Bài giải phần giải mạch P11

We find VTh using the circuit in Fig.. 15 b We obtain ZTh from the circuit in Fig... 6Ωb We find VTh at the load terminals using Fig... From Fig.d, we obtain VTh using the voltage divisi

Chapter 11, Solution 1

)t50cos(

160)

t

(

)9018030t50cos(

2)30t50sin(

20-

)t50cos(

)20)(

160()t(i)t(v)

t

(

[cos(100t 60 ) cos(60 )]W1600

)20)(

160(2

1)cos(

IV2

150jLjH

3

100j-)10)(

20)(

500(

j-C

j

1F

ω

→

µ

030

1 = ∠ ° = ∠− °=

I

)t500sin(

2.0)90t500cos(

2.0)t(

06.0j12.056.261342.0j2

3.0100j200

030

1342.0)t(

=I1 I2 0.12 j0.14 0.1844 -49.4

I

)35t500cos(

1844.0)t(

For the voltage source,

]35t500cos(

1844.0[]t500cos(

30[)t(i)t(v)t(

At , t =2s p=5.532cos(1000)cos(1000−35°)

)935.0)(

5624.0)(

532.5(

2.0[]t500cos(

30[)t(i)t(v)t(

At t =2s, p=6cos(1000)sin(1000)

)8269.0)(

5624.0)(

6(

)56.25t500cos(

1342.0[]44.63500cos(

42.13[)t(i)t(v)t(

At , t =2s p=18cos(1000−63.44°)cos(1000+26.56°)

)1329.0)(

991.0)(

18(

1342.0[]56.26t500cos(

84.26[)t(i)t(v)t(

At t =2s, p=3.602cos2(1000+25.56°)

21329.0)(

602.3(

p=

=

p 0 0636 W

Chapter 11, Solution 3

10cos(2t+30°) → 10∠30°, ω=2

2jLjH

-j2Cj

1F

25

)2j2)(

2j()2j2(

°

∠

= 1.581 11.565

2j24

3010

2j

2j2

I

For the source,

)565.11-581.1)(

3010(2

1

= I V S

5.2j5.718.43905

=

S

The average power supplied by the source = 7 5 W

For the 4-Ω resistor, the average power absorbed is

1

For the inductor,

5j)2j()236.2(2

12

For the 2-Ω resistor, the average power absorbed is

12

10j20(

2

1 j)j2(

For mesh 2,

1

2 j10)

10j5j10(

1

2 j2)

j2(

jj20

5

I I

4j

)j2(51 1

j10-2 2

1

I V S

The average power supplied=43.65cos(12.1°)= 42 68 W

For the 20-Ω resistor,

6828.1P

38.256828.12j26j

)2j2(6j1

408I

2 1

−+

P3H = P0.25F = 0

W097.522

258.2P

258.238.256828.12j26j

6jI

2 2

=

Ω Ω

)604(

Substituting (2) into (3),

0)604(8608)j2

j10

6040

608j-j10

60406042

°

∠

=+

j10

j14)608j(j10

6032j608j)2j4

°

∠

=+

)14j1)(

604(1

Applying KVL to the left-hand side of the circuit,

o

o 0.14

20

Applying KCL to the right side of the circuit,

05j105j

−+

j10

10

V V

j

5j10

8∠ °= Vo +

j1

2080

o 1

V I

1R2

6 V1 V1−V2

+

At node 2,

o o

V I

But,

j20-

2 1 o

V V

=

Hence,

40j20

-)(

5

j6)-1(37

360j6

360j

940

1R2

1P

2 2

2

Chapter 11, Solution 9

rmsV8)2)(

4(V2

61

VP

2 o

The current through the 2 -kΩ resistor is

mA1k

377(

= -37.02 6.375 -37.02 k

)754.0(1

k10Z

2 ab

mA)68t377cos(

2)22t377sin(

2)t(

2

2

102ZI

8j28

(8)(-j2)-j2

L Z

Z 0 471+j 1 882Ω

We find VTh using the circuit in Fig (b)

2j-

j64-I

8(6864R

8P

2

L

2 Th max

V

W 99 15

(b) We obtain ZTh from the circuit in Fig (c)

)3j4)(

5(2j)3j4(

||

52j

−

−+

=

−+

=

Z

=

= * Th

L Z

Z 2 5−j 1 167Ω

j100)(-j40)(80

j100)(80

=

Z

=

= * Th

L Z

Z 51 2+j 1 6Ω(b) We find VTh at the load terminals using Fig (b)

)203)(

8()203(40j100j80

=

I

6j8

)2024)(

40j(-40

8(

241040R

8P

2

L

2 Th max

V

W 5 22

From Fig.(d), we obtain VTh using the voltage division principle

3j4)3010(3j9

3j4Th

8(3

10105R

8P

2

L

2 Th max

V

W 389 1

−

=+++

++

+

−

=

∗ 8.245 j2.3Z

Z

3.2j245.87.7j245.810j8j1624j10

)8j16)(

24j10(10jZ

Th Th

V245

.8IP

V12.158j53.7166.6555.173

)8j16(40j8j1624j10

10)

8j16(V

2 2

2 Th 2

rms max

++

=+

2 o

j-j

V V V V

o 2

j-2

Substituting (1) into (2),

2 2

2 (2 j)(j) (1 j)j

j1

1

2 = −

V

5.0j5.02

j11

L Z

Z 0 5−j 0 5Ω

We now obtain VTh from Fig (b)

12

o

V V

j1

12-

o = +

V

0)

2j

o − × V +V =

V

j1

)2j1)(

12(j2)

+

=+

8(2

512R

8P

2

L

2 Th max

V

W 90

Chapter 11, Solution 16

520/14

11

F20/1,4H

1

,

x j C j j

10<0o Vo -j5 j4 VTh

-

At node 1,

2 1

2 1 1 1

425

.05

2

10

V j

V V

V V j

.004

j

V V V

V

+

−+

=

→



=+

j20 Ω

30 Ω

40 Ω -j10 Ω

(a)

10j40

-j10))(

40(20j30

)20j)(

30()10j-

||

4020j

Z

Ω+

Using current division,

3.2j1.1-)5j(10j70

20j30

10j40

8(

5000R

8

P

L

2 Th max

V

W 96 53

Chapter 11, Solution 18

We find ZTh at terminals a-b as shown in the figure below

a b

(80)(-j10)20

j20-j10)(

||

8040

||

4020j

Z

154.10j23.21

Th = +

Z

=

= * Th

j)(6)(3-j2

)j3(

||

62j-

++

=++

=

Z

561.1j049.2

To get VTh, let Z=6||(3+j)=2.049+ j0.439

By transforming the current sources, we obtain

756.1j196.8)04(

258.70R

8

P

L

2 Th max

V

W 409 3

Chapter 11, Solution 20

Combine j20 Ω and -j10 Ω to get

-j20-j10

1= V1 + V1 + V2

2

1 j4)

2j1(

1 2

1

V V V

Substituting (2) into (1),

2

2 j41.1)2j1(

=

V Z

=

= ThL

2j1(

12

2 1

°

∠

=+

−

=

818.5j9091.0

82.21j09.1092

Th V V

=

=

=

)792.6)(

8(

)893.18(R8P

2

L

2 Th max

V

W 569 6

10010j-

)30j40)(

100()30j40(

634.4j707.81

)634.4j707.31)(

50()634.4j707.31(

=

Z

73.1j5.19

Th = +

Z

=

= ThL

164

2sin2

16sin

16

1

0 0

8 =

=

rms

I

1

2 2 2

0 2 2

1t0,5)t(v

2

25dt-5)(dt52

1

1 2 1

0 2 2

32f

3

32]16016

[

3

1

dt4dt0dt)4(3

1dt)t(fT

1f

rms

3

2 1

=

++

−

=

4

1dt)10(dt54

1

2 2 2

0 2 2

1253

t5

1dtt5

1

0

3 5

0 2 2

0 2 2

rms (4t) dt 0 dt

5

1V

533.8)8(15

163

t165

1

0

3 2

V

P

2 rms 4 267 W

t(i

15

2 15

5

2 2

eff (20 2t) dt (-40 2t) dt20

1I

5

2 2

eff (100 20t t )dt (t 40t 400)dt5

1I

3 15 5

3 2 2

3

t3

tt10t1005

1I

332.33]33.8333.83[5

1

I2

=eff

2t0t)

1dt-1)(dtt4

1

2 2 2

0 2 2

42

1)

4()

2(2

1)(2

12 0

=

= ∫v t dt ∫ t dt ∫ dt

V rms

V944.2

=

rms

V

2 2 2

rms (10t ) dt 0dt

2

1I

105

t50dtt50

0

5 1

0 4 2

2t1t1020

1t010)

t

(

i

[ 10 dt (20 10t) dt 0]3

1

1

2 1

0 2 2

33.133)31)(

100(100dt)tt44(100100I

1

2 2

Chapter 11, Solution 34

472.420f

20363

t9

3

1

dt6dt)t3(3

1dt)t(fT

1f

rms

2 0 3

Chapter 11, Solution 35

5 2 5

4 2 4

2 2 2

1 2 1

0 2 2

rms 10 dt 20 dt 30 dt 20 dt 10 dt

6

1V

67.466]1004001800400100[6

34

2 2

6)10sin(

483 2

i

A487.9902

362

1664

3 2

50)(

2(jLjH

5

08.157j30jX

=

Z

)210

Apparent power = = =

160

)210

)19.79cos(

36

08.157tancoscos

)8j12)(

4j()8j12(

||

4j

4.0)6667.0j4.1(V60j92.103

50

VV30j

V20

V

30

120

2 1

2 1 1 1

o

−

−

=+

→



−+

2 2

2

4010

V V

V

V

++

(a) P j30Ω =0=P−j40Ω

W665.820/3.173

|

|2

2 2

R

V R

V

W6.034100/1.4603

|

|2

2 1

R

V V P

W86.8740/3514

|30120

|2

|

|,

3.1065

.1422

||

-j2)2j5j(

||

2j

)j4)(

2j()j4(

°

∠

=+

+

=

−+

44.0j64.1

44.0j64.0)j52.1j64.0(

Chapter 11, Solution 42

°

=θ

→

θ

=

=0.86 cos 30.683

pf

kVA798.9)683.30sin(

5sin

QSsin

S

°

=θ

=

→

θ

925

3 2

Chapter 11, Solution 45

2

6020

2 2

Reactive power =55 VAR

pf is leading because current leads voltage

Reactive power =401 2 VAR

pf is lagging because current lags voltage

Reactive power =74 54 VAR

pf is lagging because current lags voltage

Reactive power =- 961 7 VAR

pf is leading because current leads voltage

Average power =112 W Reactive power =194 VAR

)80( 2

Chapter 11, Solution 48

(a) S=P− jQ= 269−j 150 V A

(b) pf =cosθ=0.9 → θ=25.84°

31.4588)

84.25sin(

2000sin

QSsin

S

°

=θ

=

→

θ

=

48.4129cos

Qsinsin

S

59.48

=

86.396)6614.0)(

600(cosS

2 2

=

=

=

Z V

8264.01210

1000S

Pcoscos

4j

(b) 0.8 cos sin 0.6

2

6.1S

P

=θ

=V I S

1440060

j40

)120

1000j1000jQ

=

S

750j

Z

V

S=

23.23j98.30750j1000

)220

2 rms

2(

)120(2

2 2

2 rms

*

S

V S

V Z

Chapter 11, Solution 51

(a) ZT =2+(10−j5)||(8+j6)

j18

20j1102j

18

)6j8)(

5j10(2

++

=+

+

−+

2(

)16(2

SS

S

500j1000

S

2749j12009165

.0x3000j4.0x3000

S

1500j20006

.08.0

2000j2000

S

C B A

=

7494200

4200pf

749j4200I

IV

Irms = 35.55∠–55.11˚ A

Chapter 11, Solution 53

S = SA + SB + SC = 4000(0.8–j0.6) + 2400(0.6+j0.8) + 1000 + j500 = 5640 + j20 = 5640∠0.2˚

(a)

A8.2997.9388.2946.66x2I

8.2946.662

30120

2.05640V

SV

SSV

SI

rms rms

C A rms

B rms

1000j1000jQ

=

S

750j

Z

V

S=

23.23j98.30750j1000

)220

2 rms

2(

)120(2

2 2

2 rms

*

S

V S

V Z

For mesh 2, -j50=(20+j10 I2 −20I1

2

1 (2 j II

-2j5

Ij22-

1-j1j5-2

3j4

5j1-

I2 2

°

∠

=+

=

−

−+

=

−

=I I (3.5 j0.5) (2 j3) 1.5 j3.5 3.808 66.8

I3 1 2For the 40-V source,

- * 1

I V

=

= * (j50)(2 j3)2

I V

Chapter 11, Solution 56

8.1j6.02j6

)2j-)(

6(6

||

2j

−

=

j2.23.66

||

-j2)(4j

+

= (2 30 ) 0.95 47.08

2.2j6.8

2.2j6.3o

)30-)(247.0875

.4(2

12

s

= V I S

o j4)

4j5(

At node 1,

2j

2j-

1 o 1

V V V

=+

j411

24-

o = +

j411

j4)-(-24)(2

1 = +

V

The voltage across the dependent source is

o 1 o 1

2 V (2)(2V ) V 4V

4j11

)4j6)(

24-)44j2(j411

24-

−

=+

−

⋅+

=

V

)2(2

12

o 2

576j4

11

-24-4j11

)4j6)(

24-

2.0

o = =

I

mA8

20Io =

From the right portion of the circuit,

mAj7

16)mA8(3jj104

4

I

)1010(50

)1016(

2 -3 2

= I S

=

S 51 2 mVA Chapter 11, Solution 59

Consider the circuit below

240

++

=

−

o)38.0j36.0(

°

∠

=+

38.0j36.0

88o

-o 1

V I

°

∠

=+

= 3.363 -83.42

30j40

o 2

V I

Reactive power in the inductor is

L

2

2 Z I

c

2

1 Z I

Chapter 11, Solution 60

15j20))8.0(sin(cos8

.0

20j20

749.7j16))9.0(sin(cos9

.0

16j16

=+

oI 6VV

2 = −

S

kVA937.1j4))9.0(sin(cos9

.0

4j

4 2

1

I V

S =

100

10)137.1j2.5)(

2(

=

=

V

S I

104j74.22

2 = +

I

Similarly, sin(cos (0.707)) 2 j2kVA

707.0

2j

1 2

1

I V

S =

40j40-100

j

10)4j4(V

=+

= 1 2 -17.26 j144 145 96.83

o I I I

* o o

o V I2

1

=

S

VA)96.83-145)(

90100(2

1

S

=o

.0

15j

2 V I

S =

120

25.11j15

2

2

* 2

09375.0j125.0

2 = +

I

)15.0j3.0(

2 2

1 =V +I +

V

)15.0j3.0)(

09375.0j125.0(120

V

0469.0j02.120

V

843.4j10))9.0(sin(cos9

.0

10j

84.25111.11

0405.0j0837.025.82-093.0

I

053.0j2087.0

s =V +I +

V

)04.0j2.0)(

053.0j2087.0()0469.0j02.120(

V

0658.0j06.120

V

=s

V 120 06∠0 03°V

Chapter 11, Solution 63

Let S=S1+S2 +S3

929.6j12))866.0(sin(cos866

.0

12j

S

916.9j16))85.0(sin(cos85

.0

16j

S

)6.0)(

20(1 -

S

* o2

1987.22j

110

98.22j442

* o

=o

or

333.3j83.20120

400j2500V

SI

VI

923.6j615.412j8

120I

2

2 2



1010

j-C

j

1nF

1

At the noninverting terminal,

j1

4j100

100

-04

o o

V

)45t10cos(

2

4)t(

W1050

12

12

4R

V

2 2

j4)(2

s i

mAj3)j2)(4-(6

)45j4)(4(2

mA2j6

-o

°

∠+

-2 2

540

4202

1R2

Chapter 11, Solution 67

51.02

11

F1.0,6H

3,

x j C j j

ω

42510

50)

Z

j

j j

68

2052.05638.06

8

0206.0

+

+

=+

−

∠

=

mVA86

.3618mVA8.104.14)87.1606.0)(

203.0(2

j

j Z

V I V Z

Z

7.990224.0)206.0()68(12

)42(,

3 1

|

|2

o R R

S

LI2

1j0jQ

o L

1j0jQ

o c

−ω+

C

1 L j R I 2

=θ

= cos

)12j10)(

2(

)120(2

The average power absorbed =P=Re(S)= 295 1 W

(c) For unity power factor, θ1 =0°

09.354

, which implies that the reactive power due

to the capacitor is Qc =

2

1X2

VQ

c

2

c = = ω

=π

=ω

)120)(

60)(

2(

)09.354)(

2(V

Q2

Chapter 11, Solution 70

6.0sin8

.0cos

528)6.0)(

880(sinS

If the power factor is to be unity, the reactive power due to the capacitor is

VAR528Q

Qc = =

c

2 rms

V

Q2CV

C2

1X

VQ

ω

=

→

ω

=

=

=π

)220)(

50)(

2(

)528)(

2(

Chapter 11, Solution 71

67.666.0

508.08.0,

6.0,

50,

065.1067071.0

2 2

1

P

5067.66,

065.106065

98.176.18106.56735.1722

S

058.56)098.17(tan735.172)tan

058.56

Q C

rms c

)tan(tan

P

Qc = θ1− θ2

kVAR]

84.25tan(

)54.40tan(

)[

40(

kVAR84

.14

Qc =

=π

=ω

rms

c

)120)(

60)(

2(

14840V

Q

(b) θ1 =40.54°, θ2 =0°

kVAR21.34kVAR]

0)54.40tan(

)[

40(

=π

60)(

2(

34210V

V S

1 , θ =cos-1(0.96)=16.26°

2

]tan(16.26-

35tan(

[10]tantan

[P

=c

Q 4 083 kVAR

=π

=ω

rms

c

)240)(

60)(

2(

4083V

Q

(d) S2 =P2 +jQ2, P2 =P1 =10kW

kVAR917

.2083.47QQ

Q2 = 1− c = − =

kVA917.2j10

2

* 2

Chapter 11, Solution 74

(a) θ =cos-1(0.8)=36.87°

1

kVA308.0

24cos

PS

)6.0)(

30(sinS

kVA18j24

kVA105.4295.0

40cos

PS

.13sin

S

kVA144.13j40

2 = +

S

kVA144.31j64

=θ

= cos

(b) θ2 =25.95°, θ1 =0°

kVAR144

.31]0)95.25tan(

[64]tantan

[P

=π

=ω

rms

c

)120)(

60)(

2(

144,31V

Q

Chapter 11, Solution 75

5j8

576050

j80

)240

* 1

2

+

=+

=

=

Z

V S

VA91.208j13.3587j12

576070

j120

)240

)240

S

=++

68.114tan- 1

=θ

= cos

(c) Qc =P[tanθ2 −tanθ1]=18.35.88[tan(3.574°)−0]

VAR68.114

Qc =

=π

=ω

rms

c

)240)(

50)(

2(

68.114V

8j23j4

1230

3 Vo = Vo + Vo

−

−+

°

∠

∠

=+

52.23j14.36o

V

)30-3)(

19.86347.11(2

12

o

= V I S

8jLjH

-j5Cj

1F

1

(15)(-j5)-j5)

)5.4j5.1()8j6(

120

I

)5.4j5.1()5.14(2

12

12

-j3Cj

1F

12

1

=ω

)3j-)(

4(4j53j-

||

44j5

−++

=+

+

=

Z

08.2j44

08.2j44.16

20

I

)08.2j44.6()207.1(2

12

x500x100j

1C

j

1F

500,j

10x10x100jmH

2 1 2 1 2

1 2 1 o

1

V)40j6(V)40j

)VV(320

VVj

VVI

−

=

−+

−+

2 2 1

1,

4141.08443.040

j VI

S j

110

Z V

(b) 1890.625

4.6

)110

°

=θ

≅

=θ

=Scos 1559.76

Chapter 11, Solution 81

kWh consumed =4017−3246=771kWh

The electricity bill is calculated as follows :

(a) Fixed charge = $12 (b) First 100 kWh at $0.16 per kWh = $16 (c) Next 200 kWh at $0.10 per kWh = $20 (d) The remaining energy (771 – 300) = 471 kWh

,30,

600,2482.0000,

S= 1+ 2 = 1+ 2 + 1+ 2 = +

AkV22.34

|S

|

S= =(b) Q = 17.171 kVAR

220,34

600,

[tan(cos 0.865) tan(cos 0.9)] 2833VAR600

,29

)tan(tan

PQ

1 1

2 1

Q C

rms c

Chapter 11, Solution 83

(a) S VI (210 60o)(8 25o) 840 35o

2

12

W1.68835

cos840

,1

000,120

$

3 0 10 per kWh Chapter 11, Solution 85

(a) 15mH   → j2πx60x15x10− 3 = j5.655

We apply mesh analysis as shown below

I1

+

60)

120(2

* Th

)104)(

1012.4)(

2(j75Lj

Z

Ω+

80R

V I

V

2 2

2 2 2

2 2 2

)6.1()3(RX

5377.2tanR

Xtan- 1 - 1

=θ

= cos

kVAR51.7))78.0(sin(cos12

sinS

=+

=P jQ

S 9 36+j 7 51 kVA

2 2

*

*

2

10)51.7j36.9(

)210(

×+

V S

kVA94.2352cos

PS

.1239sin

S

Additional load :

kW300

kVA375cos

PS

sinS

Total load :

jQP)QQ(j)PP

2000

kVAR5

.1464225

5.1239

The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the generator is

9775.094.2352

2300S

Pcos

1

=

=

=θ

or θ=12.177°

The maximum load kVAR for this condition is

)177.12sin(

94.2352sin

S

kVAR313

.496

=

)15)(

220(

2700S

Pcos

3.1897)

09.35sin(

)15(220sin

S

When the power is raised to unity pf, θ1 =0° and Qc =Q=1897.3

=π

=ω

rms

c

)220)(

60)(

2(

3.1897V

Q

Chapter 11, Solution 92

(a) Apparent power drawn by the motor is

kVA8075.0

60cos

.52)60()80(PS

Total real power

kW8020060PPP

P= m + c+ L = + + =

Total reactive power

=+

−

=++

=Q Q Q 52.915 20 0

Total apparent power

=+

= P2 Q2

51.86

80S

7285.3pf

P

kVAR796

.2))8.0(sin(cosS

1

kVA796.2j7285.3

S

kW2.1

kVA0j2.1

2 = +

S

kW2.1)120)(

10(

kVA0j2.1

3 = +

S

.1

kVA2sin

QS

2(cosS

kVA6.1j2.1

4 = −

S

4 3 2

1 S S S S

kVA196.1j3285

=

S

Total real power = 7 3285 kW

Total reactive power = 1 196 kVAR

196.1tan- 1

=θ

1

kW700cos

S

kVAR14

.714sin

P2 = 1 =

kVA84.73695.0

700cos

PS

.230sin

S

Q

Cost of installing capacitors=$30×484.06=$ 14 , 521 80

(b) Substation capacity released =S1−S2

kVA16.26384.736

(c) Yes, because (a) is greater than (b) Additional system capacity obtained

by using capacitors costs only 46% as much as new substation and distribution facilities

* s

L = Z → R =R , X =X

Z

LC

1X

×

×π

=π

=

)1040)(

1080(2

1LC

2

1f

9 - 3

- 2 814 kHz

)10)(

4(

)6.4(R4

VP

=40 j8

ZTh

=

= * Th

L Z

)40)(

8(

)146(R

8

VP

2

Th

2 Th

W 61 66

Chapter 11, Solution 97

Ω+

=+++

=(2)(0.1 j) (100 j20) 100.2 j22

ZT

22j2.100

240Z

=

=

)22()2.100(

)240)(

100(I

100R

I