Voltage V and current I in AC circuit are given by V =50sin(50 t)volt,I = 50 sin (`50t+(pi)/(3)`)mA
The power dissipated in the ciruit is

Options

A. 5.0 W
B. 2.5 W
C. 1.25 W
D. zero

Answer

C

Explanation

Given, `V = 50 sin (50 t)V`
Maximum voltage,`V_0= 50 V`
`i = i_0 sin (50 t + pi/3) mA`
Maximum current, `i_0 = 50 mA = 50 xx 10^(-3) A`
Power dissipated, `p = (i_0)/(sqrt2) xx (v_0)/(sqrt(2))`
`= (50xx 50 xx 10^(-2))/(2)`
`= (2500 xx 10^(-2))/(2) = 1.25 W`