B.5 Übungsaufgaben zum Transformator

Lösung zur
Aufgabe 14.10.1
(Drosselspule)

  1. Windungszahl
    N   =   √--U------
          2πfA Bˆ

    =   √----------100V--------------= 450-            (B.5.1)
          2π ⋅ 50Hz ⋅ 10 ⋅ 10−4m2 ⋅ 1T
  2. Luftspaltlänge
    lL  =   RmL μ0A
                  5  −1        −7 V-s-       − 4    2
    =   1,5345 ⋅ 10 H   ⋅ 4 π ⋅ 10 A m ⋅ 10 ⋅ 10 mm
    =   0,1928mm                                             (B.5.2)
        -----------

Lösung zur
Aufgabe 14.10.2
(Drosselspule)

  1. Windungszahl
               UL
N   =   √---------
          2πfA Bˆ
        -----------221V--------------
    =   √2-π ⋅ 50Hz ⋅ 50 ⋅ 10−4m2 ⋅ 1T = 199           (B.5.3)
  2. Luftspaltlänge
    lL  =   Rm-μ0A
         2
        8,94-⋅ 105H-−-1       −7 V-s-        −4    2
    =         2      ⋅ 4π ⋅ 10  A m ⋅ 50 ⋅ 10 mm
    =  2,808mm                                             (B.5.4)
       ----------

Lösung zur
Aufgabe 14.10.3
(Transformator)

  1. Eisenverlustwiderstand
           U21N-   (230V-)2
RF e = P10 =    85W    = 622,35Ω-
    (B.5.5)

    Induktiver Hauptwiderstand

            2             2
X  =  U-1N- = --(230V--)-- = 130,56Ω
  h   Q10    405,18V ar    --------
    (B.5.6)

    Kupferwiderstände

               R
R1 =  R′2 = --K- = 0,273Ω-
             2    -------
    (B.5.7)

    Streublindwiderstände

                 XK
X σ1 = Xσ′2 = ---- = 0,342Ω-
               2
    (B.5.8)

  2. Wirkwiderstand und der Streublindwiderstand
            ′
R2  = R-2 = 0,273-=  0,0683 Ω
      ¨u2      22     --------
    (B.5.9)

    und

           X σ′   0,342
X σ2 = --22 = ---2--=  0,0854-Ω-
        ¨u       2
    (B.5.10)

  3. Sekundärspannung
              ′         ⁄      ∘
U-  =   U-2 = (221V---−-2,7-) = (110,5V ⁄ − 2,7∘)        (B.5.11)
  2      ¨u           2          -----------------

Lösung zur
Aufgabe 14.10.4
(Transformator)

  1. Leerlaufspannung
                 j37,7Ω
U-2  =   ---------------⋅ 36V = (20,58V ⁄ 17,60∘)        (B.5.12)
         20Ω +  j62,83 Ω         ----------------
  2. Lastspannung
    U-2 = 100 Ω ⋅ (0,176A ⁄-12,4∘) = (17,6V-⁄ 12,4∘)
    (B.5.13)

Lösung zur
Aufgabe 14.10.5
(Transformator)

Wirksamme Gesamtinduktivität

L  =  XG--= -330,97Ω-- = 1,05H
 G     ω    2 π ⋅ 50Hz
(B.5.14)

Gleichsinnig in Reihe geschaltet mit Gegeninduktivät

L    =   LG--−-L1-−-L2-=  1,05H-−--0,6H--−-0,15H--
  12            2                     2
     =   0,15H--                                         (B.5.15)
         -------