An ac source of angular frequency is fed across a resistor R and a capacitor C in series. The current registered is . If the frequency of source is changed to /3 (maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency

R=124Ω=3ΩI=εR2+XC2I2=εR2+9XC2∴XC=1ωCR2+9XC2R2+XC2=41⇒XCR=53

An ac source of angular frequency is fed across a resistor R and a capacitor C in series. The current registered is . If the frequency of source is changed to /3 (maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency

R=124Ω=3ΩI=εR2+XC2I2=εR2+9XC2∴XC=1ωCR2+9XC2R2+XC2=41⇒XCR=53

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An ac source of angular frequency $𝜔$ is fed across a resistor R and a capacitor C in series. The current registered is 𝐼. If the frequency of source is changed to $𝜔 / 3$ (maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency $𝜔$

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$\sqrt{\frac{3}{5}}$

$\sqrt{\frac{5}{3}}$

$\sqrt{\frac{1}{5}}$

$\sqrt{\frac{4}{5}}$

answer is A.

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Detailed Solution

$\begin{array}{l} R = \frac{12}{4} Ω = 3 Ω \\ I = \frac{ε}{\sqrt{R^{2} + X_{C}^{2}}} \\ \frac{I}{2} = \frac{ε}{\sqrt{R^{2} + 9 X_{C}^{2}}} (∴ X_{C} = \frac{1}{ω C}) \\ \frac{R^{2} + 9 X_{C}^{2}}{R^{2} + X_{C}^{2}} = \frac{4}{1} \Rightarrow \frac{X_{C}}{R} = \sqrt{\frac{5}{3}} \end{array}$