How Does AC Frequency Influence Power Dissipation in a Resistor?

[QueenFisher]

calculate the power dissipated in a pure resistor of 30 Ohms when the peak vopltage across it from a 100 Hz supply is 22V.

I did P(loss)=I^2R
Vrms=peak voltage/sqroot2=15.558 (i've just used approximate values for the moment)
R=30Ohms
I(resistor)=15.558/30=0.5186
P(loss)=0.5186^2 x 30
=8.069W

but i haven't used the frequency value anywhere, and surely it wouldn't be given if they didn't want you to use it?

 

[mohsenman]

Let's figure out the question in more detail. Power is somewhat a ambiguous term as it may mean average power or instantanious power. Your answer is exact for the former case but for the latter you may write:
1.p(t)=v(t)*i(t)
2.v(t)=R*i(t)
1 & 2->3.p(t)=((v(t))^2)/R
4.-V is peak voltage->v(t)=V*sin(2*pi*f*t+phi0)-changing time origin->v(t)=V*sin(2*pi*f*t)
3 & 4->5.p(t)=((V^2)/R)*((sin(2*pi*f*t))^2)=((V^2)/(2*R))*(1-cos(2*pi*f*t))
Notice that by averaging (5) over one period, the cosine term vanishes and the following known result emerges:
P_average=(V^2)/(2*R)