MHTCET 18 Apr 2026 Shift 2 Paper

Concept: For a series $R$-$C$ circuit, impedance depends on capacitive reactance, which varies inversely with frequency.Explanation:At angular frequency $\omega$, the capacitive reactance is:$$X_C = \frac{1}{\omega C}$$The impedance of the series combination is:$$Z = \sqrt{R^2 + X_C^2}$$So the original current is:$$I = \frac{V}{\sqrt{R^2 + X_C^2}}$$When the frequency is changed to $\frac{\omega}{3}$, the new capacitive reactance becomes:$$X_C' = \frac{1}{\left(\frac{\omega}{3}\right)C} = \frac{3}{\omega C} = 3X_C$$The new impedance is:$$Z' = \sqrt{R^2 + (3X_C)^2} = \sqrt{R^2 + 9X_C^2}$$Given that the current becomes half, $I' = \frac{I}{2}$. Since the voltage remains the same and current is inversely proportional to impedance:$$Z' = 2Z$$Substituting the impedance expressions:$$\sqrt{R^2 + 9X_C^2} = 2\sqrt{R^2 + X_C^2}$$Squaring both sides:$$R^2 + 9X_C^2 = 4(R^2 + X_C^2)$$$$R^2 + 9X_C^2 = 4R^2 + 4X_C^2$$$$5X_C^2 = 3R^2$$$$\frac{X_C^2}{R^2} = \frac{3}{5}$$Taking the square root:$$\frac{X_C}{R} = \sqrt{\frac{3}{5}} = \sqrt{0.6}$$Answer: The ratio of capacitive reactance to resistance is $\sqrt{0.6}$, which corresponds to option D.