1. **Stating the problem:** We need to find the equivalent resistance between terminals a and b in the given resistor network. 2. **Analyzing the circuit:** Break down the network into simpler series and parallel resistor combinations. 3. **Identify the 30Ω parallel resistors in the center:** Two 30Ω resistors in parallel have equivalent resistance $$R_p = \frac{30 \times 30}{30 + 30} = \frac{900}{60} = 15\,\Omega$$. 4. **Add the 30Ω resistor horizontally between the middle two 30Ω resistors:** This resistor is in series with the 15Ω calculated in step 3, so their total is $$R_s = 15 + 30 = 45\,\Omega$$. 5. **Include the two 30Ω vertical right branch resistors, repeated twice:** Each vertical 30Ω resistor is added in series, so combined resistance is $$30 + 30 = 60\,\Omega$$. 6. **Combine this 60Ω with the 45Ω from step 4 in parallel:** $$R_{center} = \frac{45 \times 60}{45 + 60} = \frac{2700}{105} \approx 25.71\,\Omega$$ 7. **Consider the 60Ω resistor on the top branch connected from node after 5Ω resistor:** This resistor is in series with the 25.71Ω resistance calculated above: $$R_{top} = 60 + 25.71 = 85.71\,\Omega$$ 8. **Calculate the series resistance after 5Ω resistor:** It is 5Ω in series with the 85.71Ω: $$R_{left} = 5 + 85.71 = 90.71\,\Omega$$ 9. **Analyze the lower branch between nodes after 15Ω and 20Ω resistors:** 10Ω horizontal resistor in series with 30Ω vertical resistor repeated twice means: Two 30Ω resistors in series: $$30 + 30 = 60\,\Omega$$. In series with 10Ω resistor: $$10 + 60 = 70\,\Omega$$ 10. **Combine the 20Ω vertical resistor in parallel with the 70Ω resistance:** $$R_{vertical} = \frac{20 \times 70}{20 + 70} = \frac{1400}{90} \approx 15.56\,\Omega$$ 11. **Add 15Ω resistor in series with previous result:** $$R_{right} = 15 + 15.56 = 30.56\,\Omega$$ 12. **Now combine left and right branches in parallel between terminals a and b:** $$R_{eq} = \frac{90.71 \times 30.56}{90.71 + 30.56} = \frac{2772.43}{121.27} \approx 22.86\,\Omega$$ **Final answer:** $$\boxed{22.86\,\Omega}$$