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, \[P=VI=-\frac{V_{0}^{2}}{\omega L}\sin \omega t\,\,\cos \omega t\,\,\,\,=-\frac{V_{0}^{2}}{2\omega L}\sin 2\omega t\]You need to login to perform this action.
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