The resonant angular frequency \( (\omega) \) of an LCR series a.c. circuit is given as
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \omega=\sqrt{\displaystyle\frac{1}{LC}} \)
At resonance in a series LCR circuit the inductive and capacitive reactances are equal:
\( \displaystyle \omega L = \frac{1}{\omega C} \)
\( \displaystyle \omega^{2} = \frac{1}{LC} \)
\( \displaystyle \omega = \sqrt{\frac{1}{LC}} \)
The resonant angular frequency does not depend on R; the resistance only affects how sharp the resonance is.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Resonance – “… So if \( \omega \) is varied, then at a particular frequency \( \omega_0 \), \( X_c = X_L \), and the impedance is minimum
\( \displaystyle Z = \sqrt{R^2 + 0^2} = R \)
This frequency is called the resonant frequency:
\( \displaystyle X_c = X_L \;\text{or}\; \frac{1}{\omega_0 C} = \omega_0 L \;\text{or}\; \omega_0 = \frac{1}{\sqrt{LC}} \qquad (7.28) \)”