For a given ac `i=i_msinomegat`, show that the average power dissipated in a resistor R over complete cycle is `1/2i_m^2R`.

Options

A. `P=(1)/(2)i_(m)^(2)R`
B. `P=(1)/(4)i_(m)^(2)R`
C. `P=(3)/(4)i_(m)^(2)R`
D. `P=i_(m)^(2)R`

Answer

A

Explanation

The instantaneous power dissipated in the resistor is
`P=i^(2)R=i_(m)^(2)Rsin^(2)omegat" "[i=i_(m)sinomegat]`
The average value of P over a cycle is
`P_(av)= lt i^(2) R gt = lt i_(m)^(2) R sin^(2) omegat gt`
Where, `P_(av)` denotes its average value and `lt . . . gt` denotes taking average of the quantity inside the bracket. since `i_(m)^(2)` and R are constant,
`P_(av)=i_(m)^(2)R lt sin^(2) omega t gt `
Using the trigonometric identtify, `sin^(2)omegat=1//2(1-cos2omegat)`,
we have ` lt sin^(2) omega t gt =(1//2)(1- lt cos2 omegatgt)` and since,
`lt cos2omegatgt=0`, we have `ltsin^(2)omegatgt=(1)/(2)`
thus, `P_(av)=(1)/(2)i_(m)^(2)R`