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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 \omega / 3 (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to ohmic resistance at the original frequency \omega will be.

Option: 1

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


Option: 2

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


Option: 3

{\frac{3}{5}}


Option: 4

{\frac{5}{3}}


Answers (1)

best_answer

According to given problem,

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

\mathrm{\text { And } \frac{1}{2}=\frac{V}{\left[R^2+(3 / C \omega)^2\right]^2}}.........................[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} \text {, i.e., } \frac{1}{C^2 \omega^2}=\frac{3}{5} R^2}

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

Posted by

shivangi.shekhar

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