Step 1: Look at how the source is connected.
The source \(v_i(t)\), the capacitor, the inductor, and the resistor are all drawn directly in parallel with each other, with no impedance in series with the ideal source. That means the voltage across every one of these branches, including the resistor, is exactly \(v_i(t)\) at every instant, no matter what the frequency is.
Step 2: Write the source explicitly.
\[
v_i(t)=12\sin(\omega t)\text{ V}
\]
so its amplitude is \(V_m=12\) V.
Step 3: Note what "resonant frequency" means here.
The resonant frequency of the parallel \(L\)-\(C\) combination is where their reactances cancel,
\[
\omega_0=\frac{1}{\sqrt{LC}}
\]
At this frequency the tank draws minimum current from a source feeding it, but here the resistor is wired straight to the ideal voltage source, not fed through the tank. So the tank's resonance does not change the voltage the resistor sees; that voltage stays fixed at \(v_i(t)\) regardless of \(\omega\).
Step 4: Compute the rms voltage across the resistor.
For a sinusoid of amplitude \(V_m\),
\[
V_{rms}=\frac{V_m}{\sqrt2}=\frac{12}{\sqrt2}\text{ V}
\]
Step 5: Compute the average power dissipated in the \(1\ \text{k}\Omega\) resistor.
\[
P_{avg}=\frac{V_{rms}^2}{R}=\frac{\left(\frac{12}{\sqrt2}\right)^2}{1000}=\frac{72}{1000}\text{ W}
\]
\[
P_{avg}=0.072\text{ W}=72\text{ mW}
\]
Step 6: Note the role of resonance.
Even though the question mentions the resonant frequency, the resistor's power does not actually depend on \(\omega\) here, because the resistor sits straight across the ideal source. Resonance only matters for the current drawn from the source itself, not for the power in the resistor.
Final Answer:
\[
\boxed{P_{avg}=72.00\text{ mW}}
\]
