Question
An a.c. source of angular frequency $\omega$ is connected across a resistor R and a capacitor C in series. The current is observed as I. Now the frequency of the source is changed to $\omega/4$, (keeping the voltage unchanged) the current is found to be $I/3$. The ratio of resistance to reactance at frequency $\omega$ is
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Solution
Sol. [Key Concept: $I = \frac{V}{\sqrt{R^{2} + X_{C}^{2}}}$; at $\omega/4$, $X_{C}$ quadruples and $I \rightarrow \ I/3$, which gives $\frac{R}{X_{C}} = \sqrt{\frac{7}{8}}$.]
As we know that current in RLC circuit is given by
$$\begin{matrix} & I = \frac{V}{Z} = \frac{V}{\left\lbrack R^{2} + \left( X_{L} - X_{C} \right)^{2} \right\rbrack^{1/2}} \\ & I = \frac{V_{0}}{\sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}}} \end{matrix}$$
As circuit is RC
$$\begin{matrix} I_{0} & \ = \frac{V_{0}}{\left( R^{2} + X_{C}^{2} \right)^{1/2}} \\ \frac{I_{0}}{3} & \ = \frac{V_{0}}{\left\lbrack R^{2} + \left( 4X_{C} \right)^{2} \right\rbrack^{1/2}} \\ R^{2} + \left( X_{C} \right)^{2} = \frac{\left( R^{2} + 16X_{C}^{2} \right)}{9} & \\ \frac{8R^{2}}{9} & \ = \frac{7X_{C}^{2}}{9} \\ \frac{R}{X_{C}} & \ = \sqrt{\frac{7}{8}} \end{matrix}$$
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