Load voltage and current are $v(t)=100\sin\omega t$ V and $i(t)=10\sin(\omega t-60^\circ)+2\sin3\omega t+5\sin5\omega t$ A. Average load power in watts is ________.
Only equal-frequency voltage and current components contribute nonzero average power. The voltage has only the fundamental, so the current harmonics average to zero in the product vi.
Fundamental rms values are 100/√2 V and 10/√2 A, with 60° displacement. Thus $P=(100/\sqrt2)(10/\sqrt2)\cos60^\circ=$ **250 W**.
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