In a series LCR circuit X_L = X_C . Which of the following is correct?

The correct answer is Option (3) → The current amplitude is maximum

When \( X_L = X_C \) the series LCR circuit is at resonance. The reactances cancel, so the impedance is the smallest it can be and equals the resistance.

    \( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = R \qquad I_0 = \frac{V_0}{Z} = \frac{V_0}{R} \)

Option (1) Current leads voltage in phase — Incorrect. The phase angle is \( \tan\phi = (X_L - X_C)/R = 0 \); the current is in phase with the voltage, so it does not lead.

Option (2) Current lags voltage in phase — Incorrect. For the same reason the current does not lag; the phase difference is zero.

Option (3) The current amplitude is maximum — Correct. \( Z \) is minimum, so \( I_0 = V_0/R \) is the maximum value.

Option (4) Current amplitude is minimum — Incorrect. The current amplitude is a maximum at resonance, not a minimum.

NCERT (2026-27):
Resonance – “… For an RLC circuit driven with voltage of amplitude v_m and frequency ω, we found that the current amplitude is given by

    \( \displaystyle i_m = \frac{v_m}{Z} = \frac{v_m}{\sqrt{R^{2} + (X_C - X_L)^{2}}} \)

with X_c = 1/ωC and X_L = ω L. So if ω is varied, then at a particular frequency ω₀, X_c = X_L, and the impedance is minimum (Z = \( \sqrt{R^{2} + 0^{2}} \) = R). This frequency is called the resonant frequency:

    \( \displaystyle X_c = X_L \ \text{or}\ \frac{1}{\omega_0 C} = \omega_0 L \)

    \( \displaystyle \text{or}\ \ \omega_0 = \frac{1}{\sqrt{LC}} \qquad (7.28) \)

At resonant frequency, the current amplitude is maximum; i_m = v_m/R.”