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Berechne für die folgende Schaltskizze den Totalstrom, den Ersatzwiderstand, den Spannungsabfall über jedem einzelnen Widerstand sowie die Stromstärke durch jeden einzelnen Widerstand.\\ Rechne mit folgenden Werten: $U_0=\SI{150}{V}$, $R_1=\SI{10}{\Omega}$, $R_2=\SI{20}{\Omega}$, $R_3=\SI{30}{\Omega}$, $R_4=\SI{40}{\Omega}$, $R_5=\SI{50}{\Omega}$. \begin{center} \begin{circuitikz} \draw (0,0) to [battery, l^=$U_0$] (0,2); \draw (0,2)--(2,2); \draw (2,2) to [european resistor, l^=$R_1$] (4,2); \draw (4,2) to [european resistor, l^=$R_2$] (6,2); \draw (6,2)--(6,0); \draw (6,0)--(2,0); \draw (2,0) to [european resistor, l^=$R_5$] (0,0); \draw (2,2) to [european resistor, l^=$R_3$] (4,0); \draw (4,2) to [european resistor, l^=$R_4$] (6,0); \end{circuitikz} \end{center}
\begin{center} \begin{circuitikz} \draw (0,-1) to [battery, l^=$U_0$] (0,2); \draw (0,2)--(2,2); \draw (2,2) to [european resistor, l^=$R_1$] (4,2); \draw (4,2.5) to [european resistor, l^=$R_2$] (6,2.5); \draw (4,1.5) to [european resistor, l^=$R_4$] (6,1.5); \draw (4,2.5)--(4,1.5); \draw (6,2.5)--(6,1.5); \draw (7,1) to [european resistor, l^=$R_5$] (9,1); \draw (2,0) to [european resistor, l^=$R_3$] (7,0); \draw (2,2)--(2,0); \draw (7,0)--(7,2)--(6,2); \draw (9,1)--(9,-1)--(0,-1); \end{circuitikz} \end{center} \begin{tabularx}{\textwidth}{X||XXXXXX} $i=$ & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline\hline $U_i$ & \cellcolor{blue!25}{\SI{150}{V}} & \SI{13.36}{V} & \SI{17.83}{V} & \SI{31.19}{V} & \SI{17.83}{V} & \SI{118.8}{V} \\ $R_i$ & \SI{63.13}{\ohm}& \cellcolor{blue!25}{\SI{10}{\ohm}} & \cellcolor{blue!25}{\SI{20}{\ohm}} & \cellcolor{blue!25}{\SI{30}{\ohm}} & \cellcolor{blue!25}{\SI{40}{\ohm}} & \cellcolor{blue!25}{\SI{50}{\ohm}} \\ $I_i$ & \SI{2.376}{A} & \SI{1.336}{A} & \SI{0.891}{A} & \SI{1.040}{A} & \SI{0.446}{A} & \SI{2.376}{A} \\ \end{tabularx}
09:24, 5. Nov. 2020 | lsg | Urs Zellweger (urs) | Current Version |
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08:19, 4. Nov. 2020 | Initial Version. | Urs Zellweger (urs) | Compare with Current |