Prüfung „Einführung in die Elektrotechnik I/II Herbst 2006“
Lösung zu Aufgabe 1:
Frage 1:
( ) x U x x 1 V U
q=
0⋅ = ⋅
( ) ( )
( ) = ⋅ ( − ) ( = − ) ⋅ Ω
⋅ +
−
⋅
⋅
⋅
−
= ⋅ R x x x x 10 k
x R x 1 R
x R x 1 x R
R
i 2 2Frage 2:
Da der invertierende Eingang Massepotential hat, ergibt sich = = R = 10 k Ω I
R U
11 AB
e
.
Die Verstärkung eines gegengekoppelten idealen OP beträgt lt. Ableitung in der Vorlesung:
R 10 v R
1 2
= −
−
=
Frage 3:
Ersatzschaltbild:
Für x = 0,5: Für x = 0,8
( ) 0 , 5 0 , 5 U 0 , 5 V
U
q= ⋅ = U
q( ) 0 , 8 = 0 , 8 ⋅ U = 0 , 8 V
( ) 0 , 5 = ( 0 , 5 − 0 , 5 ) ⋅ 10 k Ω = 2 , 5 k Ω
R
i 2R
i( ) 0 , 8 = ( 0 , 8 − 0 , 8
2) ⋅ 10 k Ω = 1 , 6 k Ω
Spannungsteilerregel:
( ) 0 , 4 V
R R 5 R , 0 U 5 , 0 U
i e
e q
AB
=
⋅ +
⋅
= ( ) 0 , 69 V
R R 5 R , 0 U 8 , 0 U
i e
e q
AB
=
⋅ +
⋅
=
( ) 0 , 5 v U ( ) 0 , 5 4 V
U
CD=
u⋅
e= − U
CD( ) 0 , 8 = v
u⋅ U
e( ) 0 , 8 = − 6 , 9 V R e
R i
U q R e =U e U a = v u ⋅ U e
R i
U q =U e U a = v u ⋅ U e
Frage 4:
Frage 5:
8 , 0 5 , 0 8 , 0 5 , 0
mittel
S S S S
S = ∩ = ∪
1,0 0,8 0,5
0 1
0 1 Füllstand
x
S 0,5
S 0,8 t
t
t 1,0
0,8 0,5
0 1
0 1 Füllstand
x
S 0,5
S 0,8 t
t
t
= 1 1
S 0,5
S 0,8
S mittel
= 1 1
S 0,5
S 0,8
S mittel
Aufgabe 2:
Frage 1:
A 53 , cos 3
U I P
N N
N , el
N
=
ϕ
= ⋅
Nm 955 , n 0 2 M P
N N
N
=
⋅
= π
Frage 2:
V I 113
U P
N N N ,
i
= =
Frage 3:
Ω
=
− = ϕ
= ⋅
+ 20 , 1
A 53 , 3
V 71 I
U cos
R U R
N
iN N N
f a
( )
H 124 , Hz 0 100
1 , 39 Hz 100 A 53 , 3
V 138 f
2 I
cos 1 L U
L
N N
2 N N
f
a
=
π
= Ω π
= ⋅
⋅ π
⋅
ϕ
−
= ⋅ +
Frage 4:
( R R ) ( 2 f ( L L ) ) 5 , 23 A
I U
2 f a 2
f a
N
a
=
+
⋅ π + +
=
Nm 1 , I 2
M I M
2
N a N
a
=
⋅
=
Frage 5:
A 93 , M 1 I M I
R N N
0
= ⋅ =
( )
( I 2 f L L ) I ( R R ) 178 V U
U
i0=
2N−
0⋅ π ⋅
a+
f 2−
0⋅
a+
f=
1 0
N iN
0 i N
0
