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: Work with squares of impedance:
Let $k = X_C/R$. Then $Z_1^2 = R^2(1 + k^2)$ and $Z_2^2 = R^2(1 + 9k^2)$.

Step 2: Use the current condition:
$I_2 = I_1/2$ means $Z_2 = 2Z_1$, so $Z_2^2 = 4Z_1^2$: $1 + 9k^2 = 4 + 4k^2$.

Step 3: Solve:
$5k^2 = 3$, so $k = \sqrt{3/5} = \sqrt{0.6}$.

Step 4: Check:
With $k^2 = 0.6$: $Z_1 = R\sqrt{1.6}$ and $Z_2 = R\sqrt{6.4} = 2R\sqrt{1.6}$. The ratio is exactly 2.

Final Answer:
Option (D). \[ \boxed{\sqrt{0.6} \text{ (D)}} \]