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. `2.5W`

Answer

A

Explanation

`V =5 cos omega t = 5 sin (omega t + pi//2)`
`I = 2 sin omegat`
phase different between `V` and I `phi = pi//2`
`P =0` .