How to calculate power from angular frequency of LRC circuit

[MeMoses]

Homework Statement

A voltage Δv = (120 V) sin ωt (in SI units) is applied across a series combination of a 2.13 H inductor, a 12.8 μF capacitor, and a 15.0 Ω resistor.

a) Determine the angular frequency, ω0 at which the power delivered to the resistor is a maximum. = 192
b) Calculate the power at that frequency.
c) Determine the two angular frequencies ω1 and ω2 at which the power delivered is one-half the maximum value. [The Q of the circuit is approximately ω0/(ω2 - ω1).] Enter the smaller one first.

Homework Equations

Not sure
P = Irms^2 * R
P = IV

z = sqrt(R + (wL - 1/(wC)))
z = V/I

The Attempt at a Solution

I got part a as 192, but I'm not sure where to go with part b. I calculated z to equal 15.0685 but how do I calculate I? Do i have to use z=V/I, but then what value do I use for V? Would it just be 120V? I'm not sure where to take this, so any help would be great. Thanks

 

[gneill]

At resonance (ω = ω_o) what is the magnitude of the imaginary component of the impedance? Hint: at resonance, |X_L| = |X_C|.

 

[MeMoses]

So the imaginary component would just be 0 and z=R? But what do I use for V?

 

[gneill]

So the imaginary component would just be 0 and z=R? But what do I use for V?

At resonance, yes. So, given that the impedance at resonance is purely real (and equal to R), what is the power dissipated?

 

[MeMoses]

Zero power is dissipated right?

 

[gneill]

Zero power is dissipated right?

Nope. The supply voltage V still sees the resistor R (since Z = R at resonance).

 

[MeMoses]

Ok, but won't I = V/R and then P=V*I? but I'm not getting the correct answer, unless I am not supposed to use 120 for V

 

[gneill]

Ok, but won't I = V/R and then P=V*I? but I'm not getting the correct answer, unless I am not supposed to use 120 for V

What value did you get for the power? Do you know what the correct value should be?

It could be that the 120V is a peak value rather than rms.

 

[MeMoses]

Yep, wasnt rms. Thanks