An electric bulb consumes only 25% of the peak power in an a.c. circuit. What is the phase difference between the circuit and the applied a.c. voltage ?

Options

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

Answer

C

Explanation

Average Power = `(E_(0)I_(0)cos theta)/(2)`
and peak power = `P_(0) = E_(0)I_(0)`
It gives that, `(E_(0)I_(0)cos theta)/(2) = (P_0)/4 = (E_(0)I_(0))/(4)`
`:. cos theta = 1/2 :. theta = 60^(@) = (pi)/3`.