A sinusoidal voltage with peak value 220 V and variable frequency is…

The correct answer is Option (2) → (A), (B) and (D) only

At resonance in a series LCR circuit \( X_L = X_C \), so \( \omega_0 = 1/\sqrt{LC} \) and the impedance equals R.

    \( L = 4\,\mathrm{mH} = 4\times10^{-3}\,\mathrm{H} \)
    \( C = 100\,\mathrm{mF} = 0.1\,\mathrm{F} \)

    \( \displaystyle \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{4\times10^{-3} \times 0.1}} = \frac{1}{\sqrt{4\times10^{-4}}} = \frac{1}{2\times10^{-2}} = 50\,\mathrm{rad\,s^{-1}} \)

(A) Resonant angular frequency \( (\omega_0) = 50\,\mathrm{rad/s} \) — Correct. This is the value found above.

(B) Resonant frequency \( (\nu_0) = 25/\pi\,\mathrm{Hz} \) — Correct.

    \( \displaystyle \nu_0 = \frac{\omega_0}{2\pi} = \frac{50}{2\pi} = \frac{25}{\pi}\,\mathrm{Hz} \)

(C) Resonant frequency \( (\nu_0) = 50/\pi\,\mathrm{Hz} \) — Incorrect. This is \( \omega_0/\pi \), twice the resonant frequency; the factor 2 in \( \nu_0 = \omega_0/2\pi \) is missed.

(D) The rms current at resonance is 15.5 A — Correct. At resonance \( Z = R = 10\,\Omega \).

    \( \displaystyle I_{\mathrm{rms}} = \frac{V_0/\sqrt{2}}{R} = \frac{220}{1.414 \times 10} = 15.56\,\mathrm{A} \approx 15.5\,\mathrm{A} \)

Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27):
Power in AC Circuit: The Power Factor – “… Example 7.8 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH, and C = 796 μF.”

Power in AC Circuit: The Power Factor – “… Example 7.9 Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition.”