Chapter Ten Solutions - University of pramamoo/CourseWork/ECE250ClassMaterials/solutions...12 π≥ π π π ∴π π ... SOLUTIONS Engineering Circuit Analysis, ... CHAPTER TEN (Phasor Analysis) SOLUTIONS Engineering Circuit Analysis,

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  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    1. (a) (b) (c)

    33 3

    3

    rad rad

    2 10T 4(7.5 2.1)10 21.6 10 , 290.9 rad/s

    21.6

    ( ) 8.5sin (290.9 ) 0 8.5sin (290.9 2.1 10 )

    0.6109 2 5.672 or 325.0

    ( ) 8.5sin (290.9 325.0 )

    = = = =

    = + = + = + = = +

    t

    f t t

    f t t

    8.5sin (290.9 325.0 ) 8.5

    cos(290.9 235 ) 8.5cos (290.9 125 )

    t

    t t

    + =+ =

    8.5cos ( 125 )cos 8.5sin125

    sin 4.875 cos 290.9 6.963sin 290.9+ +

    = +

    t

    t t t

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    2. (a) (b) (c) (d)

    10cos 4sin ACos ( ), A 0, 180 180

    A 116 10.770, A cos 10, A sin 4 tan 0.4, 3 quad

    21.80 201.8 , too large 201.8 360 158.20

    d

    t t wt + + + > <

    = = = = = = = = =

    200cos (5 130 ) Fcos5 G sin 5 F 200cos130 128.56

    G 200sin130 153.21

    + = + = = = =

    t t t

    sin10 5( ) 5cos10 3sin10 0, 0 1 , 10 1.0304,

    cos10 30.10304 ; also, 10 1.0304 , 0.4172 ; 2 : 0.7314

    = = = =

    = = + =

    ti t t t t s t

    tt s t t s s

    0 10ms, 10cos100 12sin100 ; let 10cos100 t =12sin100 t

    10tan100 t = , 100 0.6947 2.211ms 0 2.211ms

    12

    < <

    = = < = =

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    45. At = 4 rad/s, the 1/8-F capacitor has an impedance of j/C = -j2 , and the 4-H inductor has an impedance of jL = j16 . (a) (b)

    (8 16) (2 2) 16(3 1)abOC : Z

    10 14 10 14

    2.378 1.7297

    in

    j j j

    j j

    j

    + += =+ +

    = 16 32

    abSC : Z (8 2) (2 16)8 2 2 16

    2.440 1.6362

    = + = + +

    =

    in

    in

    j jj j

    j j

    Z j

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    46. f = 1 MHz, = 2f = 6.283 Mrad/s 2 F -j0.07958 = Z1 3.2 H j20.11 = Z2 1 F -j0.1592 = Z3 1 H j6.283 = Z4 20 H j125.7 = Z5 200 pF -j795.8 = Z6

    The three impedances at the upper right, Z3, 700 k, and Z3 reduce to j0.01592

    Then we form Z2 in series with Zeq: Z2 + Zeq = j20.09 .

    Next we see 106 || (Z2 + Zeq) = j20.09 .

    Finally, Zin = Z1 + Z4 + j20.09 = j26.29 .

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    47. 2 H 2, 1F 1 Let I 1 0 A

    V 2V I I 0.5 1 1

    V 2 (1 1) ( 1) 1 1

    1 0 1 1 1V 0.5 0.55

    V 1 1 1 1

    1Now 0.5 2 , 0.5S 2H

    2

    = = = + = + = + + = +

    = = = +

    =

    L c in L

    in

    inin

    j j

    j v j

    j j j j

    jj

    j j

    s jj

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    48. (a)

    (b) ,106

    R 21.25

    = = in ab

    (c)

    500, Z 5 10 1 5 9

    1 5 9 9Y Y 500C

    5 9 106 106

    9C 169.81 F

    53,000

    inRLC

    inRLC c

    j j j

    j

    j

    = = + = + = = = =

    +

    = =

    ,

    9 11000 Y

    53 5 20 0.5

    0.012338 0.12169S or 0.12232 84.21 S

    = = ++

    = +

    in ab j j j

    j

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    49. (a)

    (b) (c) (d)

    6 2

    4 6 2

    6 26

    4 6 2

    4 2 6 4 2

    4 2

    0.1R 550 : Z 500

    100 0.001

    50,000 0.6 100 0.001Z

    100 0.001 100 0.001

    5 10 0.0006 (60 50 )Z

    10 10

    5 10 0.006R 550 5.5 10

    10 10

    5.5 10 5 10 10

    0.5 10

    in in

    in

    in

    in

    j

    j

    j j

    j j

    j

    = = ++

    + = +

    + + =+

    + = = +

    + =

    = 6 2 10 50.5 10 , 10 , 10 rad/s = =

    6 4 24 6 2

    2 5 10

    5 10 105 5

    10X 50 0.5 10 0.5 10 10

    10 10

    0, 2 10 10 0

    2 10 4 10 4 1010 10 rad/s

    2

    = = = + +

    = + =

    = = =

    in

    3

    6 4 2

    8 2

    6 4 23

    8 2

    6 4 2 6 6 2

    6 6 2

    100 0.001 50,000 0.6G 1.8 10 : Y

    50,000 0.6 50,000 0.6

    5 10 6 10 (50 6 )

    25 10 0.36

    5 10 6 101.8 10

    25 10 0.36

    5 10 6 10 4.5 10 648 10

    0.5 10 48 10 102.06

    + = = +

    + + = + + =

    + + = +

    = =

    in in

    j j

    j j

    j

    Krad/s

    48 2

    4 6 2

    6 2 4

    6

    10B 1.5 10

    25 10 0.36

    10 37.5 10 54 10

    54 10 10 37.5 10 0,

    100 8110 52.23 and 133.95krad/s

    108 10

    = = +

    = + + =

    = =

    in

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    50.

    (a) 11 131

    I 0.1 30V 20 23.13 V 20 V

    Y (3 4)10 = = = =

    + j

    (b) 2 1 2V V V 20V= = (c) (d)

    32 2 2

    3 1 2

    33 33

    3

    I Y V (5 2)10 20 23.13 0.10770 1.3286 A

    I I I 0.1 30 0.10770 1.3286 0.2 13.740 A

    I 0.2 13.740V 44.72 77.18 V V 44.72V

    Y (2 4)10

    j

    j

    = = + = = + = + =

    = = = =

    1 3V V V 20 23.13 44.72 77.18 45.60 51.62

    V 45.60V

    = + + + = =

    in

    in

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    51.

    (a)

    (b)

    1

    1

    1 1

    50 F 20 Y 0.1 0.05

    1 1000 1Y R 8 4

    1000 C 0.1 0.05RC

    1R 8 and C 250 F

    4

    in

    in

    j j

    j jjj

    = +

    = = = +

    = = =

    11

    1 1 11

    12000 : 50 F 10 Y 0.1 0.1

    500R

    C

    500R 5 5 R 5 , C 100 F

    C

    inj jj

    j j

    = = + =

    = = =

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    52. (a)

    Gin Bin

    0 1 2 5 10 20

    0 0.0099 0.0385 0.2 0.5 0.8 1

    0 0.0099 0.1923 0.4 0.5 0.4 0

    2

    2

    2

    2 2

    10 10Z 1

    10Y

    10 10

    10Y

    100

    10G , B

    100 100

    in

    in

    in

    in in

    j

    j j

    j j

    j j

    j

    += + =

    +

    + =+

    = =+ +

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    53.

    1 1 2 1 21 1 2 1 2

    1 2

    2 1 2 1 2

    2 1 2 1 2 1 2

    2

    V V V5 , 75 5V 3V 3V 5V 5V

    3 5 3

    (5 2) V 2V 75

    V V V V10

    3 5 6

    10V 10V 6V 6V 5V 300 4V (5 4) V 300

    5 2 75

    4 300 1500 600 300 1200V

    5 2 2 17 30 8

    4 5 4

    v vj j j j j j

    j j

    j j j

    v

    j j

    j j j j j j

    j j

    j jj j j

    j j

    = + + = + +

    + = + + =

    + + + = + =

    = = =

    +

    60034.36 23.63 V

    25 30

    j

    j=

    (1)

    (2)

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    54.

    3I 5(I I ) 0 2I 5I 0

    3(I 5) 5(I I ) 6(I 10) 0

    5I (9 5) I 60 15

    0 5

    60 15 9 5 75 300I

    2 5 15 18

    5 9 5

    13.198 154.23 A

    B B D B D

    D D B D

    B D

    B

    j j j

    j j

    j j j

    j

    j j jj j j

    j j

    = + =+ + + =

    + =

    += =

    =

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    55.

    1 2

    1 2

    20cos1000 V, 20sin1000 V

    V 20 0 V, V 20V

    0.01H 10 , 0.1mF 10

    20 200, 0.04 2 2 0,

    10 25 10

    V 25(2 2) 70.71 45 V

    ( ) 70.71cos(1000 45 ) V

    s s

    s s

    x x xx

    x

    x

    v t v t

    j

    j j

    v v v jv j

    j j

    j

    v t t

    = = = =

    + + + = + =

    = = =

  • CHAPTER TEN (Phasor Analysis) SOLUTIONS

    Engineering Circuit Analysis, 6th Edition Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

    56. (a)

    (b)

    3 2 2

    1

    1

    3

    Assume V 1V V 1 0.5V, I 1 0.5mA

    V 1 0.5 (2 0.5) ( 0.5) 0.75 1.5V

    I 0.75 1.5mA, I 0.75 1.5 2 0.5 2.75 2 mA

    V 0.75 1.5 1.5 (2.75 2) ( 0.5)

    1000.25 2.875V V 34.65 94.

    0.25 2.875

    in

    in

    j j

    j j j j

    j j j j

    j j j j

    jj j

    +

    = = = = + = = = + = = +

    = = =

    97 V

    3 3

    2 2 12

    2 2 21 1

    2 2 3 2 3 3

    2

    0.5 Assume V 1V I 1A,

    V 1 X, I 1 X, I 2 X

    V 1 X (2 X)( X) 1 X 3X, I 1 X 3X, I 3 X 4

    V 1 X 3X 4X X 3X 1 5X (X 6X) X 6X 0

    X 6, X 6, Z 2.449K

    in

    in

    c

    j jx

    j