Question

In an AC circuit having \(R\) and \(C\), angular frequency \(\omega\) has peak current \(I\). If angular frequency becomes \(\omega/4\), the peak current becomes \(I/3\). Find the ratio of resistance and reactance at frequency \(\omega\).

• A. \(\sqrt{\frac{8}{7}}\)
• B. \(\sqrt{\frac{2}{7}}\)
• C. \(\sqrt{\frac{7}{8}}\)
• D. \(\sqrt{\frac{9}{8}}\)

Answer 1

The Correct Option is C

Concept: Current in an \(RC\) circuit: \[ I = \frac{V_0}{\sqrt{R^2 + X_C^2}} \] where \[ X_C = \frac{1}{\omega C} \] Step 1: Current at frequency \(\omega\) \[ I_0 = \frac{V_0}{\sqrt{R^2 + X_C^2}} \] Step 2: When frequency becomes \(\omega/4\) \[ X_C' = \frac{1}{(\omega/4)C} = 4X_C \] Thus current \[ \frac{I_0}{3} = \frac{V_0}{\sqrt{R^2 + (4X_C)^2}} \] Step 3: Form ratio \[ \frac{I_0}{I_0/3} = \frac{\sqrt{R^2 + 16X_C^2}}{\sqrt{R^2 + X_C^2}} \] \[ 3 = \sqrt{\frac{R^2 + 16X_C^2}{R^2 + X_C^2}} \] Step 4: Solve \[ 9(R^2 + X_C^2) = R^2 + 16X_C^2 \] \[ 8R^2 = 7X_C^2 \] \[ \frac{R}{X_C} = \sqrt{\frac{7}{8}} \]