Power Factor In Ac Circuits

According to NCERT Class 12, Chapter 7 (Alternating Current), the average power dissipated in an AC circuit is $P=V_{rms}{I}_{rms}\cos \varphi$, where $\cos \varphi$ is the power factor accounting for the phase difference between voltage and current; only the in-phase component of current transfers real power. Here $\cos 60^{\circ }=0.5$, so $P=200\times 5\times 0.5=500\text{W}$. The option 1000 W ignores the power factor entirely, giving only the apparent power $V_{rms}{I}_{rms}$. The option 866 W wrongly uses $\sin 60^{\circ }$ instead of $\cos 60^{\circ }$, which represents reactive rather than real power. The option 250 W squares the power factor incorrectly. A sanity check shows the real power must be less than the apparent power of 1000 VA, and 500 W is exactly half, consistent with a power factor of 0.5. Reactive components store and return energy each cycle without net dissipation, so they do not contribute to average power. The distinction between apparent power, measured in volt-amperes, and real power, measured in watts, is central here: the supply must deliver 1000 VA of apparent power even though only 500 W does useful work. Engineers seek to improve the power factor toward unity, often by adding capacitors to inductive loads, so that a greater fraction of the delivered current does real work and transmission losses are minimised. A power factor of 0.5 as in this problem is comparatively poor, indicating a strongly reactive load where half the apparent power circulates uselessly between the source and the reactive elements.