To solve this problem, we need to calculate the power loss in the AC circuit.$\\$Given:$\\$• Inductance $L = 20 \,$mH$= 20 \times 10^{-3} \,$H$\\$• Capacitance $C = 50 \, \mu$F$= 50 \times 10^{-6} \,$F$\\$• Resistance $R = 40 \, \Omega \\$• Source voltage $V = 10 \sin(340t) \\$The angular frequency $\omega$ is given by the equation of the source:$\\ \omega = 340 \,$rad/s$\\$First, calculate the inductive reactance $X_L$ and capacitive reactance $X_C: \\ X_L = \omega L = 340 \times 20 \times 10^{-3} = 6.8 \, \Omega \\ X_C = \frac{1}{\omega C} = \frac{1}{340 \times 50 \times 10^{-6}} = 58.82 \, \Omega \\$The total impedance $Z$ of the series circuit is given by:$\\ Z = \sqrt{R^2 + (X_L - X_C)^2} \\ Z = \sqrt{40^2 + (6.8 - 58.82)^2} \\ Z = \sqrt{1600 + (-52.02)^2} \\ Z = \sqrt{1600 + 2704.08} \\ Z = \sqrt{4304.08} \\ Z \approx 65.6 \, \Omega \\$The RMS voltage $V_{rms}$ is:$\\ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.07 \,$V$\\$The RMS current $I_{rms}$ is:$\\ I_{rms} = \frac{V_{rms}}{Z} = \frac{7.07}{65.6} \approx 0.108 \,$A$\\$The power factor $\cos\phi$ is:$\\ \cos\phi = \frac{R}{Z} = \frac{40}{65.6} \approx 0.61 \\$The power loss (average power) in the circuit is given by:$\\ P = V_{rms} \cdot I_{rms} \cdot \cos\phi \\ P = 7.07 \times 0.108 \times 0.61 \\ P \approx 0.51 \,$W$\\$Therefore, the power loss in the AC circuit is $0.51 \,$W$. \\$This corresponds to Option 3.