UPSC NDA I 2024 General Ability Paper

Concept:For infinite resistance networks, the equivalent resistance can be found by assuming the network beyond the first stage has the same equivalent resistance $R_{eq}$.Explanation:Let $R_{eq}$ be the equivalent resistance between points A and B in the infinite ladder.Each resistor has resistance $R = 4\,\Omega$.Consider the first part of the network: a resistor from A to node X, and then from X to B there are two parallel paths – one direct resistor of $4\,\Omega$ and the remaining infinite network which also has resistance $R_{eq}$.Thus the resistance from X to B is the parallel combination: $\frac{4 \times R_{eq}}{4 + R_{eq}}$.Adding the series resistor from A to X gives total resistance: $4 + \frac{4 R_{eq}}{4 + R_{eq}}$.Since the network is infinite, this total must equal $R_{eq}$ itself.Set up the equation: $R_{eq} = 4 + \frac{4 R_{eq}}{4 + R_{eq}}$.Multiply both sides by $(4+R_{eq})$: $R_{eq}(4+R_{eq}) = 4(4+R_{eq}) + 4R_{eq}$.Simplify: $R_{eq}^2 + 4R_{eq} = 16 + 4R_{eq} + 4R_{eq}$ → $R_{eq}^2 + 4R_{eq} = 16 + 8R_{eq}$.Bring all terms to one side: $R_{eq}^2 - 4R_{eq} - 16 = 0$.Solve using quadratic formula: $R_{eq} = \frac{4 \pm \sqrt{16 + 64}}{2} = \frac{4 \pm \sqrt{80}}{2} = \frac{4 \pm 4\sqrt{5}}{2} = 2 \pm 2\sqrt{5}$.Resistance cannot be negative, so take the positive value: $R_{eq} = 2 + 2\sqrt{5}\,\Omega$.Answer: $2+2\sqrt{5}\,\Omega$ (Option B)