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)/(6)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. `(pi)/(2)`

Answer

B

Explanation

`P=(1)/(2)V_(0)i_(0)cos phiimpliesP=P_("Peak.cosphi")`
`implies(1)/(2)(P_("peak"))cos theta`
`impliescospi=(1)/(2)impliesphi=(pi)/(3)`