Question

An a.c. source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current is observed as $I$. Now the frequency of the source is changed to $\omega/4$, (keeping the voltage unchanged) the current is found to be $I/3$. The ratio of resistance to reactance at frequency $\omega$ is:

• A. $\sqrt{6/7}$
• B. $\sqrt{3/5}$
• C. $\sqrt{7/8}$
• D. $\sqrt{3/4}$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to set up equations for current in an RC circuit at two different frequencies and solve for the ratio $R/X_C$.
Step 2: Key Formula or Approach:
1. $I = V / Z$.
2. Impedance $Z = \sqrt{R^2 + X_C^2}$.
3. Capacitive reactance $X_C = 1/(\omega C)$.
Step 3: Detailed Explanation:
Case 1: At frequency $\omega$.
$I = \frac{V}{\sqrt{R^2 + X_C^2}} \implies I^2 = \frac{V^2}{R^2 + X_C^2}$ ...(1)
Case 2: At frequency $\omega/4$.
New reactance $X_C' = \frac{1}{(\omega/4) C} = 4 X_C$.
Current $I/3 = \frac{V}{\sqrt{R^2 + (4X_C)^2}} \implies \frac{I^2}{9} = \frac{V^2}{R^2 + 16 X_C^2}$ ...(2)
Divide (1) by (2):
\[ 9 = \frac{R^2 + 16 X_C^2}{R^2 + X_C^2} \]
\[ 9R^2 + 9X_C^2 = R^2 + 16 X_C^2 \]
\[ 8R^2 = 7X_C^2 \implies \frac{R^2}{X_C^2} = \frac{7}{8} \]
\[ \frac{R}{X_C} = \sqrt{\frac{7}{8}} \]
Step 4: Final Answer:
The ratio is $\sqrt{7/8}$.