average power delivered?

Question

A 2.46 ?F ideal capacitor and a 300 ? resistor are connected in series with an AC source of 120 V (rms) and 60Hz. Find the average power delivered to the circuit. Answer: 3.45 W

Pave= Iv cos () 120 cos() v=IZ Z=Square(r^2+ (wl-xc)^2)

I think this is what you do but i can not get the right answer

Answer 1

P( ave) = Pa + jPr

P(ave)= V I
I= V/Z
P(ave)= V^2/Z =
Z= R +j/(Cw) where w=2 pi f

P(ave) = V^2 / Z or
P(ave) = V^2 ( R - j/(Cw))/(R^2 + 1/(Cw)^2) )

I don't know what did you meant by 2.46 ?F and 300 ?