Tài liệu Bài giải mạch P9 docx

We obtain I by applying the principle of current division twice... First, we convert the circuit into the frequency domain... Since the left portion of the circuit is twice as large as t

(a) angular frequency ω = 10 3 rad/s

(a) v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°)

(b) i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)

i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°)

Thus, i(t) leads v(t) by 20°

Thus, y(t) leads x(t) by 9.24°

Chapter 9, Solution 7

If f(φ) = cosφ + j sinφ,

)(j)sinj(cosjcosj-sin

d

df

φ

=φ+φ

=φ+φ

−

°

∠+ j2 =

°

∠

°

∠53.13-5

4515

+ j2 = 3∠98.13° + j2 = -0.4245 + j2.97 + j2

= -0.4243 + j4.97

(b) (2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57°

j4)-j)(3(2

20-8+

°

∠

+j125-

10

°

∠26.57-11.18

20-8

+

14425

)10)(

12j5-+

−

= 0.7156∠6.57° − 0.2958

− j0.71

= 0.7109 + j0.08188 − 0.2958 − j0.71

4j3

−

+ = 2 +

6425

)8j5)(

4j3(+

++

= 2 +

89

3220j24j

2j1 = 4∠-10° +

°

∠

°

∠63

63.43-236.2

∠

504809

20-6108

=

064.3j571.2863.8j5628.1

052.2j638.53892.1j879.7

−

−+

−+

+

=

799.5j0083.1

6629.0j517.13

81.2-533.13

4j3-

−

+ =

25144

)5j12)(

4j3(-

+

++

2 1

zz

zz

−

+

=

)j15(

)j1(9+

+

115

j)-15)(

j1(9-

= -0.6372 – j0.5575

−

+ =

25144

)5j12)(

4j3(-

+

++

2 1

zz

zz

−

+

=

)j15(

)j1(9+

+

115

j)-15)(

j1(9-

.246

9.694424186

)5983.1096.16)(

8467

(

)8060)(

8056.13882.231116

62

(

j j

j j

j j

+

−

=+

+

−

−+

+

(c) (−2+ j4)2 (260− j120) =−256.4− j200.89

(a)

j1-5-

3j26j10

+

−+

16

10-4-3020

0jj1

j1j1

j1j

0jj1

The phasor form is 5∠-100°

(c) 4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°)

The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87°

Chapter 9, Solution 17

(a) Let A = 8∠-30° + 6∠0°

= 12.928 – j4 = 13.533∠-17.19°

a(t) = 13.533 cos(5t + 342.81°)

= 3.881 + j42.33 = 42.51∠84.76°

b(t) = 42.51 cos(120πt + 84.76°)

(c) Let C = 4∠-90° + 3∠(-10° – 90°)

= -j4 – 0.5209 – j2.954 = 6.974∠265.72°

v2 = 10 cos(40t + 53.13°)

(c) i1(t) = 2.8 cos(377t – π/3)

(d) I2 = -0.5 – j1.2 = 1.3∠247.4°

)t(

i2 = 1.3 cos(10 3 t + 247.4°) Chapter 9, Solution 19

(a) 3∠10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5

= -1.376 + j3.021 = 3.32∠114.49°

Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t +

114.49°)

(b) 4∠-90° + 3∠-45° = -j40 + 21.21 – j21.21

= 21.21 – j61.21

= 64.78∠-70.89°

Therefore, 40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°)

(c) Using sinα = cos(α − 90°),

20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699

= 6.7101 – j6.641 = 9.44∠-44.7°

Therefore, 20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°)

= 9.44 cos(400t – 44.7°)

(a) V =4∠−60o −90o −5∠40o =−3.464− j2−3.83− j3.2139=8.966∠−4.399o

Hence,

)399.4377cos(

966

(b) I =10∠0o + jω8∠20o −90o, ω =5, i.e I =10+40∠20o =49.51∠16.04o

)04.165cos(

51

324.8)

(b) G=8∠−90o +4∠50o =2.571− j4.9358=5.565∠−62.49o

)49.62cos(

565.5)

i.e H =0.25∠−90o +0.125∠−180o =−j0.25−0.125=0.2795∠−116.6o

)6.11640cos(

2795.0)

v( ) 4 2 ( )

10

o V

j

V V j

V

F =10 + 4 − 2 , ω =5, =20∠−30

ωω

o j

j V

j V j

V

F =10 + 20 − 0.4 =(10− 19.6)(17.32− 10)=440.1∠−92.97

)97.925cos(

1.440)

(a) v(t) = 40 cos(ωt – 60°)

(b) V = -30∠10° + 50∠60°

= -4.54 + j38.09 = 38.36∠96.8°

010

j = ∠ ° ω=ω

+ V

V

10)j1( − =

V

°

∠

=+

=

−

= 5 j5 7.071 45j

9010(20j

45

ω++

454j

V

°

∠

=+

°

∠

= 3.43 -110.96

3j5

80-20

V

Therefore, v(t) = 3.43 cos(4t – 110.96°)

(a)

2,

45-432jωI+ I= ∠ ° ω=

°

∠

=+ j4) 4 -453

°

∠

13.535

45-4j43

45-4ITherefore, i(t) = 0.8 cos(2t – 98.13°)

(b)

5,

2256jj

°

∠

56.26708.6

2253

j6

225

I

Therefore, i(t) = 0.745 cos(5t – 4.56°)

Chapter 9, Solution 26

2,

01j2

ω++

ωI I I

12j

122

I

°

∠

=+

= 0.4 -36.87

5.1j2

10-110j

10050

ω+

.380(

)t(v)

10(j

1C

j

1

6 -

×

=ω

65-60

Z

V I

Therefore, i(t) = 30 cos(500t – 155°) A

60-65

I

V Z

Since V and I are in phase, the element is a resistor with R = 6.5 Ω

V = 180∠10°, I = 12∠-30°, ω = 2

Ω+

0180

I

V

Z

One element is a resistor with R = 11.49 Ω

The other element is an inductor with ωL = 9.642 or L = 4.821 H

Chapter 9, Solution 33

2 L

vo =

LC

1C

1

ω

=ω

105(

V

2j)1)(

2(jL

2j-)25.0)(

2(j

1C

j

1

=

=ω

−

2

2j2j2j2

2j

Let Z be the input impedance at the source

2010

100200mH

100 → jωL= j x x − 3 = j

500200

1010

11

F

x x j C

→

ωµ

1000//-j500 = 200 –j400

1000//(j20 + 200 –j400) = 242.62 –j239.84

o j

Z =2242.62− 239.84=2255∠−6.104

mA896.361.26104.62255

1

5(j

1C

j

1

=

=ω

Let Z1 =-j,

5j2

10j5j2

)5j)(

2(5j

2

Z Z

Z I

=+

+

+

8j5

20j)2(5j2

10jj-

5j2

10jx

I

Therefore, ix(t)= 2.12 sin(5t + 32°) A

(a) -j2

)6/1)(

3(j

1C

j

1F

6

1

=

=ω

j4

2j-

I Hence, i(t) = 4.472 cos(3t – 18.43°) A

)12/1)(

4(j

1C

j

1F

12

1

=

=ω

→



12j)3)(

4(jLjH

050

Z

V I Hence, i(t) = 10 cos(4t + 36.87°) A

= (50 0 ) 41.6 33.69j12

8

12j

V Hence, v(t) = 41.6 cos(4t + 33.69°) V

Chapter 9, Solution 39

10j810j5j

)10j-)(

5j(8)10j-

||

5j

−+

=+

20j10

8

040

j

10j-

5j

I

Therefore, i1(t)= 6.248 cos(120πt – 51.34°) A

=)t(

i2 3.124 cos(120πt + 128.66°) A

(a) For ω=1,

j)1)(

1(jLjH

20j-)05.0)(

1(j

1C

j

1F

05

40j-j)20j-

||

2

−+

=+

04j0.802

1.98

04

V I

Hence, io(t)= 1.872 cos(t – 22.05°) A

(b) For ω=5,

5j)1)(

5(jLjH

4j-)05.0)(

5(j

1C

j

1F

05

4j-5j)4j-

||

25

−+

=+

04j4

1.6

04

V I

Hence, io(t)= 0.89 cos(5t – 69.14°) A

(c) For ω=10,

10j)1)(

10(jLjH

2j-)05.0)(

10(j

1C

j

1F

05

4j-10j)2j-

||

210

−+

=+

049

j1

04

V I

Hence, io(t)= 0.4417 cos(10t – 83.66°) A

,

=ω

j)1)(

1(jLjH

j-)1)(

1(j

1Cj

1F

1j-1)j-

||

)j1(

=

Z

j2

10s

j2

)10)(

j1()j1()j1)(

200(j

1C

j

1F

×

=ω

→

µ

20j)1.0)(

200(jLjH

1

20j40j2-1

j100-j10050

)(50)(-j100-j100

70

20j)060(20j403020j

20jo

V

Thus, vo(t)= 17.14 sin(200t + 90°) V

or vo(t)= 17.14 cos(200t) V

2j)1)(

2(jLjH

5.0j-)1)(

2(j

1Cj

1F

=+

−

−

5.1j1

5.1j15.0j2j

5.0j2j

200(jLjmH

j-)105)(

200(j

1C

j

1mF

×

=ω

→



4.0j55.010

j35.0j25.0j3

12j

14

1

−

=

++

−

=

−++

=

Y

865.0j1892.14.0j55.0

11

°

∠

=+

°

∠

=+

°

∠

865.0j1892.6

065

06

Z I

Thus, i(t) = 0.96 cos(200t – 7.956°) A

We obtain I by applying the principle of current division twice o

2j-

1 =

j2-2

j4-j42

||

-j2)(4j

Z

j1

j10-)05(3j12j-

2j-

2 1

1

Z I

=+

10-j1

j10-j-1

j-j2

2

-j2-2

j-)1.0)(

10(j

1C

j

1F

1

10(jLjH

2

2j4

8j2j

j1.60.8

s 2 1

Z Z

Z I

)405)(

43.63789.1(o

I

Thus, io(t)= 2.325 cos(10t + 94.46°) A

First, we convert the circuit into the frequency domain

=++

−

+

−+

63.52854.10

5626

.8j588.42

54

j2010j

)4j20(10j2

We can solve this using nodal analysis

4338.0i

4.94338.020

j30

29.24643.15I

29.24643.1503462.0j12307.0

402V

402)01538.0j02307.005.0j1.0(V

020j3020j10

x x 1 1

°+

+

−

=

−++

Chapter 9, Solution 49

4j1

)j1)(

2j(2)j1(

||

2j2

+

−+

=

−+

I

j1

2jj12j

2j

105

x =0 ∠ °=

I

4j

j12

j

j1

j1T

Using the current dividing rule:

V)50t100cos(

50

v

5050I20

V

505.2405.2j40510j2010j

10jI

x

x x

2(j

1C

j

1F

1

2(jLjH

5

The current I through the 2-Ω resistor is

4j32j5j1

s = −+

I

Therefore,

=)t(

is 50 cos(2t – 53.13°) A

Chapter 9, Solution 52

5.2j5.2j1

5j5j5

25j5

s s

s 2 1

=

)5.2j5.2

5.2(j5

430

−

+

=+

)j5)(

308

46,

7692.01532.0210

42

2

j

x j Z j

j

x j

)2308.01538.9()10//(

)

8

163.0878.66062.0726.4

A 64.28721.83575.188.6

3060Z

3060

o

o o

Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below

V s

−

V 1 +

V

)j1(4

V

)j1(62 1

s =V +V = −

V

=s

j

48j

o

I

j8j4

-)8j((-j0.5)j4

-)8j(1

I

5.0j8

j8

-j0.52

I

)8j(12j20

)8j(2

j-2

j812j20

1j2

3

j26-4

3 → jωL= j

30/

3015

30)15///(

j j

j x j j

)06681.02(033.0)

06681.02//(

j j

j

j Z

Chapter 9, Solution 57

2H

2 → jωL= j

j C

2.1j6.2j22j

)j2(2j1)j2//(

2j1

−+

−+

=

−+

=

S 1463.0j3171.0Z1

(a) -j2

)1010)(

50(jCjmF

×

=ω

→



5.0j)1010)(

50(jLjmH

)2j1(

||

15.0j

Z

2j2

2j15.0j

−+

=

Z

)j3(25.05.0j

50(jLjH

2

20j-)101)(

50(j

1C

j

1mF

×

=ω

→



For the parallel elements,

20j-

110j

120

11

p

++

=

Z

10j10

||

)2j1(6

Z

)4j2()2j1(

)4j2)(

2j1(6

+

−+

=

Z

5385.1j308.26

=

−+

+

=+

−++

=(25 j15) (20 j50)//(30 j10) 25 j15 26.097 j5.122 51.1 j9.878

Z

All of the impedances are in parallel

3j1

15j

12j1

1j1

11

Z

4.0j8.0)3.0j1.0()2.0j-)4.0j2.0()5.0j5.0(1

eq

−

=

−++

−++

1eq

Chapter 9, Solution 62

2mH → jωL= j(10×103)(2×10-3)= j20

100j-)101)(

1010(j

1C

j

1F

×

×

=ω

→

µ

50 Ω j20 Ω

+

+ − V +

01

=

V

)50)(

2()100j20j50)(

01(

V

80j15010080j50

in in

V

First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with the corresponding delta

5.22j1020

450j200z

,333.13j3015

j

450j200

z

45j2010

300j150j200

z

3 2

−+

Z

821.3j70.21)16j10

(

z

938.8j721.833

.29j40

)16j10)(

333.13j30()16j10

(

z

1 T

3

2

Chapter 9, Solution 64

A7.104527.14767.1j3866.0Z

9030

I

5j192

j6

)8j6(10j4

=

||

)6j4(2

Z

2j7

)4j3)(

6j4(2

+

−+

10120T

1705

j60

)10j40)(

5j20()10j40(

||

)5j20(

9060T

I 2 j10 Ω

I I

I

j12

2j85j60

10j40

+

=+

+

=

I I

I

j12

j45j60

5j20

−

=+

−

=

2 1

ab -20I j10I

I I

V

145

j)(150)-12

(j12

150-ab

+

=+

=

)76.9725.4)(

24.175457.12(

10(j

1C

j

1F

5

×

=ω

→

µ

)80j60(

||

20j60

Z

60j60

)80j60)(

20j(60

−+

10(j

1C

j

1F

×

=ω

→

µ2060

||

)10j40(

||

2050j-

Z

10j60

)10j40)(

20(50j-

++

1

Z

Y 0.0197∠74.56° S = 5.24 + j18.99 mS

1j3

12j5

12j-

14

11

o

+

=+

=

Y

6.1j8.05

)2j1)(

4(2j1

Y

)6.0j8.0()333.0j()1(6.0j8.0

13

j-

11

11

o

+++

=

−++

Y

773.4j4378.05j

Y

97.22

773.4j4378.05.0773.4j4378.0

12

11

eq

−+

=+

+

=

Y

2078.0j5191.01

2078.0j5191.0eq

Make a delta-to-wye transformation as shown in the figure below

c b

9j75

j15

)10j15)(

10(15j1010j5

)15j10)(

10j-

+

−

=++

)15j10)(

5(

)10j-)(

5(

||

)2

||

)5.3j5.6(9j7

Z

5.4j5.13

)8j7)(

5.3j5.6(9j7

−+

We apply a wye-to-delta transformation

j4 Ω

Z eq -j2 Ω

2j22

j

4j2j2

Z

j12

2j2

Z

j2-2j-

2j2

Z

8.0j6.13j1

)j1)(

4j()j1(

||

4j

)j1)(

1()j1(

=

Z

6.0j2.2

12

j2-

12j-

11

Transform the delta connections to wye connections as shown below

j2 Ω

j2 Ω j2 Ω

b

6j-18j-

102020

)20)(

20(

50

)10)(

20(

50

)10)(

20(3R

44)j6(j2

||

)82j(j2

Z

j4)(4

||

)2j8(j24

Z

j2-12

)4jj2)(4(8

j24ab

−+

++

=

Z

4054.1j567.3j24

Transform the delta connection to a wye connection as in Fig (a) and then

transform the wye connection to a delta connection as in Fig (b)

j2 Ω

j2 Ω j2 Ω

b

8.4j-10j

486j8j8j

)6j-)(

8j(

−+

64-10j

)8j)(

8j(

Z

=+

++

+++ )(4 ) (4 )( ) (2 )( )

2

6.9j4.46)4.6j)(

8.4j2()4.6j)(

8.4j4()8.4j4)(

8.4j2

25.7j5.14

.6j

6.9j4.46

Z

688.6j574.38.4j4

6.9j4.46

6.9j4.46

.12j574.3

)88.61583.7)(

906(

j7.25)-j4)(1.5

||

12j

||

4j-

||

)

||

6j

)5673.2j7494.0(

||

)3716.3j7407.0(

20j20-40

j20

)20j20)(

20j()20j20(

+

=+

=

Z

)j1(3

13j6

3j1)01(12j24

12j4

+

=+

Z

Z V

=+

3

j)j1(3

1j1

j20

j20

20j

)tcos(ω = ω + °

This is achieved by the RL circuit shown below, as explained in the previous problem

1X

π

=ω

=

394.9566116R

|Z

|XX

1fX

V V

−

=

)1020)(

102)(

2(

1C

1

×

×π

=ω

=

))53.979(tan-90(979.35

979.3j3.979

5

-j3.979

2 2

)51.38-90(83.1525

979.3i

+

=

V V

Therefore, the phase shift is 51.49° lagging

1f

2π =

=ω

=

×π

=π

=

)1020)(

5)(

2(

1RC

)]

6(8[)6(8

−++

−+

=

−+

=

X j R

X j R X j R

Z

i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X

Equating real and imaginary parts:

8R = 5R + 40 which leads to R=13.33Ω 6R-XR =30-5 which leads to X=4.125Ω

j30

)60j30)(

30j()60j30(

||

30j

+

+

=+

=+

+

=+

31j43

)21j43)(

10j()40(

)01)(

21.80028.9(

Z

Z V

+

=+

=

03.2685.47

)77.573875.0)(

87.81213.21(21j43

21j3

1

5

22

j1

2j60

j30

60j

V V

V

+

=+

=

)6.1131718.0)(

56.268944.0(

V Therefore, the phase shift is 140.2°

(b) The phase shift is leading

=ω

→

 j L j(2 )(60)(200 10 ) j75.4mH

)0120(4.75j50R

4.75j4

.75j50R

4.75j

i

++

=+

=

69.2688.167

)0120)(

904.75()0120(4.75j150

4.75jo

V

=o

V 53.89∠63.31° V

45.5647.90)

0120(4.75j50o

V

=o

V 100∠33.55° V

(c) To produce a phase shift of 45°, the phase of Vo = 90° + 0° − α = 45°

Hence, α = phase of (R + 50 + j75.4) = 45°

For α to be 45°, R + 50 = 75.4 Therefore, R = 25.4 Ω

Chapter 9, Solution 81

Let Z1 =R1,

2 2

1R

ω+

=

x x

1R

ω+

=

2 1

=ω

+

2

2 1

3 x

1R

R

RCj

1R

R

RCC

1R

RC

2 3

1 x 2

1

3 x

R

R

s 2

R

R

s 1 2

Let

s 1

1

||

Rω

=

Z , Z2 =R2, Z3 =R3, and Zx =Rx + jωLx

1CRjRC

j

1R

CjR

s 1 1

s 1

s 1

ω+

RRR

1CRjRRLj

1

3 2 1

s 1 3 2 x

Equating the real and imaginary components,

1

3 2

R R

R =

)CR(R

RR

1

3 2

s 3 2

L =Given that R1 =40kΩ, R2 =1.6kΩ, R3 = k4 Ω, and Cs =0.45µF

=Ω

=Ω

=

40

)4)(

6.1(R

RRR

1

3 2

1R

ω+

=

4 4

1

||

Rω

=

jCR

Rj-1CRj

R

4 4

4 4

3 2 3 2

4

2 4 2

4 4 1 4

C

jRRR1

CR

)jCR(RRj-

ω

−

=+

ω

+ω

Equating the real and imaginary components,

3 2 2

4

2 4 2 4

1CR

RR

=+

(1)

2

3 2

4

2 4 2

4

2 4 1

C

R1CR

CRR

ω

=+ω

ω

(2) Dividing (1) by (2),

2 2 4

4

CRC

R

1

ω

=ω

4 4 2 2

2

CRCR

1

=ω

4 4 2

2C R CR

1f

2π =

=ω

4 2 4

2 R C C R

2

1 f

π

=

Chapter 9, Solution 86

84j-

195j

1240

1

++

=

Y

0119.0j01053.0j101667

=

=

2.183861.4

100037

.1j1667.4

10001

Y

Z

Z = 228∠-18.2° Ω

102)(

2(

j-50

Cj

Z

)1010)(

102)(

2(j80Lj

Z

66.125j80

111

1

Z Z Z

66.125j80

179

.39j50

1100

11

+

+

−+

=

Z

)663.5j605.3745.9j24.1210(10

1 -3

−+

++

1C

1π

=ω

Lf2L

would cause the capacitive impedance to double, while ω = π would cause the inductive impedance to halve Thus,

40j12015j40j

=

Z

Z = 120 – j65 Ω

Cj

1R

||

Ljin

−ω+

ω

+

=ω+ω+

=

C

1LjR

RLjCL

Cj

1LjR

Cj

1RLjin

Z

2 2

in

C

1LR

C

1LjRRLjCL

−ω+

−ω

=

Z

To have a resistive impedance, Im(Zin)=0 Hence,

0C

1LC

LR

−ω

C

1LC

R2

ω

−ω

=ω

1LCC

2 2 2

ω

+ω

=

(1) Ignoring the +1 in the numerator in (1),

0145jX

R80

s

++

°

∠

=++

I

80

V

jXR80

)145)(

80(50

++

jXR80

)0145)(

jXR()jXR(

°

∠+

=+

V

jXR80

)145)(

jXR(110

++

+

From (1) and (2),

jXR

80110

5

11)80(jXR

30976X

From (1),

23250

)145)(

80(jXR

53824X

RR160

6400+ + 2 + 2 =

47424X

RR

||

RCj

LRjC

j-

ω+ω

=

Z

2 2 2

2 2

2

LR

LRjRLC

j-

ω+

ω+ω

+ω

=

To have a resistive impedance, Im(Zin)=0

Hence,

0LR

LRC

1-

2 2 2

2

=ω+

ω+ω

2 2 2

2

LR

LRC

1

ω+

ω

=ω

2 2

2 2 2

LR

LR

Cω

ω+

=

where ω=2πf =2π×107

)109)(

1020)(

104

(

)10400)(

104

(109

12 14

2 4

×

×

×π

×

×π+

×

nF72

169

2

π

π+

x Y

Z

1048450

75100

6

(b) γ = ZY = 100∠75o x450∠48o x10− 6 =0.2121∠61.5o

Z

20j

0115