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 is

• A. \(\sqrt{6}\)
• B. \(\sqrt{0.3}\)
• C. \(\sqrt{3}\)
• D. \(\sqrt{0.6}\)

Hint:

Lowering the frequency to one third makes X_C three times larger. Current halving means impedance doubles.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In an RC series circuit the impedance is \(Z = \sqrt{R^2 + X_C^2}\), with \(X_C = \dfrac{1}{\omega C}\). With the same voltage, the current is inversely proportional to the impedance.

Step 2: Key Formula or Approach:
At frequency \(\omega\): \(Z_1 = \sqrt{R^2 + X_C^2}\). At \(\omega/3\), the reactance becomes \(3X_C\), so \(Z_2 = \sqrt{R^2 + 9X_C^2}\).

Step 3: Detailed Explanation:
The current halves, so \(Z_2 = 2Z_1\):
\[ R^2 + 9X_C^2 = 4(R^2 + X_C^2) \]
\[ 9X_C^2 - 4X_C^2 = 4R^2 - R^2 \]
\[ 5X_C^2 = 3R^2 \]
\[ \frac{X_C}{R} = \sqrt{\frac35} = \sqrt{0.6} \]
Option (A) \(\sqrt6\) and (C) \(\sqrt3\) are too large. Option (B) \(\sqrt{0.3}\) would arise from taking \(5X_C^2 = 1.5R^2\).

Final Answer:
The ratio \(X_C/R\) is \(\sqrt{0.6}\), option (D). \[ \boxed{\sqrt{0.6} \text{ (D)}} \]