## ac frequency question

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?
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 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)