# 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

1. (a) (b) (c)

33 3

3

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

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

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

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

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

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

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

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,

108 10

= = +

= + + =

= =

in

• CHAPTER TEN (Phasor Analysis) SOLUTIONS

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

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

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

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

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

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

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