In an AC circuit, the power factor:

The correct answer is Option (2) → is one, when the circuit contains an ideal resistor only

The power factor is \( \cos\phi \), where \( \phi \) is the phase difference between voltage and current. The average power is \( P = V_{\mathrm{rms}} I_{\mathrm{rms}} \cos\phi \).

Option (1) is zero when the LCR series circuit is at resonance — Incorrect. At resonance \( X_L = X_C \), so the circuit is purely resistive, \( \phi = 0 \) and the power factor is one, not zero.

Option (2) is one, when the circuit contains an ideal resistor only — Correct. In a purely resistive circuit the voltage and the current are in phase, so \( \phi = 0 \) and \( \cos\phi = 1 \).

Option (3) is one, when the circuit contains an ideal inductor only — Incorrect. In an ideal inductor the current lags the voltage by \( 90^\circ \), so \( \cos\phi = 0 \) and no average power is used.

Option (4) is one, when the circuit contains an ideal capacitor only — Incorrect. In an ideal capacitor the current leads the voltage by \( 90^\circ \), so the power factor is again zero.

NCERT (2026-27):
Power in AC Circuit: The Power Factor – “… Case (i) Resistive circuit: If the circuit contains only pure R, it is called resistive. In that case \( \phi = 0 \), \( \cos\phi = 1 \). There is maximum power dissipation.
Case (ii) Purely inductive or capacitive circuit: … Therefore, \( \cos\phi = 0 \), and no power is dissipated even though a current is flowing in the circuit.
Case (iv) Power dissipated at resonance in LCR circuit: At resonance \( X_c - X_L = 0 \), and \( \phi = 0 \). Therefore, \( \cos\phi = 1 \) and \( P = I^{2}Z = I^{2}R \).”