UPSC NDA II 2018 General Ability Solved Paper

Concept:The equivalent resistance is found by simplifying series and parallel combinations step by step from the given circuit.Explanation:First, identify the two resistors connected in parallel between the left junction and the top node.Their equivalent, $R_{12}$, is given by $\frac{1}{R_{12}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$, so $R_{12} = \frac{R}{2}$.Next, $R_{12}$ is in series with the third resistor $R$ (the one on the right side of the top loop).Their combined resistance, $R_{123} = R_{12} + R = \frac{R}{2} + R = \frac{3R}{2}$.Now $R_{123}$ is in parallel with the fourth resistor $R$ (the bottom one between A and B).The net equivalent resistance $R_{\text{net}}$ satisfies $\frac{1}{R_{\text{net}}} = \frac{1}{R_{123}} + \frac{1}{R} = \frac{2}{3R} + \frac{1}{R} = \frac{5}{3R}$.Thus $R_{\text{net}} = \frac{3}{5}R$.Answer:$\boxed{\frac{3}{5}R}$