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An ac 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 $\omega/3$ (but maintaining the same voltage), the current in the circuit is found to be halved. Then the ratio of reactance to resistance at the original frequency $\omega$
Option: 1
$\sqrt {3/5 }$

Option: 2
$\sqrt 5/3$

Option: 3
$\sqrt 2/3$

Option: 4
$\sqrt 3/2$

Answers (1)

best_answer

As we learned

Peak current -

${i}'_{0}= \frac{V_{0}}{z}= \frac{V_{0}}{\sqrt{R^{2}+X_{c}^{2}}}= \frac{V_{0}}{\sqrt{R^{2}+\frac{1}{4\pi ^{2}\nu ^{2}c^{2}}}}$

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According to given problem,

$I = \frac{V}{Z} = \frac{V }{[ R ^2 + (1/C\omega)^2]^{1/2}}$ ... (1)

and $I/2 = \frac{V }{[ R ^2 + (3/C\omega)^2]^{1/2}}$ ...(2)

Substituting the value of I from Equation (1) in (2),

$4 ( R^2 + \frac{1}{C^2 \omega ^2 }) = R^2 + \frac{9}{C^2 \omega ^2 }$ .

i.e., $\frac{1}{C^2\omega ^2 } = 3/5 R^2$

So that $\frac{X}{R } = \frac{(1/C\omega)}{R} = \frac{(3/5 R^2)^{1/2} }{R } = \sqrt {\frac{3}{5}}$

Posted by

Gaurav

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