The potential differences `V` and the current `i` flowing through an instrument in an `AC` circuit of frequency `f` are given by `V=5 cos omega t` and `I=2 sin omega t` amperes (where `omega=2 pi f`). The power dissipated in the instrument is

Options

A. zero
B. `10W`
C. `5W`
D. `2.5 W`

Answer

A

Explanation

`V=5 cos omega t=5sin(omega t+(pi)/2)` and `i=2 sin omegat`
Power `=V_(rms)xxi_(rms)xxcos varphi=0`
(Since `varphi=(pi)/2`, therefore `cos varphi=cos((pi)/2)=0`)