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 impedance in the L-C-R circuit.
In a series L-C-R circuit, the total impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2}, \]
where \( X_L \) is the inductive reactance, \( X_C \) is the capacitive reactance, and \( R \) is the resistance. The potential difference across the capacitor \( V_C \) can be found using the voltage divider rule.

Step 2: Voltage divider rule.
The voltage across the capacitor is given by:
\[ V_C = \frac{X_C}{Z} V, \]
where \( V \) is the total voltage across the series circuit, and \( Z \) is the total impedance.

Step 3: Use the given relationship \( X_L - X_C = R \).
We are given that \( X_L - X_C = R \). Therefore, the impedance simplifies to:
\[ Z = \sqrt{R^2 + R^2} = \sqrt{2} R. \]

Step 4: Finding the r.m.s. value of potential difference.
The r.m.s. value of the potential difference is given by:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}}. \]
Thus, the potential difference across the capacitor is:
\[ V_C = \frac{X_C}{Z} V_{\text{rms}} = \frac{X_C}{R} V_{\text{rms}} = \frac{V_0}{X_L}. \]
Final Answer:
Thus, the r.m.s. value of the potential difference across the capacitor is:
\[ \boxed{\frac{V_0}{X_L}}. \]