1. **Stating the problem:** We are given a network of resistors with values $R_1=8\Omega$, $R_2=18\Omega$, $R_3=9\Omega$, $R_4=20\Omega$, $R_5=5\Omega$, $R_6=1\Omega$, and $R_7=2\Omega$ connected between points A, B, C, D, E, F, G, H, K. The goal is to find the equivalent resistance $R_{center}$ at the center of the network or between specific points as implied. 2. **Analyzing the circuit:** The resistors form a polygonal network with connections: - $R_1$ between A and B - $R_2$ between K and B - $R_3$ between B and G - $R_4$ between C and D - $R_5$ between B and E - $R_6$ between E and F - $R_7$ between G and F 3. **Simplify the network step-by-step:** - First, note that $R_5$, $R_6$, and $R_7$ form a path from B to F via E and G. - Combine $R_6$ and $R_7$ in series: $$R_{67} = R_6 + R_7 = 1 + 2 = 3\Omega$$ - Now, $R_5$ is in series with $R_{67}$ through points B-E-F-G, but since $R_5$ is between B and E, and $R_{67}$ is between E and F-G, the total resistance from B to F through E and G is: $$R_{5,67} = R_5 + R_{67} = 5 + 3 = 8\Omega$$ 4. **Consider $R_3$ between B and G:** - $R_3 = 9\Omega$ is between B and G. - Since G connects to F through $R_7$, and F connects to E through $R_6$, and E connects to B through $R_5$, the path B-E-F-G forms a loop with $R_3$. 5. **Calculate the parallel combination of $R_3$ and $R_{5,67}$ between B and G:** $$\frac{1}{R_{BG}} = \frac{1}{R_3} + \frac{1}{R_{5,67}} = \frac{1}{9} + \frac{1}{8} = \frac{8 + 9}{72} = \frac{17}{72}$$ $$R_{BG} = \frac{72}{17} \approx 4.24\Omega$$ 6. **Now consider $R_1$ between A and B and $R_2$ between K and B:** - $R_1 = 8\Omega$ (A-B) - $R_2 = 18\Omega$ (K-B) 7. **Assuming points A and K are connected externally or to a common node, the resistors $R_1$, $R_2$, and $R_{BG}$ meet at B. If we want the equivalent resistance looking into B from A and K, we combine $R_1$ and $R_2$ in parallel with $R_{BG}$:** First, combine $R_1$ and $R_2$ in parallel: $$\frac{1}{R_{1,2}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{8} + \frac{1}{18} = \frac{9 + 4}{72} = \frac{13}{72}$$ $$R_{1,2} = \frac{72}{13} \approx 5.54\Omega$$ 8. **Finally, combine $R_{1,2}$ in series with $R_{BG}$ to find the total equivalent resistance at the center:** $$R_{center} = R_{1,2} + R_{BG} = 5.54 + 4.24 = 9.78\Omega$$ **Answer:** The equivalent resistance at the center of the network is approximately $9.78\Omega$.