An ideal 1/3 H inductor is connected to a 50 Hz sinusoidal voltage source. The energy stored is stated as 6 J. Find the maximum power delivered to the inductor in W.
The official A treats the stated six joules as cycle-average magnetic energy. With sinusoidal current, average energy is LIrms²/2 and instantaneous power amplitude is ωLIrms²=2ωEavg. Thus Pmax=2×(2π×50)×6=1200π W. If six joules meant maximum stored energy instead, Pmax would be 600π W, B. The source does not specify average versus peak energy; the official grading is retained with its necessary interpretation. Average real power in an ideal inductor remains zero.
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