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find the req and vs using series, parallel and delta wye

To find the equivalent resistance (\(R_{\text{eq}}\)) and the source voltage (\(V_s\)) in the circuit you provided, we'll proceed step by step by analyzing the series and parallel combinations of resistors, and potentially using a delta-wye transformation if necessary.

### 1. Identifying Key Sections of the Circuit:
- The current source provides a current of \( I = 2 \, \text{A} \).
- Resistors are present in a combination of series and parallel arrangements.
  
We will now simplify this circuit step-by-step:

### 2. Parallel Combinations:
Look at the right-most part of the circuit:

- The 1 Ω and 2 Ω resistors are in parallel.
- The middle branch consists of two 1 Ω resistors in series (equivalent to 2 Ω), which are in parallel with another 1 Ω resistor.
- The bottom branch has two resistors in series, a 3 Ω and a 0.5 Ω resistor (equivalent to 3.5 Ω).

Let's calculate the equivalent resistance for the parallel branches.

#### a. Top Parallel Branch:
For the 1 Ω and 2 Ω resistors in parallel:
\[
\frac{1}{R_{\text{top}}} = \frac{1}{1 \, \Omega} + \frac{1}{2 \, \Omega} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}
\]
\[
R_{\text{top}} = \frac{2}{3} \, \Omega
\]

#### b. Middle Parallel Branch:
For the two 1 Ω resistors in series, the total resistance is \(1 \, \Omega + 1 \, \Omega = 2 \, \Omega\).

This 2 Ω resistor is in parallel with another 1 Ω resistor:
\[
\frac{1}{R_{\text{middle}}} = \frac{1}{2 \, \Omega} + \frac{1}{1 \, \Omega} = \frac{1}{2} + 1 = \frac{3}{2}
\]
\[
R_{\text{middle}} = \frac{2}{3} \, \Omega
\]

#### c. Bottom Branch:
The bottom branch consists of a 3 Ω and 0.5 Ω resistor in series:
\[
R_{\text{bottom}} = 3 \, \Omega + 0.5 \, \Omega = 3.5 \, \Omega
\]

### 3. Combine Parallel Branches:
Now, these three equivalent resistances (\(R_{\text{top}} = \frac{2}{3} \, \Omega\), \(R_{\text{middle}} = \frac{2}{3} \, \Omega\), and \(R_{\text{bottom}} = 3.5 \, \Omega\)) are in parallel:
\[
\frac{1}{R_{\text{combined}}} = \frac{1}{\frac{2}{3} \, \Omega} + \frac{1}{\frac{2}{3} \, \Omega} + \frac{1}{3.5 \, \Omega}
\]
\[
\frac{1}{R_{\text{combined}}} = \frac{3}{2} + \frac{3}{2} + \frac{1}{3.5}
\]
\[
\frac{1}{R_{\text{combined}}} = 3 + \frac{1}{3.5} \approx 3 + 0.2857 \approx 3.2857
\]
\[
R_{\text{combined}} \approx \frac{1}{3.2857} \approx 0.3044 \, \Omega
\]

### 4. Combine with Series Resistors:
Now, add the resistors in series with this combination:

- There's a 1 Ω resistor in series at the top.
- There's a 0.5 Ω resistor at the bottom.
  
The total resistance is:
\[
R_{\text{eq}} = 1 \, \Omega + R_{\text{combined}} + 0.5 \, \Omega
\]
\[
R_{\text{eq}} \approx 1 \, \Omega + 0.3044 \, \Omega + 0.5 \, \Omega = 1.8044 \, \Omega
\]

### 5. Source Voltage (\(V_s\)):
Using Ohm's law:
\[
V_s = I \times R_{\text{eq}} = 2 \, \text{A} \times 1.8044 \, \Omega = 3.6088 \, \text{V}
\]

### Final Results:
- The equivalent resistance \(R_{\text{eq}} \approx 1.804 \, \Omega\).
- The source voltage \(V_s \approx 3.61 \, \text{V}\).

Would you like any further clarifications or additional details on any specific step?

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Here are 5 related questions to deepen your understanding:
1. What happens to the equivalent resistance when more resistors are added in parallel?
2. How would the source voltage change if the current increased to 3 A?
3. What is the difference between series and parallel connections in terms of current and voltage?
4. How do delta-wye transformations simplify complex circuits?
5. Can you calculate the total power dissipated in this circuit?

**Tip:** In parallel circuits, the total resistance is always lower than the smallest resistor in the network.

Find the Req and Vs using series, parallel, and delta-wye transformations.

Find the equivalent resistance (Req) and source voltage (Vs) in a complex circuit by simplifying series and parallel resistors and using delta-wye transformations. This guide explains each step clearly and applies key formulas, including Ohm's Law and the delta-wye transformation theorem. Suitable for advanced high school or college students.

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