To solve this problem, we need to calculate the power loss in an RLC series circuit.$\\$Given:$\\$• Inductance $L = 20$ mH$= 20 \times 10^{-3}$ H$\\$• Capacitance $C = 100 \mu F = 100 \times 10^{-6}$ F$\\$• Resistance $R = 50 \Omega \\$• Source voltage $V = 10 \sin 314t \\$The angular frequency $\omega$ is given by the source:$\\ \omega = 314$ rad/s$\\$The impedance $Z$ of the RLC circuit is given by:$\\ Z = \sqrt{R^2 + (X_L - X_C)^2} \\$Where $X_L$ is the inductive reactance and $X_C$ is the capacitive reactance:$\\ X_L = \omega L = 314 \times 20 \times 10^{-3} = 6.28 \, \Omega \\ X_C = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} = 31.85 \, \Omega \\$Substitute $X_L$ and $X_C$ into the impedance formula:$\\ Z = \sqrt{50^2 + (6.28 - 31.85)^2} \\ Z = \sqrt{50^2 + (-25.57)^2} \\ Z = \sqrt{2500 + 653.92} \\ Z = \sqrt{3153.92} \\ Z \approx 56.15 \, \Omega \\$The RMS voltage $V_{rms}$ is given by:$\\ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 7.07 \,$V$\\$The RMS current $I_{rms}$ is:$\\ I_{rms} = \frac{V_{rms}}{Z} = \frac{7.07}{56.15} \\ I_{rms} \approx 0.126 \,$A$\\$The power loss (average power) in the resistor is given by:$\\ P = I_{rms}^2 \times R \\ P = (0.126)^2 \times 50 \\ P = 0.015876 \times 50 \\ P \approx 0.79 \,$W$\\$Therefore, the power loss in the circuit is $0.79 \,$W$. \\$This corresponds to Option 3.