A series combination of resistor 'R' and capacitor 'C' is connected to an a.c. source of angular frequency ' ω '. Keeping the voltage same, if the frequency is changed to ((ω)/(3)) the current becomes half of the original current. Then the ratio of capacitive reactance and resistance is (a) √(6) (b) √(0.3) (c) √(3) (d) √(0.6)

Answer: (d). I ∝ (1)/(√(R² + XC²)); at (ω)/(3), XC becomes 3XC. Current halves: √(R² + 9XC²) = 2√(R² + XC²) R² + 9XC² = 4R² + 4XC² ⇒ 5XC² = 3R² XC/R = √(3/5) = √(0.6) So the answer is (d) √(0.6).

A series combination of resistor 'R' and capacitor 'C' is connected… | MHT CET 2026 (18 Apr, Shift 2)

Q8MHT CET 2026 · 18 APR · SHIFT 2Medium

RL and RC Series Circuits

A series combination of resistor 'R' and capacitor 'C' is connected to an a.c. source of angular frequency ' $\omega$ '. Keeping the voltage same, if the frequency is changed to $\left(\frac{\omega}{3}\right)$ the current becomes half of the original current. Then the ratio of capacitive reactance and resistance is

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Answer: (d)

$I \propto \frac{1}{\sqrt{R^2 + X_C^2}}$ ; at $\frac{\omega}{3}$ , $X_C$ becomes $3X_C$ .
Current halves: $\sqrt{R^2 + 9X_C^2} = 2\sqrt{R^2 + X_C^2}$
$R^2 + 9X_C^2 = 4R^2 + 4X_C^2 \Rightarrow 5X_C^2 = 3R^2$
$\frac{X_C}{R} = \sqrt{\frac{3}{5}} = \sqrt{0.6}$
So the answer is (d) $\sqrt{0.6}$ .

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RL and RC Series Circuits question from MHT CET 2026 (18 Apr, Shift 2)

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