A series LCR circuit with L = 5.0 H, C = 80 μF and R = 50 Ω is connected to a variable frequency 220 V ac source. The angular frequency and the corresponding r.m.s. current in the circuit at resonance is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 50 rad/sec, 4.4 A
At resonance the inductive and capacitive reactances are equal and cancel, so the impedance is just R and the current is maximum.
\( \displaystyle \omega_0 = \frac{1}{\sqrt{LC}} \)
\( \displaystyle \omega_0 = \frac{1}{\sqrt{5.0\,\mathrm{H} \times 80\times10^{-6}\,\mathrm{F}}} \)
\( \displaystyle \omega_0 = \frac{1}{\sqrt{4\times10^{-4}}}\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle \omega_0 = \frac{1}{0.02}\,\mathrm{rad\,s^{-1}} \)
\( \omega_0 = 50\,\mathrm{rad\,s^{-1}} \)
The 220 V of an ac source is its rms value, and at resonance Z = R.
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{R} = \frac{220\,\mathrm{V}}{50\,\Omega} \)
\( I_{\mathrm{rms}} = 4.4\,\mathrm{A} \)
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 – …
