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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: $U_0=\SI{60}{V}$, $R_1=\SI{40}{\ohm}$, $R_2=\SI{40}{\ohm}$, $R_3=\SI{20}{\ohm}$, $R_4=\SI{30}{\ohm}$, $R_5=\SI{30}{\ohm}$. \begin{center} \begin{circuitikz} \draw (0,0) to [battery, l^=$U_0$] (0,2); \draw (0,2) to [european resistor, l^=$R_1$] (2,2); \draw (2,2) to [european resistor, l^=$R_2$] (4,2); \draw (4,2)--(6,2); \draw (6,2)--(6,0); \draw (4,0) to [european resistor, l^=$R_3$] (6,0); \draw (2,2) to [european resistor, l^=$R_4$] (4,0); \draw (2,0)--(4,0); \draw (0,0) to [european resistor, l^=$R_5$] (2,0); \end{circuitikz} \end{center}
\begin{center} \begin{circuitikz} \draw (0,-1) to [battery, l^=$U_0$] (0,1); \draw (0,1) to [european resistor, l^=$R_1$] (3,1); \draw (3,0)--(3,2); \draw (3,2) to [european resistor, l^=$R_2$] (5,2); \draw (5,2) to [european resistor, l^=$R_3$] (7,2); \draw (3,0) to [european resistor, l^=$R_4$] (7,0); \draw (7,0)--(7,2); \draw (7,1) to [european resistor, l^=$R_5$] (9,1); \draw (0,-1)--(9,-1)--(9,1); \end{circuitikz} \end{center} \begin{tabularx}{\textwidth}{X||XXXXXX} $i=$ & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline\hline $U_i$ & \cellcolor{blue!25}{\SI{60}{V}} & \SI{26.67}{V} & \SI{8.904}{V} & \SI{4.452}{V} & \SI{13.33}{V} & \SI{20}{V} \\ $R_i$ & \SI{90}{\ohm}& \cellcolor{blue!25}{\SI{40}{\ohm}} & \cellcolor{blue!25}{\SI{40}{\ohm}} & \cellcolor{blue!25}{\SI{20}{\ohm}} & \cellcolor{blue!25}{\SI{30}{\ohm}} & \cellcolor{blue!25}{\SI{30}{\ohm}} \\ $I_i$ & \SI{0.667}{A} & \SI{0.667}{A} & \SI{0.223}{A} & \SI{0.223}{A} & \SI{0.444}{A} & \SI{0.667}{A} \\ \end{tabularx}
09:32, 5. Nov. 2020 | lsgtabelle | Urs Zellweger (urs) | Current Version |
09:12, 4. Nov. 2020 | Initial Version. | Urs Zellweger (urs) | Compare with Current |