A series LCR circuit connected to an AC source with voltage of the source \( v = v_m\sin\omega t \).
If ‘q’ is the charge on the capacitor and ‘i’ is the current, from Kirchhoff’s loop rule :
\( L\displaystyle\frac{di}{dt} + iR + \frac{q}{c} = V \)
The current in the circuit is given by \( i = I_m\sin(\omega t + \phi) \) where \( \phi \) is the phase difference between the voltage across the source and current in the circuit.
We know \( V_{Rm} = I_m R \); \( V_{Lm} = I_m X_L \); \( V_{Cm} = I_m X_C \); and
\( X_L = \omega L \); \( X_C = \displaystyle\frac{1}{\omega C} \)
Total impedance in the circuit regulates current. At resonance frequency of the LCR circuit current in the circuit is maximum.
In a series LCR circuit \( L = 4\,\mathrm{H} \), \( C = 100\,\mu\mathrm{F} \) and \( R = 25\,\Omega \). The circuit is connected to a variable frequency 220 V source.
The source angular frequency \( (\omega) \) which drives the circuit in resonance is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 50\,\mathrm{rad\,s^{-1}} \)
Resonance occurs when \( X_L = X_C \), that is \( \omega L = 1/(\omega C) \). This gives the resonant angular frequency
\( \displaystyle \omega_0 = \frac{1}{\sqrt{LC}} \)
\( L = 4\,\mathrm{H} \)
\( C = 100\,\mu\mathrm{F} = 100\times10^{-6}\,\mathrm{F} = 10^{-4}\,\mathrm{F} \)
\( \displaystyle \omega_0 = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{2\times10^{-2}} = 50\,\mathrm{rad\,s^{-1}} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
