A current `I=3 sin omega t` ampere flows through a bulb. The P.D. across the bulb is given by `V=4 cos omega t` volt. The power dissipated in the bulb is

Options

A. 12 W
B. 6 W
C. zero W
D. 3 W

Answer

C

Explanation

`V = 4 cos (omega t) = 4 sin (omega t +(pi)/2)`
Phase diff. between V and I is `(pi)/2`
`:.` Power factor = `cos(pi)/2 = 0`.