In an A.C. circuit, voltage applied is $V = 220 \sin 100\, t$. If the impedance is 110 and phase difference between current and voltage is 60°, the power consumption is equal to

110 W

$ V_{rms} = \frac{V_{peak}}{root(2)} = 220/\sqrt 2$

$ I_{rms} = \frac{V_{rms}}{Z} = \sqrt 2$

$P = V_{rms}I_{rms} cos\phi = \frac{220}{\sqrt 2} \sqrt 2 cos 60^o = 110W$