Match List – I with List – II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | Resonant frequency | (I) | \( \displaystyle\frac{\omega L}{R} \) |
| (B) | Quality factor | (II) | \( 2\pi fL \) |
| (C) | \( X_L \) | (III) | \( \displaystyle\frac{1}{2\pi\sqrt{LC}} \) |
| (D) | \( X_C \) | (IV) | \( 1/2\pi fC \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
These are standard results for a series LCR circuit.
(A) Resonant frequency — this matches (III) \( \frac{1}{2\pi\sqrt{LC}} \). At resonance \( X_L = X_C \), that is \( \omega L = 1/\omega C \), which gives \( \omega_0 = 1/\sqrt{LC} \) and \( f_0 = \frac{1}{2\pi\sqrt{LC}} \).
(B) Quality factor — this matches (I) \( \frac{\omega L}{R} \). The quality factor measures the sharpness of resonance, \( Q = \frac{\omega_0 L}{R} \).
(C) \( X_L \) — this matches (II) \( 2\pi fL \). The inductive reactance is \( X_L = \omega L = 2\pi fL \).
(D) \( X_C \) — this matches (IV) \( 1/2\pi fC \). The capacitive reactance is \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC} \).
NCERT (2026-27):
Summary – …
