The power loss in a series LCR circuit occurs only across the resistor and is given by $P = I_{rms}^2 R$ .

1. Identify parameters: $L = 20 \text{ mH} = 20 \times 10^{-3} \text{ H}$, $C = 50 \mu\text{F} = 50 \times 10^{-6} \text{ F}$, $R = 40 \Omega$. From $V = 10 \sin 340t$, peak voltage $V_m = 10 \text{ V}$ and angular frequency $\omega = 340 \text{ rad/s}$ .
2. Calculate Reactances: $X_L = \omega L = 340 \times 20 \times 10^{-3} = 6.8 \Omega$ . $X_C = \frac{1}{\omega C} = \frac{1}{340 \times 50 \times 10^{-6}} \approx 58.82 \Omega$ .
3. Calculate Impedance ($Z$): $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{40^2 + (58.82 - 6.8)^2} = \sqrt{1600 + (52.02)^2} \approx \sqrt{1600 + 2706} \approx 65.62 \Omega$.
4. Calculate RMS Current: $I_{rms} = \frac{V_{rms}}{Z} = \frac{V_m}{\sqrt{2} Z} = \frac{10}{1.414 \times 65.62} \approx 0.108 \text{ A}$ .
5. Calculate Power: $P = I_{rms}^2 R \approx (0.108)^2 \times 40 \approx 0.46 \text{ W}$. Among the given options, $0.51 \text{ W}$ is the closest value.