A series LCR circuit with \( L = 5.0\,\mathrm{H} \), \( C = 80\,\mu\mathrm{F} \) and \( R = 40\,\Omega \) is connected to a variable frequency 200 V source. The amplitude of current at the resonant frequency is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7.1 A
At resonance \( X_L = X_C \), so the reactances cancel and the impedance is just the resistance, \( Z = R \).
The supply voltage of an a.c. source is quoted as its rms value unless stated otherwise, so \( V_{\mathrm{rms}} = 200\,\mathrm{V} \) and the peak voltage is \( \sqrt{2} \times 200\,\mathrm{V} \).
\( \displaystyle I_0 = \frac{V_0}{R} = \frac{\sqrt{2} \times 200}{40} = 5\sqrt{2}\,\mathrm{A} \approx 7.1\,\mathrm{A} \)
The values of \( L \) and \( C \) only fix the resonant frequency; they do not affect the current at resonance.
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 – …
