In an circuit, V and I are given by `V=150sin(150t) V` and `I =150sin(150t+(pi)/(3))A`. The power dissipated in the circuit is

Options

A. 106 W
B. 150 W
C. 5625 W
D. zero

Answer

C

Explanation

Compare C = 150 sin (150 t) with
`V = V_(0)sin omega t`, we get, `V_(0)=150 V`
Compare `I = 150 sin (150 t+(pi)/(3))` with
`I = I_(0)sin(omega t+ phi)`, we get
`I_(0)=150A, phi = (pi)/(3)=60^(@)`
The power dissipated in ac circuit is
`P=(1)/(2)V_(0)I_(0)cos phi=(1)/(2)xx150xx150xx cos 60^(@)`
`=(1)/(2)xx150xx150xx(1)/(2)=5625 W`