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Alternating Current — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsAlternating Current1 Mark
MCQ
An ac source of angular frequency $\omega$ is fed across a resistor $r$ and a capacitor $C$ in series. The current registered is $l$. If now the frequency of source is changed to $\omega / 3$ (but maintaining the same voltage), the current in then circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency $\omega$
  • ✓
    $\sqrt{\frac{3}{5}}$
  • B
    $\sqrt{\frac{1}{5}}$
  • C
    $\sqrt{\frac{2}{5}}$
  • D
    $\sqrt{\frac{4}{5}}$
✓

Answer

Correct option: A.
$\sqrt{\frac{3}{5}}$
At angular frequency $\omega$, the current in $R C$ circuit is given by$i_{r m s}=\frac{V_{r m s}}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}$
Also $\frac{i_{r m s}}{2}=\frac{V_{r m s}}{\sqrt{R^2+\left(\frac{1}{\frac{\omega}{3} C}\right)^2}}=\frac{V_{r m s}}{\sqrt{R^2+\frac{9}{\omega^2 C^2}}}$
From equation (i) and (ii) we get $3 R^2=\frac{5}{\omega^2 C^2} \Rightarrow \frac{\frac{1}{\omega C}}{R}=\sqrt{\frac{3}{5}} \Rightarrow \frac{X_C}{R}=\sqrt{\frac{3}{5}}$

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