Ac frequency question

1. Nov 8, 2005

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?

2. May 5, 2010

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)