1. **Problem statement:** Determine the equivalent resistance $R_{AB}$ between terminals A and B for an infinite resistive ladder with repeating units consisting of a 20Ω resistor on the top row, a 10Ω vertical resistor, and a 30Ω resistor on the bottom row. 2. **Assume the equivalent resistance of the infinite ladder is $R$.** Since the network is infinite and repeating, adding one more unit to the ladder does not change the overall resistance $R$. 3. **Analyze one unit appended to the ladder:** - The top resistor is 20Ω. - The vertical resistor is 10Ω. - The bottom resistor is 30Ω. The portion to the right of the first unit has resistance $R$. 4. **Write the circuit relation:** Looking from terminals A-B: - The 30Ω resistor is in series with $R$ (the rest of the ladder), giving total $30 + R$ Ω. - The 10Ω resistor is in parallel with this series branch. 5. **Calculate the parallel combination of 10Ω and $(30 + R)$:** $$ R_p = \frac{10 \times (30+R)}{10 + 30 + R} = \frac{10(30+R)}{40 + R} $$ 6. **Add the 20Ω resistor in series on top:** The total resistance $R$ is then: $$ R = 20 + R_p = 20 + \frac{10(30+R)}{40 + R} $$ 7. **Solve for $R$:** Multiply both sides by $40 + R$: $$ R(40+R) = 20(40+R) + 10(30 + R) $$ Expand: $$ 40 R + R^2 = 800 + 20 R + 300 + 10 R $$ Simplify the right side: $$ 40 R + R^2 = 1100 + 30 R $$ Bring all terms to one side: $$ R^2 + 40 R - 30 R - 1100 = 0 $$ $$ R^2 + 10 R - 1100 = 0 $$ 8. **Use quadratic formula:** $$ R = \frac{-10 \pm \sqrt{10^2 - 4 \times 1 \times (-1100)}}{2} = \frac{-10 \pm \sqrt{100 + 4400}}{2} = \frac{-10 \pm \sqrt{4500}}{2} $$ Since resistance must be positive, select the positive root: $$ R = \frac{-10 + 30\sqrt{5}}{2} = -5 + 15\sqrt{5} $$ 9. **Final approximate value:** $$ R \approx -5 + 15 \times 2.236 = -5 + 33.54 = 28.54\ \Omega $$ **Answer:** The equivalent resistance between terminals A and B is approximately $28.54\ \Omega$.