контрольная работа / 1.2 вариант №84

1.2.

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:

W

p

L1 = 1,36

C1

a

e/3

L3 = 10,9

k

/

1 = 3,25

e1

f

L3

3 = 9,46

R2

i1

R2 = 65

L1

m

f = 700

d

//

1

(

)

e1

e3

= 141cos

ωt

− 90

,B

h

e′

//

C3

e′′

= 0

b

1

e′

= 0

. 1.29.

3

(

)

3

= 282 sin

ωt

− 50

e′′

,B

p

:

XC1

ω = 2π f

= 2π ×700 » 4398 ,23

a

I3

:

k

/

X

= wL = 4398,23 × 1,36 ´ 10−3 » 5,98

L1

E1

XL3

1

f

R2

X L3 = wL3 = 4398,23 × 10,9 ´ 10−3 » 47 ,94

I1

XL1

I2

m

X C1 =

1

=

1

» 69,96

d

//

−6

wC1 4398,23 × 3,25 ´ 10

E3

h

X C 3

=

1

=

1

» 24,03

wC3

b

XC3

4398,23 × 9,46 ´ 10−6

) :

+ i3 = 0

i1 − i2

di

1

L

1

+

i dt + R i = e′

dt

C

1

1

2 2

1

1

di3

1

i dt = e′′

R i + L

+

dt

C

2 2

3

3

3

3

) :

I1 − I2 + I3 = 0

(

X

− X

C1 )

I + R I

= E′

j

L1

1

2

2

1

R I

+ j

(

X

L3

− X

C3 )

I

= E′′

2

2

3

3

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2)

! ,

"

.

,

!

"

!. # "

.

(ωt - 90

) = 141sin(ωt ),B

¢

= 141cos

¢

=

141

j0

» 99,7 B

e1

E1

e

2

(ωt - 50

),B

¢¢

= 282 sin

¢¢

282

e

− j50

»

199,4e

− j50

» (128,17 - j152,75) B

e3

E3 =

2

Y1 =

1

=

1

» j0,01563

Z1

- j63,98

Y 2 =

1

=

1

» 0,015385

Z 2

65

Y 3 =

1

=

1

» - j0,041824

Z3

j23,91

#ϕb = 0 &

E1Y1

+ E3Y 2

99,7 ×

(

- j152,75

)(

- j0,041824

)

jp =

¢

¢¢

=

j0,01563 + 128,17

»

Y1 + Y 2 + Y 3

j0,01563 + 0,015385 - j0,041824

»

-6,388616 - j3,802271

»

7,434493e− j149,24

= 244,73e − j89,67

» (1, 41

- j244,73) B

0,015385 - j0,026194

0,030378e− j59,57

j68,12

-1, 41+ j244,73 + 99,7

I1 =

-jp + E1¢

=

» 263,73e

» 4,1221e

j158,12 » (-3,8252 + j1,5362) A

- j63,98

Z1

63,98e− j90

244,73e

− j89,67

I2

=

jp

=

» 3,7651e − j89,67 » (0,0217 - j3,765)

A

Z 2

65

+ E¢¢

-j

j35,97

-1,41 + j244,73 + 128,17 - j152,75 » 156,616e

p

3

I3 =

=

»

Z3

j23,91

23,91e j90

» 6,5502e − j54,03 » (3,8473 - j5,3012)

A

PW

= Re

- 244,73e

− j89,67

)× 4,1221e

− j158,12

»

= Re Ubp I1

= Re (jb - jp )× I1

(0

» Re 1008,8 × e− j67,79

» Re[381,33 - j933,95] = 381,33

= 0 ; (& ;

ja

B

jp = ja

+ jXC1I1 = 0 + j69,96 × (-3,8252 + j1,5362) » (-107, 473 - j267,611)

jm = jp + jX L3I3 = -107,473 - j267,611 + j47,94 × (3,8473 - j5,3012) » (146,667 - j83,171) B

(

)

(

)

j

h

= j

m

3

=

+ j69,579

B

- E¢¢ = 146,667 - j83,171 -

128,17 - j152,75

18, 497

B

jb

= jh

- jXC3I3 = 18,497 + j69,579 - j24,03 × (3,8473 - j5,3012) » (-108,891 - j22,872)

- jX L1I1 = -108,891 - j22,872 - j5,98 × (-3,8252 + j1,5362) » (-99,705 + j0,003)

j f

= jb

B

ja

= j f

+ E1¢ = -99,705 + j0,003 + 99,7 = (-0,005 + j0,003) » 0

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&

)

, " :

j

80

70

ϕh

60

I ×30, A

− jX

C3

I

3

50

1

40

30

−E′′

20

3

ϕa

+

10

ϕ f

0

E1′

10

20

30

ϕb

− jX L1I 1

40

50

60

70

jXC1I 1

80

90

ϕm

100

I 2×30, A

110

−R I

2

120

2

130

140

150

160

I 3×30, A

170

180

190

jX L3I 3

200

210

220

230

240

250

260

270

280

ϕ p

120 110

100

90

80

70

60

50

40

30

20

10

0

10

20

30

40

50

60

70

80

90

100

110

120

130

140 150

U

= E′

− U ′

(5.1)

1

pb

U ′pb – p b 3.

U ′pb

Y

1 = 0 .

199,4e− j50

(

)

¢

¢¢

¢¢

- j0,041824

U ¢

=

E1Y1

+ E3Y 2

=

E3Y 2

=

»

pb

Y1 + Y 2

+ Y 3

Y 2 + Y 3

0,015385 - j0,041824

» 199,4e− j50 × 0,041824e− j90

= 187,14e − j70,2 »

(

63,39 - j176,08

)

0,044564e− j69,8

(5.1)

(

)

(

)

1

pb

» 179,78e j78,35 B

U

=

E¢ -U ¢

= 99,7 -

63,39

-

j176,08

=

36,31 +

j176,08

Z

Z 2

Z 3

jXL1

Z 2 × Z 3

65 × j23,91

65 × 23,91e j90

Z = jX L1 +

= j5,98 +

» j5,98

+

»

Z 2 + Z

3

65 + j23,91

69,258e j20,2

.

» j5,98 + 22,44 × e j69,8

» j5,98 + 7,748 + j21,06 » (7,748 + j27,04) » 28,128e j74,01

» 6,3915 × e j4,34 » (6,3732 + j0,4837)

A

I

= U

=

179,78e j78,35

1

Z

28,128e j74,01

1 )

I1 (xC1 ) =

E

(5.2)

Z - jX C1

!

y = j X C1

- j Z = -90 - 74,01 » -164,01 .

I1

(xC1 ) =

E

=

E

(6)

Z − jX C1

Re(Z )

2

+ (Im(Z ) − X C1 )

2

, ) I1 ( X C1 )

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:

I1, A

24