Исследование линейных электрических цепей в системе "Electronics Workbench". Быковская Л.В - 22 стр.

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Uc Z I⋅:= Uc 18= Ul 18=
Cхема замещения пассивного двухполюсника
ω 2 π⋅ 1000⋅:=
R 200:= Xl i ω⋅ L⋅:= Xc
1
i ω⋅ C⋅
:=
R 200= Xl 75.398i= Xc 159.155i−=
X
Xl Xc⋅()
Xl Xc+()
:= X 143.272i= X 143.272=
Z
RX⋅()
RX+
:= Z 67.827 94.683i+= Z 116.471= arg Z()
180
π
⋅ 54.384=
Y
1
Z
:= Y510
3−
× 6.98i 10
3−
×−=
g
R
Z
()
2
:= g 0.015= b
X
Z
()
2
:= b 0.011=
E1 15:= I1
E1
Z
:= I1 0.075 0.105i−= I2 Y E1⋅:= I2 0.075 0.105i−=
I1 0.129= I2 0.129=
Активное сопротивление
R 200:=
E18e
i
30π⋅
180
⋅
⋅:=
ZR:=
I
E
Z
:= I 0.078 0.045i+= I 0.09=
Ur R I⋅:= Ur 15.588 9i+= Ur 18=
Катушка индуктивности
L1210
3−
⋅:= E18e
i
30π⋅
180
⋅
⋅:= ω 2 π⋅ 1500⋅:= ω 9.425 10
3
×=
Ziω⋅ L⋅:=
I
E
Z
:= I 0.08 0.138i−= I 0.159=
Ul Z I⋅:= Ul 15.588 9i+= Ul 18=
Конденсатор
C110
6−
⋅:= E18:= ω 2 π⋅ 1500⋅:= ω 9.425 10
3
×=
Z
1
i ω⋅ C⋅
()
:=
I
E
Z
:= I 0.17i= I 0.17=
Рисунок 25
22
     Активное сопротивление
                                                                 ⋅
                                                               30π
                                                          i⋅
     R := 200                                                  180
                                             E := 18 ⋅ e
     Z := R
             E
      I :=                               I = 0.078 + 0.045i                       I = 0.09
             Z

      Ur := R ⋅ I                        Ur = 15.588 + 9i                         Ur = 18


      Катушка индуктивности
                                                                              ⋅
                                                                            30π
                                                                       i⋅
                             −3                                             180                                                                 3
      L := 12 ⋅ 10                                        E := 18 ⋅ e                      ω := 2 ⋅ π ⋅ 1500                   ω = 9.425 × 10

      Z := i ⋅ ω ⋅ L
                 E
       I :=                                  I = 0.08 − 0.138i                     I = 0.159
                 Z

       Ul := Z ⋅ I                           Ul = 15.588 + 9i                      Ul = 18

       Конденсатор

                                 −6                                                                                                      3
       C := 1 ⋅ 10                                    E := 18                        ω := 2 ⋅ π ⋅ 1500                ω = 9.425 × 10
                             1
       Z :=
                 ( i ⋅ ω ⋅ C)
                     E
         I :=                                 I = 0.17i                             I = 0.17
                     Z

         Uc := Z ⋅ I                          Uc = 18                               Ul = 18


       Cхема замещения пассивного двухполюсника                                                                ω := 2 ⋅ π ⋅ 1000
                                                                                                 1
       R := 200                                      Xl := i ⋅ ω ⋅ L                 Xc :=
                                                                                              i⋅ ω ⋅ C

       R = 200                                       Xl = 75.398i                    Xc = −159.155i
                  ( Xl ⋅ Xc)
       X :=                                          X = 143.272i                     X = 143.272
                 ( Xl + Xc)

                 ( R ⋅ X)                                                                                                      180
       Z :=                                      Z = 67.827 + 94.683i                       Z = 116.471            arg( Z) ⋅         = 54.384
                 R+ X                                                                                                          π
                             1                                       −3                −3
              Y :=                                   Y = 5 × 10             − 6.98i × 10
                             Z
                                 R                                                                        X
             g :=                                     g = 0.015                                b :=                       b = 0.011
                         (   Z       )   2
                                                                                                      (   Z   )2
                                                     E1
       E1 := 15                              I1 :=             I1 = 0.075 − 0.105i                    I2 := Y ⋅ E1             I2 = 0.075 − 0.105i
                                                     Z
                                                                I1 = 0.129                                                     I2 = 0.129

                                                                                  Рисунок 25




22

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