контрольная работа / 2.3 вариант №23

2.3.

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.

L,

,

,

Um,

R , !

0,55

0,44

0,184

33,4

27

" #

$ %,

$ :

U

m

4a

1

1

u1(t) =

+

sin (α )sin (ωt ) +

sin (3α )sin (3ωt ) +

sin (5α )sin (5ωt ) =

2

32

πα

52

U

4

Um

π

1

π

m

2

=

+

sin

π

sin (ωt ) + 1 sin 3

sin (3ωt ) +

sin 5

sin (5ωt ) =

2

π

π

6

25

6

6

9

6

U

m

12U

m

1

1

1

=

+

sin (ωt ) +

× sin (3ωt ) +

sin (5ωt )

2

9

50

,

π 2

2

π

α =

a =

Um

6 ;

2

ω =

2π

=

2π

» 34147,746

T

0,184 ´10−3

X (1)

=

1

=

1

» 66,556

C

ωC

34147,746 × 0, 44 ´10−6

X (3)

=

1

=

1

» 22,185

C

3

×ωC

3 × 34147,746 × 0, 44 ´10−6

X (5)

= 5 ×ωL = 5 × 34147,746 × 0,55 ´10−3 » 93,906

L

X (5)

=

1

=

1

» 13,311

C

5 ×ωC

5 × 34147,746 × 0, 44 ´10−6

I1 ( jω )

jX L

I2 ( jω)

U1 ( jω)

I4 ( jω)

U2 ( jω)

R

I5 ( jω )

− jXC

I3

( jω )

− jXC

( !:

U2 ( jω) = R I2 ( jω) = − jXC I3 ( jω)

I2

( jω) =

U2 ( jω)

R

I3

( jω) =

U2 ( jω)

- jXC

( 1- ) *:

U2 ( jω)

U2 ( jω)

1

R - jXC

I4 ( jω) = I2 ( jω) + I3 ( jω) =

+

=

1

+

U2 ( jω) =

U2 ( jω)

R

- jXC

R

- jXC

- jXC R

( !:

X L (R - jXC )

U4 ( jω) = jX L I4 ( jω) = jX L ×

R - jXC

U2 ( jω) =

U2 ( jω)

- jXC R

- XC R

( 2- ) *:

(R

)

(R

)

U ( jω) = U

( jω) + U

( jω) =

X

L

− jX

C

( jω) +U

( jω) =

X

L

− jX

C

( jω) =

U

+1 U

−XC R

−XC R

1

4

2

2

2

2

=

XC R − X L

(R − jXC )

U2 ( jω) =

R ( XC

− X L ) + jX L XC

U2

( jω)

XC R

XC R

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K ( jω) =

U2 ( jω)

=

U2 ( jω)

=

XC R

U ( jω)

R ( X

C

- X

L

) + jX

L

X

R ( X

C

- X

L

) + jX

L

X

C

1

C

U2 ( jω)

XC R

# XL

( ,

C

# +, .

. . R& + # ' L,

#

' ,

U (0)

= U (0)

=

Um

= 33,4 = 16,7

B

2

1

2

2

- 1- :

(1)

X (1) R

K

( jω) =

C

=

R (XC(1) - X L(1) )+ jX L(1) XC(1)

66,556 × 27

» 1,0004e− j44,1

= 27 × 66,556 -18,781 + j18,781× 66,556

(

)

U

(1)

( jω) = U (1) ( jω)K (1)

( jω) =

12Um

× 1 × K (1) ( jω) = 6 × 33, 4 ×1,0004e− j44,1 » 20,3129e− j44,1 B

2m

1m

π 2

2

π 2

- 3- :

(3)

X

(3) R

K

( jω) =

C

=

R (XC(3) - X L(3) )+ jX L(3) XC(3)

=

22,185 × 27

» 0,3856e− j126,42

27 × (22,185 - 56,344) + j56,344 × 22,185

U

(3)

( jω) = U (3) ( jω)K (3) ( jω) =

12Um

1 × K (3) ( jω) = 12 × 33, 4 × 0,3856e− j126,42

» 1,7399e− j126,42 B

2m

1m

π 2

9

9π 2

- 5- ( )

(5)

X

(5) R

K

( jω) =

C

=

R (XC(5) - X L(5) )+ jX L(5) XC(5)

13,311× 27

» 0,1432e− j150,13

= 27 × 13,311 - 93,906 + j93,906 ×13,311

(

)

U

(5)

( jω) = U (5) ( jω)K (5) ( jω) =

12Um

1

× K (5) ( jω) = 12 × 33,4 × 0,1432e− j150,13 » 0,1163e− j150,13 B

2m

1m

π 2

50

50 ×π 2

21

u1

, B

20

19

18

17

16

15

14

13

12

11

10

9

8

7

6

5