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To determine the voltage \( V_s \) and the equivalent resistance \( R_{eq} \) of this circuit, we need to analyze it step by step.

### Step 1: Identify resistors in parallel and series

- The top-left branch consists of two resistors, 1Ω and 1Ω, in series.
  \[
  R_1 = 1\Omega + 1\Omega = 2\Omega
  \]
- The middle branch has resistors 1Ω and 1Ω in series.
  \[
  R_2 = 1\Omega + 1\Omega = 2\Omega
  \]
- The bottom branch has two resistors, 3Ω and 0.5Ω in series.
  \[
  R_3 = 3\Omega + 0.5\Omega = 3.5\Omega
  \]
- The rightmost branch has a combination of a 1Ω and a 2Ω in series.
  \[
  R_4 = 2\Omega + 1\Omega = 3\Omega
  \]

### Step 2: Simplify the circuit
Now, notice that \( R_1 \), \( R_2 \), \( R_3 \), and \( R_4 \) are all in parallel. Therefore, the equivalent resistance \( R_{eq} \) of the parallel combination is given by:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}
\]
Substituting the values:
\[
\frac{1}{R_{eq}} = \frac{1}{2\Omega} + \frac{1}{2\Omega} + \frac{1}{3.5\Omega} + \frac{1}{3\Omega}
\]
Let's calculate this value.

### Step 3: Calculate the voltage source \( V_s \)
Once we have the equivalent resistance, we can find the voltage \( V_s \) using Ohm's law, \( V = IR \). Given that the current \( I = 2 \, A \), and knowing the equivalent resistance, we can find \( V_s \) using:
\[
V_s = I \times R_{eq}
\]

I'll compute these values next.The equivalent resistance \( R_{eq} \) of the circuit is approximately \( 0.62 \, \Omega \).

The voltage \( V_s \) of the source is approximately \( 1.24 \, V \).

Would you like to ask any further questions or need more details? Here are some related questions to explore:

1. How is the equivalent resistance affected by changes in one of the branches?
2. How would the current be distributed among the parallel branches?
3. What would happen if one of the resistors was removed or changed?
4. Can you verify the power dissipation in each resistor?
5. How does Kirchhoff's current law apply to this circuit?

**Tip:** Always simplify complex circuits by identifying series and parallel combinations to calculate the equivalent resistance easily.

Find the voltage source (Vs) and the equivalent resistance (Req) in the given circuit.

Learn how to calculate the voltage source (Vs) and equivalent resistance (Req) in a complex circuit with parallel and series resistors. This problem involves key concepts in circuit analysis, including Ohm's Law and resistor combination rules. Suitable for Grades 11-12 students.

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