Применение MathCAD в решении задач электротехники. Ч.1 Линейные электрические цепи

Iab2m :=

Uab2m

Iab2m

= 1.84

arg(Iab2m) = -72.724×deg

Zab2

Ibc2m :=

Ubc2m

Ibc2m

= 2.395

arg(Ibc2m) = 118.859×deg

Zbc2

Ica2m :=

Uca2m

= 3.057

arg(Ica2m) = -150×deg

Ica2m

Zca2

Ia2m := Iab2m - Ica2m

Ia2m

= 3.202

arg(Ia2m) = -4.09×deg

Ib2m := Ibc2m - Iab2m

Ib2m

= 4.214

arg(Ib2m) = 113.83×deg

Ic2m := Ica2m - Ibc2m

Ic2m

= 3.921

arg(Ic2m) = -112.352×deg

1

Pab2 := Re

Uab2m×Iab2m

= 140.447

2

1

Pbc2 := Re

Ubc2m×Ibc2m

= 218.003

2

1

Pca2 := Re

Uca2m×Ica2m

= 317.647

2

V2 := ( 0

Ua2m

0 Ub2m

0 Uc2m Ua2m Ub2m Uc2m )T

I2 := ( 0

Ia2m

0

Ib2m 0

Ic2m 0

Iab2m

0 Ibc2m

0 Ica2m )T

150

100

50

Re(V2)

Re(I2) ×20

150

100

50

0

- 50

- 100

- 150

- 50

- 100

Im(V2) , Im(I2) ×20

3. Расчет схемы для 3-ой гармоники

1

2

2

2

Uf :=

× (

Ua1m

)

+

(

Ua2m

)

+ (

Ua3m

)

= 180.693

2

1

2

2

Ul := 3×

× (

Ua1m

)

+ (

Ua2m

) = 306.92

2

1

2

2

Iab :=

× (

Iab1m

)

+

(

Iab2m

)

= 3.222

2

1

2

2

Ibc :=

× (

Ibc1m

)

+

(

Ibc2m

)

= 2.923

2

1

2

2

Ica :=

× (

Ica1m

)

+

(

Ica2m

)

= 4.514

2

1

2

2

Ia :=

× (

Ia1m

)

+ (

Ia2m

)

= 6.971

2

1

2

2

Ib :=

× (

Ib1m

)

+ (

Ib2m

)

= 3.707

2

1

2

2

Ic :=

× (

Ic1m

)

+

(

Ic2m

)

= 6.893

2