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:
The topic of this question is Alternating Current (AC circuits).
We need to deduce the ratio of resistance to capacitive reactance by forming an equation based on the changes in current at two different frequencies.
Step 2: Key Formula or Approach:
1. Impedance: $Z = \sqrt{R^2 + X_C^2}$.
2. Capacitive Reactance: $X_C = \frac{1}{\omega C}$.
3. Ohm's Law: $I = \frac{V}{Z}$.
Step 3: Detailed Explanation:
At initial frequency $\omega$:
$I^2 = \frac{V^2}{R^2 + X_C^2} \implies V^2 = I^2(R^2 + X_C^2)$.
At frequency $\omega/4$, the new reactance becomes $X_C' = \frac{1}{(\omega/4)C} = 4X_C$.
The new current squared is:
$\left(\frac{I}{3}\right)^2 = \frac{V^2}{R^2 + (4X_C)^2} \implies \frac{I^2}{9} = \frac{V^2}{R^2 + 16X_C^2}$.
Equating the expressions by eliminating $V^2$:
\[ 9(R^2 + X_C^2) = R^2 + 16X_C^2 \]
\[ 9R^2 + 9X_C^2 = R^2 + 16X_C^2 \implies 8R^2 = 7X_C^2 \]
\[ \frac{R}{X_C} = \sqrt{\frac{7}{8}} \]
Step 4: Final Answer:
The ratio of resistance to reactance is $\sqrt{7/8}$.