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

The current in the RC circuit at frequency \(\omega\) is given by

\(I = \dfrac{V}{\sqrt{R^2 + X_C^2}}\)

where \(X_C = \dfrac{1}{\omega C}\) is the capacitive reactance.

When the frequency is changed to \(\omega/4\), the new reactance becomes \(X_C' = \dfrac{1}{(\omega/4)C} = 4X_C\).

The new current is given as \(I/3\), so

\(\dfrac{I}{3} = \dfrac{V}{\sqrt{R^2 + (4X_C)^2}}\)

Substituting the expression for \(I\), we get

\(\dfrac{V}{3\sqrt{R^2 + X_C^2}} = \dfrac{V}{\sqrt{R^2 + 16X_C^2}}\)

Squaring both sides, we obtain

\(\dfrac{1}{9(R^2 + X_C^2)} = \dfrac{1}{R^2 + 16X_C^2}\)

\(9R^2 + 9X_C^2 = R^2 + 16X_C^2\)

\(8R^2 = 7X_C^2\)

\(\dfrac{R^2}{X_C^2} = \dfrac{7}{8}\)

\(\dfrac{R}{X_C} = \sqrt{\dfrac{7}{8}}\)

Answer: \(\sqrt{\dfrac{7}{8}}\)