Wissensdatenbank Wirtschaftsrecht: history "TutoriumE3L1"

Revision history for TutoriumE3L1

Revision [61261]

Last edited on 2015-11-11 16:10:13 by Jorina Lossau

Deletions:

~~{{files}}~~

Revision [61260]

Edited on 2015-11-11 16:09:41 by Jorina Lossau

Additions:

Warum errechnet sich die komplexe Scheinleistung aus dem Produkt der komplexen Spannung und dem konjugiert, komplexen Strom beziehungsweise aus dem Produkt des konjugiert, komplexen Leitwertes und der Spannung zum Quadrat aber dem Produkt der (nicht konjugierten) Impedanz und dem Strom zum Quadrat? Warum das konjugieren in manchen Fällen?
Im Verbraucherzählpfeilsystem ist die Blindleistung an einer Kapazität negativ, an einer Induktivität positiv. Der Imaginärteil der Scheinleistung muss also an einem Kondensator negativ sein. Der Strom an einer Kapazität eilt der Spannung aber immer vor, hat also einen positiven Imaginärteil. Wird konjugiert stimmt das Ergebnis.
Der Imaginärteil X der Impedanz Z ist bei einer Kapazität bereits negativ, der Imaginärteil B der Admittanz Y dagegen positiv. Umgekehrt gilt gleiches für die Induktivität.

Revision [61259]

Edited on 2015-11-11 15:53:34 by Jorina Lossau

Additions:

S = P + jQ = U* I* = Z* I² = Y* * U²

Revision [61258]

Edited on 2015-11-11 15:40:59 by Jorina Lossau

Additions:

QK,10 = -2π 50Hz 2,781pF*(230V(1-0,1))²
QK,10 = -37µVAr
QK1 = -2π 50Hz 2,781pF*(230V(1-0,9997545))²
QK1 = -2,8pVAr

Revision [61257]

Edited on 2015-11-11 15:32:17 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung44.jpg" width="150" class="left"}}~~

Revision [61256]

Edited on 2015-11-11 15:31:07 by Jorina Lossau

Additions:

QE10 = -2π50Hz 25pF(0,1*230V)²
QE10 = -4,2µVAr
QE1 = -2π 50Hz 25pF*(0,9997545*230V)²
QE1 = -0,42 mVAr
QK = -BK(U-U2)², BK = wCK
QK =
{{image url="anpassung44.jpg" width="150" class="left"}}

Deletions:

~~QE10 = -2π~~

Revision [61255]

Edited on 2015-11-11 15:14:35 by Jorina Lossau

Additions:

QE10 = -2π

Revision [61254]

Edited on 2015-11-11 15:05:48 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung43.jpg" width="200" class="left"}}~~

Revision [61253]

Edited on 2015-11-11 15:05:04 by Jorina Lossau

Additions:

QE =
{{image url="anpassung43.jpg" width="200" class="left"}}

Revision [61252]

Edited on 2015-11-11 12:17:01 by Jorina Lossau

Additions:

QE = -BEU2², BE =wCE

Revision [61251]

Edited on 2015-11-11 12:04:56 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung42.jpg" width="150" class="left"}}~~

Revision [61250]

Edited on 2015-11-11 12:04:10 by Jorina Lossau

Additions:

P1 =
{{image url="anpassung42.jpg" width="150" class="left"}}
= 13µW

Revision [61249]

Edited on 2015-11-11 11:52:34 by Jorina Lossau

No Differences

Revision [61248]

Edited on 2015-11-11 11:52:03 by Jorina Lossau

Additions:

{{image url="anpassung40.jpg" width="150" class="left"}}, U = U1 = 230V
P10 =
{{image url="anpassung41.jpg" width="150" class="left"}}
= 4,8 mW

Deletions:

~~{{image url="anpassung40.jpg" width="150" class="left"}}~~

Revision [61247]

Edited on 2015-11-11 11:35:28 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung40.jpg" width="100" class="left"}}~~

Revision [61246]

Edited on 2015-11-11 11:34:53 by Jorina Lossau

Additions:

P = (U-U2²)/R1
{{image url="anpassung40.jpg" width="100" class="left"}}

Revision [61245]

Edited on 2015-11-11 11:26:33 by Jorina Lossau

Additions:

{{image url="anpassung39.jpg" width="100" class="left"}} = -BU² = -wCU²

Deletions:

~~{{image url="anpassung39.jpg" width="100" class="left"}}~~

Revision [61244]

Edited on 2015-11-11 11:25:11 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung39.jpg" width="150" class="left"}}~~

Revision [61243]

Edited on 2015-11-11 11:24:28 by Jorina Lossau

Additions:

Q =
{{image url="anpassung39.jpg" width="150" class="left"}}

Revision [61186]

Edited on 2015-11-10 21:45:26 by Jorina Lossau

Additions:

{{image url="anpassung38.jpg" width="150" class="left"}} = 0,53 mW

Deletions:

~~{{image url="anpassung38.jpg" width="100" class="left"}} = 0,53 mW~~

Revision [61183]

Edited on 2015-11-10 21:44:14 by Jorina Lossau

Additions:

{{image url="anpassung37.jpg" width="100" class="left"}} = 0,53 mW
PE,1 =
{{image url="anpassung38.jpg" width="100" class="left"}} = 0,53 mW

Deletions:

~~{{image url="anpassung37.jpg" width="100" class="left"}}~~

Revision [61178]

Edited on 2015-11-10 21:38:10 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung36.jpg" width="100" class="left"}}~~

Revision [61177]

Edited on 2015-11-10 21:34:33 by Jorina Lossau

Additions:

PE,10 =
{{image url="anpassung36.jpg" width="100" class="left"}}

Revision [61176]

Edited on 2015-11-10 21:26:42 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung35.jpg" width="150" class="left"}}~~

Revision [61175]

Edited on 2015-11-10 21:25:42 by Jorina Lossau

Additions:

PE =
{{image url="anpassung35.jpg" width="150" class="left"}}

Revision [61174]

Edited on 2015-11-10 21:22:00 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung34.jpg" width="100" class="left"}}~~

Revision [61173]

Edited on 2015-11-10 21:21:08 by Jorina Lossau

Additions:

PE =
{{image url="anpassung34.jpg" width="100" class="left"}}

Revision [61172]

Edited on 2015-11-10 20:50:34 by Jorina Lossau

Additions:

=GU^2 = U^2/R

Revision [61171]

Edited on 2015-11-10 20:47:43 by Jorina Lossau

Additions:

P =
{{image url="anpassung33.jpg" width="100" class="left"}}

Deletions:

~~P = {{image url="anpassung33.jpg" width="600" class="left"}}~~

Revision [61170]

Edited on 2015-11-10 20:44:26 by Jorina Lossau

Additions:

P = {{image url="anpassung33.jpg" width="600" class="left"}}

Revision [61169]

Edited on 2015-11-10 20:34:14 by Jorina Lossau

Additions:

||Berechnen Sie die Leistungen an den Widerständen und Kapazitäten, wenn eine Spannung von U = 230 V an der Spitze des Tastkopfes anliegt.

Revision [61166]

Edited on 2015-11-10 20:15:57 by Jorina Lossau

Additions:

|| ℓ10 in Grad ||0||0||0||0||
||H1(f, CK = 2,781pF) ||0,9997545||0,9997545||0,9997545||0,7623251||
||ℓ1(CK = 2,781pF) in Grad ||0||-0,0022065||-0,2206444||-35,848693||
||H1(f,CK = 0) ||0,9997545||0,9997545||0,9997471||0,7920504||
||ℓ1(CK = 0) in Grad ||0||-0,0022065||-0,2206508 ||-37,604201||

Revision [61160]

Edited on 2015-11-10 19:57:43 by Jorina Lossau

Additions:

|| f in Hz ||0||10^3||10^5||2*10^7||
|| H10(f) ||0,1||0,1||0,1||0,1||
|| H10(f) ||0,1||0,1||0,1||0,1||

Revision [61128]

Edited on 2015-11-10 10:00:27 by Jorina Lossau

Additions:

{{image url="anpassung32.jpg" width="600" class="left"}} .

Deletions:

~~{{image url="anpassung32.jpg" width="250" class="left"}} .~~

Revision [61127]

Edited on 2015-11-10 09:59:32 by Jorina Lossau

Additions:

f ={0; 10^3; 10^5; 2*10^7} Hz. Kommt es auch zu einer Phasenverschiebung? Wenn ja, berechnen Sie diese für die gleichen Frequenzen f .
{{image url="anpassung32.jpg" width="250" class="left"}} .

Revision [61126]

Edited on 2015-11-10 09:31:04 by Jorina Lossau

Additions:

wenn keine Abschwächung gewählt ist für:

Revision [61125]

Edited on 2015-11-10 09:30:12 by Jorina Lossau

Additions:

{{image url="anpassung30.jpg" width="250" class="left"}} .
{{image url="anpassung31.jpg" width="100" class="left"}} .

Deletions:

{{image url="anpassung30.jpg" width="300" class="left"}} .
{{image url="anpassung31.jpg" width="300" class="left"}} .

Revision [61124]

Edited on 2015-11-10 09:28:42 by Jorina Lossau

Additions:

CK = 2,781 pF
Berechnen Sie das Teilerverhältnis
{{image url="anpassung31.jpg" width="300" class="left"}} .

Revision [61123]

Edited on 2015-11-10 09:21:14 by Jorina Lossau

Additions:

{{image url="anpassung30.jpg" width="300" class="left"}} .

Deletions:

~~{{image url="anpassung30.jpg" width="400" class="left"}} .~~

Revision [61122]

Edited on 2015-11-10 09:18:59 by Jorina Lossau

Additions:

{{image url="anpassung30.jpg" width="400" class="left"}} .

Revision [61121]

Edited on 2015-11-10 09:13:05 by Jorina Lossau

Additions:

Auf welchen Wert muss die Kompensationskapazität CK eingestellt werden, um einen Frequenzunabhängigen Spannungsteiler zu erhalten?

Revision [61120]

Edited on 2015-11-10 09:09:47 by Jorina Lossau

Additions:

{{image url="anpassung29.jpg" width="400" class="left"}} .

Revision [61119]

Edited on 2015-11-10 08:55:45 by Jorina Lossau

Additions:

Gegeben ist ein Oszilloskop mit einem Eingangswiderstand von RE= 998;6 kΩ und einer Eingangskapazität von CE = 25 pF. Dazu gab es einen Tastkopf an den zwischen 1-facher und 10-facher Abschwächung umgeschaltet werden kann.
Der Widerstand für 10-fache Abschwächung beträgt R10 = 8;978 MΩ. Ohne Abschwächung beträgt der Widerstand noch R1 = 245;23 Ω. Skizzieren Sie einen Schaltplan für angeschlossenen Tastkopf (der tastkopf ist mit einer einstellbaren Kompensationskapazität ausgestattet).
.
.

Revision [61118]

Edited on 2015-11-10 08:46:47 by Jorina Lossau

Additions:

**3 Frequenzkompensierter Spannungsteiler**

Revision [61117]

Edited on 2015-11-10 08:40:04 by Jorina Lossau

Additions:

Revision [61116]

Edited on 2015-11-10 08:25:48 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung27.jpg" width="300" class="left"}}~~

Revision [61115]

Edited on 2015-11-10 08:24:32 by Jorina Lossau

Additions:

Bode - Diagramm der Zusammenschaltung:
{{image url="anpassung27.jpg" width="300" class="left"}}

Revision [61114]

Edited on 2015-11-10 08:19:33 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung26.jpg" width="200" class="left"}}~~

Revision [61113]

Edited on 2015-11-10 08:18:34 by Jorina Lossau

Additions:

Revision [61112]

Edited on 2015-11-10 08:06:17 by Jorina Lossau

Additions:

Wie würde der Amplituden- und Phasengang aussehen, wenn Sie je einen der entworfenen Tief- und einen der Hochpässe, getrennt durch einen idealen OPV mit der Verstärkung 1, in Reihe schalten? Skizzieren Sie die Anordnung, den Amplituden- und Phasengang.

Revision [61040]

Edited on 2015-11-07 15:10:10 by Jorina Lossau

Additions:

Rneu = 2 π fu =3,4 Ω

Revision [61029]

Edited on 2015-11-07 11:31:17 by Jorina Lossau

Additions:

Weil große Induktivitäten teuer sind, solle eine um den Faktor 1000 kleinere Induktivität verwendet werden. Wie groß muss dann der neue Widerstand gewählt werden?

Revision [61028]

Edited on 2015-11-07 11:29:22 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung25.jpg" width="50" class="left"}}~~

Revision [61027]

Edited on 2015-11-07 11:23:31 by Jorina Lossau

Additions:

Revision [61026]

Edited on 2015-11-07 11:15:26 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung24.jpg" width="200" class="left"}}~~

Revision [61024]

Edited on 2015-11-07 11:13:02 by Jorina Lossau

Additions:

Es handelt sich jetzt um einen RL-Hochpass mit:
R2 = 3;39 kΩ
L2 =
{{image url="anpassung24.jpg" width="200" class="left"}}
= 54 H

Revision [61022]

Edited on 2015-11-07 10:52:52 by Jorina Lossau

No Differences

Revision [61018]

Edited on 2015-11-07 10:46:34 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung23.jpg" width="90" class="left"}}~~

Revision [61017]

Edited on 2015-11-07 10:45:26 by Jorina Lossau

Additions:

Revision [61016]

Edited on 2015-11-07 10:43:41 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung22.jpg" width="70" class="left"}}~~

Revision [61015]

Edited on 2015-11-07 10:43:17 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung22.jpg" width="50" class="left"}}~~

Revision [61014]

Edited on 2015-11-07 10:42:39 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung22.jpg" width="400" class="left"}}~~

Revision [61013]

Edited on 2015-11-07 10:41:47 by Jorina Lossau

Additions:

R2 =
{{image url="anpassung22.jpg" width="400" class="left"}}
= 3,39 Ω

Revision [61012]

Edited on 2015-11-07 10:36:15 by Jorina Lossau

Additions:

Es handelt sich um einen RC-Hochpass mit:
fu = 10 Hz
C2 = 4;7µF

Revision [61004]

Edited on 2015-11-06 22:01:51 by Jorina Lossau

Additions:

Entwerfen Sie eine RC-Schaltung, die den Anforderungen gerecht wird, Skizzieren Sie den Schaltplan und geben Sie die Werte der Bauteile an. (Verwenden Sie dafür wieder oben genannte Kondensatoren)

Revision [61003]

Edited on 2015-11-06 21:58:26 by Jorina Lossau

Additions:

Revision [61002]

Edited on 2015-11-06 21:57:07 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung20.jpg" width="200" class="left"}}~~

Revision [61001]

Edited on 2015-11-06 21:55:38 by Jorina Lossau

Additions:

Abbildung 2: Bode - Diagramm:
{{image url="anpassung20.jpg" width="200" class="left"}}

Revision [61000]

Edited on 2015-11-06 21:48:31 by Jorina Lossau

Additions:

Es soll nun zusätzlich eine Schaltung entworfen werden, die den Anforderungen aus Abb. 2 gerecht wird.

Revision [60999]

Edited on 2015-11-06 21:47:02 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung19.jpg" width="50" class="left"}}~~

Revision [60998]

Edited on 2015-11-06 21:34:33 by Jorina Lossau

Additions:

Revision [60997]

Edited on 2015-11-06 21:32:50 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung18.jpg" width="200" class="left"}}~~

Revision [60996]

Edited on 2015-11-06 21:21:27 by Jorina Lossau

Additions:

Es handelt sich jetzt um einen RL-Tiefpass mit:
fo = 1 kHz
R1= 33;9 Ω
L1 =
{{image url="anpassung18.jpg" width="200" class="left"}}
= 5,4 mH

Revision [60967]

Edited on 2015-11-04 16:46:38 by Jorina Lossau

Additions:

Wie muss die Schaltung aussehen, wenn Sie den gleichen Widerstand verwenden wollen und statt des Kondensators eine Luftspule verwendet werden soll? Skizzieren Sie den Schaltplan und berechnen Sie die nötige Induktivität der Luftspule.

Revision [60966]

Edited on 2015-11-04 16:33:54 by Jorina Lossau

No Differences

Revision [60965]

Edited on 2015-11-04 16:33:23 by Jorina Lossau

Additions:

Revision [60964]

Edited on 2015-11-04 16:32:00 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung16.jpg" width="400" class="left"}}~~

Revision [60963]

Edited on 2015-11-04 16:30:57 by Jorina Lossau

Additions:

Revision [60962]

Edited on 2015-11-04 16:13:46 by Jorina Lossau

Additions:

Skizzieren Sie den Schaltplan der notwendigen Schaltung und geben Sie die Werte der Bauteile an.
Es handelt sich um einen RC-Tiefpass mit: fo = 1 kHz
C1= 4;7µF

Revision [60961]

Edited on 2015-11-04 16:08:02 by Jorina Lossau

Additions:

Abbildung 1: Bode - Diagramm

Revision [60960]

Edited on 2015-11-04 16:06:17 by Jorina Lossau

Additions:

Revision [60959]

Edited on 2015-11-04 16:04:05 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung14.jpg" width="250" class="left"}}~~

Revision [60958]

Edited on 2015-11-04 16:03:07 by Jorina Lossau

Additions:

Revision [60957]

Edited on 2015-11-04 15:54:56 by Jorina Lossau

Additions:

a = 10 log(1/2)
dB = 20 log(1/√2)
dB = - 3,0103 dB
**2.2 Aufgaben**
Es soll eine Schaltung mit einer Übertragungsfunktion wie in Abb. 1 erstellt werden. Es sind zahlreiche Kondensatoren mit einer Kapazität von C = 4;7µF vorhanden. Darüber hinaus stehen unzählige Widerstände zur Verfügung.

Revision [60954]

Edited on 2015-11-04 14:47:07 by Jorina Lossau

Additions:

Ist u2< u1spricht man von Dämpfung (a < 0). Für u2 = u1 ist a = 0. Für die Bandbreite eines Signals werden die Frequenzen betrachtet für die gilt:
P2 = 1/2
P1 bzw. u2 = (1/√2)u1

Revision [60953]

Edited on 2015-11-04 14:30:29 by Jorina Lossau

Additions:

a = 20log (u2/u1) dB oder u2 = u1*10^(0,05 a/db)

Revision [60952]

Edited on 2015-11-04 14:23:18 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung13.jpg" width="200" class="left"}}~~

Revision [60951]

Edited on 2015-11-04 14:22:30 by Jorina Lossau

Additions:

Revision [60950]

Edited on 2015-11-04 14:09:36 by Jorina Lossau

No Differences

Revision [60949]

Edited on 2015-11-04 14:08:55 by Jorina Lossau

Additions:

Etwas anders geschrieben:
a/dB =
{{image url="anpassung12.jpg" width="200" class="left"}}

Revision [60945]

Edited on 2015-11-03 19:31:06 by Jorina Lossau

Additions:

10 dB =
{{image url="anpassung11.jpg" width="200" class="left"}}

Deletions:

~~10 dB = {{image url="anpassung11.jpg" width="100" class="left"}}~~

Revision [60944]

Edited on 2015-11-03 19:29:20 by Jorina Lossau

Additions:

In der Elektrotechnik ist das Dezibel gebräuchlicher:
10 dB = {{image url="anpassung11.jpg" width="100" class="left"}}

Revision [60943]

Edited on 2015-11-03 19:26:57 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung10.jpg" width="500" class="left"}}~~

Revision [60942]

Edited on 2015-11-03 19:26:05 by Jorina Lossau

Additions:

Bel (Formelzeichen B)ist eine Pseudoenheit für den Zehnerlogarithmus des Verhältnisses zweier Leistungs- bzw. Energiegrößen:
{{image url="anpassung10.jpg" width="500" class="left"}}

Deletions:

~~Bel (Formelzeichen B)ist eine Pseudoenheit für den Zehnerlogarithmus des Verhältnisses~~

Revision [60941]

Edited on 2015-11-03 19:19:24 by Jorina Lossau

Additions:

**2.1 Dämpfungsangaben**
Für die Darstellung des Phasengangs ist es sinnvoll, die f -Achse logarithmisch zu skalieren. Für den Amplitudengang ist eine doppellogarithmische Skalierung sehr sinnvoll.
Eine gängige Einheit ist Dezibel (dB), nach Alexander Graham Bell. Neben dem dB gibt es noch Neper (wird mit ln berechnet).
Bel (Formelzeichen B)ist eine Pseudoenheit für den Zehnerlogarithmus des Verhältnisses

Revision [60940]

Edited on 2015-11-03 19:17:48 by Jorina Lossau

Additions:

--------------------------------------------------------------------
**2. Frequenzabhängigkeit**
Vor den Aufgaben eine kleine Ergänzung zu Dämpfungs- und Pegelangaben.

Revision [60939]

Edited on 2015-11-03 19:16:41 by Jorina Lossau

Additions:

SL = (3,82 + j 2,20) kVA; SL = PL + jQL; Si = SL

Revision [60938]

Edited on 2015-11-03 16:52:09 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung9jpg" width="500" class="left"}}~~

Revision [60937]

Edited on 2015-11-03 16:51:07 by Jorina Lossau

Additions:

Revision [60936]

Edited on 2015-11-03 16:44:43 by Jorina Lossau

Additions:

QL = 2,20 kVAr; Qi = QL

Revision [60935]

Edited on 2015-11-03 16:09:39 by Jorina Lossau

No Differences

Revision [60934]

Edited on 2015-11-03 16:08:27 by Jorina Lossau

Additions:

PL = 3,82 kW; Pi = PL
{{image url="anpassung8.jpg" width="500" class="left"}}

Revision [60933]

Edited on 2015-11-03 15:49:06 by Jorina Lossau

Additions:

**a)**
**b)**
PL =
{{image url="anpassung7.jpg" width="500" class="left"}}

Deletions:

a)
b)

Revision [60932]

Edited on 2015-11-03 15:35:49 by Jorina Lossau

Additions:

SL = (5,09 + j2,94) kVA; SL = PL + jQL; Si = S*L = (5,09 - j2,94) kVA
b)

Deletions:

~~SL = (5,09 + j2,94) kVA; SL =~~

Revision [60931]

Edited on 2015-11-03 15:31:39 by Jorina Lossau

No Differences

Revision [60930]

Edited on 2015-11-03 12:01:06 by Jorina Lossau

Additions:

SL = (5,09 + j2,94) kVA; SL =

Revision [60929]

Edited on 2015-11-03 11:47:10 by Jorina Lossau

No Differences

Revision [60928]

Edited on 2015-11-03 11:45:51 by Jorina Lossau

Additions:

QL = 2,94 kVAr; Qi = -QL = -2,94 kVAr

SL =
{{image url="anpassung6.jpg" width="500" class="left"}}

Revision [60927]

Edited on 2015-11-03 11:38:44 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung1.jpg" width="350" class="left"}}~~

Revision [60926]

Edited on 2015-11-03 11:31:25 by Jorina Lossau

No Differences

Revision [60925]

Edited on 2015-11-03 11:30:53 by Jorina Lossau

Additions:

PL = 5,09 kW; Pi = PL
QL =
{{image url="anpassung5.jpg" width="500" class="left"}}

Revision [60924]

Edited on 2015-11-03 11:22:22 by Jorina Lossau

Additions:

{{image url="anpassung1.jpg" width="350" class="left"}}
PL =
{{image url="anpassung4.jpg" width="500" class="left"}}

Deletions:

{{image url="anpassung1.jpg" width="400" class="left"}}
PL = {{image url="anpassung4.jpg" width="200" class="left"}}

Revision [60923]

Edited on 2015-11-03 11:20:27 by Jorina Lossau

Additions:

a)
PL = {{image url="anpassung4.jpg" width="200" class="left"}}

Revision [60905]

Edited on 2015-11-03 10:00:37 by Jorina Lossau

Additions:

Berechnen Sie alle Leistungen wenn Uq = 200 V und f= 50 Hz ist.

Revision [60904]

Edited on 2015-11-03 09:57:08 by Jorina Lossau

No Differences

Revision [60903]

Edited on 2015-11-03 09:54:12 by Jorina Lossau

Additions:

Die Blindwiderstände der Innenimpedanz sind:
{{image url="anpassung3.jpg" width="200" class="left"}}

Revision [60902]

Edited on 2015-11-03 09:49:08 by Jorina Lossau

Additions:

Deletions:

~~{{image url="anpassung2.jpg" width="400" class="left"}}~~

Revision [60901]

Edited on 2015-11-03 09:48:18 by Jorina Lossau

Additions:

{{image url="anpassung1.jpg" width="400" class="left"}}
Um was für eine Last handelt es sich? Geben Sie die Werte der Induktivität bzw. Kapazität an.
Es ist eine induktive Last mit
{{image url="anpassung2.jpg" width="400" class="left"}}
||

Deletions:

~~{{image url="anpassung1.jpg" width="400" class="left"}}||~~

Revision [60898]

Edited on 2015-11-03 09:35:15 by Jorina Lossau

Additions:

||**1 Leistungsanpassung**
Es ist eine Last mit ZL (f = 50 Hz) = 3 Ohm exp (j*pi/6) gegeben. Welchen Wert muss die Innenimpedanz Zi der Quelle annehmen, dass an ZL
{{image url="anpassung1.jpg" width="400" class="left"}}||

Deletions:

||1 Leistungsanpassung
Es ist eine Last mit ZL (f=50 Hz) = 3 Ohm exp (j*pi/6) gegeben. Welchen Wert muss die Innenimpedanz Zi der Quelle annehmen, dass an ZL
{{image url="anpassung1.jpg" width="100" class="left"}}||

Revision [60897]

Edited on 2015-11-03 09:26:18 by Jorina Lossau

Additions:

{{files}}
||1 Leistungsanpassung
Es ist eine Last mit ZL (f=50 Hz) = 3 Ohm exp (j*pi/6) gegeben. Welchen Wert muss die Innenimpedanz Zi der Quelle annehmen, dass an ZL
a) die maximale Wirkleistung
b) die maximale Scheinleistung
umgesetzt wird. Geben Sie Zi jeweils in algebraischer Form und Exponentialform an.
{{image url="anpassung1.jpg" width="100" class="left"}}||

Deletions:

1 Leistungsanpassung
{{image url="anpassung1.jpg" width="100" class="left"}}

Revision [60860]

Edited on 2015-11-02 23:01:52 by Jorina Lossau

Additions:

1 Leistungsanpassung
{{image url="anpassung1.jpg" width="100" class="left"}}

Deletions:

{{image url="E3L11.jpg" width="650" class="center"}}
{{image url="E3L12.jpg" width="650" class="center"}}
{{image url="E3L13.jpg" width="650" class="center"}}
{{image url="E3L14.jpg" width="650" class="center"}}
{{image url="E3L15.jpg" width="650" class="center"}}
{{image url="E3L61.jpg" width="650" class="center"}}
{{image url="E3L71.jpg" width="650" class="center"}}
{{image url="E3L81.jpg" width="650" class="center"}}

Revision [43857]

Edited on 2014-08-28 11:50:47 by Jorina Lossau

Deletions:

~~{{files}}~~

Revision [43856]

Edited on 2014-08-28 11:50:07 by Jorina Lossau

Additions:

Deletions:

Revision [43855]

Edited on 2014-08-28 11:49:05 by Jorina Lossau

Additions:

{{image url="E3L11.jpg" width="650" class="center"}}
{{image url="E3L12.jpg" width="650" class="center"}}
{{image url="E3L13.jpg" width="650" class="center"}}
{{image url="E3L14.jpg" width="650" class="center"}}
{{image url="E3L15.jpg" width="650" class="center"}}
{{image url="E3L16.jpg" width="650" class="center"}}
{{image url="E3L71.jpg" width="650" class="center"}}
{{image url="E3L18.jpg" width="650" class="center"}}

Deletions:

{{image url="E3L1.jpg" width="650" class="center"}}
{{image url="E3L2.jpg" width="650" class="center"}}
{{image url="E3L3.jpg" width="650" class="center"}}
{{image url="E3L4.jpg" width="650" class="center"}}
{{image url="E3L5.jpg" width="650" class="center"}}
{{image url="E3L6.jpg" width="650" class="center"}}
{{image url="E3L7.jpg" width="650" class="center"}}
{{image url="E3L8.jpg" width="650" class="center"}}

Revision [43854]

Edited on 2014-08-28 11:47:36 by Jorina Lossau

Additions:

{{files}}
{{image url="E3L2.jpg" width="650" class="center"}}
{{image url="E3L3.jpg" width="650" class="center"}}
{{image url="E3L4.jpg" width="650" class="center"}}
{{image url="E3L5.jpg" width="650" class="center"}}
{{image url="E3L6.jpg" width="650" class="center"}}
{{image url="E3L7.jpg" width="650" class="center"}}
{{image url="E3L8.jpg" width="650" class="center"}}

Revision [43851]

The oldest known version of this page was created on 2014-08-28 11:22:46 by Jorina Lossau

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