1. **State the problem:** Find the equivalent resistance $R_{eq}$ of the given resistor network. 2. **Analyze the circuit:** - The top branch has resistors 2Ω, 3Ω, and 4Ω in series. - From the junction between 2Ω and 3Ω, a 6Ω resistor goes down. - From the junction between 3Ω and 4Ω, a 4Ω resistor goes down. - The 6Ω resistor connects at the bottom with a 1Ω resistor. - The 4Ω resistor connects at the bottom with a 3Ω resistor in series with a 5Ω resistor. - The 1Ω resistor and the 3Ω resistor at the bottom are in series. 3. **Step 1: Combine series resistors at the bottom right:** - The 3Ω and 5Ω resistors are in series: $$R_{3+5} = 3 + 5 = 8\ \Omega$$ 4. **Step 2: Combine the 4Ω resistor (from junction) with $R_{3+5}$ in parallel:** $$R_{4||8} = \frac{1}{\frac{1}{4} + \frac{1}{8}} = \frac{1}{\frac{2}{8} + \frac{1}{8}} = \frac{1}{\frac{3}{8}} = \frac{8}{3} \approx 2.67\ \Omega$$ 5. **Step 3: Combine the 1Ω and 3Ω resistors at the bottom left in series:** $$R_{1+3} = 1 + 3 = 4\ \Omega$$ 6. **Step 4: Combine the 6Ω resistor with $R_{1+3}$ in series:** $$R_{6+4} = 6 + 4 = 10\ \Omega$$ 7. **Step 5: Combine $R_{6+4}$ with the 2Ω resistor in parallel?** Actually, the 6Ω resistor is connected from the junction between 2Ω and 3Ω down to the 1Ω resistor, so the 6Ω and 1Ω+3Ω are in series (already combined as 10Ω). This 10Ω is connected in parallel with the 4Ω resistor from the junction between 3Ω and 4Ω combined with 3Ω+5Ω series (already combined as 2.67Ω). So the two parallel branches are 10Ω and 2.67Ω. 8. **Step 6: Combine the two parallel branches (10Ω and 2.67Ω):** $$R_{parallel} = \frac{1}{\frac{1}{10} + \frac{1}{\frac{8}{3}}} = \frac{1}{\frac{1}{10} + \frac{3}{8}} = \frac{1}{\frac{4}{40} + \frac{15}{40}} = \frac{1}{\frac{19}{40}} = \frac{40}{19} \approx 2.11\ \Omega$$ 9. **Step 7: Add the 2Ω and 3Ω resistors in series with $R_{parallel}$ and the 4Ω resistor at the end:** - The top branch is 2Ω + 3Ω + 4Ω = 9Ω, but since the 6Ω and 4Ω branches are connected at the junctions, the equivalent resistance is: $$R_{eq} = 2 + \left(\frac{1}{\frac{1}{3 + R_{parallel}} + \frac{1}{4}}\right)$$ But from the problem description, the 2Ω, 3Ω, and 4Ω are in series, with the 6Ω and 4Ω branches connected at the junctions between these resistors. We already accounted for the parallel branches, so the equivalent resistance is: $$R_{eq} = 2 + \left(\frac{1}{\frac{1}{3 + R_{parallel}} + \frac{1}{4}}\right) + 4$$ 10. **Step 8: Calculate $3 + R_{parallel}$:** $$3 + 2.11 = 5.11\ \Omega$$ 11. **Step 9: Calculate the parallel of $5.11$ and $4$:** $$R_{parallel2} = \frac{1}{\frac{1}{5.11} + \frac{1}{4}} = \frac{1}{0.1957 + 0.25} = \frac{1}{0.4457} \approx 2.24\ \Omega$$ 12. **Step 10: Add the 2Ω resistor at the start and the 4Ω resistor at the end:** $$R_{eq} = 2 + 2.24 + 4 = 8.24\ \Omega$$ **Final answer:** $$\boxed{R_{eq} \approx 8.24\ \Omega}$$