Show that the current in a pure resistor is in phase with the ac voltage across it and hence S.T. average power dissipation in a resistor is `i^(2)R` where I is r.m.s value of ac.

Explanation

Let I be the current in a resistor voltage drop a cross the resistor will be iR, supplying KVL to the closed loop. We write `v_(m)sinomegat=iR`.
i.e., `i=((v_(m))/(R))sinomegat`
i.e., `i=i_(m)sinomegat` where `i_(m)=(v_(m))/(R)`
Hence current is in phase with the applied voltage.
By definition of power `p=i^(2)R`
i.e., `p=i_(m)^(2)Rsinomegat`
Average power, `barp=(:i^(2)R:)=(:i_(m)^(2)Rsin^(2)omegat:)`
Since, `(:sin^(2)omegat:)=(1)/(2)`
`barp=i_(m)^(2)R(:sin^(2)omegat:)`
i.e. `barp=(1)/(2)i_(m)^(2)R`