Lösung zu Aufgabe 1:
Frage 1:
( 1 ) U ( 1 99 % ) 250 kV 2 , 5 kV
U
K= − η ⋅ = − ⋅ = Frage 2:
A kV 400
250 MW 100 U
I
K= P = =
Frage 3:
Ω
⋅ =
=
⋅
= 3 , 125
A 400 2
V 2500 I
U 2 R 1
K K K
Frage 4:
2 K
K
286 mm
R A l A
R = ρ ⋅ l � ≥ ρ ⋅ =
U
0R
iA
B U
0R
iA
B
Frage 2:
V 58 , 0 V k 10 1 , 9 560 U 560 R R
U R
V2 1
2
0
⋅ =
Ω + Ω
= Ω + ⋅
=
Ω Ω =
+ Ω
Ω
⋅
= Ω +
= ⋅ 527 , 5
560 k
1 , 9
560 k 1 , 9 R R
R R R
2 1
2 1 i
Frage 3:
A 6 , 50 34
5 , 527
V 6 , 0 V 58 , 0 R
R U I U
B i
0 BE 0
B
= − µ
Ω + Ω
= − +
= −
Ω µ =
⋅
⋅
= π
= ω 39 , 8
F 1 Hz 4000 2
1 C
X
C1
Ω
−
=
−
= jX j 39 , 8
Z
C CFrage 2:
�� �
�� �
�
� =
Ω
= Ω Ω =
= Ω
� =
=
⋅ 0 , 9
50 x 45 80 , 50 0
8 , 39 R x X X R
x
C CFrage 3:
ℑ ℑ
ℜ 50 Ω j50 Ω
-j50 Ω -50 Ω
ℜ 50 Ω j50 Ω
-j50 Ω -50 Ω
ℑ ℑ
ℜ 50 Ω j50 Ω
-j50 Ω -50 Ω
ℜ 50 Ω j50 Ω
-j50 Ω
-50 Ω
L j R Z
LR1=
1+ ω Frage 2:
( )
(
2)
22 2
2 2
2 2 2
CR
1 CR
CR j 1 R CR j 1
R C
j R 1
C j R 1
Z + ω
ω
−
= ⋅ ω
= + + ω
⋅ ω
=
Frage 3:
= 0 ϕ Frage 4:
{ ZLR1+ Z
CR2} = 0 ℑ
C L
C R 1 L
22
0
⋅
−
=
ω
Ω
− =
= −
−
= −
�
=
−
−
⋅
−
k mA 1
5
V 1 V 6 V 12 I
U U R U
0 U U I R U
D RS DS V D
RS DS D D V
Frage 2:
mA 5 I I
S=
D=
Ω
=
=
= 200
mA 5
V 1 I R U
S S S
Frage 3:
V 97 , V 2 mA 17
mA 5 V 2
2 , S 2
I U 2
U
GS th D⋅
2=
+
⋅ = +
=
−Ω + =
= 400 k
I U R U
2 RG
GS RS 2 G
mA 68 , k 0 10
V 8 , 6 R I U
P Z
P
=
= Ω
=
15 6,8 1 0,68 4,88
V Z
V
Z P
U U V V
R k
I I mA mA
− −
= = = Ω
+ +
Frage 2:
1
v
U= (Spannungsfolger s. VL)
Die Spannung kann im Bereich 0 ≤ U
a≤ 6 , 8 V eingestellt werden.
Frage 3:
z. B. durch Einsatz einer Verstärkerstufe anstelle des Spannungsfolgers.
ZD
P OPV R
VU
VU
+U
aI
ZI
PU
ZR
2R
1ZD
P OPV R
VU
VU
+U
aI
ZI
PU
ZR
2R
1m 796 kA Am 10 Vs 4
T 1 H B
0 7
=
⋅ π µ =
=
− δ
δ
Frage 2:
A 59 , w 1
l 2 I H ⋅ ⋅ =
=
δ δFrage 3:
Vs 12 , 0 A B w ⋅ ⋅ =
=
Ψ
δFrage 4:
mH 4 , A 75 59 , 1
Vs 12 , 0
L Ψ I = =
=
Vs 25 , A 0 200
Nm 50 I
M 2 k
K
K
= =
π = Φ
Frage 2:
Ω
=
=
= 50 m
A 200
V 10 I
R U
K K a
Frage 3:
1 a 1
0
7 , 64 s 458 min
2 Vs 25 , 0
V 12 k
n U =
−=
−π
= ⋅
= Φ Frage 4:
Nm 60 U M
M U
KK a 12 ,
K
= ⋅ =
2b 12a 22a
3a 13c 23a
4c 14b 24c
5a 15a 25b
6b 16b 26b
7c 17a 27c
8b 18c 28c
9b 19c 29b
10a 20c 30a
Aufgabe 9 (V2)
1a 11a 21a
2b 12a 22b
3c 13c 23b
4c 14c 24b
5c 15b 25c
6b 16a 26b
7b 17a 27a
8c 18c 28a
9a 19b 29b
10a 20c 30c
Aufgabe 9 (Klausur 4)
1b 11 nicht lösbar 21a
2b 12a 22a
3b 13a 23a
4b 14 nicht lösbar 24c
5a 15a 25b
6a 16b 26b
7a 17a 27c
8c 18c 28c
9c 19c 29b
10b 20c 30a
D (Dreieck) Frage 2:
Ω
⋅ =
⋅ =
= 0 , 77
3 A 300
V 400 3
I Z U
N N NY
( ) = °
=
ϕ
Narccos 0 , 8 36 , 87
⋅
°Ω
=
j36,87N
0 , 77 e Z
Frage 3:
kVA 3 , 69 3
V 400 A 300 3
U
S
N,StrangI
N N⋅ =
⋅ =
=
kW 4 , 55 8 , 0 kVA 3 , 69 cos
S
P
N,Strang=
N,Strang⋅ ϕ
N= ⋅ =
Aufgabe 9b (Klausur 5) Frage 1:
% f 2
n p s f
1 N 1
N
− ⋅ =
= (nur p = 2 ergibt eine sinnvolle Lösung!) Frage 2:
( 1 s ) 3 400 V 69 , 3 A 0 , 85 ( 100 % 2 % ) 40 kW cos
I U 3 P
P
N=
el,n⋅ η
N= ⋅
N⋅
N⋅ ϕ
N⋅ −
N= ⋅ ⋅ ⋅ ⋅ − = Frage 3:
Nm min 260
1470 2
min 60 s kW 40 n
2
M P
1N N
N
=
⋅ π
⋅
⋅ =
= π
−Frage 4:
Nm 2 676
02 , 0
1 , 0 1 , 0
02 , Nm 0 260 2
s s s
M s
M
NKipp Kipp
N N
Kipp
=
� �
� �
�
� +
⋅
=
� �
�
�
� �
�
� +
⋅
=
Frage 5:
Nm 134 1
1 , 0 1 , 0
1
Nm 676 2 1 s s
1 M M 2
Kipp Kipp
Kipp
K