Lösungen Klausur Version 1 und 4
Aufgabe 1:
� ( ) = Ω
⋅ π
⋅ Ω
⋅
⋅ =
⋅ π ρ
=
⋅ ρ
=
−−
6 , 37
2 m 10
m 1 , m 0 10 2 5
d l A
R l
24 7
2
� = ρ ⋅ = ⋅ = ⋅ = 6 , 83 Ω 97
, 0
01 , R 1 A
´ A
l
´ R l
´ A
´
´ l
R
2Aufgabe 2:
� k = 2 z = 3
(k-1) = 1 Knotengleichungen z - (k-1) = 2 Maschengleichungen
� ( )
− °°
µ =
⋅
⋅ π
⋅
⋅
= ⋅
= ω
−
⋅
=
j900 j C
C C
C
15 , 9 V e
F 10 Hz 50 2 j
e mA 50 C
j jX I I
U
Aufgabe 3:
� Ω
⋅
− ω =
+
= f Hz
10 592 , j 1 C 30
j R 1 Z
5
� f 5305 Hz
Hz f
10 592 , j 1 30
5
⇒ =
⋅ Ω
= Ω
( ) ( )
45 4 1
arctan 1 j
1 R f
Z
g Zπ
−
=
°
−
− =
=
⇒ ϕ
−
⋅
=
Aufgabe 4:
U = 10 V
R = 100 Ω
U S = 0,7 V I D
R S = 10 Ω
D (Ideale Diode) U = 10 V
R = 100 Ω
U S = 0,7 V I D
R S = 10 Ω
D (Ideale Diode)
� aus Maschenumlauf: 84 , 5 mA
10 100
V 7 , 0 V 10 R R
U I U
S S
D
=
Ω + Ω
= − +
= −
Aufgabe 5:
� aus Vorlesungsskript: Verstärkung eines invertierenden OPV:
� 2 10 f Hz j
Hz f 10 2 nF 100 f 2 k j 10
k 100 Z
v Z
32
e g
U
π ⋅ ⋅ −
⋅
⋅
= π
⋅
− π Ω
= Ω
=
−−
für f g müssen Real- und Imaginärteil der Verstärkung gleich sein:
Hz 159 10 Hz
2 f 1 1 Hz f 10
2
3 3=
⋅
= π
⇒
=
⋅
⋅
π
− −Aufgabe 6:
� für u e > U 2 gilt u a ≈ -15 V
� für u e < U 2 gilt u a ≈ +15 V
� Spannungsteilerregel: ⇒ = Ω
Ω
= − R 24 k
k 1 R
R V
15 V 4 , 14
2 2
2
Aufgabe 7:
� F = B ⋅ I ⋅ l = 0 , 5 T ⋅ 1 A ⋅ 0 , 01 m = 5 mN
� 100 5 mN m
2 m 01 , mN 0 5 2 2 w F d 2
F = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =
� ⋅ = °
=
=
ϕ 1 , 05 bzw . 60 , 1 m
N 048 , 0
rad m mN 5 c M
Aufgabe 8:
� ( X
a+ X
f) = 2 π ⋅ f
N⋅ ( L
a+ L
f) = 2 π ⋅ 50 Hz ⋅ 300 mH = 94 , 25 Ω
� ( ) ( ) ( 20 ) ( 94 , 25 ) 2 , 39 A
V 230 X
X R
R I U
2 2
2 f a 2 f a
N
K
=
Ω +
Ω
= +
+ +
=
� A
30 Vs , I 3 ' k
k k . bzw Vs 89 , 7 k . bzw Vs 255 , 2 1 I k 2 M k
K K K K
K K
K
K
Φ =
= Φ
⋅
⋅
= Φ π =
⇒ Φ π ⋅
= Φ
� Reihenschluss: 3 Nm 0 , 525 Nm
A 39 , 2
A M 1
I M I I
~ M
2
K 2
K
2
⋅ =
=
⋅
=
⇒
� 7 , 89 Vs 3 , 30 Vs
39 , 2
1 2
k I k I I
~
k
KK N
N
= ⋅ =
π
⋅ Φ
=
⇒ Φ Φ
( )
( U X X I ) ( ( R R ) I ) ( 230 V ) ( 94 , 25 1 A ) ( 20 1 A ) 189 , 8 V
U
iN=
2N−
a+
f 2⋅
2N−
a+
f⋅
N=
2− Ω ⋅
2− Ω ⋅ =
1 1
N iN N
i
