At the resonant frequency, the current amplitude in a series LCR circ…

The correct answer is Option (2) → maximum

In a series LCR circuit the current amplitude is \( i_m=v_m/Z \), with the impedance \( Z=\sqrt{R^2+(X_C-X_L)^2} \). At the resonant frequency \( \omega_0 \) the two reactances are equal, \( X_C=X_L \).

Option (1) minimum — Incorrect. Away from resonance the term \( (X_C-X_L)^2 \) adds to \( R^2 \). So the impedance is larger there and the current is smaller. At resonance the impedance is least, so the current is greatest.

Option (2) maximum — Correct. With \( X_C=X_L \), the impedance falls to its minimum value \( Z=R \), so the current amplitude reaches its largest value \( i_m=v_m/R \).

Option (3) zero — Incorrect. A zero current would need an infinite impedance. At resonance the impedance is only R, its smallest value.

Option (4) infinite — Incorrect. The impedance never falls below the resistance R. So the current amplitude cannot go above \( v_m/R \). It could become infinite only if R were zero.

NCERT (2026-27):
Resonance – “… So if ω is varied, then at a particular frequency ω₀, X_c = X_L, and the impedance is minimum \( \left(Z=\sqrt{R^2+0^2}=R\right) \). This frequency is called the resonant frequency:

    \( \displaystyle X_c=X_L\ \ \text{or}\ \ \frac{1}{\omega_0C}=\omega_0L\ \ \text{or}\ \ \omega_0=\frac{1}{\sqrt{LC}} \qquad (7.28) \)

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