Übungsaufgaben zu Schwingkreise

Lösung zur Aufgabe 8.7.1 (Tiefpass): Widerstand
$R = --1---- 2πfgC 1 = ------3------------−6--= 1,59k-Ω- (B.2.1) 2π ⋅ 10 Hz ⋅ 0,1 ⋅ 10 F$

Lösung zur Aufgabe 8.7.2 (Hochpass): Kapazitätswert
$C = --1---- 2πfgR 1 = --------------------= 132,6nF-- (B.2.2) 2π ⋅ 1200Hz ⋅ 1000 Ω$
Betragsverhältnis
$U2- ------1-------- U = ∘ -------1----- 1 1 + (2πf1CR)2 --------------1--------------- = ∘ ---------------1------------ 1 + (2π⋅500Hz⋅132,6⋅10−9F⋅1000Ω)2 = 0,385-- (B.2.3) ------$

Lösung zur Aufgabe 8.7.3 (Schwingkreis): Kapazität
$C = ----1---- (2πfr)2L 1 = -----4----2---------= 25,33nF-- (B.2.4) (2π10 Hz ) ⋅ 0,01H$
Widerstand
$∘ --- R = bXk--= b- ⋅ L- fr fr ---C---------- 103Hz ∘ 0,01H = --4----⋅ ---------−9-- = 62,83Ω- (B.2.5) 10 Hz 25,33 ⋅ 10 F -------$

Lösung zur Aufgabe 8.7.4 (Schwingkreis)

Widerstand

R = Z1 = 12,5Ω- ------

(B.2.6)

Induktivität

∘ --------- Z2 − R2 L = ----2---2- ω2 − ω1 ∘ -----ω2----------- (25Ω )2 − (12,5Ω )2 = ------−1---(500s−1)2--= 40,59mH--- (B.2.7) 300s − 300s−1

Kapazität

--1- ---------1----------- C = ω21L = (500s−1)2 ⋅ 0,04059H = 98,55μF--

(B.2.8)

Lösung zur Aufgabe 8.7.5 (Schwingkreis)

Widerstand

-1- R = Z1 = Y1 = 25,0Ω--

(B.2.9)

Kapazität

∘ --2------ --Y2-−-G2-- C = ω21 ∘ω2-−--ω2------------- (0,08S )2 − (0,04S )2 = ---------------−1-2-- = 129,9μF (B.2.10) 300s− 1 − (503000ss−)1-- ---------

Induktivität

1 1 L = -2---= -----−1-2----------−-6--= 30,79mH---- ω1C (500s ) ⋅ 129,9 ⋅ 10 F ----------

(B.2.11)

Lösung zur Aufgabe 8.7.6 (Schwingkreis)

Resonanzfrequenz

∘-R1-+--R2---- ω = ωr = R2LC − L2 ---2----------

(B.2.12)

Lösung zur Aufgabe 8.7.7 (Schwingkreis)

Resonanzfrequenz

┌ ------------ ││ 2 L- ω = ω = ∘ --R1-−-C---- r LCR22 − L2 --------------

(B.2.13)

Lösung zur Aufgabe 8.7.8 (Schwingkreis)

Resonanzfrequenz
$1 f = fr = ---∘------------2 2π LC − (RC ) 1 = ---∘------------------------------------------- 2π 0,01H ⋅ 100 ⋅ 10− 9F − (100Ω ⋅ 100 ⋅ 10−9F )2 = 5,31kHz (B.2.14) ---------$
Gesamtwiderstand
$Z = 1--= ---1--- -r Yr Re{Y-} ( 1 )2 R2-+-X2C- -ωC--- ---1---- = R = R + R = R + R (ωC )2 = 100 Ω + -----------------1---------------- 100 Ω ⋅ (2π ⋅ 5310Hz ⋅ 100 ⋅ 10− 9F)2 = 998,0Ω (B.2.15) -------$

Lösung zur Aufgabe 8.7.9 (Schwingkreis)

Frequenz
$1 1 f1 = --√------= --√--------------−6-- 2π L2C 2π 0,2H ⋅ 10 ⋅ 10 F = 112,5Hz-- (B.2.16)$
Frequenz
$┌│ ----L-- -1-│∘ 1-+-L12- f2 = 2π CL1 ┌│ ------------------ 1 │∘ 1 + 0,1H- = --- ------−-60,2H----- 2π 10 ⋅ 10 F ⋅ 0,1H = 194,9Hz-- (B.2.17)$

Lösung zur Aufgabe 8.7.10 (Schwingkreis): Kapazität
$1 1 C = -2--- = --------------2-------- ω1L1 (2π ⋅ 4000Hz ) ⋅ 0,01H = 158,3nF-- (B.2.18)$
Induktivität
$L2 = --1--− L1 ω22C 1 = --------------2-----------−9- − 0,01H (2π ⋅ 3000Hz ) ⋅ 158,3 ⋅ 10 F = 7,78mH-- (B.2.19)$