An a.c. source of angular frequency ' ω ' is fed across a resistor ' R ' and a capacitor ' C ' in series. The current registered is I. If now the frequency of source is changed to (ω)/(3) (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency ' ω ' will be (a) √(2/5) (b) √(1/5) (c) √(4/5) (d) √(3/5)

Answer: (d). At (ω)/(3), XC → 3XC; current halves so √(R² + 9XC²) = 2√(R² + XC²). 5XC² = 3R² XC/R = √(3/5) So the answer is (d) √(3/5).

An a.c. source of angular frequency ' ω ' is fed across a resistor '… | MHT CET 2021 (22 Sep, Shift 1)

Q41MHT CET 2021 · 22 SEP · SHIFT 1Medium

RL and RC Series Circuits

An a.c. source of angular frequency ' $\omega$ ' is fed across a resistor ' $R$ ' and a capacitor ' $C$ ' in series. The current registered is I. If now the frequency of source is changed to $\frac{\omega}{3}$ (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency ' $\omega$ ' will be

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Answer: (d)

At $\frac{\omega}{3}$ , $X_C \to 3X_C$ ; current halves so $\sqrt{R^2 + 9X_C^2} = 2\sqrt{R^2 + X_C^2}$ .
$5X_C^2 = 3R^2$
$\frac{X_C}{R} = \sqrt{\frac{3}{5}}$
So the answer is (d) $\sqrt{\frac{3}{5}}$ .

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