An AC voltage of $V=220\sqrt{2}\sin \left(100\pi t+\frac{\pi }{2}\right)$ is applied across a DC voltmeter, its reading will be

The potential difference $V$ across and the current $I$ flowing through an instrument in an AC circuit are given by 

$V=5\cos \omega t \mathrm{V}$

$I=2\sin \omega t \mathrm{A}$

The power dissipated in the instrument is