1.60 + j1.20 MVA Old situation per phase: S old = 1.60 + j1.20 MVA New situation: S new = 1.60 cos -1 ( 0.9 ) 0.9 = 1.60 + j 0.77 kVA = Pold + j ( Qold + QC ) QC = 0.77 - 1.20 = -0.43 MVAR But, 2 QC = - CYVan = - 2 CYVab 3 CY = 161.7 F Problem 10.61 Find C in the network shown such that the total load has a power factor of 0.9 leading. + 4.6 kV rms Balanced three-phase source 60 Hz _ C C C Balanced three-phase load 6 MVA 0.8 pf lagging Suggested Solution pf new = 0.9 leading f = 60 Hz Original complex power per phase: S old = 236.87 = 1.6 + j1.2 MVA Corrected complex power per phase: S new = 1.6 - cos -1 ( 0.9 ) 0.9 = 1.6 - j 0.77 MVA Qold + QC = Qnew QC = -0.77 - 1.20 = -1.97 MVAR But, 2 QC = - CYVan = - 2 CYVab 3 CY = 740.9 F Problem 10.62 Find C in the network shown such that the total load has a power factor of 0.92 leading. + 34.5 kV rms Balanced three-phase source 60 Hz _ C C C Balanced three-phase load 20 MVA 0.707 pf lagging Suggested Solution Vab = 34.5 kV rms f = 60 Hz S3 = 20 MVA pf old = 0.707 lagging pf new = 0.92 leading Old complex power per phase: S old = 20 cos -1 ( 0.707 ) = 4.71 + j 4.71 MVA 3 New complex power per phase: S new = 4.71 - cos -1 ( 0.92 ) = 4.71 - j 2.01 MVA 0.92 2 QC = - CVab = Qnew - Qold = -6.72 MVAR C = 15.0 F Problem 10.63 Find C in the network shown such that the total load has a power factor of 0.92 lagging. + 34.5 kV rms Balanced three-phase source 60 Hz _ C C C Balanced three-phase load 20 MVA 0.707 pf lagging Suggested Solution Vab = 34.5 kV rms f = 60 Hz S3 = 20 MVA pf old = 0.707 lagging pf new = 0.92 lagging Old complex power per phase: S old = 20 cos -1 ( 0.707 ) = 4.71 + j 4.71 MVA 3 New complex power per phase: S new = 4.71 cos -1 ( 0.92 ) = 4.71 + j 2.01 MVA 0.92 2 QC = - CVab = Qnew - Qold = -2.70 MVAR C = 6.0 F Problem 10FE-1 A wye-connected load consists of a series RL impedance. Measurements indicate that the rms voltage across each element is 84.85 V. If the rms line current is 6 A, find the total complex power for the threephase load configuration. Suggested Solution VL = 84.85 + j84.85 = 12045 V 2 84.85 2 84.85 ST = 3 ( 6 ) + j ( 6) 6 6 = 1527.3 + j1527.3 = 216045 VA or ST = 3 (12045 )( 6 ) = 216045 VA Problem 10FE-2 A balanced three-phase delta-connected load consists of an impedance of 12 + j12 . If the line voltage at the load is measured to be 230 V rms, find the magnitude of the line current and the total real power absorbed by the three-phase configuration. Suggested Solution I = 230 = 13.55 A rms 12 + j12 I L = 3 I = 23.47 A rms Ptotal = 3 VL I L cos 45 = 6.61 kW or Ptotal = 3 I R = 3 (13.55 ) (12 ) = 6.61 kW 2 2 Problem 10FE-3 Two balanced three-phase loads are connected in parallel. One load with a phase impedance of 24 + j18 is connected in delta, and the other load has a phase impedance of 6 + j 4 and is connected in wye. If the line-to-line voltage is 208 V rms, determine the line current. Suggested Solution Assume 2080 V rms For : I = IL = 2080 = 6.933 - 36.87 A rms 24 + j18 ( 3 ) ( - 30)( 6.933 - 36.87) = 12 - 66.87 A rms For Y: 208 - 30 3 IL = = 16.64 - 63.69 A rms 6 + j4 Total line current: I L = 12 - 66.87 + 16.64 - 63.69 = 12.09 - j 25.96 = 28.64 - 65 A rms or Convert Y 8 + j6 Ztotal = ( 6 + j 4 )(8 + j 6 ) 24 + j 68 = = 4.1935.06 ( 6 + j 4 ) + (8 + j 6 ) 14 + j10 Then 208 - 30 3 IL = = 28.66 - 65.06 A rms 4.1935.06 Problem 10FE-4 The total complex power at the load of a three-phase balanced system is 2430 kVA . Find the real power per phase. Suggested Solution S = 24, 00030 VA = 20, 785 + j12, 000 VA P = 20, 785 = 6.928 kW 3 Problem 12.1 Find the Laplace Transform of the function f (t ) = te - t (t - 1) Suggested Solution F ( s ) = te - at (t - 1)e - st dt 0 let g (t ) = te - at F ( s ) = g (t ) (t - 1)e - st dt = g (1)e- s (1) = e- ( s + a ) 0 Problem 12.2 Find the Laplace transform of the function f (t ) = te - a (t -1) (t - 1) Suggested Solution f (t ) = te - a (t -1) (t - 1) let g (t ) = te - ( a -1)t F ( s ) = g (t ) (t - 1)e - st dt = g (1)e- s (1) = e- s 0 F ( s ) = e- s Problem 12.3 If f (t ) = e - at cos( t ) show s+a F (s) = ( s + a ) 2 + ( ) 2 Suggested Solution F ( s ) = e - at e - st cos( t )dt = e- ( a + s ) [ 0 0 e - j t + e - j t ]dt 2 1 F ( s ) = [ e - ( s + a - j ) t dt + e- ( s + a + j )t dt ] 0 2 0 1 -1 1 e - ( s + a - j )t | - e- ( s + a + j )t | ] = [ 0 0 s + a + j 2 s + a - j 1 1 -1 = [ + ] 2 s + a - j s + a + j 1 s + a + j + s + a - j ] = [ 2 ( s + a - j )( s + a + j ) s+a F (s) = ( s + a ) 2 + ( ) 2 F (s) = s+a ( s + a ) 2 + ( ) 2 Problem 12.4 Find F(s) if f (t ) = e - at sin( t )u (t - 1) Suggested Solution f (t ) = e - at sin( t )u (t - 1) F ( s ) = e - s L[e - a (t +1) sin( )(t + 1)] = e - ( s + a ) L[e - at sin (t + 1)] = e - ( s + a ) L[e - at (sin t cos + cos t sin )] = e-( s+a ) [ cos ( s + a ) sin + ] 2 2 ( s + a ) + ( ) ( s + a ) 2 + ( ) 2 cos ( s + a ) sin + F ( s) = e- ( s + a ) 2 2 2 2 ( s + a ) + ( ) ( s + a ) + ( ) Problem 12.5 If f (t ) = t cos( t )u (t - 1) find F ( s ) Suggested Solution f (t ) = t cos( t )u (t - 1), F ( s ) = e - s L[(t + 1) cos (t + 1)] = e - s L[(t + 1)(cos t cos - sin t sin )] = e - s L[t cos t cos + cos t cos - t sin t sin - sin t sin ] L[cos t cos ] = s cos s2 + 2 s sin L[sin t sin ] = 2 s +2 -d s cos ( s 2 + 2 ) cos - s cos (2 s ) ( 2 ) = -[ L[t cos t cos ] = ] ds s + 2 ( s 2 + 2 )2 sin (2 s) -d sin L[t sin t sin ] = ( 2 ) = -[ 2 ] 2 ds s + ( s + 2 )2 F (s) = e- s [ 2 s 2 cos - ( s 2 + 2 ) cos 2 s sin s cos sin + 2 + + ] (s 2 + 2 )2 (s + 2 )2 s 2 + 2 s 2 + 2 2 s 2 cos - ( s 2 + 2 ) cos 2 s sin s cos sin + 2 + + ] (s 2 + 2 )2 (s + 2 )2 s 2 + 2 s 2 + 2 F (s) = e- s [ Problem 12.6 Find F(s) if f (t ) = te - at u (t - 4) Suggested Solution f (t ) = te - at u (t - 4) F ( s ) = e -4 s L[(t + 4)e - a ( t + 4) ] = e -4( s + a ) L[(t + 4)e- at ] = e -4( s + a ) L[te- at + 4e - at ] = e -4( s + a ) [ 1 4 + ] 2 ( s + a) ( s + a) F ( s ) = e -4( s + a ) [ 1 4 + ] 2 ( s + a) ( s + a) Problem 12.7 Use the shifting Theorem to determine L { f ( t )} where f (t ) = [e - (t - 2) - e-2(t - 2) ]u (t - 2) Suggested Solution f (t ) = [e - (t - 2) - e-2(t - 2) ]u (t - 2) The shifting Theorem states L[ f (t - t0 )u (t - t0 )] = e -t0 s F ( s) L[ f (t - t0 )u (t - t0 )] = e -t0 s F ( s) let g (t ) = (e - e )u (t ) so 1 1 G ( s) = - s +1 s + 2 F ( s ) = e -2 s G ( s ) -t -2 t F ( s ) = e -2 s [ 1 1 - ] s +1 s + 2 e -2 s F (s) = ( s + 1)( s + 2) Problem 12.8 Use the shifting Theorem to determine L{f(t)} where f (t ) = [e - ( t - 2) - e - ( t -1) ]u (t - 1) Suggested Solution f (t ) = [e - ( t - 2) - e - ( t -1) ]u (t - 1) let g (t ) = (t + e - t )u (t - 1) so G ( s) = 1 1 + 2 s +1 s F (s) = e- sG(s) F ( s ) = e -2 s [ 1 1 + 2] s +1 s F ( s ) = e -2 s [ 1 1 + 2] s +1 s Problem 12.9 Use Property Number 5 to find L{f(t)} if f (t ) = e - at u (t - 1) Suggested Solution f (t ) = e - at u (t - 1) let g (t ) = e - at Thm _ 5 : L[ g (t )u (t - t0 )] = e - t0 s L[ g (t + t0 )] so L[e - at u (t - 1)] = e - sL[e - a (t +1) ] = e- ( s + a ) L[e- at ] sin ce 1 L[e - at ] = s+a -( s+a ) e F (s) = s +1 F ( s) = e- ( s + a ) s +1 Problem 12.10 Use Property Number 6 to find L{f(t)} if f (t ) = te - at u (t - 1) Suggested Solution f (t ) = te - at u (t - 1) let g (t ) = tu (t - 1) Thm _ 6 : L[e - at g (t )] = G ( s + a ) 1 1 + ) = G(s) s2 s 1 1 F ( s) = G ( s + a) = e- ( s + a ) ( + ) 2 ( s + a) ( s + a) L[ g (t )] = L[tu (t - 1)] = e - s L[t - 1] = e - s ( F ( s) = G ( s + a) = e- ( s + a ) ( 1 1 + ) 2 ( s + a) ( s + a) Problem 12.11 Given the following functions F(s), find f(t), 4 ( s + 1) + ( s + 2) 10 s F (s) = ( s + 1) + ( s + 4) F (s) = Suggested Solution A. F (s) = for s = -1 4 ( s + 1) + ( s + 2) 4 = 4 = k1 s+2 for s = -2 4 = -4 = k2 s +1 so F (s) = 4 4 - s + 2 s +1 f (t ) = (4e - t - 4e -2t )u (t ) f (t ) = (4e - t - 4e -2t )u (t ) B. F (s) = 10 s ( s + 1) + ( s + 4) for s = -1 10 s -10 = = k1 3 s+4 for s = -4 10 s 40 = = k2 s +1 3 so F (s) = -10 / 3 40 / 3 - s +1 s+4 -10 -t 40 -4t f (t ) = ( e + e )u (t ) 3 3 f (t ) = ( -10 -t 40 -4t e + e )u (t ) 3 3 Problem 12.12 Given the following functions F(s), find f(t) s + 10 ( s + 4)( s + 6) 24 F (s) = ( s + 2)( s + 8) F (s) = Suggested Solution A. F (s) = for s = -4 s + 10 ( s + 4)( s + 6) s + 10 = 3 = k1 s+6 for s = -6 s + 10 = -2 = k2 s+4 so F (s) = 3 2 - s+4 s+6 f (t ) = (3e -4t - 2e -6t )u (t ) f (t ) = (3e -4t - 2e -6t )u (t ) B. F (s) = 24 ( s + 2)( s + 8) for s = -2 24 = 6 = k1 s +8 for s = -8 24 = -4 = k2 s+2 so F (s) = 4 4 - s + 2 s +8 f (t ) = (4e -2t - 4e -8t )u (t ) f (t ) = (4e -2t - 4e -8t )u (t ) Problem 12.13 Given the following functions F(s), find f(t) if F (s) = s +1 s ( s + 2)( s + 3) s2 + s + 1 F (s) = s ( s + 2)( s + 1) Suggested Solution A. F (s) = s +1 s ( s + 2)( s + 3) s +1 = 1/ 6 = k1 ( s + 2)( s + 3) s +1 1 for s = -2, = = k2 s ( s + 3) 2 s +1 -2 for s = -3, = = k3 s ( s + 2) 3 1/ 6 1/ 2 2 / 3 so F ( s ) = + - s s+2 s+3 2 1 1 and f (t ) = + e -2t - e-3t u (t ) 3 6 2 for s = 0, B. 2 1 1 f (t ) = + e -2t - e-3t u (t ) 3 6 2 F (s) = s2 + s + 1 s ( s + 2)( s + 1) s2 + s + 1 = 1/ 2 = k1 for s = 0, ( s + 2)( s + 3) s2 + s +1 = -1 = k2 for s = -1, s ( s + 2) s2 + s + 1 3 = = k3 for s = -2, s ( s + 2) 2 1/ 2 -1 3 / 2 + + so F ( s ) = s s +1 s + 2 1 3 and f (t ) = ( + e - t + e -2t )u (t ) 2 2 1 3 f (t ) = ( + e - t + e -2t )u (t ) 2 2 Problem 12.14 Given the following functions F(s), find f(t). s 2 + 5s + 4 ( s + 2)( s + 4)( s + 6) ( s + 3)( s + 6) F (s) = s ( s 2 + 8s + 12) F (s) = Suggested Solution A. s 2 + 5s + 4 ( s + 2)( s + 4)( s + 6) ( s + 1)( s + 4) ( s + 1) A B F (s) = = = + ( s + 2)( s + 4)( s + 6) ( s + 2)( s + 6) s + 2 s + 6 -1 A = F ( s ) ( s + 2) s =-2 = 4 5 B = F ( s ) ( s + 6) s =-6 = 4 F (s) = F (s) = 5 / 4 1/ 4 - s+6 s+2 5 1 f (t ) = ( + e -6t - e-2t )u (t ) 4 4 5 1 f (t ) = ( + e -6t - e-2t )u (t ) 4 4 B. F (s) = ( s + 3)( s + 6) s ( s 2 + 8s + 12) ( s + 3)( s + 6) ( s + 3) A B = = + s ( s + 2)( s + 6) s ( s + 2) s s + 2 3 A = F ( s) s s =0 = 2 -1 B = F ( s ) ( s + 2) s =-2 = 2 F (s) = F (s) = 3 / 2 1/ 2 3 / 2 - + s s+2 s+2 3 1 f (t ) = ( - e -2t )u (t ) 2 2 3 1 f (t ) = ( - e -2t )u (t ) 2 2 Problem 12.15 Given the following functions F(s), find f(t). s 2 + 7 s + 12 ( s + 2)( s + 4)( s + 6) ( s + 3)( s + 6) F (s) = s ( s 2 + 10s + 24) F (s) = Suggested Solution A. s 2 + 7 s + 12 ( s + 2)( s + 4)( s + 6) ( s + 3)( s + 4) ( s + 3) A B F (s) = = = + ( s + 2)( s + 4)( s + 6) ( s + 6)( s + 2) s + 6 s + 2 F (s) = F (s) = 1/ 4 3 / 4 + s+2 s+6 1 3 f (t ) = ( e -2t + e-6t )u (t ) 4 4 1 3 f (t ) = ( e -2t + e-6t )u (t ) 4 4 B. F (s) = F (s) = ( s + 3)( s + 6) s ( s 2 + 10s + 24) A B ( s + 3)( s + 6) ( s + 3) = = + ( s )( s + 4)( s + 6) ( s + 6)( s + 2) s s + 4 3 1 f (t ) = ( + e -4t )u (t ) 4 4 F (s) = 3 / 4 1/ 4 + s s+4 3 1 f (t ) = ( + e -4t )u (t ) 4 4 Problem 12.16 Given the following functions F(s) find f(t) s 2 + 7 s + 12 ( s + 2)( s + 4)( s + 6) 10( s + 2) F (s) = 2 ( s + 4s + 5) F (s) = Suggested Solution A. F (s) = s 2 + 7 s + 12 ( s + 2)( s + 4)( s + 6) Matlab code EDU syms s t EDU ilaplace((s^2+7*s+12)/(s+2)/(s+4)/(s+6)) ans = 3/4*exp(-6*t)+1/4*exp(-2*t) or 3 1 f (t ) = ( e -6t + e-2t ) 4 4 3 1 f (t ) = ( e -6t + e-2t ) 4 4 B. F ( s) = 10( s + 2) ( s 2 + 4s + 5) Matlab code ilaplace((s+3)*(s+6)/(s*(s^2+10*s+24))) ans = 1/4*exp(-4*t)+3/4 or 3 1 f (t ) = ( + e -4t ) 4 4 3 1 f (t ) = ( + e -4t ) 4 4 Problem 12.17 Given the following functions F(s) find f(t) 10 ( s + 2 s + 2) 10( s + 2) F (s) = 2 ( s + 4s + 5) F (s) = 2 Suggested Solution A. F (s) = 10 ( s + 2s + 2) k1 k1 10 F (s) = 2 = + s + 2s + 2 s + 1 - s + 1 + 2 -5 5 + s +1- s +1+ f ( ) = 10e - t cos(t - 90o )u (t ) f ( ) = f ( ) = 10e - t cos(t - 90o )u (t ) B. F ( s) = F (s) = so 10( s + 2) ( s 2 + 4s + 5) k1 k1 10( s + 2) = + 2 s + 4s + 5 s + 2 - s + 2 + 5 5 + s + 2 - s + 2 + f ( ) = 10e -2t cos(t )u (t ) f ( ) = f ( ) = 10e -2t cos(t )u (t ) Problem 12.18 Given the following functions F(s) find inverse Laplace functions. 10 ( s + 2 s + 2) 10( s + 2) F (s) = 2 ( s + 4s + 5) F (s) = 2 Suggested Solution A F (s) = 10 ( s + 2s + 2) 2 F (s) = k1 k1 10( s + 2) = + 2 s + 4s + 5 s + 2 - s + 2 + for s = -1 + 10( s + 1) = 5 = k1 s +1+ so 5 5 F (s) = + s +1- s +1+ f (t ) = 10e -6 cos tu (t ) B f (t ) = 10e -6 cos tu (t ) F ( s) = F (s) = 10( s + 2) ( s 2 + 4s + 5) k3 s +1 k k2 = 1+ + s ( s + 4s + 5) s s + 2 - s + 2 + 2 for s=0 s +1 = 1/ 5 = k1 2 s + 4s + 5 for s = -2 + s +1 = 0.31 - 108.43o = k2 s( s + 2 + ) F (s) = 1/ 5 0.31 - 108.43o 0.31108.43o + + s s + 2 + s + 2 + 1 f (t ) = ( + 0.62e -2t cos(t - 108.43o ))u (t ) 5 1 f (t ) = ( + 0.62e -2t cos(t - 108.43o ))u (t ) 5 Problem 12.19 Given the following functions F(s), find f(t) s ( s + 6) ( s + 3)( s 2 + 6s + 18) ( s + 4)( s + 8) F (s) = s ( s 2 + 8s + 32) F (s) = Suggested Solution A. F (s) = s ( s + 6) ( s + 3)( s 2 + 6s + 18) s ( s + 6) s ( s + 6) = 2 ( s + 3)( s + 6s + 18) ( s + 3)( s + 3 + 3 )( s + 3 - 3 ) k k2 k3 = 1 + + s + 3 s + 3 + 3 s + 3 - 3 at F (s) = s = -3 s ( s + 6) = -1 = k1 ( s + 3)( s 2 + 6s + 18) at s = -3 + 3 s ( s + 6) = 1 = k2 ( s + 3)( s 2 + 6s + 18) so F (s) = 1 1 -1 + + s + 3 s + 3 + 3 s + 3 - 3 f (t ) = (e -3t + 2e -3t cos 3t )u (t )) f (t ) = (e -3t + 2e -3t cos 3t )u (t )) B. F (s) = ( s + 4)( s + 8) s ( s 2 + 8s + 32) ( s + 4)( s + 8) s ( s + 6) = 2 ( s )( s + 8s + 32) ( s + 3)( s + 3 + 3 )( s + 3 - 3 ) k k2 k2 = 1+ + s s + 4 + 4 s + 4 - 4 at F (s) = s=0 ( s + 4)( s + 8) = 1 = k1 ( s )( s 2 + 8s + 32) at s = -4 + 4 ( s + 4)( s + 8) -1 = = k2 2 ( s )( s + 8s + 32) 2 so /2 1 1/ 2 F (s) = + + s s + 4 + 4 s + 4 - 4 f (t ) = (1 + e -4t cos(4t - 90o )u (t )) f (t ) = (1 + e -4t cos(4t - 90o )u (t )) Problem 12.20 Given the following functions F(s) find f(t) 6 s + 12 ( s + 4 s + 5)( s 2 + 4s + 8) s ( s + 2) F (s) = 2 s + 2s + 2 F (s) = 2 Suggested Solution 6s + 12 (( s + 2) + 12 )(( s + 2) 2 + 22 ) k k k k = 1 1 + 1 1 + 2 2 + 2 2 s + 2 - j1 s + 2 + j1 s + 2 - j 2 s + 2 + j 2 F (s) = 2 k11 = F ( s )( s + 2 - j1) |s =-2+ j1 = 10o k11 = F ( s )( s + 2 - j 2) |s =-2- j 2 = -10o f (t ) = [2e -2t cos(t ) - 2e-2t cos(2t )]u (t ) f (t ) = [2e -2t cos(t ) - 2e -2t cos(2t )]u (t ) s 2 + 2s + 2 - 2 2 = 1- 2 2 ( s + 2 s + 2) ( s + 2s + 2) 2 F (s) = 1 - ( s + 1) 2 + (1) 2 F (s) = f (t ) = [ (t ) - 2e -t sin(t )]u (t ) f (t ) = [ (t ) - 2e - t sin(t )]u (t ) Problem 12.21 Use Matlab to solve Problem 12.19 Suggested Solution F (s) = s ( s + 6) ( s + 3)( s 2 + 6s + 18) Matlab Code >> syms s t >> ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18))) ans = -exp(-3*t)+2*exp(-3*t)*cos(3*t) >> f (t ) = 2e -3t cos(3t ) - e -3t f (t ) = 2e -3t cos(3t ) - e -3t F (s) = ( s + 3)( s + 8) s ( s 2 + 8s + 32) Matlab Code >> syms s t >> ilaplace((s+4)*(s+8)/(s*(s^2+8*s+32))) ans = 1+exp(-4*t)*sin(4*t) >> f (t ) = 1 + e -4t sin(4t ) f (t ) = 1 + e -4t sin(4t ) Problem 12.22 Given following functions F(s), find f(t) s +1 F (s) = 2 s ( s + 2) s+3 F (s) = ( s + 1) 2 ( s + 4) Suggested Solution A. F (s) = at s=0 s +1 1 = k1 = s+2 2 d s +1 1 [ ] = k2 = ds ( s + 2) 4 at s = -2 s +1 -1 = k3 = 2 4 (s) so F (s) = 1/ 2 1/ 4 1/ 4 + - s2 s s+4 1 1 1 -2t f (t ) = ( + t - e )u (t ) 4 2 4 k k k s +1 = 1+ 2+ 3 2 s ( s + 4) s s s+2 2 1 1 1 f (t ) = ( + t - e -2t )u (t ) 4 2 4 B. F (s) = ( s + 4)( s + 8) s ( s 2 + 8s + 32) f (t ) = 1 + e -4t sin(4t ) F (s) = at s = -1 s+3 2 = k1 = s+4 3 d s+3 1 [ ] = k2 = ds ( s + 4) 9 at s = -4 s+3 1 = k3 = 2 ( s + 1) 9 so F (s) = 2/3 1/ 9 1/ 9 + - 2 s +1 s + 4 ( s + 1) 1 2 1 f (t ) = ( e - t + te - t - e -4t )u (t ) 9 3 9 k k1 k s+3 = + 2 + 3 2 2 s +1 s + 4 ( s + 1) ( s + 4) ( s + 1) 1 2 1 f (t ) = ( e- t + te- t - e-4t )u (t ) 9 3 9 Problem 12.23 Given the following functions F(s), find f(t) s +8 F (s) = 2 s ( s + 6) 1 F (s) = 2 s ( s + 1) 2 Suggested Solution A. F (s) = at s=0 s +8 4 = k1 = s+6 3 d s +8 -1 [ ] = k2 = ds ( s + 6) 18 at s = -6 s +8 1 = k3 = 2 18 ( s) so F (s) = 4 / 3 1/18 1/18 - + s2 s s+6 4 1 1 -6t f (t ) = ( t - + e )u (t ) 3 18 18 k k k s +8 = 1+ 2+ 3 2 s s+6 s ( s + 6) s 2 4 1 1 f (t ) = ( t - + e -6t )u (t ) 3 18 18 B. F (s) = k3 k k k s +8 = 1+ 2+ + 4 2 2 2 s ( s + 1) s +1 s ( s + 1) s 2 at s=0 1 = k1 = 1 ( s + 1) 2 d 1 [ ] = k2 = -2 ds ( s + 1) 2 at s = -1 1 1 = k3 = 18 ( s)2 d 1 [ ] = k2 = 1 ds ( s ) 2 so 1 2 1 1 F (s) = 2 - - + s s ( s + 1) 2 s + 1 f (t ) = (t - 2 + te - t + 2e- t )u (t ) f (t ) = (t - 2 + te- t + 2e - t )u (t ) Problem 12.24 Given the following functions F(s), find f(t) F (s) = s+4 ( s + 2) 2 s+6 F (s) = s ( s + 1) 2 Suggested Solution F (s) = k1 k s+4 = + 2 2 2 s+2 ( s + 2) ( s + 2) at s = -2 s + 2 = 2 = k1 d [ s + 2] = k2 = 1 ds so F (s) = 2 1 + ( s + 2) 2 s + 2 f (t ) = (2te -2t + e -2t )u (t ) f (t ) = (2te -2t + e -2t )u (t ) B. F (s) = at s=0 s+6 = k1 = 6 ( s + 1) 2 at s = -1 s+6 = k2 = -5 s d s+6 [ ] = k3 = -6 ds s so F (s) = 6 5 6 - - 2 s ( s + 1) s +1 f (t ) = (6 - 5te - t - te - t )u (t ) k k k2 s+6 = 1+ + 3 2 2 s ( s + 1) s +1 s ( s + 1) f (t ) = (6 - 5te- t - te - t )u (t ) Problem 12.25 Given the following functions F(s) find f(t) F (s) = s2 ( s + 1) 2 ( s + 2) s 2 + 9 s - 20 s ( s + 4)3 ( s + 5) F (s) = Suggested Solution A. F (s) = at s = -1 s2 = k1 = 1 ( s + 2) at s = -2 s2 = k3 = 4 ( s + 1) 2 d s2 [ ] = k2 = -3 ds ( s + 1) 2 so F (s) = 1 3 4 - + ( s + 1) 2 s + 1 s + 2 k k1 k s2 = + 2 + 3 2 2 s +1 s + 2 ( s + 1) ( s + 2) ( s + 1) f (t ) = (te - t - 3e - t + 4te-2t )u (t ) f (t ) = (te - t - 3e- t + 4te-2t )u (t ) B. F (s) = at s=0 k k k2 s 2 + 9s - 20 1 = = 1+ + 3 3 2 2 s ( s + 4) ( s + 5) s ( s + 4) s ( s + 4) s+4 1 = k1 = 1/10 ( s + 4) 2 at s = -4 1 = k2 = -1/ 4 s d 1 [ ] = k3 = -1/16 ds s so F (s) = 1/16 1/ 4 1/16 - - 2 s s+4 ( s + 4) 1 t -4 t 1 - 4 t f (t ) = ( - e - e )u (t ) 16 4 16 1 t 1 f (t ) = ( - e-4t - e -4t )u (t ) 16 4 16 Problem 12.26 Find f(t) if F(s) is given by expression F (s) = s ( s + 1) ( s + 2)3 ( s + 3) Suggested Solution F (s) = s ( s + 1) A B C D = + + + ( s + 2)3 ( s + 3) s + 3 ( s + 3)3 ( s + 3) 2 s + 2 A = F ( s )( s + 3) |s =-3 = -6 B = F ( s )( s + 2) 2 |s =-2 = 2 -6 2 C D + + + C + 2D = 7 3 8 4 2 -6 + 2+C + D C + D =1 F (-1) = 0 = 2 D = 6, C = -5 -6 2 5 6 + - + F (s) = 3 2 s + 3 ( s + 3) ( s + 3) s+2 F (0) = 0 = f (t ) = (t 2e -2t - 5te -2t + 6e -2t - 6e -3t )u (t ) f (t ) = (t 2e-2t - 5te -2t + 6e-2t - 6e-3t )u (t ) Problem 12.27 Find f(t) if F(s) is given by F (s) = 12( s + 2) s 2 ( s + 1)( s 2 + 4s + 8) Suggested Solution F (s) = at s=0 12( s + 2) = 3 = k1 ( s + 1)( s 2 + 4s + 8) d 12( s + 2) [ ] = k2 = -3 ds ( s + 1)( s 2 + 4 s + 8) at s = -1 12( s + 2) 12 = k3 = ( s 2 )( s 2 + 4 s + 8) 5 at s = -2 + 2 12( s + 2) 1 = k4 = - 26.56o 3 s 2 ( s + 1) so 1 1 - 26.56o 26.56o 3 3 12 / 5 3 3 F (s) = 2 - + + + s s +1 s + 2 - 2 s + 2 + 2 s 12 - t 2 -2t f (t ) = [-3 + 3t + e + e cos(2t - 26.56o )]u (t ) 5 3 k k1 k2 k4 k4 + + 3 + + s 2 s s + 1 s + 2 - 2 s + 2 + 2 f (t ) = [-3 + 3t + 12 - t 2 -2t e + e cos(2t - 26.56o )]u (t ) 5 3 Problem 12.28 Use Matlab to solve Problem 12.25 Suggested Solution s2 ( s + 1) 2 ( s + 2) Matlab Code F (s) = >> syms s t >> >> ilaplace(s^2/((s+1)^2*(s+2))) ans = (-3+t)*exp(-t)+4*exp(-2*t) >> f (t ) = te- t - 3e - t + 4e-2t f (t ) = te- t - 3e - t + 4e-2t s 2 + 9s + 20 s ( s + 4)3 ( s + 5) Matlab Code >> syms s t >> ilaplace((s^2+9*s+20)/(s*(s+4)^3*(s+5))) F (s) = ans = -1/4*t*exp(-4*t)-1/16*exp(-4*t)+1/16 >> f (t ) = 0.0625{1 - e -4t - 4te-4t } f (t ) = 0.0625{1 - e -4t - 4te-4t } Problem 12.29 Find the inverse Laplace transform of the following functions e- s F (s) = s +1 F (s) = 1 - e -2 s s 1 - e- s s+2 F (s) = Suggested Solution a. F (s) = let G( s) = 1 g (t ) = e - t u (t ) s +1 f (t ) = g (t - 1) f (t ) = e - (t -1) u (t - 1) e- s s +1 b. F (s) = let G( s) = 1 g (t ) = u (t ) s f (t ) = g (t ) - g (t - 2) f (t ) = u (t ) - u (t - 2) 1 - e -2 s s c. F (s) = let G( s) = 1 g (t ) = e -2t u (t ) s+2 f (t ) = g (t ) - g (t - 1) f (t ) = e -2t u (t ) - e-2(t -1) u (t - 1) 1 - e- s s+2 f (t ) = e - (t -1) u (t - 1) f (t ) = u (t ) - u (t - 2) f (t ) = e-2t u (t ) - e -2(t -1) u (t - 1) Problem 12.30 Find the inverse Laplace transform of the following function ( s + 1)e - s F (s) = s ( s + 2) F (s) = 10e -2 s ( s + 1)( s + 3) Suggested Solution F (s) = at s=0 s +1 = k1 = 1/ 2 s+2 at s = -2 s +1 = 1/ 2 = k2 s 1/ 2 1/ 2 F (s) = e- s [ ] + s s+2 1 1 f (t ) = u (t - 1) + e -2(t -1) u (t - 1) 2 2 F (s) = k k 10e -2 s = e -2 s [ 1 + 2 ] ( s + 1)( s + 3) s +1 s + 3 k k ( s + 1)e - s = e- s [ 1 + 2 ] s ( s + 2) s s+2 at s = -1 10 = k1 = 5 s+3 at s = -3 10 = k2 = -5 s +1 5 5 F ( s ) = e -2 s [ ] - s +1 s + 3 f (t ) = [5e- (t - 2) - 5e-3(t - 2) ]u (t - 2) f (t ) = 1 1 u (t - 1) + e-2(t -1) u (t - 1) 2 2 f (t ) = [5e - (t - 2) - 5e -3(t - 2) ]u (t - 2) Problem 12.31 Find the inverse Laplace transform f(t) if F(s) is se - s F (s) = ( s + 1)( s + 2) Suggested Solution F (s) = let G( s) = s A B = + ( s + 1)( s + 2) s + 1 s + 2 A = G ( s )( s + 1) |s =-1 = -1 2 1 - g (t ) = [2e -2t - e- t ]u (t ) s + 2 s +1 f (t ) = g (t - 1) f (t ) = [2e -2(t -1) - e - (t -1) ]u (t - 1) se - s ( s + 1)( s + 2) f (t ) = [2e -2(t -1) - e - (t -1) ]u (t - 1) B = G ( s )( s + 2) |s =-2 = 2 G( s) = Problem 12.32 Find f(t) if F(s) is given by the following functions s 2 + 2s + 3 ( s + 2)e -4 s F ( s ) = e -2 s [ ] F (s) = 2 s ( s + 1)( s + 2) s ( s + 1) Suggested Solution A. F ( s ) = e -2 s [ at s=0 s 2 + 2s + 3 3 = k1 = ( s + 1)( s + 2) 2 at s = -1 s 2 + 2s + 3 = k2 = -2 s ( s + 2) at s = -2 3 s 2 + 2s + 3 = k3 = 2 s ( s + 1) so 3/ 2 3/ 2 -2 ] F ( s ) = e -2 s [ + + ( s + 1) ( s + 2) s 3 3 f (t ) = [ - 2e- (t - 2) + e -2(t - 2) ]u (t - 2) 2 2 B. F (s) = k k k ( s + 2)e -4 s = e -4 s [ 1 + 2 + 3 ] s 2 ( s + 1) s 2 s ( s + 1) k3 k k s 2 + 2s + 3 ] = e -2 s [ 1 + 2 + ] s ( s + 1)( s + 2) s ( s + 1) ( s + 2) at s=0 s+2 = k1 = 2 s +1 d s+2 [ ] = -1 = k 2 ds s + 1 at s = -1 s+2 = k3 = 1 s +1 so F ( s ) = e -4 s [ 2 1 1 ] - + 2 s s ( s + 1) f (t ) = [2(t - 4) - 1 + e- (t - 4) ]u (t - 4) 3 3 f (t ) = [ - 2e- (t - 2) + e -2(t - 2) ]u (t - 2) 2 2 f (t ) = [2(t - 4) - 1 + e- (t - 4) ]u (t - 4) Problem 12.33 Find f(t) if F(s) is given by following functions F (s) = e- s [ 2( s + 1) ] ( s + 3)( s + 2) F (s) = 10( s + 2)e -2 s ( s + 4)( s + 1) Suggested Solution A. F (s) = e- s [ at s = -2 2( s + 1) = k1 = -2 ( s + 3) at s = -3 2( s + 1) = k2 = 4 ( s + 2) so F ( s ) = e -2 s [ -2 4 + ] s+2 s+3 f (t ) = [-2e - (t -1) + 4e -3(t -1) ]u (t - 1) k k 2( s + 1) ] = e- s [ 1 + 2 ] ( s + 3)( s + 2) s+2 s+3 B F (s) = at s = -1 10( s + 2) 10 = k1 = ( s + 4) 3 at s = -4 10( s + 2) = k2 = 4 ( s + 1) so 10 / 3 4 + F ( s ) = e -2 s [ ] s +1 s + 4 10 20 f (t ) = [ e- (t - 2) + e -4(t - 2) ]u (t - 2) 3 3 k k 10( s + 2)e -2 s = e -2 s [ 1 + 2 ] ( s + 4)( s + 1) s +1 s + 4 f (t ) = [-2e- (t -1) + 4e-3(t -1) ]u (t - 1) 10 20 f (t ) = [ e- (t - 2) + e -4(t - 2) ]u (t - 2) 3 3 Problem 12.34 Solve the following differential equations using Laplace Transform dx(t ) + 3x(t ) = e -2t dt dx(t ) + 3 x(t ) = 2u (t ) dt Suggested Solution A. dx(t ) + 3x(t ) = e -2t , x(0) = 1 dt dx(t ) 1 1 s+3 sX ( s ) - x(0) = X ( s )( s + 3) = +1 = dt s+2 s+2 s+2 1 X (s) = x(t ) = e -2t u (t ) s+2 B. dx(t ) + 3x(t ) = 2u (t ), x(0) = 2 dt dx(t ) 2 2( s + 1) sX ( s ) - 2 + 3 X ( s ) X ( s )( s + 3) = + 2 = dt s s 2 x(t ) = [1 + 2e -3t ]u (t ) 3 x(t ) = e -2t u (t ) 2 x(t ) = [1 + 2e -3t ]u (t ) 3 Problem 12.35 Solve the following differential equations using Laplace transform d 2 y (t ) 2dy (t ) + + y (t ) = e-2t , y (0) = y '(0) = 0 A. dt dt 2 d 2 y (t ) 4dy (t ) B. + + 4 y (t ) = u (t ), y (0) = 0; y '(0) = 1 dt dt 2 Suggested Solution A. d 2 y (t ) 2dy (t ) + + y (t ) = e-2t , y (0) = y '(0) = 0 2 dt dt 1 s+2 1 A B C Y (s) = ; C = Y ( s )( s + 2) |s =-2 = 1 = + + ( s + 1) 2 ( s + 2) ( s + 1) 2 ( s + 1) ( s + 2) s 2Y ( s ) + 2sY ( s ) + Y ( s ) = Y ( s )[ s 2 + 2s + 1] = A = Y ( s )( s + 1) 2 |s =-1 = 1 1 1 = 1 + B + B = -1 2 2 1 1 1 Y (s) = - + ( s + 1) 2 ( s + 1) ( s + 2) Y (0) = y (t ) = (te - t - e - t + e -2t )u (t ) B. d 2 y (t ) 4dy (t ) + + 4 y (t ) = u (t ), y (0) = 0; y '(0) = 1 dt 2 dt s 2Y ( s ) - sy '(0) - y (0) + 4 sY ( s ) - 4 y (0) + 4Y ( s ) = 1 s2 + 1 +s= s s 2 s +1 A B C Y (s) = = + + s ( s + 2) 2 s ( s + 2) 2 s + 2 1 A = Y ( s ) s |s = 0 = 4 B = Y ( s )( s + 2) 2 |s =-2 = -2.5 Y ( s )[ s 2 + 4s + 4] = 1 5 3 Y (-1) = -2 = - - + C C = 4 2 4 1 1 10 3 + ] Y (s) = [ - 4 s ( s + 2) 2 s + 2 1 y (t ) = [1 - 10te -2t + 3e-2t ]u (t ) 4 1 s y (t ) = (te - t - e - t + e-2t )u (t ) 1 y (t ) = [1 - 10te -2t + 3e -2t ]u (t ) 4 Problem 12.36 Use Laplace transform to find y(t) if dy (t ) + 5 y (t ) + 4 y ( x)dx = u (t ), y (0) = 0, t > 0 dt 0 t Suggested Solution In Laplace terms 4 1 sY ( s ) + 5Y ( s ) + Y ( s ) = 5 s 2 Y ( s )[ s + 5s + 4] = 1 so Y (s) = so Y (s) = 1/ 3 1/ 3 - s +1 s + 4 1 1 y (t ) = [ e - t - e -4t ]u (t ) 3 3 k k 1 1 = = 1 + 2 k1 = 1/ 3; k2 = -1/ 3 s + 5s + 4 ( s + 1)( s + 4) s + 1 s + 4 2 1 1 y (t ) = [ e- t - e -4t ]u (t ) 3 3 Problem 12.37 Solve the integrodifferential equations using Laplace transforms. t dy (t ) + 2 y (t ) + y ( )d = 1 - e -2t , y (0) = 0, t > 0 dt 0 Suggested Solution In Laplace terms 1 1 1 sY ( s ) + 2Y ( s ) + Y ( s ) = - s s s+2 1 1 1 Y ( s )[ s + 2 + ] = - s s s+2 so Y (s) = at s = -1 2 = 2 = k1 s+2 d 2 [ ] = -2 = k2 ds s + 2 at s = -2 2 = 2 = k3 ( s + 1) 2 so F (s) = 2 2 2 - + 2 ( s + 1) s +1 s + 2 k k1 k + 2 + 3 2 ( s + 1) s +1 s + 2 f (t ) = [2te- t - 2e - t + 2e -2t ]u (t ) f (t ) = [2te- t - 2e- t + 2e -2t ]u (t ) Problem 12.38 Determine the y(t) in the following equation if all initial conditions are zero. d 3 y (t ) d 2 y (t ) dy (t ) +4 +3 = 10e -2t 3 dt dt dt 2 Suggested Solution d 3 y (t ) d 2 y (t ) dy (t ) +4 +3 = 10e -2t 3 dt dt 2 dt All intial conditions are zero. 10 s+2 A B C D 10 Y (s) = = + + + s ( s + 2)( s + 1)( s + 3) s s + 2 s + 1 s + 3 A = Y ( s ) s |s = 0 = 5 / 3 s 3Y ( s ) + 4 s 2Y ( s ) + 3sY ( s ) = B = Y ( s )( s + 2) |s =-2 = 5 C = Y ( s )( s + 1) |s =-1 = -5 D = Y ( s )( s + 3) |s =-3 = -5 / 3 Y (s) = 5/3 5 5 5/3 + - - s s + 2 s +1 s + 3 5 5 y (t ) = ( + 5e -2t - 5e - t - e -3t )u (t ) 3 3 5 5 y (t ) = ( + 5e -2t - 5e - t - e -3t )u (t ) 3 3 Problem 12.39 Find f(t) using convolution if F(s) is 1 F (s) = ( s + 1)( s + 2) Suggested Solution F (s) = let 1 s +1 1 F2 ( s ) = s+2 f1 (t ) = e- t u (t ) F1 ( s ) = f 2 (t ) = e -2t u (t ) f (t ) = e - (t - ) e-2 d = e - t e - d =e- t [e - ]t0 0 0 t t 1 ( s + 1)( s + 2) f (t ) = e- t [1 - e - t ] f (t ) = (e- t - e-2t )u (t ) f (t ) = (e - t - e -2t )u (t ) Problem 12.40 Use convolution to find f(t) if 1 F (s) = ( s + 1)( s + 2) 2 Suggested Solution F (s) = let 1 ( s + 2) 2 1 F2 ( s ) = s +1 f1 (t ) = te -2t u (t ) F1 ( s ) = f 2 (t ) = e - t u (t ) f (t ) = e - (t - ) e -2 d = e - t e- d =e - t [-e - - e - ]t0 0 0 t t 1 ( s + 1)( s + 2) 2 f (t ) = e [-e - te + 1] f (t ) = (e - t - te -2t - e-2t )u (t ) -t -t -t f (t ) = (e - t - te -2t - e-2t )u (t ) Problem 12.41 Determine the intial and final value of F(s) given by the expression F (s) = 2( s + 2) s ( s + 1) F (s) = F (s) = 2( s 2 + 2s + 6) s ( s + 1) 2s 2 s ( s + 1)( s 2 + 2 s + 2) Suggested Solution A. F (s) = for 2( s + 2) ]= 2 s ( s + 1) 2( s + 2) ]= 4 t , lim s 0 sf ( s ) = lim s 0 [ s ( s + 1) B. t = 0+ , lim s sf ( s ) = lim s [ F (s) = for t = 0+ , lim s sf ( s ) = lim s [ t , lim s 0 sf ( s ) = lim s 0 [ 2( s 2 + 2 s + 6) ]= 2 s ( s + 1) 2( s 2 + 2 s + 6) s ( s + 1) 2( s + 2) s ( s + 1) 2( s 2 + 2 s + 6) ]= 0 s ( s + 1) C. F (s) = for t = 0+ , lim s sf ( s ) = lim s [ t , lim s 0 sf ( s ) = lim s 0 [ 2s 2 ]= 2 s ( s + 1)( s 2 + 2s + 2) 2s 2 s ( s + 1)( s 2 + 2 s + 2) 2s 2 ]= 0 s ( s + 1)( s 2 + 2s + 2) Problem 12.42 Find the initial and final value of the time function f(t) if F(s) is given as F (s) = 10( s + 2) ( s + 1)( s + 3) F (s) = F (s) = ( s 2 + 2s + 4) ( s + 1)( s 3 + 4s 2 + 8s + 10) 2s ( s 2 + 2 s + 2) Suggested Solution A. F (s) = for 10( s + 2) ] = 10 ( s + 1)( s + 3) 10( s + 2) ]= 0 t , lim s 0 sf ( s ) = lim s 0 [ ( s + 1)( s + 3) B. t = 0+ , lim s sf ( s ) = lim s [ F (s) = for t = 0+ , lim s sf ( s ) = lim s [ t , lim s 0 sf ( s ) = lim s 0 [ ( s 2 + 2s + 4) ]= 0 ( s + 1)( s 3 + 4 s 2 + 8s + 10) ( s 2 + 2s + 4) ( s + 1)( s 3 + 4s 2 + 8s + 10) 10( s + 2) ( s + 1)( s + 3) ( s 2 + 2s + 4) ]= 0 ( s + 1)( s 3 + 4s 2 + 8s + 10) C. F (s) = for t = 0+ , lim s sf ( s ) = lim s [ t , lim s 0 sf ( s ) = lim s 0 [ 2s 2 ]= 2 s ( s + 1)( s 2 + 2s + 2) 2s ( s + 2s + 2) 2 2s 2 ]= 0 s ( s + 1)( s 2 + 2s + 2) Problem 12.43 Find the final values of the time function f(t) if F(s) is given as F (s) = 10( s + 1) ( s + 2)( s + 3) F (s) = 10 s + 4s + 4 2 Suggested Solution A. F (s) = for 10( s + 1) ] = 10 ( s + 2)( s + 3) 10( s + 1) t , lim s 0 sf ( s ) = lim s 0 [ ]= 0 ( s + 2)( s + 3) B. 10 F (s) = 2 s + 4s + 4 for 10 t = 0+ , lim s sf ( s ) = lim s [ 2 ]= 0 s + 4s + 4 10 t , lim s 0 sf ( s ) = lim s 0 [ 2 ]= 0 s + 4s + 4 t = 0+ , lim s sf ( s ) = lim s [ 10( s + 1) ( s + 2)( s + 3) Problem 12.44 In the network in fig, the switch opens at t =0. Use laplace transform to find I(t) for t>0. t=0 3H 3H 12V 6 i(t) 12 Suggested Solution t=0 + Vl 3H 12V 6 i(t) 12 12 6 I(s) 3H for t<0 VL = 0V and i (0- ) = i (0+ ) = 1A 18I ( s ) + 3sI ( s ) - 3i (0) = 0 I ( s )[3s + 18] = 3 1 I (s) = s+6 i (t ) = e -6t u (t ) i (t ) = e -6t u (t ) Problem 12.45 The switch in the circuit opens at t=0. Find I(t) for t>0 using Laplace transforms. 2 4 t=0 3 12V 2H Suggested Solution 2 R1 4 t=0 3 R3 12V R2 + VL 2 4 R4 t=0 3 12V iL(o) 6 3 4 i (t) 2H 2 di (t ) =0 dt 9 I ( s ) + 2 I ( s ) - 2i (0+ ) = 0 9 I ( s) = [ s + ] = 4 2 4 I ( s) = 9 s+ 2 3i (t ) + 4i (t ) + 2i (t ) + 2 i (t ) = 4e 9 - t 2 u (t ) A i (t ) = 4e 9 - t 2 u (t ) A Problem 12.46 In the circuit in fig, the switch moves from position 1 to 2 at t =0. Use Laplace transforms to find v(t) for t>0. 1 t=0 2 12V + v(t) 100F 3K 6K Suggested Solution 1 t=0 2 12V R1 + v(t) 100F R2 R1 + V(o) 12V - R2 R2 + V(t) - V (t ) Cdv(t ) + =0 R dt V (s) + CV ( s ) - Cv (0) = 0 R 1 4 V ( s )[ s + ] = v (0) V ( s ) = RC s+5 v(t ) = 4e-5t u (t )V v(t ) = 4e -5t u (t )V Problem 12.47 In the network the switch closes at t=0. Use Laplace transforms to find Vc(t) for t>0. 1 12V t=0 1 2H Vc(t) 0.5F + Suggested Solution 1 12V t=0 1 2H iL 0.5F + t 12 - Vc(t ) = 1 dVc(t ) Vc( x)dx + C L dt 0 s 12 1 - Vc( s ) = Vc( s ) + Vc( s) Vc(t) s 2s 2 12 1 s = Vc( s )[1 + + ] s 2s 2 12 2s + 1 + s 2 ] = Vc( s )[ s 2s 24 24 Vc( s ) = = 2s + 1 + s 2 ( s + 1) 2 Vc(t ) = 24te- t u (t )V Vc(t ) = 24te - t u (t )V Problem 12.48 In the network the switch opens at t=0. Use Laplace transforms to find iL(t) for t>0. 3 iL(t) t=0 1A 1H 0.5F Suggested Solution 3 iL(t) t=0 1A 0.5F 1H - ic + Vc(t) For _ t > 0 L t diL 1 = 3iC + iC (t )dt dt C0 and iL + iC = 1 iC = 1 - iL now, 1 - I L ( s) ) sI L ( s ) = 3 - 3I L ( s ) + 2( s 2 2 I L ( s )[ s + ] = 3 + s s 3s + 2 A B = + A = 4, B = -1 I L ( s) = 2 s + 3s + 2 s + 2 s + 1 4 1 - I L ( s) = s + 2 s +1 iL (t ) = (4e -2t - e- t )u (t ) iL (t ) = (4e -2t - e- t )u (t ) Problem 12.49 In the network the switch opens at t=0. Use Laplace transforms to find V0(t) for t>0. 3K t=0 4K + Vc(t) 2K + V(t) - 12V - Suggested Solution 3K t=0 4K + + Vc(t) 12V 4K + V(t) 0.1mF 2K - 2K - V(t) Vc(0- ) = 12( 2+4 ) = 8v = Vc(0+ ) 2+4+3 8 v0 (0+ ) = vc (0+ ) = v 3 1 i (t )dt - vc (0) = 0 0.1m 0 t 6ki (t ) + I ( s ) 0.8m - =0 s s 0.8 4/3 I ( s) = mA = 1 0.6s + 1 s+ 0.6 4 - t / 0.6 i (t ) = e u (t )mA 3 8 V0 (t ) = 2ki (t ) = e- t / 0.6V 3 0.6 I ( s ) + i (t ) = 4 - t / 0.6 e u (t )mA 3 Problem 12FE-1 The output function of a network is expressed using Laplace transforms in the following form. V0 ( s ) = 12 s ( s + 1)( s + 2) Find the output as a function of time v0(t). Suggested Solution V0 ( s ) = 12 A B C = + + s ( s + 1)( s + 2) s s + 1 s + 2 12 |s = 0 = A = 6 ( s + 1)( s + 2) 12 |s =-1 = B = -12 s ( s + 2) 12 |s =-2 = C = 6 s ( s + 1) V0 (t ) = (6 - 12e- t + 6e-2t )u (t )V V0 (t ) = (6 - 12e- t + 6e-2t )u (t )V Problem 12FE-2 The Laplace transform function representing the output voltage of a network is expressed as V0 ( s ) = 120 s ( s + 10)( s + 20) Determine the time domain function and the value of the v0(t) at t=100mSec. Suggested Solution V0 ( s ) = 120 A B C = + + s ( s + 10)( s + 20) s s + 10 s + 20 120 |s = 0 = A = 0.6 ( s + 10)( s + 20) 120 |s =-10 = B = -1.2 s ( s + 20) 120 |s =-20 = C = 0.6 s ( s + 10) V0 (t ) = (0.6 - 1.2e -10t + 0.6e-20t )u (t )V V0 (t ) |t = 0.1sec = 0.24V V0 (t ) |t = 0.1sec = 0.24V Problem 12FE-3 The Laplace transform function for the output voltage of a network is expressed in the following form V0 ( s ) = 12( s + 2) s ( s + 1)( s + 3)( s + 4) Determine the final value i.e. vo ( t ) as t , of this voltage. Suggested Solution V0 ( s ) = and V0 () = sV0 ( s ) V0 () = 12( s + 2) |s = 0 = 2v s ( s + 1)( s + 3)( s + 4) 12( s + 2) s ( s + 1)( s + 3)( s + 4) V0 = 2v ...
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