контрольная работа / 2.1 вариант №64

ω = 2π f = 2π ×700 » 4398 ,23

:

X

L1

= wL

= 4398,23 × 1,36 ´ 10−3 » 5,98

1

X

L3

= wL

= 4398,23 × 5,46 ´ 10−3 » 24,01

3

X C1

=

1

=

1

» 69,96

wC1

× 3,25 ´ 10

−6

4398,23

I ′

I ′

I

Z1

Z 2

I

Z

1

1

1

2

2

2

U1

Z 3

U

2

Z

E

1/

2 /

! #-$ % :

U

= H

I + H U

2

1

11 1

12

I2 = H 21I1 + H 22U2

!T-! $$ :

A = 1 +

Z1

= 1 + - j63,98 » 1 - j0,98431 » 1,40316e− j44,55

Z3

65

B = Z1 + Z 2

+

Z1 Z 2

= - j63,98 + j24,01 +

- j63,98 × j24,01

» (23,633 - j39,97) » 46,434e− j59,41

Z 3

65

C =

1

=

1

» 0,015385 C

Z 3

65

D = 1 +

Z 2

= 1 +

j24,01

» 1 + j0,36938 » 1,06604e j20,27

Z 3

65

B

46,434e

− j59,41

H11 =

=

» 43,5575e− j79,68

D

1,06604e

j20,27

H12 =

D

=

1

» 0,938051e− j20,27

D

1,06604e

j20,27

H 21 =

-1 =

-1

» 0,938051e j159,73

j20,27

D

1,06604e

H 22 =

C

=

0,015385

» 0,014432e− j20,27

D

1,06604e

j20,27