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AC Circuits

Question

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 now the frequency of source is changed to $\frac{ω}{3}$ (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 $ω$ is

Moderate

A

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

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

$\sqrt{\frac{2}{3}}$
D

$\sqrt{\frac{3}{2}}$

Solution

According to given problem,
$I = \frac{V}{Z} = \frac{V}{{[R^{2} + {(\frac{1}{C ω})}^{2}]}^{\frac{1}{2}}}$ ……..(1)

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

substituting the value of I from eq.(1) in (2),

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

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

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

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