In a series LCR ac circuit, off resonance, the current

The correct answer is Option (4) → may lead or lag behind the generator voltage

In a series LCR circuit, the phase angle between the current and the source voltage is \( \tan\phi=(X_C-X_L)/R \). As the frequency increases, \( X_L=\omega L \) rises and \( X_C=1/\omega C \) falls.

Option (1) is always in phase with the generator voltage — Incorrect. The current is in phase only at resonance, where \( X_L=X_C \) and \( \phi=0 \). Off resonance, \( X_C-X_L\ne0 \), so \( \phi\ne0 \).

Option (2) always lags behind the generator voltage — Incorrect. The current lags only when \( X_L\gt X_C \), that is, above the resonant frequency. Below it, the current leads.

Option (3) always leads the generator voltage — Incorrect. The current leads only when \( X_C\gt X_L \), that is, below the resonant frequency. Above it, the current lags.

Option (4) may lead or lag behind the generator voltage — Correct. Off resonance, which reactance is larger depends on the frequency. Below \( \omega_0 \), the circuit is capacitive and the current leads. Above \( \omega_0 \), it is inductive and the current lags.

NCERT (2026-27):
Phasor-Diagram Solution – “If X_C > X_L, φ is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage. If X_C < X_L, φ is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage.”