In a circuit, the instantaneous values of alternating current and voltages in a circuit is given by
`I = (1)/(sqrt2) sin (100 pi t) A` and
`E = (1)/(sqrt2) sin (100 pi t + (pi)/(3)) V`.
The average power in watts consumed in the circui is

Options

A. `1/4`
B. `(sqrt3)/(4)`
C. `1/2`
D. `1/8`

Answer

D

Explanation

Given `i=(1)/(sqrt2) sin (100pit)` ampere
Compare it with `i=i_0 sin (omega t), `we get
`i_0 = (1)/(sqrt2)A`
Given `e = (1)/(sqrt2) sin (100 pi t + pi/3) `volt
Compare it with `e = e_0 sin (omega t + phi), ` we get
`e_0 = (1)/(sqrt2) V, phi = pi/3`
`therefore i_(rms) = (i_0)/(sqrt2) = (1/sqrt2)/(sqrt2) A = 1/2A " "e_(rms) = (e_0)/(sqrt2) = (1/sqrt2)/(sqrt2) V = 1/2 V`
Average power consumed in the circuit ,
`P = i_(rms)e_(rms) cos phi`
` = (1/2)(1/2) cos pi/3 = (1/2)(1/2)(1/2) = 1/8 W`