question_answer

In an AC circuit, voltage \[V={{V}_{0}}\,\sin \omega t\] and inductor L is connected across the circuit. Then the instantaneous power will be 

A) \[\frac{V_{0}^{2}}{2\omega L}\sin \omega t\]               

B) \[\frac{-V_{0}^{2}}{2\omega L}\sin \omega t\]           

C) \[\frac{-V_{0}^{2}}{2\omega L}\sin 2\omega t\]         

D) \[\frac{V_{0}^{2}}{\omega L}\sin 2\omega t\]

Correct Answer: C

Solution :

[c] \[V=L\frac{dI}{dt}\] \[\Rightarrow \,\,\,\frac{dI}{dt}=\frac{{{V}_{0}}}{L}\sin \omega t\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,I=-\frac{{{V}_{0}}}{\omega L}\cos \omega t\] Instantaneous power, \[P=VI=-\frac{V_{0}^{2}}{\omega L}\sin \omega t\,\,\cos \omega t\,\,\,\,=-\frac{V_{0}^{2}}{2\omega L}\sin 2\omega t\]