https://r2.erweima.ai/mathbot/image/1726745375294it1ou1ge.jpg

The problem in the image consists of a basic electrical circuit with specific values for voltage (\( U_L \)), and resistances (\( R_1 \) and \( R_a \)), along with three questions concerning the working point, resistor adjustments, and current changes due to voltage variation. Let's break it down step by step:

### Given:
- \( U_L = 10 \, \text{V} \) (source voltage)
- \( R_1 = 1 \, \Omega \) (resistor 1)
- \( R_a = 4 \, \Omega \) (resistor 2)

---

### a) **Bestimmen Sie grafisch den Arbeitspunkt.** (Determine the working point graphically.)

The "working point" in an electrical circuit typically refers to the operating current \( I \) for a given voltage. The circuit appears to be a simple series circuit. To find the current \( I \), we use **Ohm's Law**:
\[
I = \frac{U_L}{R_{\text{total}}}
\]
where \( R_{\text{total}} = R_1 + R_a \).

Calculating the total resistance:
\[
R_{\text{total}} = 1 \, \Omega + 4 \, \Omega = 5 \, \Omega
\]

Now, calculate the current \( I \):
\[
I = \frac{10 \, \text{V}}{5 \, \Omega} = 2 \, \text{A}
\]
So, the working point is a current of 2 A.

---

### b) **Welchen Wert muss \( R_a \) annehmen, damit ein Strom von \( I = 6 \, \text{A} \) fließt?**  
(What value must \( R_a \) take for a current of \( I = 6 \, \text{A} \)?)

Using the same Ohm's Law:
\[
I = \frac{U_L}{R_{\text{total}}}
\]
Rearranging to find \( R_{\text{total}} \) when \( I = 6 \, \text{A} \):
\[
R_{\text{total}} = \frac{U_L}{I} = \frac{10 \, \text{V}}{6 \, \text{A}} = \frac{5}{3} \, \Omega \approx 1.67 \, \Omega
\]

Since \( R_{\text{total}} = R_1 + R_a \), we can solve for \( R_a \):
\[
R_a = R_{\text{total}} - R_1 = 1.67 \, \Omega - 1 \, \Omega = 0.67 \, \Omega
\]

So, \( R_a \) must be approximately \( 0.67 \, \Omega \) to have a current of 6 A.

---

### c) **Die Quellenspannung wird auf \( U_L = 12 \, \text{V} \) angehoben. Auf welchen Wert steigt der Strom \( I \)?**  
(The source voltage is raised to \( U_L = 12 \, \text{V} \). What is the new value of the current \( I \)?)

With the new voltage \( U_L = 12 \, \text{V} \), we can again use Ohm's Law with the original total resistance \( R_{\text{total}} = 5 \, \Omega \):
\[
I = \frac{12 \, \text{V}}{5 \, \Omega} = 2.4 \, \text{A}
\]

Thus, the current increases to 2.4 A.

---

Let me know if you need more details or have further questions!

### Relative questions:
1. What is the significance of the working point in an electrical circuit?
2. How does changing \( R_a \) affect the overall resistance and current in the circuit?
3. Why do we use Ohm's Law to calculate the current in a series circuit?
4. How would increasing the source voltage further (e.g., to 15 V) affect the circuit's current?
5. How can the graphical method for determining the working point be illustrated?

### Tip:
Always check whether a circuit is in series or parallel when applying Ohm's Law, as this affects how you calculate total resistance.

Gegeben ist ein Grundstromkreis mit UL = 10 V, R1 = 1 Ohm und Ra = 4 Ohm.

 a) Bestimmen Sie grafisch den Arbeitspunkt.
 b) Welchen Wert muss Ra annehmen, damit ein Strom von I = 6 A fließt?
 c) Die Quellenspannung wird auf UL = 12 V angehoben. Auf welchen Wert steigt der Strom I?

This problem involves analyzing a basic series circuit using Ohm's Law. It covers the calculation of the working point, determining the value of resistance Ra for a specific current, and how a change in source voltage affects the current. Key concepts include Ohm's Law and series circuit resistance.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation