In an ac circuit the current lags behind the applied voltage. Its pow…

The correct answer is Option (3) → draws a leading wattless current that neutralizes the lagging wattless component of the current

When the current lags the voltage by \( \phi \), it can be split into two components. One is the power component \( I_p \), in phase with the voltage. The other is the wattless component \( I_q \), at right angles to it. Only \( I_p \) carries power. The smaller \( I_q \) is, the closer the current comes to being in phase with the voltage and the closer \( \cos\phi \) comes to 1.

Option (1) dissipates extra power, so the power component of the current increases — Incorrect. An ideal capacitor dissipates no average power, since its current is \( \pi/2 \) out of phase with the voltage across it. The power taken by the circuit stays the same.

Option (2) draws a current in phase with the voltage, which adds to the power component of the current — Incorrect. The current in a capacitor is never in phase with the voltage across it. It leads by \( \pi/2 \). So it adds nothing to the power component and only changes the wattless part of the total current.

Option (3) draws a leading wattless current that neutralizes the lagging wattless component of the current — Correct. The current in the capacitor leads the voltage by \( \pi/2 \), opposite to the lagging component \( I_q \). With a suitable capacitance the two neutralize each other. The supply current then reduces to \( I_p \), in phase with the voltage, and the power factor approaches 1.

Option (4) draws a lagging wattless current, which brings the total current closer in phase with the voltage — Incorrect. A capacitor current leads the voltage. A lagging current is what an inductor draws. A lagging current added to an already lagging circuit current would increase the angle of lag and lower the power factor further.

NCERT (2026-27):
Example 7.7 – “… Let us resolve I into two components. I_p along the applied voltage V and I_q perpendicular to the applied voltage. I_q as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. I_P is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I_q by an equal leading wattless current I′_q. This can be done by connecting a capacitor of appropriate value in parallel so that I_q and I′_q cancel each other and P is effectively I_p V.”