Задача К8 вариант№20

8 ( 20)

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ω

2−3

= ω

2

= ω =

VC

=

VA

;

CP = AC + AP;

VC

=

VA

VC = AC + AP ;

3

CP

AP

AC + AP

AP

VA

AP

VC

VC

1

-VA

VA

VC -V

= AC +1

AC =

-1

=

VC

-1

1

=

VC

;

ω

2−3

=

=

.

VA

AP

AP VA

AP

VA

AC r2 ×VA

AP

r2

!

V = ω I (r3 + r4 ) = 90 ×(15 + 25) = 3600 ;

V = ω1r1 = -40 ×50 = -2000 .

= V −V

= −2000 − 3600 = -560 ; ω

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ω

2

−3

2

−3

= ω

2

= ω

3

= -560 .

r2

10

3 4 ( . 2 )

( ) .

" :

V = -V -VAB = -V -ω2−3r3;

V

= -V

- r ω

= -3600 -15×(-560) = 4800 ;

ω

4

= V = 4800 = 192 .

3

3

r4

25

i

=

ω 2 −ω I

; i =

r1

ω 2 −ω I

=

r1

ω

=

r1

(ω

-ω

) + ω

,

(1)

2

1

I

I

12

ω1

-ω I

12

r2

ω1 -ω I

r2

r2

i12 – 1 2 (

) ; , i34 > 0, – .

i

=

ω 4 −ω I

;

i = -

r3

ω 4 −ω I

= -

r3

ω

= -

r3

(ω

-ω

) + ω

,

(2)

4

3

I

I

34

ω 3 -ω I

34

r4

ω 3 -ω I

r4

r4

i34 – 3 4 (

); , i34 < 0, – $ .

&, $ (1) (2)

()")

; « + »

, « – » — .

* , :

ω

= ω 3

=

r1

(ω1 -ω I ) + ω I

=

50

×(-40 - 90) + 90

= -560 .

2

r2

10

+ « + »

, 2 – ,

.

ω

= -

r3

(ω 3

-ω I ) + ω I

= -

15

×(-560 - 90) + 90

= 480 .

4

r4

25

+ « – » , II 4

.

: ω 2 = ω 3 = -560 ;

ω 4 = 480 .