Question

A series combination of resistor 'R' and capacitor 'C' is connected to an a.c. source of angular frequency 'ω'. Keeping the voltage same, if the frequency is changed to \(\frac{ω}{3}\) the current becomes half of the original current. Then the ratio of capacitive reactance and resistance at the former frequency is

• A. \(\sqrt{0.6}\)
• B. \(\sqrt{6}\)
• C. \(\sqrt{3}\)
• D. \(\sqrt{2}\)

Hint:

When frequency changes, capacitive reactance is inversely proportional to frequency. For a constant voltage, the current ratio gives impedance ratio. Square and solve for \(X_c/R\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
An RC series circuit with AC source of constant voltage. At frequency ω, current is I. At frequency ω/3, current becomes I/2. We need \(\frac{X_c}{R}\) at the original frequency ω.

Step 2: Key Formula or Approach:
Capacitive reactance \(X_c = \frac{1}{ωC}\). Impedance \(Z = \sqrt{R^2 + X_c^2}\). Current \(I = V/Z\).
At new frequency ω/3, \(X_c' = \frac{1}{(ω/3)C} = 3X_c\). New impedance \(Z' = \sqrt{R^2 + 9X_c^2}\). Given \(I' = I/2\), so \(Z' = 2Z\).

Step 3: Detailed Explanation:
\[ \sqrt{R^2 + 9X_c^2} = 2 \sqrt{R^2 + X_c^2} \] Square both sides: \(R^2 + 9X_c^2 = 4(R^2 + X_c^2) = 4R^2 + 4X_c^2\).
Thus \(9X_c^2 - 4X_c^2 = 4R^2 - R^2\) ⇒ \(5X_c^2 = 3R^2\) ⇒ \(\frac{X_c^2}{R^2} = \frac{3}{5} = 0.6\).
Hence \(\frac{X_c}{R} = \sqrt{0.6}\).

Step 4: Final Answer:
Option (A).