A series LCR circuit is predominantly inductive if

The correct answer is Option (2) → \( X_L\gt X_C \)

The sign of the phase angle \( \phi \) tells which reactance dominates. For a series LCR circuit, \( \tan\phi=(X_C-X_L)/R \), where \( \phi \) is the phase angle between the source voltage and the current.

Option (1) \( X_L=X_C \) — Incorrect. Then \( \tan\phi=0 \). The two reactances balance, so the impedance is just R. The current is in phase with the voltage. This is resonance, where the circuit is neither inductive nor capacitive.

Option (2) \( X_L\gt X_C \) — Correct. With \( X_C\lt X_L \), \( \phi \) is negative, the circuit is predominantly inductive and the current lags the source voltage.

Option (3) the resistance R is zero — Incorrect. The resistance only sets the size of the phase angle. Whether the circuit is inductive or capacitive depends on which of \( X_L \) and \( X_C \) is larger.

Option (4) \( X_L\lt X_C \) — Incorrect. With \( X_C\gt X_L \), \( \phi \) is positive and the circuit is predominantly capacitive, so the current leads the voltage.

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.”