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  • An ac source of angular frequency  is fed across a resistor R and a capacitor C in series. The current regist

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\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

Option: 1

\mathrm{\sqrt{\frac{3}{5}}}


Option: 2

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


Option: 3

\mathrm{\frac{3}{5}}


Option: 4

\mathrm{\frac{5}{3}}


Answers (1)

best_answer

According to given Problem

\mathrm{\begin{aligned} & I=\frac{V}{Z}=V /\left[R^2+\left(1 / C \omega^2\right)\right]^{1 / 2} \\ & \text { and } \frac{1}{2}=\frac{V}{\left[R^2+(3 / \mathrm{C} \omega)^2\right]^2} \\ & \end{aligned}}-----------(1) & (2)

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

\mathrm{4\left(R^2+\frac{1}{C^2 \omega^2}\right)=R^2+\frac{9}{C^2 \omega^2} \quad \text { i.e., } \frac{1}{C^2 \omega^2}=\frac{3}{5} R^2}

So that \mathrm{\frac{X}{R}=\frac{(1 / C \omega)}{R}=\frac{\left[(3 / 5) R^2\right]^{1 / 2}}{R}=\sqrt{\frac{3}{5}}}

 

Posted by

Devendra Khairwa

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