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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 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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. 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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 threephase system containing both phase voltages and line voltages if Van = 10045 V rms . Label all magnitudes and assume an abcphase 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 = 100195 V rms
Real Vca = 173.2165 V rms Vbn = 100 75 V rms Vbc = 173.2 45 V rms Problem 10.2
A positivesequence threephase 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 abcsequence balanced threephase 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 = 208120 V rms Problem 10.4
Sketch a phasor representation of a balanced threephase system containing both phase voltages and line voltages if Van = 12060 V rms . Label all magnitudes and assume an abcphase 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 = 208150 V rms Vbn = 120 60 V rms Problem 10.5
A positivesequence threephase 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 threephase system containing both phase voltages and line voltages if Vab = 20845 V rms . Label all phasors and assume an abcphase 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 positivesequence balanced threephase wyeconnected source with a phase voltage of 100 V supplies power to a balanced wyeconnected 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 positivesequence balanced threephase wyeconnected 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 abcsequence balanced threephase wyeconnected source supplies power to a balanced wyeconnected 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 threephase balanced wyewye system, the source is an abcsequence 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 abcsequence set of voltages feeds a balanced threephase wyewye 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 threephase wyewye system, the source is an abcsequence 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 abcsequence of voltages feeds a balanced threephase wyewye 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 threephase wyewye system, the source is an abcsequence 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 abcphase sequence balanced threephase 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 threephase wyewye 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 threephase wyewye 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 threephase wyewye 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 threephase wyewye 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 threephase wyewye system, the source is an abcsequence 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 threephase wyewye system, the source is an abcsequence set of voltages and Van = 12040 V rms. If the aphase 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 abcsequence set of voltages feeds a balanced threephase wyewye 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 threephase balanced positivesequence system a deltaconnected 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 208220 V rms 20820 V rms 8 0.4 0.4 j0.8 j0.8 j6 B 208100 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 threephase deltawye system the source had an abcphase 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 abcphasesequence threephase balanced wyeconnected 60Hz source supplies a balanced deltaconnected load. The phase impedance in the load consists of a 20  resistor in series with a 50mH 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 abcphasesequence threephase balanced wyeconnected source supplies a balanced deltaconnected 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 wyedelta 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 abcphasesequence threephase balanced wyeconnected 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 wyeconnected 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 abcphasesequence threephase balanced wyeconnected source supplies a balanced deltaconnected 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 abcphasesequence threephase balanced wyeconnected 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 threephase wyedelta system, the source has an abcphase 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 threephase wyedelta system, the source has an abcphase 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 threephase balanced deltadelta system, the source has an abcphase 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 threephase deltadelta system, the source has an abcphase 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 threephase 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 threephase system, the abcphasesequence 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 threephase system, the source is a balanced wye with an abcphase 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 threephase system, the source is a balanced wye with an abcphase 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 threephase system the source, which has an abcphase 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 threephase system, the source has an abcphase sequence and is connected in delta. There are two parallel wyeconnected 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 threephase system, the source has an abcphase 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 threephase system, the source has an abcphase 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 threephase deltaconnected 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 abcphasesequence 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 wyeconnected load. Suggested Solution
Vab = 44060 V rms Van = 440 3 ( 60  30 ) = 25430 V rms Equivalent singlephase (aphase) 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 abcsequence wyeconnected source having a phasea 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 threephase balanced wyewye 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 threephase wyewye 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 threephase loads. The plants receive their power from a balanced threephase 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 threephase 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 threephase 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 threephase 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 threephase loads. The power lines to the shopping center represent a threephase 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 threephase 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 threephase 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 threephase 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 Perphase 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 threephase 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 threephase 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 singlephase users and threephase users. The utility must provide threephase users, typically industries, with all three phases. However, singlephase users, residential and light commercial, are connected to only one phase. To reduce cable costs, all singlephase users in a neighborhood are connected together. This means that even if the threephase users present perfectly balanced loads to the power grid, the singlephase loads will never be in balance, resulting in current flow in the neutral connection. Consider the 60Hz, abcsequence network shown. With a line voltage of 41630 V rms , phase a supplies the singlephase users on A Street, phase b supplies B Street and phase c supplies C Street. Furthermore, the threephase 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 Threephase 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 threephase abcsequence wyeconnected 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 threephase abcsequence wyeconnected 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 threephase source 60 Hz _ C C C Balanced threephase 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 threephase source 60 Hz _ C C C Balanced threephase 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 threephase source 60 Hz _ C C C Balanced threephase 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 threephase source 60 Hz _ C C C Balanced threephase 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 10FE1
A wyeconnected 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 10FE2
A balanced threephase deltaconnected 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 threephase 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 10FE3
Two balanced threephase 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 linetoline 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 10FE4
The total complex power at the load of a threephase 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)  e2(t  2) ]u (t  2) Suggested Solution f (t ) = [e  (t  2)  e2(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  e3t u (t ) 3 6 2 for s = 0,
B. 2 1 1 f (t ) = + e 2t  e3t 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  e2t )u (t ) 4 4 5 1 f (t ) = ( + e 6t  e2t )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 + e6t )u (t ) 4 4 1 3 f (t ) = ( e 2t + e6t )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 + e2t ) 4 4 3 1 f (t ) = ( e 6t + e2t ) 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 )  2e2t 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  e4t )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 + 4te2t )u (t ) f (t ) = (te  t  3e t + 4te2t )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 ) = (  e4t  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 2e2t  5te 2t + 6e2t  6e3t )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 + 4e2t
f (t ) = te t  3e  t + 4e2t 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  4te4t }
f (t ) = 0.0625{1  e 4t  4te4t } 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 )  e2(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 ) = e2t 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)  5e3(t  2) ]u (t  2) f (t ) = 1 1 u (t  1) + e2(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) + 4e3(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 ) = e2t , 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 ) = e2t , 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 + 3e2t ]u (t ) 4 1 s y (t ) = (te  t  e  t + e2t )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  ) e2 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  e2t )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  e2t )u (t ) t t t f (t ) = (e  t  te 2t  e2t )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 ) = 4e5t 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 12FE1
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 + 6e2t )u (t )V V0 (t ) = (6  12e t + 6e2t )u (t )V Problem 12FE2
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 + 0.6e20t )u (t )V V0 (t ) t = 0.1sec = 0.24V V0 (t ) t = 0.1sec = 0.24V Problem 12FE3
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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