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Convert the wye connection Ri—Rj — Rh to its equivalent delta... b 480 mW 4.53 a Minimize the number of equations to write and solve by using the mesh current method... Thus, when the lo

Trang 1

2.34 a) i = 385 mA,so a warning sign should be

posted and precautions taken

b) Use the following resistors: 390 ft, 47 (1,

and 220 ft

2.35

2.36 a) Pa r m = 59.17 W;/> l e g = 29.59 W;

P t m n k = 7.40 W

b) ^ m = 1414.23 s; r leg = 7071.13 s;

'trunk = 70.677.37 S

c) All values are much greater than a few minutes

2.37 a) 40 V

b) No, 12 V/800 ft = 15 mA will cause a shock

2.38 3000 V

Chapter 3

3.1 a) 6 kfl and 12 kft, 9 kfl and 7 kft; simplified

circuit is

10 V

b) 3 kft, 5 kft, and 7 kft; simplified circuit is

3 mA

c) 300 ft, 400 ft and 500 ft; simplified circuit is

1200 Q

200 mV

3.2 a) 10 ft and 40 ft, 10011 and 25 ft; simplified

circuit is

60 V

b) 9 kfl and 18 kft, and 6 kfl; simplified circuit is

50 mA

c) 600 ft, 200 ft, and 300 ft; simplified circuit is

250 fl

0.2 A

3.3

3.4

3.15

3.16

3.18 3.23

3.24

a) b) c) a) b) c) a) b) c) a) b)

21.2 ft

10 kft

1600 ft

30 ft

5 kft

80 fl

66 V 1.88 W, 1.32 W 17,672 ft, 12,408 ft

1200 fl, 300 ft

1 W

1875 ft, 3750ft, 7.5 k f a)

b) c) d) e) a)

b) c) d)

36 V

2 A 0.96 A

24 V 6.4 V

25 mA

250 V

50 V

25 mA 4.167 mA

Trang 2

3.31 a) 49,98012

b) 498012

c) 23012

d) 5ft

3.35 a) R,„ = 50fl;

(25/12)

3.53

3.56

3.58

3.72

3.73

50 + (25/12) (k 25 b)

2500 mcas

c) Yes

3.51 a) 1500 (1

b) 28.8 mA

c) 750 a , 276.48 mW

d) 100012,92.16 mW

23.2 V, 21 V

a) A-connected R2—R3—R4 becomes Y-connected

512—2012—4ft; equivalent resistance is 3312

b) Y-connected R2—R4—R>, becomes A-connected

10012 —8012—20 ft; equivalent resistance

is 3312

c) Convert the delta connection R 4 —R s —Rf, to

its equivalent wye Convert the wye connection

Ri—Rj — Rh to its equivalent delta

9012

R { = 1.0372ft, R 2 = 1.1435 ft,/?3 = 1.212,

R 4 = 1.1435 ft, R 5 = 1.0372 ft, R a = 0.0259 ft,

R b = 0.006812, R c = 0.006812 R L{ = 0.025912

Pdiss = 624W = Pd d

3.74 a) R] = 0.426912, R2 = 0.4617 ft, R3 = 0.48 ft,

R4 = 0.4617ft, R5 = 0.4269 ft, Ra = 0.0085 ft,

Rh = 0.0022 ft, rtc = 0.002212, Rd = 0.008512

b) i, = 26.51 A, /?/?, = 300 W or 200 W/m;

i2 = 25.49 A, i\R2 = 300 W or 200 W/m;

ib = 52 A, ilRh = 6 W or 200 W/m;

Pd d= 1548 W =P{Sss

Chapter

4.1 a)

b)

c)

d)

e)

f)

g)

4

11

10

9

s

6

4

6

4.2 a) 8 b) 3 c) 4 d) Avoid the topmost mesh and the leftmost mesh, which both contain current sources

4.3 a) 2 b) 5 c) 7 d) 1,4,7 4.4

4.8

4.9 4.13

4.17

4.19 4.24 4.26

4.27 4.33

4.34

4.38 4.39 4.42

4.44

4.48

a) 5 b) 3

c) -i s + ij + U = 0; —/1 + /4 + /3 = 0;

/5 — / 2 — /3 = 0 d) 2

e) /?,/! + tf3/3 - R 2 i 2 = 0;

/?3* 3 + R 5 i 5 - R4I4 = 0

120 V, 96 V

4 V a) - 6 8 A, 2.7 A, - 9 5 A, 2.5 A, - 1 2 A b) 3840W

a) 8800W b) 8800W

750 W 3.2 V a) -37.5 V, 75 W b) -37.5 V, 75 W c) Part (b), because there are fewer equations to write and solve

- 2 0 V a) 5.6 A, 3.2 A , - 2 4 A b) -8.8 A , - 1 6 A, 7.2 A a) 17,940 W

b) 17,940 W 259.2 W

2700 W a) 162.92 W b) 518.52 W c) Power delivered equals power absorbed

a) 2 raA b) 304 mW c) 0.9 mW

740 W

Trang 3

4.51 a) 5.7 A, 4.6 A, 0.97 A , - 1 1 A, 3.63 A

b) 5 > d e v = £ / ^ = 1319.685W

4.52 a) The constraint equations are easier to

formu-late in the node voltage method, making it the

preferred method

b) 480 mW

4.53 a) Minimize the number of equations to write and

solve by using the mesh current method

b) 4 m W

c) No, since the mesh current method still minimizes

the number of equations to write and solve

d) 200 mW

4.59 a) - 1 m A

b) - 1 mA

4.60 a) -0.85 A

b) -0.85 A

4.63 60 V source, positive at the top, in series with a

1011 resistor

4.64 1 mA current source, with current flowing from top

to bottom, in parallel with a 3.75 O resistor

4.71 a) 51.3 V

b) - 5 %

4.74 160 V source, positive at the bottom, in series with

a 56.4 kft resistor

4.77 8 ft (The voltage source is zero because there are

no independent sources in the circuit.)

4.83 2.5 H and 22.5 O

4.87 a) 6 f t

b) 24 W

4.91 a) 50 V

b) 250W

4.96 30 V

4.105 v { = 39.583 V,v 2 = 102.5 V

4.106 t»i = 37.5 V,v 2 = 105 V

4.107 V] = 52.0833 V,i>2 = 117.5 V

Chapter 5

input

non-inverting

input

positive power supply output

negative power supply

b) The input resistance; i n = 0

c) The open-loop voltage gain; (v p - v n ) = 0

d) ^ = 9 V

5.2 - 1 mA 5.3 a) - 1 5 V (saturates) b) - 1 0 V

c) - 4 V

d) 7 V

e) 15 V (saturates)

f) -1.08 V < v a < 4.92 V 5.8 a) Many possible designs; one uses a single 3.3 kft

input resistor and three series-connected 3.3 kft resistors in the feedback path

b) ±15 V

5.9 a) 0 < or < 0.40

b) 556.25 At A

5.11 0<i?f<60kft 5.12 a) Inverting summing amplifier b) - 6 V

c) - 0 5 V < v „ < 2 V

5.14 a) 14 V

b) 3.818 V < v a < 9.273 V

5.17 a) Non-inverting amplifier

b) 2v s

c) - 6 V < v s < 4 V

5.18 a) 10.54 V

5.25

5.26 5.28

5.33 5.34

5.43

5.45

b) -4.55 V < v g < 4.55 V c) 181.76 kft

a) - 1 5 1 V b) 34.3 kft c) 250 kft

20 kft a) 16 V

b) - 4 2 V < v b < 3 8 V

19.93 kft < R x < 20.07 kft a) 24.98

b) -0.04 c) 624.5 a) -19.9844 b) 736.1 jaV c) 5003.68 ft d) - 2 0 , 0 V, 5000 ft a) 13.49

b) 999.446 mV, 999.834 mV c) 387.78/AV

d) 692.47 pA e) 13.5,1V,0V,0A

Trang 4

5.49 a) 2 k f l

b) 12 mfl

Chapter 6

6.1 »f(mV)

e) v(V)

6.4

4

3

2

1

0

(

-1

- 2

- 3

- 4

l /*""~3

l

4

1.4

1.2

1 F

0.8

0.6

0.4

0.2

-0

0 1

a) / = 0

i = 50/ A

/ = 0.5 - 5 0 / A

/ = 0

b) v = 0

v = lV

v = -\V

v = 0

p = 0

p = 50/ W

/? = 50/ - 0.5 W

/7 = 0

w = 0

w = 25/2 J

/(s)

/(s)

4 / < 0

0 < / < 5 ms

5 < / < 10 ms

10 ms < / / < 0

0 < / < 5 ms

5 < / < 10 ms

10 ms < / / < 0

0 < / < 5 ms

5 < / < 10 ms

10 ms < / / < 0

0 < / < 5 ms u> = 25/2 - 0.5/ + 0.0025 J 5 < / < 10 ms

to = 0 10 ms < /

6.16 a) - 5 0 X 104/ + 15 V

b) 106/V

c) 1.6 X 106/ - 12 V

d) 52 V

6.17

/(jxs)

t (ms)

6.21 8 H

6.25 a) -5e~ Al A

b) -4e~4 ? - 6 A c) - e_ 4 i + 6 A d) 40 J

e) 400 J f) 360 J 9) 1(4)(-6)2 + j(16)(6)2 = 360 J (checks)

6.26 2 ^cF, initial voltage is 25 V 6.31 a) - 2 0 e "2 5' V

b) -16<T25' + 21 V c) -4<T25/ - 21 V d) 320/AJ e) 2525/AJ f) 2205/xJ

g) \(2 X 10"6)(21)2 + |(8 X 1()"6)(-21)2 =

2205 /xJ

6.39 a) 16 ^ + 32/2 = 2-±

at at

b) -16<Tf + 32e_2f + 32e-' - 32e~2' = 16e"

c) 34<T' - 4e~2tV

d) 30 V, which is consistent with the circuit's behavior

6.44 a) 2721.6 mJ

b) 2721.6 mJ c) 518.4 mJ d) 518.4 mJ

Trang 5

6.45 a) - 4 5 A

b) No

6.48 a) 50mH,2.4

b) 0.2 X 10~6Wb/A,0.2 X 1 0_ 6W b / A

6.49 0.8 nWb/A, 1.2 nWb/A

6.53 There is no difference in the output voltage for

these two circuits

Chapter

7.4

7.5

7.7

7.23

7.26

7.35

7.36

a)

b)

c)

d)

e)

a)

b)

c)

d)

e)

f)

9)

h)

i)

i)

k)

I)

a)

b)

c)

d)

a)

b)

c)

a)

b)

c)

a)

b)

a)

b)

c)

d)

7

5 mA, 15 mA

5 mA, - 5 mA

5,,-20.000/ m A

-5e-2 0-, ) 0 0'mA

The current in a resistor can change

instantaneously

0 A

160mA

65 mA

160 mA

225 mA

0 A

160e"2n"' mA

OV

-3.2 V

OV

-32e~ 2m V

225 - 160e"200' mA

2 A

20 ms

2e~ 5i)t A, -160e~5lM V, -144<T50' V

69.92%

1.6e~5()fmA,32<r5()' + 8V, -8<T50' + 8 V

800/d

160/xJ,640/J

2 4 « - « " raA

- Se -5om + 8 0 V

2880/d

- 2 - 3 e -5 0 0 , ,' A , 4 8 - 4 8 e -5 , K , ( ,' V

6 0 V , 0 V

- 5 mA

0.333 mA

5/AS

0.333 -5.333<r2 ( , a ( , 0 , )'mA

7.37

7.51

7.53 7.55

7.56

7.68

7.69

7.71 7.78 7.85 7.87

7.95

7.96 7.103

7.104

7.105

a) b) a) b) c) d)

e)

5 + 15e-1000'A

5 0 - 4 5 0 < T1 ( l o a ,V

3.4 mA

- 6 0 + 90e~20,)0' V

a)

b)

c)

d) e) f) a) b) c) d) a) b)

c)

d) e) a) b)

c)

d)

e)

50 V

- 2 4 V

0.1 fis

-18.5 A

- 2 4 + 74<Tlt,7'V -18.5e-1 ( ) 7'A

90 V

- 6 0 V

1000/AS 916.3/AS

4 - 4 < T2 0' A 80e"20fV 2.4 - 2.4e~20' A 1.6 - 1.6<T20'mA Yes

40 - 40e"500(" mA

1 0 e - 5 0 o o , v

24 - 24e"5000' mA Yes

-5.013 V

-5V, 0 < t < 5 s; -5£>-(U('" 5) V, 5 s < t <

83.09 ms a) b)

2.25

272.1 /AS

- 1 6 0 0 / + 8 V , - 1 5 + l l e -2 0 0' V ,

23

80 a) b) a) b) a) b) c)

- 1600* - l l e "2 0 0' V

ms 1.091 MO 0.29 s 8.55 flashes/min 559.3 kH 24.32 flashes/min 99.06 mA

$43.39 per year

CO

Trang 6

Chapter 8

8.1 a) -10,000 rad/s,-40,000 rad/s

b) overdamped

c) 3125 ft

d) -16,000 +/12,000 rad/s,

-16,000 - /12,000 rad/s

e) 2500ft

8.7 a) 25 nF

b) 2500ft

c) 75 V

d) 30 mA

8.8

8.9

8.18

8.19

8.20

8.29

8.30

8.31

8.50

e) -8000, (30 cos 6000? + 71.25 sin 6000?) mA

a) 8 kft, 40 H, 625 rad/s, 500 rad/s

O.&T250* -0.8<T10()"fmA,

02e- 25{)[ - 32e- xmt mA

a) l k f t , l / x F , 6 0 0 0 V / s , 8 V

b) (-3000? + 2)^ 5 ( l l l , mA

8.11 a) 500 rad/s, 400 rad/s, 1.5625 H,4/xF,

- 1 5 mA, 60 mA

b) 18.75e"200' - 18.75<r8()0'V

c) 75<T200' - 7 5 ^8 0 0' mA

800/

d) - 6 0 e -2 0 0' + 15(TftUUf mA

5e~ 2im + \Qe-* m) <V

\5e~ 25m cos 3122.5? + 721e~ 2mi sin 3122.5? V

15e- 4m] 'V

60 - 105e"W)00' cos 6000? - 90e"800<)rsin 6000? mA

60 - 750,000?e"104' - 105eT104' mA

60-80ff-8 n 0 r + 2Off-32WfV

8.51 60 - 120,000?e -2000(

- 60e 2000( V

8.52 60 - 60e>-200(" cos 1500? - 80<r20(K)/ sin 1500? V

8.63 a) 0 < ? < 0.5" s:

v n (t) = l0t 2 V,v lA (t) = -1.6? V;

0.5 + < ? < ?sal:

v 0 (t) = -5?2 + 15? - 3.75 V,

v ol (t) = 0.8? - 1.2 V

b) 3.5 s

8.64 0 < ? < 0.5 ~ s:

v a {t) = 10 - 20e"' + Hfe~*V,

v oi (t) = -0.8 + 0.8e~aV,

0.5+ < ? < ?sal:

Vo {t) = - 5 + 19.42ff"(t-a5) - 12.87e_ 2 ( f-a 5 )V,

8.68 a) 55.23fis

b) 262.42 V c) ?max = 53.63 /xs, v(t max ) = 262.15 V

8.69 a) 40 mJ

b) -27,808.04 V c) 568.15 V

Chapter 9 9.1 a) 80 V

b) 500 Hz c) 3141.59 rad/s d) -0.5236 rad e) -30°

f) 2 ms g) 166.67 ^ s h) -80sinl0007r?V i) 333.33/AS

j) 166.67jiis

9.4 a) 600 Hz

b) 1.67 ms c) 10V d) 6 V e) -53.13°,-0.9273 rad f) 662.64/AS

g) 245.97^s

2

a) -195.72<rl066-67f + 200cos(800? - 11.87°) mA b) -195.72^- 1066-67/mA,

200 cos(800? - 11.87°) mA c) 28.39 mA

d) 200 mA, 800 rad/s, -11.87°

e) 36.87°

9.11 a) 111.8 cos(500? - 3.43°)

b) 102.99 cos(377? + 40.29°) c) 161.59 cos(100? - 29.96°) d) 0

9.13 a) 502,654.82 rad/s

b) 90°

c) -39.79 ft

d) 0.05/AF

e) -/39.79 ft

9.14 a) 400 Hz

b) -90°

9.8

9.9

Trang 7

9.85

9.15

9.24

9.28

9.29

9.32

9.37

9.45

9.46

9.49

9.55

9.59

9.60

9.64

9.76

9.77

9.83

9.84

c) 5fi

d) 1.99 mH

e) /5 ft

AAA /~W~lT\

j V V v

600/20° v f J

7-b) 8.32 /76.31° A

^ -/100 ft

c) 8.32cos(8000f + 76.31°) A

a) 200 /36.87° mS

b) 160 mS

c) 120 mS

d) 10 A

500 rad/s

42.43 cos(50,000f + 45°) V

42.43 cos(2000? + 45°) V

2/3 ft

227.68 / -18.43° V, (3.6 + /10.8)(1

2 / - 3 6 8 7 ° A, (100 - /50) ft

1 0 / - 4 5 ° A, (1.6 + /3.2) ft

188.43/-42.88° V

/80 = 80 / 9 0 ° V

36 cos 2000 fV

56.57 cos(10,000f - 4 5 ° ) V

a) 0.3536

b) 2 A

a) 5 cos(5000f - 36.87°) A, 1 cos(5000f - 180°) A

b) 0.5

c) 9 m J , 1 2 m J

512,000/60° ft

I

V = ( V / 2 ) - 1 R

o v in / m v a) 247.11/1.68° V

b) - 3 2 ft, 241.13/1.90° V

c) -26.90 ft

9.88 a) l { = 24 / 0 ° A, 12 = 2.04 / 0 ° A,

I3 = 2 1 9 6 / 0 ° A, I4 = 19.40/0° A,

I5 = 4 6 / 0 ° A, I6 = 2 5 5 / 0 ° A b) 0.42 / 0 ° A

9.89 a) 0 A

b) 0.436/0° A c) Yes; when the loads are equal, no power is lost

to the neutral line, so the cost of power is lower

Chapter 10 10.1 a) 409.58 W (abs), 286.79 VAR (abs)

b) 103.53 W (abs), -386.37 VAR (del) c) - 1 0 0 0 W (del),-1732.05 VAR (del) d) - 2 5 0 W (del), 433.01 VAR (abs)

10.2 a) Yes b) Yes 10.15 a) 15.81 V(rms) b) 62.5 W 10.18 a) 6.4 W, 4.8 VAR, 8 VA b) 6.4 W

c) 4.8 VAR 10.26 a) 0.96 lagging, 0.28; 0.8 leading,-0.6;

0.6 leading, - 0 8 b) 0.74 leading,-0.67

10.27 a) 1.875 + /0.625 ft

b) 0.9487 lagging

10.44 a) 20 + /20 ft b) 20 W c) With 22 ft and 1 mH, the load impedance is

22 + /5 ft and the load power is 17.7 W

10.47 a) 360 mW b) 4000ft,0.1/AF

c) 443.1 mW; yes d) 450 mW e) 4000 ft, 66.67 nF

f) Yes 10.48 a) 4123.1 ft, 0.1/xF, 443.18 mW b) Yes

c) Yes 10.64 90 W 10.65 a) 10 b) 250W 10.66 a) 28.8 ft b) 28.8 ft c) Yes

Trang 8

10.67 a) P L =

PH =

V 1 R\ + R 2

v\Ri + R2)

R1R2

PM = Yl

Ri

V 2 V 2

Pi

(Yl _ vL\(¥l\

(p L p ¥ )yp,)

PM - PL

b) 1125W

10.68 36ft, 24 ft

Chapter 11

11.2 a) acb

b) abc

11.3 a) Balanced, negative phase sequence

b) Balanced, positive phase sequence

c) Balanced, negative phase sequence

d) Balanced, positive phase sequence

e) Unbalanced, due to unequal amplitudes

f) Unbalanced, due to unequal phase angle

separation

11.7 v AB = 13,198.23 cos a>t V,

vBC = 13,198.23 cos(wr + 120°) V,

vCA = 13,198.23 cos(w/ - 120°) V

11.9 a) 15.24 A(rms)

b) 6583.94 V(rms)

11.11 a) I aA = 5 / - 3 6 8 7 ° A, IbB = 5/83.13° A,

IcC - 5 / - 1 5 6 8 7 ° A

b) Va b = 216.51 / - 3 0 ° V , Vb c = 216.51 / 9 0 ° V,

Vca = 2 1 6 5 1 / - 1 5 0 ° V

c) VA N = 122.23/-1.36° V,

VB N = 122.23/118.64° V,

VCN = 122.23/-121.36° V

d) VA B = 211.72/-31.36° V,

VB C = 211.72/88.64° V,

VC A = 211.72/-151.36° V

11.12 a)

1 + / 3 H

20/If

V(

150° / + \

rms) \-J

aA

39 - /33 n

b) 0.4 / - 1 7 3 1 3 ° A(rms)

c) 35.39/176.63° V(rms)

11.13 21.64/121.34° V(rms)

11.16 159.5 /29.34° V(rms)

11.22 6120/36.61° VA

11.24 a) 1833.46 / 2 2 ° VA

b) 519.62 V(rms)

11.43 a) W 2 ~ Wx = VL/L[cos (d - 30°) - cos (6 +

30°)] = 2V L I L $in0sm30 o = V L I L sm6

Thus, V 3 ( W2 - Wx) = VwLILsm 0 = QT

b) 2592 VAR, -2592 VAR, 3741.23 VAR, -4172.80 VAR

11.44 197.26 W, 476.64 W 11.52 a)

1.70 MVA

1.2MVAR

1.2 MW

b)

1.2 MW 11.53 a) 16.71 AIF

b) 50.14/aF 11.56 |Vab| = 12,548.8 V, so the voltage is below the acceptable level of 13 kV Thus, when the load at the substation drops to zero, the capacitor bank must be switched off

11.57 PL(before) = 81.66 kW, PL ( a f t e r ) = 40.83 kW

Chapter 12 12.2

12.3

12.7

12.9

12.10

12.14

a) (t + \0)u(t + 10) - 2tu(t) + (t - 10)«(f - 10)

b) -8(t + 3)u(t + 3) + 8(7 + 2)u(t + 2) + 8(r + l)«(f + 1) 8(f l)u(t 1 ) -8(/ - 2)u{t -2) + -8(/ - 3}M(/ - 3)

a) 5t[u{t) u{t 2)] + 10[«(f 2 ) -u(t - 6)] + (-5? + A0)[-u(t - 6) - -u(t - 8)] b) 10 sin irt[u{t) - u(t - 2)]

c) 4t[u(t) - u{t - 5)]

a) 1.0 b) 0 c) oo a) 26 b) 2.25

2/9

at SC °

3 ) f + c,2

b) * + J

c) 2 d) check

Trang 9

12.17 a)

b)

c)

d)

e)

12.22 a)

b)

12.40 a)

(.v + a)2

CO

2 i ~>

co cos 0 +

s 2 +

1

->

sr

sinh 0 +

(r

-1

v(.v + a)

1

.v sin 6 ->

of

y[cosh 0]

- 1)

13.9 a)

12.41

12.42

12.50

12.55

12.56

s(s + a)

[«?"' + 5e'2' + 2e-4']u(t)

b) [6 + 4e~2t + 2e~4t + e~("]u(t)

c) [4e'1 + 20cT'cos(2f + 36.87°)]M(?)

d) [490 + 250e"7'cos(r - 163.74°)]«(r)

a) [20/ - 4 + 4e-*]u(t)

b) [250 - lOOte'1 - 250e~f]u(t)

c) [30r - 8 + K)e 3lcos(t + 36.87°) ]u(t)

d) [20 - 2 5 f V - I5te~' - 20e"']«(0

e) [16 + S9A4te~ 2t cos (t + 26.57°) +

I13.14e_2rcos0 + 98.13°) ]u(t)

c) 5'(t) - 105(f) + [3(k'_5/ + 20e~10f]«(f)

a) / ( ( ) ' ) = 8,.f(oo) = 0

b) /(0^) = 13,/(00) = 6

c) / ( 0+) = 20, / ( o o ) = 0

d) / ( 0+) = 250, / ( o o ) = 490

0.947

588

Chapter 13

13.4 a) — 8 X 107s

13.6 a)

.v2 + 40,000* + 256 X 106

b) Zero at 0; poles at -8000 rad/s and

-32,000 rad/s

s 2 + 8000* + 25 x 106

b) Zeros at -4000 + /3000 rad/s and

-4000 - /30()() rad/s; pole at 0

16 x Uf

-^vw—

5000 (2

SI

150 V-s

b)

-150s

+

V„ 2.5 s a

(s + 400)(.v + 1600)

c) (SO*-*** - 200e-l60()f)«(r)V

A/W-+ 10012

b)

c) 13.12 a)

5 x 10s n

s '

V

o.ob' n

75.v2 + 812,500.v + 6875 X

sis1 + 104v + 5 X 107)

/ + \ 137.5 /

1.25 mV

-Hv ^_7

106

[137.5 + 8().04^ 50()()'cos(5000/ + 141340) ] K ( 0

ii,

V-s

R

b)

c)

•48(s + 8000) .v2 + 8000* + 25 X 106

2.4(.v + 4875) .v2 + 8000s + 25 X 106

d) [80e"-")00fcos(3000f + 126.87°) ]u(t)V e) [2.5e- 4mt cos(3(mt - 1 6 2 6 ° ) ] H ( 0 A

13.21 a) [35 + 5.73e~' cos(7r + 167.91 ° ) ] H ( I ) V

b) Compare solution at t = 0 and t = oo to circuit at t = 0 and t = oo

13.22 a) [10 - Kk>""a5'cos0.5/]u(f)A b) 7.07e_OA cos(0.5r - 45°)w(r) V

c) Compare solutions at t = 0 and t = oo to circuit at t = 0 and t = oo

Trang 10

13.29 a) 652 + 6s - 18 - 9 r - 30.v 18

s(s + 2)(5 + 3)' s(s + 2)(5 + 3)

b) Initial values: 6 A, - 9 A;

final values: - 3 A, - 3 A

c) [ - 3 + 3e~2' + 6e-3,]u(t)A,

[ - 3 - 3e~ 2 ' - 3e" 3 ']u(t)A

13.34 63.25*"ia*cos(50> + 71.57°)w(0 mA

240(5 + 40)

13.39 a) —

' s(s + 20)(5 + 80)

b) Initial value is 0 final value is 6 A

c) (6 - 4e~m - 2e~m)u(t) A

13.40 a) (-2e~ m + 2 e ^)' ) « ( 0 m A

b) (2< -2()(

2e- h l>(0 mA

13.42 a) 480(5 + 2.5)

5(5 + 4)(5 + 6)

b) [50 + 90e~4' - \4Qe~ ( "]u(t)V

13.50 a)

b)

c)

d)

e)

250

5 + 250

5

5 + 250

V

5 + 8000

8000

5 + 8000

, no zeros, pole at -250 racl/s

, zero at 0, pole at -250 rad/s

, zero at 0, pole at -8000 rad/s

, no zeros, pole at -8000 rad/s

100

, no zeros, pole at -500 rad/s

5 + 500

13.62 (e - l)e"?V

13.63 (1 - e)e-'V

13.77 16.97 cos (3/ + 8.13°) V

5(5 + 30,000)

13.79 a)

1 (5 + 5000)(5 + 8000)

b) (5e-5l)m - 4Ae-*m')u(t)V

13.80 a) = 1 * 2

; (5 + 400)(5 + 1000)

b) 13.13 cos(400f - 156.8°) V

13.87 a) 0.8 A

b) 0.6 A

c) 0.2 A

d) -0.6 A

e) Q.6e~ zxm '\t(t)A

f) - 0 6 e -2 x l% ( f ) A

g) -1.6 x 1(T38(0 - 72QOe'2xliftu(t)V

13.88 a) 80 V

b) 20 V

c) 0 V

d) 325(f)juA

e) 16 V f) 4V g) 20 V

13.92 a) i 2(0~) = *2<0+) = 0 A ;

1440TT( 122.92 V 2 5 - 3 0 0 0 T T \ / 2

300 V 2

5 + 1475 w

va = 252.89<Tl475jr/ + 172.62 cos(1207it + 6.85°) V y„(0+) = 424.26 V

c) V„ = 122.06/6.85° V(rms)

d) »„ (V)

t (ms)

13.93 a) - 2 0 5 8 ^- 1 4 7 5^ + 172.62 cos( 12()77-/ - 83.15°) V b)

v„ (V)

t (ms)

c) Voltage spikes in Problem 13.92 but not here

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