Lösungen zur Klausur 1 28.02.2009 Aufgabe 1:
(100°C)=100Ω⋅(1+0,004K− ⋅(100°C−20°C))=132Ω
R 1
(100°C)=R(100°C)⋅I=132Ω⋅1mA=0,132Ω U
Vorteile: lineare Kennlinie Aufgabe 2:
Ω
− =
= Ω
=
= 6M
I U R U
; M I 4
R U
11 31 1 1 11
31 2
( ) ( )
π⋅ +
⋅
= + ω
+
⋅ +
⋅
⋅
= 0,416
s 1000 t sin V 915 , 0 C j
1 R R
R R
R R
R R t
u t u
1 2
1 2 1
2 1
2 1
2 32
( )
π⋅ +
⋅ +
= +
= 0,416
s 1000 t sin V 915 , 0 V 4 t u U u3 31 32
Aufgabe 3:
Hz 5033 LC
1 2
f0 1 ⋅ =
= π
Hz L 4775
R 2 f 1 ⋅ =
= π Δ
Aufgabe 4:
( )
V 74 , S 4
I U 2
U
U 2 U
I S
D th
GS
2 th GS D
⋅ = +
⇒ =
−
⋅
=
Ja, denn UDS=UV−RD⋅ID=9,5V>UGS−Uth=2,24V
Aufgabe 5:
( ) 1V
s dt t RC u 0 1 u u
t
0 e C
a= − ⋅∫ =− ⋅
ua → -15 V; der theoretisch gegen -∞ strebende Wert wird durch die Betriebsspannung begrenzt.
Aufgabe 6:
Subtrahierer: ( ( ) )
1 2 2 e
a R
U R t u
u = − ⋅
U2 entspricht Gleichanteil ue: U2 = 5 V
ua entspricht 3-fachem WS-Anteil ue, R2 = 3 R1 = 300 kΩ
Lösungen zur Klausur 1 28.02.2009 Aufgabe 7:
( ) = ⋅δ
δ
⋅
= ⋅
δ I
2 100 2
I I w , H
( )
mm A 000 I . m 50 A
I , H
1 = ⋅δ
δ
−
( ) ( ) ( )
( )
⋅ δ π⋅ δ =
⋅δ
⋅
⋅
⋅ π δ =
⋅
⋅ ⋅ µ
= δ
⋅ µ
= δ
= δ
δ
− δ
mm A
I 10 2 T
I , B
I 2 100 m A
s 10 V 2 4
I I w
, H I , B I ,
B Fe 0 0 7
( ) ( ) ( ) 2
2 2 2 4 7
2 0
mag I
2 100 m 10 m A
s 10 V 2 4
I w V A
2 I , H I , 2 B I ,
W ⋅ ⋅
⋅ δ
⋅ π
=
⋅ δ ⋅
⋅
=µ δ ⋅
⋅
⋅ δ
= δ
− δ −
δ
( )
⋅ δ
⋅
⋅ π
δ = − mm
A 10 I VAs
I ,
Wmag 4 2
( )
2
0 mag
2 I A w
I W ,
F
δ
⋅
⋅ ⋅
⋅ µ
− δ =
∂
= ∂
δ δ
A 6 , 100 35
m 10 2 2 m 10 s V 10 4
m A N 100 w
2 A F 2 I
3 3
2 4 7
0
⋅ =
⋅ ⋅
⋅
⋅ π
= ⋅ δ
⋅ ⋅
⋅
⋅µ
=
−
−
−
− δ
Aufgabe 8:
s V 144 , min 0 1000 4 n , n 2
k U 1
0
a = ⋅ =
=
Φ −
A R 12
I U
a a
K = =
m N 275 , 0 2 I
MK k ⋅ K = π
= Φ
Aufgabe 9 (Klausur 1)
1b 11c 21b
2c 12a 22a
3c 13b 23b
4a 14a 24b
5a 15c 25a
6a 16c 26a
7c 17c 27b
8c 18b 28b
9c 19a 29b
10a 20a 30a
