Question

An alternating e.m.f. having voltage \( V = V_0 \sin \omega t \) is applied to a series L-C-R circuit. Given \( X_L - X_C = R \), the r.m.s. value of the potential difference across the capacitor will be

• A. \( V_0 \frac{R}{X_L} \)
• B. \( \frac{V_0}{X_C} \)
• C. \( \frac{V_0}{X_L} \)
• D. \( \frac{V_0}{\sqrt{R^2 + (X_L - X_C)^2}} \)

Hint:

In an L-C-R circuit, the r.m.s. potential difference across the capacitor can be found using the voltage divider rule, taking into account the impedance of the entire circuit.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the r.m.s. potential difference across the capacitor in a series LCR circuit where the magnitude of the difference between inductive and capacitive reactance is equal to the resistance.
Step 2: Key Formula or Approach:
1. Peak voltage \( V_0 \) and r.m.s. voltage \( V_{rms} = \frac{V_0}{\sqrt{2}} \).
2. Impedance \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).
3. r.m.s. Current \( I_{rms} = \frac{V_{rms}}{Z} \).
4. Voltage across capacitor \( V_{C,rms} = I_{rms} \cdot X_C \), where \( X_C = \frac{1}{\omega C} \).
Step 3: Detailed Explanation:
Given \( |X_L - X_C| = R \).
Substitute this into the impedance formula:
\[ Z = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]
Now, find the r.m.s. current:
\[ I_{rms} = \frac{V_0 / \sqrt{2}}{R\sqrt{2}} = \frac{V_0}{2R} \]
The r.m.s. potential difference across the capacitor is:
\[ V_{C,rms} = I_{rms} \cdot X_C = \left( \frac{V_0}{2R} \right) \left( \frac{1}{\omega C} \right) = \frac{V_0}{2R \omega C} \]
Step 4: Final Answer:
The r.m.s. value of potential difference across the capacitor is \( \frac{V_0}{2R \omega C} \).