Lösung Aufgabe 1:
1. = ρ ⋅ = Ω ⋅ = 0 , 5 Ω
mm 1
m 1 m 5 mm , A 0
R l
22
2. R wird größer
3. R 101 , 1 %
% 4 , 99
% 5 , R 100
´
R = ⋅ = ⋅
Lösung Aufgabe 2:
1. I = I
W+ jI
b= I
N⋅ ( cos ϕ
N− j sin ϕ
N) = ( 1 , 61 − j 1 , 64 ) A 2.
R U = U
I = I w + jI b
L R C U = U
I = I w + jI b
L
C
3. 22 , 7 F
V 230 Hz 50 2
A 64 , 1 U
C I
b= μ
⋅
⋅ π
= ⋅
⋅
= ω
Lösung Aufgabe 3:
1. = ( − ) Ω
⋅
⋅ π + ⋅ Ω ω =
+
=
−30 j 160 . 000
F 10 Hz 1000 2
j 30 1 C j R 1
Z
92. 5 , 3 MHz
F 10 30 2
1 RC
1 2 f 1 C
R 1
9=
⋅ Ω
⋅
= π π ⋅
= ω ⇒
=
−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
Lösung Aufgabe 4:
1.
0 5 10
I D mA
U DS V 4,4
3,2 3,6 15
10 5
U GS /V
4,0
0 5 10
I D mA
U DS V 4,4
3,2 3,6 15
10 5
U GS /V
4,0
2. = − = − = 769 Ω
mA 5 , 6
V 5 V 10 I
U R U
0 DS
0 DS V D
3. = Ω
μ
= −
= − 640 k
A 10
V 6 , 3 V 10 I
U R U
10 0 GS V 1
Ω μ =
=
= 360 k
A 10
V 6 , 3 I
R U
10 0 GS 2
Lösung zu Aufgabe 5:
1. 2 , 2 pF
m 10
m 10 25 m 85 pF , d 8
C A
42 6
0
⋅ =
⋅
=
⋅ ε
=
−−V 678 , pF 0 2 , 2
pC 5 , 1 C
U = Q = =
2. 0 , 610 V
mm 1 , 0
mm 09 , 0 pF 2 , 2
pC 5 , 1
´ C
´ Q
U = = ⋅ =
3. v U = 1; er wird eingesetzt um den Kondensator wg. R e → ∞ nicht mit einem Strom zu
belasten
Lösung zu Aufgabe 6:
1. u a (t) kann entweder +U V oder -U V annehmen.
2.
a(
e a)
2 1
1 e
2 1
2
2 u u
3 u 1 R R u R R R
u R ⋅ = ⋅ ⋅ +
+ + + ⋅
+
=
3.
+U
v-U
vt u
e(t) +U
v-U
vt
u
e(t)
Lösung zu Aufgabe 7:
1. 157 mH
m 10 2 2
m 10 5 m A
s 10 V 256 , 1 l 1000
2 w A
L
32 4 6
2 0
2
=
⋅
⋅
⋅ ⋅
⋅
⋅
⋅ =
⋅ μ
⋅
=
− − −δ δ
2. L I 78 , 5 mJ
2
W
mag= 1 ⋅ ⋅
2=
3. In den beiden Luftspalten
Lösung zu Aufgabe 8:
1. 0 , 8 Nm
min 1200 2
min 60 s W 100 n
2
M P
1N N
N
=
⋅ π
⋅
= ⋅
⋅ π
= ⋅
−2. P
V= P
el− P
N= U
N⋅ I
N− P
N= 14 V ⋅ 10 A − 100 W = 40 W
( ) = Ω
=
= 0 , 4
A 10
W 40 I
R P
2 2N V a
3. 35 A
4 , 0
V 14 R I U
a N
K
=
= Ω
=
Nm 8 , 2 Nm 8 , A 0 10
A M 35
I
M I
NN K
K
= ⋅ = ⋅ =
Lösung zu Aufgabe 9a (Klausur 5):
1. Y-Schaltung
2. (
ϕ) ⋅ = Ω ⋅
°= ⋅
⋅ ⋅
=
jarccoscos jarccos0,6 j53,13NY NY
N
e 3 , 46 e
A 115 3
V e 690
I 3
Z U
N3. S
N= 3 ⋅ U
NY⋅ I
NY⋅ e
−jarccos(
cosϕN) = 3 ⋅ 690 V ⋅ 115 A ⋅ e
−jarccos0,6= 13 , 7 kVA ⋅ e
−j53,13°kW
25 , 8 6 , 0 A 115 V 690 3 cos
I U 3
P
N= ⋅
NY⋅
NY⋅ ϕ
N= ⋅ ⋅ ⋅ =
Lösung zu Aufgabe 9b (Klausur 5):
1. 260 Nm
min 1470 2
min 60 s kW 40 n
2
M P
1N N
N
=
⋅ π
⋅
= ⋅
⋅ π
= ⋅
−2. 69 , 3 A
85 , 0 V 400 3
min 1470
min kW 1500
40 cos
U 3
n P n cos
U 3
P U
3 I S
1 1
N N
N 0 N
N N
N , el N
N , el
N
=
⋅
⋅
⋅ ϕ =
⋅
⋅
⋅ ϕ =
⋅
= ⋅
= ⋅
−−
3. 98 %
min 1500
min 1470 n
n
1 1
0 N
N