% \ifthenelse{\equal{\toPrint}{Lösung}}{% \newcounter{acounter} \setcounter{acounter}{1} \newcounter{bcounter} \newcounter{bsubcounter} \setcounter{bcounter}{12} \setcounter{bsubcounter}{0} \section{Ergebnisse} %% Blatt 12 \textbf{\arabic{acounter} {Blitzableiter}} %\textbf{\arabic{acounter} Blitzableiter} \textbf{\arabic{acounter} {Blitzableiter}} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $u=-133{,}7\,\volt$\\ \textbf{\arabic{acounter} Drahtschleife} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $u=-36{,}39\,\milli\volt$\\ \textbf{\arabic{acounter} Metallstab} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $|B_z|=0{,}398\,\frac{\volt\second}{\square\metre}$\\ \textbf{\arabic{acounter} Spannungsverlauf} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $f=40\,\hertz$\\ $\overline{|u|}=6\,\volt$\\ $U=7{,}30\,\volt$\\ $F=1{,}22$\\ $P=0{,}533\,\watt$\\ \textbf{\arabic{acounter} Phasenanschnitt} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $U=261\,\volt$\\ \textbf{\arabic{acounter} Rechteckspannung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $u_1=1{,}5\,\volt \qquad u_2=-3{,}5\,\volt\\ U=2{,}29\,\volt$\\ %% Blatt 13 \textbf{\arabic{acounter} Scheinersatzwiderstände} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter} \setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ $R_r=520\,\ohm\\ L_r=60\,\milli\henry G_p=693\,\ohm\\ L_p=239\,\milli\henry\\ \varphi'_r=23{,}5\,\degree\\ \varphi'_p=-37{,}6\,\degree$\\ \textbf{\arabic{acounter} Verbraucherleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $p(t)=1079\,\watt+1318\,\volt\ampere\cdot \sin(2\omega t+0{,}96)\\ S=1318\,\volt\ampere\\ P=1079\,\watt\\ Q=-756\,\mathrm{var}$\\ \textbf{\arabic{acounter} Blindleistungskompensation} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $I_N=20{,}63\,\ampere\\ \varphi_N=32,52\degree\\ C=153{,}6\,\micro\farad$\\ \textbf{\arabic{acounter} Energieübertragung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $X_C=-25\,\ohm\\ P_{VR_L}=46{,}7\,\watt\\ P_W=583\,\watt $\\ \textbf{\arabic{acounter} Wechselstrommotor} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $I=15{,}5\,\ampere\\ C=80{,}6\,\micro\farad\\ I'=12{,}1\,\ampere $\\ \clearpage \textbf{\arabic{acounter} Parallelschaltung von L und C} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $i_L(t_2)=2{,}92\,\ampere $\\ \textbf{\arabic{acounter} Werte $R_L$ und $L$ einer Spule} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $I=1\,\ampere\\ R_L=12{,}5\,\ohm\\ L=0{,}219\,\henry $\\ %% Blatt 14 \textbf{\arabic{acounter} Zeigerdiagramm} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ $\underline{U}_e=15\,\volt\cdot e^{+j90\degree}$\\ \textbf{\arabic{acounter} Gesamtwiderstand} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ Induktiv\\ \textbf{\arabic{acounter} Brückenschaltung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{U}_a=130\,\volt\cdot e^{j83\,\degree}$\\ \textbf{\arabic{acounter} Zeigerdiagramm Netzwerk} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $R_1=13{,}89\,\kilo\ohm$\\ \textbf{\arabic{acounter} Blind- Wirk- und Scheinleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{Z}=84{,}85\,\ohm\cdot e^{-j45\,\degree}=(60-j60)\,\ohm\\ \uline{I}=2{,}71\,\ampere\cdot e^{j45\,\degree}=(1{,}916+j1{,}916)\,\ampere\\ \uline{I}_{R_C}=1{,}21\,\ampere\cdot e^{-j18{,}4\,\degree}=(1{,}150-j0{,}383)\,\ampere\\ \uline{I}_{C}=2{,}42\,\ampere\cdot e^{+j71{,}6\,\degree}=(0{,}766+j2{,}30)\,\ampere\\ \uline{U}_{R_C}=242\,\volt\cdot e^{-j18{,}4\,\degree}\\ \uline{U}_{R_L}=54{,}2\,\volt\cdot e^{j45\,\degree}\\ \uline{U}_L=54.2\,\volt\cdot e^{j135\,\degree}\\ S=623\,\volt\ampere\\ P=447\,\watt\\ Q=-447\,\volt\ampere r $\\ %% Blatt 15 \textbf{\arabic{acounter} Komplexe Wechselstromrechnung Netzwerk Strom} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ $\uline{I_2}=53{,}92\,\milli\ampere\cdot e^{j176{,}32\,\degree}$\\ \textbf{\arabic{acounter} Übergang Zeitabhängige zu Komplexen Größen} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{I}_L=34{,}45\,\milli\ampere\cdot e^{j50{,}83\,\degree}\\ i_L(T)=-25{,}79\,\milli\ampere$\\ \textbf{\arabic{acounter} Leitwert} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $R=500\,\ohm\\ B_C=2{,}175\,\milli\siemens$\\ \textbf{\arabic{acounter} Strom L-R-C} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{I}=0{,}4058\,\ampere\cdot e^{j67{,}4\,\degree}=(0{,}156+j0{,}375)\,\ampere $\\ \textbf{\arabic{acounter} Überlagerungsmethode} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{I}_C=(578{,}41+j279{,}99)\,\milli\ampere=642{,}6\,\milli\ampere\cdot e^{+j25{,}38\,\degree}$\\ \textbf{\arabic{acounter} Momentan Leistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $u_R(T)=-114{,}1\,\milli\volt\\ u_L(T)=-50{,}5\,\milli\volt\\ u_C(T)=119{,}7\,\milli\volt\\ p(T)=0{,}427\,\milli\watt$\\ \clearpage %% Blatt 16 \textbf{\arabic{acounter} CLR Netzwerk} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ $S=243{,}32\,\milli\volt\ampere\\ P=141{,}50\,\milli\watt\\ Q=-197{,}93\,\milli\,\var$\\ \textbf{\arabic{acounter} Wirkleistung vs. Blindleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ --\\ \textbf{\arabic{acounter} Wirkleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $R_v=17{,}24\,\ohm\\ C=19{,}14\,\micro\farad\\ P_{v\text{,}max}=14{,}5\,\milli\watt$ \textbf{\arabic{acounter} Abgebbare Wirkleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $F_ \% =-78{,}8\, \%$\\ \textbf{\arabic{acounter} Wirkleistung Spannungsquelle} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $P=-16{,}31\,\milli\watt$\\ \textbf{\arabic{acounter} Dualitätskonstante} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $L_2=100\,\milli\henry\\ C_2=5\,\micro\farad$ \textbf{\arabic{acounter} Dualitätskonstante verlustbehaftete Bauelemente} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $R_G=1\,\milli\henry\\ G_R=2\,\milli\siemens\\ C_L=100\,\nano\farad$\\ \textbf{\arabic{acounter} Vierpol Y-Parameter} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{Y}_L=\frac{1}{j\omega\cdot L}=-j\frac{1}{\omega\cdot L}\\ \uline{Y}_{11}=\frac{1}{3}\cdot \uline{Y}_L\\ \uline{Y}_{12}=-\frac{1}{6}\cdot \uline{Y}_L\\ \uline{Y}_{22}=\uline{Y}_{11}=\frac{1}{3}\cdot \uline{Y}_L=\\ \uline{Z}_{11}=\uline{Z}_{24}= 4\cdot j\omega\cdot L\\ \uline{Z}_{12}=\uline{Z}_{21}= 2\cdot j\omega\cdot L$\\ \textbf{\arabic{acounter} Spannung Vierpol} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{U}_0=130{,}4\,\milli\volt$\\ %% Blatt 17 \textbf{\arabic{acounter} Stromortskurve} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ $\uline{I}=236\,\milli\ampere\cdot e^{(j35{,}2\,\degree)}=(193+j136)\,\milli\ampere$\\ \textbf{\arabic{acounter} Leitwerts-, Widerstandsortskurve} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ -- \textbf{\arabic{acounter} Ortskurve} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $f_g=828{,}9\,\hertz\\ a=-4{,}9\,\deci\bel$\\ \textbf{\arabic{acounter} Stromortskurve} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $I_{max}=5{,}55\,\milli\ampere $\\ \textbf{\arabic{acounter} Widerstandstransformation} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $X_L{_p}=12{,}2\,\kilo\ohm\\ X_{C_s}=-5{,}6\,\kilo\ohm\\ {X_{L_s}}=+2{,}6\,\kilo\ohm\\ X_{C_p}=-4{,}2\,\kilo\ohm $\\ \clearpage \textbf{\arabic{acounter} Brückenschaltung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{U}_{ab}=+j100\,\volt=100\,\volt\cdot e^{+j90\degree} $ \textbf{\arabic{acounter} Wechselstrombrücke} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $ \uline{Z}_2=(1+j0{,}5)\,\kilo\ohm\\ R_2=1{,}25\,\kilo\ohm\\ L_2=2{,}5\,\milli\henry$\\ \textbf{\arabic{acounter} Wechselstrombrücke} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $X=+\frac{R^2}{X_L}$\\ %% Blatt 18 \textbf{\arabic{acounter} Übertrager im Leerlauf} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ $Z_1=23\,\ohm\cdot e^{j77{,}44\,\degree}\\ P=500\,\watt\\ Q=2245\,var\\ \omega L_1=22{,}45\,\ohm\\ \omega M=10\,\ohm\\ \omega L_2=4{,}45\,\ohm$\\ \textbf{\arabic{acounter} Übertrager mit kapazitiver Last} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{U}_2=0{,}398\,\volt\cdot e^{-j174{,}3\,\degree}\\ \uline{U}_2=0{,}42\,\volt\cdot e^{+j18{,}4\,\degree}\\ P_2=178\,\micro\watt $\\ \textbf{\arabic{acounter} Übertrager mit Verbindung zum Eingang} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{U}_{ab}=200\,\volt\cdot e^{j73{,}74\,\degree}\\ \uline{U}_S=317\,\volt\cdot e^{j79{,}8\,\degree}\\ \uline{I}=0{,}689\,\ampere\cdot e^{-j8{,}43\,\degree}\\ \uline{U}_{ab}=159\,\volt\cdot e^{+j71{,}7\,\degree}\\ P=12{,}2\,\watt$\\ \textbf{\arabic{acounter} Impedanzmatrix} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{Z}_{11}=(730+j100)\,\ohm\\ \uline{Z}_{12}=\uline{Z}_{21}=1200\,\ohm\\ \uline{Z}_{22}=(2000-j200)\,\ohm $\\ \textbf{\arabic{acounter} Netztransformator} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $U_2=105{,}1\,\volt\\ I_1=118\,\milli\ampere\\ \widehat{B}=1{,}095\,\tesla$\\ \textbf{\arabic{acounter} 3-Phasen Spannungssystem} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{I}_N=10{,}6\,\ampere\cdot e^{j14{,}7\,\degree} $\\ \textbf{\arabic{acounter} 3-Phasen System mit unsymmetrischem Verbraucher} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $P=-335{,}44\,\watt $\\ \textbf{\arabic{acounter} Strangströme 3-Phasen System mit unsymmetrischem Verbraucher} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{I}_C=4\,\ampere\cdot e^{j120\,\degree}=(-2+j3{,}46)\,\ampere\\ \uline{I}_{RL}=4\,\ampere\cdot e^{j113{,}1\,\degree}=(-1{,}57+j3{,}68)\,\ampere\\ \uline{I}_1=0{,}48\,\ampere\cdot e^{-j153\,\degree}\\ \uline{I}_2=3{,}66\,\ampere\cdot e^{j109{,}1\,\degree}\\ \uline{I}_3=4{,}03\,\ampere\cdot e^{j66{,}1\,\degree}\\ P=1280\,\watt $ \clearpage %% Blatt 19 \textbf{\arabic{acounter} Resonanzfrequenz Zweipol} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): } $f_{res}=899\,\hertz\\ Z=37{,}5\,\ohm $\\ \textbf{\arabic{acounter} RLC-Reihenschwingkreis} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $R=10\,\ohm\\ C=3{,}47\,\micro\farad\\ L=35{,}4\,\milli\henry$\\ \textbf{\arabic{acounter} Effektivwert und Klirrfaktor} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $I_{\mu}=7{,}36\,\milli\ampere\\ k_{\mu}=27{,}7\%\\ I_0=7{,}88\,\milli\ampere\\ k_0=25{,}8\%$\\ \textbf{\arabic{acounter} Klirrfaktor} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $k_a=21{,}5\%\\ U_a=2{,}49\,\volt$\\ \textbf{\arabic{acounter} Momentanspannung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $u_C(t=T=2\,\micro\second)=1{,}315\,\volt $\\ \textbf{\arabic{acounter} Nichtlinears Bauelement} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $k_i=19{,}61\%$\\ \textbf{\arabic{acounter} Wirkleistung Zweipol} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $P=573\,\watt$\\ %% Blatt 20 \textbf{\arabic{acounter} Ringspule} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): } $U=37{,}73\,\milli\volt$\\ \textbf{\arabic{acounter} Netzwerk Wirk- und Blindanteil} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ $\uline{I}=5{,}25\milli\ampere\cdot e^{j18{,}5\,\degree}=(4{,}98+j1{,}67)\milli\ampere\\ \uline{I}_{\textrm{w}}=5{,}14\,\milli\ampere\cdot e^{j30\,\degree}\\ \uline{I}_b=1{,}05\,\milli\ampere\cdot e^{-j60\,\degree}\\ P_K=21{,}6\,\milli\watt\\ Q_K=4{,}4\,\milli\var\\ P_I=14{,}5\,\milli\watt\\ Q_I=-11{,}5\,\milli\var $\\ \textbf{\arabic{acounter} Gleichungen in Matrizenschreibweise} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ \scriptsize \begin{align*} &\left( \begin{array}{ccc} (0{,}8706+j1{,}6824) & (0-j2) & (-0{,}4706+j0{,}1176) \\ (0-j2) & (0{,}5+j2{,}0) & (0{,}5+j0) \\ (-0{,}4706-j0{,}1176) & (0{,}5+j0) & (1{,}4706+j1{,}8823) \\ \end{array} \right)\,\milli\siemens \left( \begin{array}{c} \uline{U}_{10} \\ \uline{U}_{20} \\ \uline{U}_{30} \\ \end{array} \right)= \left( \begin{array}{c} (6-j0)\\ (13+j5)\\ (0-j5)\\ \end{array} \right)\,\milli\ampere \end{align*}\\ \clearpage % }{}%