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This is a typical problem involving a combination of series and parallel resistors. Let's break it down step by step.

### Circuit Information:
- \( R_1 = 40 \, \Omega \)
- \( R_2 = 70 \, \Omega \)
- \( R_3 = 150 \, \Omega \)
- \( U = 36 \, \text{V} \)

The circuit has:
- A resistor \( R_1 \) in series with a parallel combination of \( R_2 \) and \( R_3 \).
- You are asked to calculate the following:
  - The voltage \( U_1 \) across resistor \( R_1 \),
  - The current \( I_1 \) through the series part,
  - The branch currents \( I_2 \) and \( I_3 \) through \( R_2 \) and \( R_3 \), respectively.

### Step-by-Step Solution:

1. **Equivalent Resistance of \( R_2 \) and \( R_3 \):**
   Since \( R_2 \) and \( R_3 \) are in parallel, we can calculate their equivalent resistance \( R_{\text{eq}} \) as:
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{R_2} + \frac{1}{R_3}
   \]
   Substituting the values:
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{70} + \frac{1}{150}
   \]
   Solving for \( R_{\text{eq}} \).

2. **Total Resistance \( R_{\text{total}} \):**
   Once we have \( R_{\text{eq}} \), the total resistance of the circuit is the sum of \( R_1 \) and \( R_{\text{eq}} \):
   \[
   R_{\text{total}} = R_1 + R_{\text{eq}}
   \]

3. **Current \( I_1 \) Through the Series Resistor \( R_1 \):**
   The total current \( I_1 \) can be calculated using Ohm's Law:
   \[
   I_1 = \frac{U}{R_{\text{total}}}
   \]

4. **Voltage Drop Across \( R_1 \) ( \( U_1 \) ):**
   The voltage drop across \( R_1 \) is given by:
   \[
   U_1 = I_1 \times R_1
   \]

5. **Currents \( I_2 \) and \( I_3 \):**
   The voltage across the parallel branch is \( U - U_1 \), so the currents \( I_2 \) and \( I_3 \) can be found using Ohm's Law:
   \[
   I_2 = \frac{U - U_1}{R_2}
   \]
   \[
   I_3 = \frac{U - U_1}{R_3}
   \]

I'll proceed to calculate these values.Here are the calculated results:

1. **Equivalent Resistance of \( R_2 \) and \( R_3 \)**:  
   \( R_{\text{eq}} = 47.73 \, \Omega \)

2. **Total Resistance of the Circuit**:  
   \( R_{\text{total}} = 87.73 \, \Omega \)

3. **Current Through Resistor \( R_1 \) ( \( I_1 \) )**:  
   \( I_1 = 0.41 \, \text{A} \)

4. **Voltage Drop Across \( R_1 \) ( \( U_1 \) )**:  
   \( U_1 = 16.41 \, \text{V} \)

5. **Current Through Resistor \( R_2 \) ( \( I_2 \) )**:  
   \( I_2 = 0.28 \, \text{A} \)

6. **Current Through Resistor \( R_3 \) ( \( I_3 \) )**:  
   \( I_3 = 0.13 \, \text{A} \)

Would you like further details or have any questions?

Here are five related questions for further practice:
1. What would happen to the total current if \( R_1 \) was increased?
2. How would the circuit's behavior change if \( R_2 \) and \( R_3 \) were swapped?
3. How is power distributed across each resistor in this circuit?
4. What happens to \( U_1 \) if \( R_3 \) is removed from the circuit?
5. How can you calculate the power dissipated in \( R_1 \)?

**Tip**: Always remember that in a series circuit, the current is the same through all components, but the voltage is divided according to the resistances.

Gegeben ist die folgende Schaltung mit R_1 = 40 Ω, R_2 = 70 Ω, R_3 = 150 Ω und U = 36 V. Berechnen Sie die Spannung U_1, den Strom I_1 sowie die Teilströme I_2 und I_3.

In this example, we calculate the voltage U_1 across a 40 Ω resistor, the current I_1 through a series resistor, and the branch currents I_2 and I_3 in a parallel circuit with resistors R_2 = 70 Ω and R_3 = 150 Ω, using a total voltage of 36 V. Detailed steps involve calculating equivalent resistances, current using Ohm's law, and voltage drops across each resistor.

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