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: Set up the impedances.
At frequency $\omega$, the reactance is $X_c$ and the impedance is $Z = \sqrt{R^2 + X_c^2}$. At $\tfrac{\omega}{3}$ the reactance triples to $3X_c$, so $Z' = \sqrt{R^2 + 9X_c^2}$.

Step 2: Use the current drop.
The current halves, so the impedance doubles: $Z' = 2Z$.

Step 3: Square and simplify.
$R^2 + 9X_c^2 = 4(R^2 + X_c^2)$, which gives $5X_c^2 = 3R^2$, so $\tfrac{X_c^2}{R^2} = \tfrac{3}{5} = 0.6$.

Step 4: Take the root.
$\tfrac{X_c}{R} = \sqrt{0.6}$. \[ \boxed{\sqrt{0.6}} \]