# Voltage Source across Resistor

V(t) = Vm sin ωt

$$i\left( t \right)=\frac{v\left( t \right)}{R}$$.

$$i\left( t \right)=\frac{{{V}_{m}}}{R}\sin \omega t$$.

Where, $${{I}_{m}}=\frac{{{V}_{m}}}{R}$$.

i(t) = Im sin ωt

And P(t) = v(t) i(t)

P(t) = Vm sin ωt Im sin ωt

$$P\left( t \right)={{V}_{m}}\frac{{{I}_{m}}}{2}~\left( 1-\cos 2~\omega t \right)$$.

Hence, average power is

$${{P}_{avg}}=\frac{1}{2\pi }~\underset{0}{\overset{2\pi }{\mathop \int }}\,p\left( t \right)d\omega t$$.

$${{P}_{avg}}=\frac{{{V}_{m}}{{I}_{m}}}{2}=\frac{{{V}_{m}}}{\sqrt{2}}\frac{{{I}_{m}}}{\sqrt{2}}={{V}_{rms}}~{{I}_{rms}}$$.It is clear that if current or voltage waveform has a frequency of 50 Hz then power waveform has a frequency of 100Hz.