In an a.c. Circuit the voltage applied is `E=E_(0) sin (omega)t`. The resulting current in the circuit is `I=I_(0)sin((omega)t-(pi/2))`. The power consumption in the circuit is given by

Options

A. `0.5V_(0)l_(0)W`
B. `0.707V_(0)l_(0)W`
C. `1.919V_(0)l_(0)W`
D. zero

Answer

D

Explanation

Given, `V=V_(0)sinomegat and l=l_(0) sin(omegat-pi//2)` It is obvious that current lags behind the voltage by `pi//2` in phase.
`therefore `Power consumed in the circuit
`P=(1)/(2)V_(0)l_(0)cosphi`
But `phi=pi//2=90^(@)`
`therefore P=(1)/(2)V_(0)II_(0)cos90^(@)=0" "(becausecos90^(@)=0)`