Thank you for the clarification. If $R_2$ and $R_3$ are in series, then the calculation changes as follows:

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(a) Berechnen Sie den Gesamtwiderstand $R_{\text{ges}}$ der Schaltung.

1. Series Combination of $R_2$ and $R_3$: $R_{23} = R_2 + R_3$

2. Parallel Combination of $R_{23}$ and $R_4$: $R_{234} = \frac{R_{23} \cdot R_4}{R_{23} + R_4} = \frac{(R_2 + R_3) \cdot R_4}{R_2 + R_3 + R_4}$

3. Total Resistance $R_{\text{ges}}$ (series combination of $R_1$ and $R_{234}$): $R_{\text{ges}} = R_1 + R_{234}$ Substituting $R_{234}$, we get: $R_{\text{ges}} = R_1 + \frac{(R_2 + R_3) \cdot R_4}{R_2 + R_3 + R_4}$

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(b) Wie groß sind der Strom $I$ und die Spannung $U_4$?

1. Total Current $I$ using Ohm's Law: $I = \frac{U_0}{R_{\text{ges}}}$

2. Voltage $U_4$ Across $R_4$: Since $R_{234}$ (which includes $R_4$) is in series with $R_1$, the voltage drop across $R_{234}$ is: $U_{234} = I \cdot R_{234}$ Thus, the voltage across $R_4$ specifically is the same as $U_{234}$, because $R_4$ is part of the parallel combination: $U_4 = I \cdot R_4$

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(c) Berechnen Sie die Spannung $U_5$ in Abhängigkeit vom Gesamtwiderstand $R_{\text{ges}}$.

The voltage $U_5$ across the section with $R_2$, $R_3$, and $R_4$ can be expressed in terms of $R_{\text{ges}}$ as follows:

1. The voltage $U_5$ is simply the voltage drop across $R_{234}$: $U_5 = U_0 - (I \cdot R_1)$
2. Substitute $I = \frac{U_0}{R_{\text{ges}}}$ into the equation: [ U_5 = U_0 - \left(\frac{U_0}{R_{\text{ges}}} \cdot R_1\right) = U_0 \left(1 - \frac{R_1