1. **Problem Statement:** Find the resultant resistance between points A and B in the given resistor network. 2. **Understanding the Network:** The network has resistors: 5 Ω (vertical from A), 10 Ω (diagonal from A), 5 Ω (horizontal from lower junction), 2 Ω (vertical from central junction), 4 Ω (horizontal from central junction), and 3 Ω (diagonal to B). 3. **Step 1: Identify series and parallel connections.** - The 2 Ω, 4 Ω, and 3 Ω resistors form a triangle from the central junction to B. - The 5 Ω (vertical) and 5 Ω (horizontal) resistors connect from A to the central junction. - The 10 Ω resistor connects directly from A to the right junction. 4. **Step 2: Simplify the triangle of 2 Ω, 4 Ω, and 3 Ω resistors.** - These three resistors form a delta (triangle) between the central junction and B. - We can convert this delta to a star (Y) to simplify. 5. **Delta to Star conversion formulas:** $$R_1 = \frac{R_a R_b}{R_a + R_b + R_c},\quad R_2 = \frac{R_b R_c}{R_a + R_b + R_c},\quad R_3 = \frac{R_a R_c}{R_a + R_b + R_c}$$ where $R_a=2$, $R_b=4$, $R_c=3$. 6. **Calculate sum:** $$S = 2 + 4 + 3 = 9$$ 7. **Calculate star resistors:** $$R_1 = \frac{2 \times 4}{9} = \frac{8}{9}$$ $$R_2 = \frac{4 \times 3}{9} = \frac{12}{9} = \frac{4}{3}$$ $$R_3 = \frac{2 \times 3}{9} = \frac{6}{9} = \frac{2}{3}$$ 8. **Step 3: Replace delta with star resistors:** - Now the network has star resistors $\frac{8}{9}$ Ω, $\frac{4}{3}$ Ω, and $\frac{2}{3}$ Ω connected to the central junction. 9. **Step 4: Combine resistors in series and parallel:** - The 5 Ω (horizontal) resistor is in series with $\frac{8}{9}$ Ω. - The 5 Ω (vertical) resistor is in series with $\frac{4}{3}$ Ω. - The $\frac{2}{3}$ Ω resistor connects to B through the 10 Ω resistor from A. 10. **Calculate series resistances:** $$R_{series1} = 5 + \frac{8}{9} = \frac{45}{9} + \frac{8}{9} = \frac{53}{9}$$ $$R_{series2} = 5 + \frac{4}{3} = \frac{15}{3} + \frac{4}{3} = \frac{19}{3}$$ 11. **Step 5: Combine $R_{series1}$, $R_{series2}$, $\frac{2}{3}$, and 10 Ω appropriately:** - $R_{series1}$ and $R_{series2}$ are connected to the central junction. - The $\frac{2}{3}$ Ω resistor and 10 Ω resistor form a path to B. 12. **Assuming $\frac{2}{3}$ Ω and 10 Ω are in series:** $$R_{series3} = \frac{2}{3} + 10 = \frac{2}{3} + \frac{30}{3} = \frac{32}{3}$$ 13. **Now, $R_{series1}$, $R_{series2}$, and $R_{series3}$ are in parallel between A and B:** $$\frac{1}{R_{eq}} = \frac{1}{\frac{53}{9}} + \frac{1}{\frac{19}{3}} + \frac{1}{\frac{32}{3}} = \frac{9}{53} + \frac{3}{19} + \frac{3}{32}$$ 14. **Calculate common denominator and sum:** - Common denominator is $53 \times 19 \times 32 = 32224$ - Convert each fraction: $$\frac{9}{53} = \frac{9 \times 19 \times 32}{32224} = \frac{5472}{32224}$$ $$\frac{3}{19} = \frac{3 \times 53 \times 32}{32224} = \frac{5088}{32224}$$ $$\frac{3}{32} = \frac{3 \times 53 \times 19}{32224} = \frac{3027}{32224}$$ 15. **Sum:** $$\frac{5472 + 5088 + 3027}{32224} = \frac{13587}{32224}$$ 16. **Therefore:** $$R_{eq} = \frac{32224}{13587} \approx 2.37\ \Omega$$ **Final answer:** The resultant resistance between points A and B is approximately $2.37$ Ω.