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

For series RC circuit:

current $$I=\frac{V}{Z}$$, where

$$Z=\sqrt{R^2+X_C^2},\quad X_C=\frac{1}{\omega C}$$

At frequency ω\omegaω:

$$Z_1=\sqrt{R^2+X^2}$$

$$(where\ X=\frac{1}{\omega C}​)$$

At frequency $$\omega/4$$:

$$X'=\frac{1}{(\omega/4)C}=4X$$

$$Z_2=\sqrt{R^2+(4X)^2}=\sqrt{R^2+16X^2}$$

Given:

$$\frac{I_2}{I_1}=\frac{1/3}{1}=\frac{1}{3}\Rightarrow\frac{Z_1}{Z_2}=\frac{1}{3}$$

$$\frac{\sqrt{R^2+X^2}}{\sqrt{R^2+16X^2}}=\frac{1}{3}$$

Square:

$$\frac{R^2+X^2}{R^2+16X^2}=\frac{1}{9}$$

$$9(R^2+X^2)=R^2+16X^2$$

$$9R^2+9X^2=R^2+16X^2$$

$$8R^2=7X^2$$

$$\frac{R}{X}=\sqrt{\frac{7}{8}}$$