A lamp consumes only `50%` of peak power in an `a.c.` circuit. What is the phase difference between the applied voltage and the circuit current

Options

A. `pi/3` rad
B. `pi/6` rad
C. `pi/4` rad
D. `pi/2` rad

Answer

A

Explanation

We know that, `cosphi=eta`
(`becauseeta=50%` where `phi` is the phase difference)
`cosphi=50%`
`cosphi=(1)/(2)`
`cosphi=cos60^(@)implies theta=(pi)/(3)`