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To find the equivalent resistance of the circuit, we need to simplify the series and parallel combinations of resistors step-by-step.

### Step-by-Step Solution:

1. **Identify Series and Parallel Combinations:**

   - **Series:** Resistors that are connected end-to-end, sharing only a single connection point with each other.
   - **Parallel:** Resistors that are connected across the same two points.

2. **Simplify Parallel Combinations:**
   - First, look at the resistors that are in parallel:
     - The \(10 \, \Omega\) resistor on the left and the \(2 \, \Omega\) resistor are in parallel.
     - The \(10 \, \Omega\) resistor on the right and the \(4 \, \Omega\) resistor are also in parallel.

   The formula for combining two resistors in parallel, \( R_1 \) and \( R_2 \), is:
   \[
   R_{\text{parallel}} = \frac{R_1 \times R_2}{R_1 + R_2}
   \]

   **Calculations:**

   - For the left parallel combination:
     \[
     R_{\text{left parallel}} = \frac{10 \times 2}{10 + 2} = \frac{20}{12} = \frac{5}{3} \, \Omega
     \]

   - For the right parallel combination:
     \[
     R_{\text{right parallel}} = \frac{10 \times 4}{10 + 4} = \frac{40}{14} = \frac{20}{7} \, \Omega
     \]

3. **Simplify Series Combinations:**
   - Now add these parallel results to the series resistors connected to them:
     - The \(5 \, \Omega\) and \(R_{\text{left parallel}}\) are in series with \(3 \, \Omega\).
     - The \(R_{\text{right parallel}}\) is in series with \(1 \, \Omega\) and \(5 \, \Omega\).

   **Calculations:**

   - Total resistance of the left branch:
     \[
     R_{\text{left series}} = 5 + \frac{5}{3} + 3 = 8 + \frac{5}{3} = \frac{24}{3} + \frac{5}{3} = \frac{29}{3} \, \Omega
     \]

   - Total resistance of the right branch:
     \[
     R_{\text{right series}} = \frac{20}{7} + 1 + 5 = \frac{20}{7} + \frac{7}{7} + \frac{35}{7} = \frac{62}{7} \, \Omega
     \]

4. **Combine the Two Branches in Parallel:**

   Since these two equivalent resistances are now in parallel, the total equivalent resistance \( R_{\text{eq}} \) of the circuit is:
   \[
   \frac{1}{R_{\text{eq}}} = \frac{1}{\frac{29}{3}} + \frac{1}{\frac{62}{7}}
   \]

   **Calculations:**

   - Reciprocal of \( R_{\text{eq}} \):
     \[
     \frac{1}{R_{\text{eq}}} = \frac{3}{29} + \frac{7}{62}
     \]

   - Finding a common denominator:
     \[
     \frac{1}{R_{\text{eq}}} = \frac{6}{58} + \frac{7}{62} = \frac{31}{62}
     \]

   - Therefore, \( R_{\text{eq}} \) is:
     \[
     R_{\text{eq}} = \frac{62}{31} = 2 \, \Omega
     \]

### Conclusion:
The equivalent resistance of the circuit is **2 ohms**.

Would you like further details or have any questions?

Here are five related questions to expand on this topic:
1. How do you calculate the current flowing through each branch in this circuit?
2. What would be the effect on the total resistance if another resistor was added in series?
3. Can you explain how the voltage is distributed across each resistor in this circuit?
4. What happens to the total resistance if one of the resistors in parallel is removed?
5. How can Kirchhoff's Voltage and Current Laws be applied to this circuit for more complex analysis?

**Tip:** When working with circuits, always check if resistors are in series or parallel, as this greatly affects how you calculate total resistance.

Learn how to find the equivalent resistance of a circuit using series and parallel combinations of resistors. This detailed step-by-step solution covers identifying series and parallel resistors, applying the parallel resistance formula, and calculating the reciprocal to find the total resistance. Suitable for Grades 10-12.

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