From the given voltage equation $V = 10\sin(340t)$, the peak voltage is $V_m = 10$ V and the angular frequency is $\omega = 340$ rad/s. The inductive reactance is $X_L = \omega L = 340 \times 20 \times 10^{-3} = 6.8 \, \Omega$. The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{340 \times 50 \times 10^{-6}} = \frac{1}{0.017} \approx 58.82 \, \Omega$. The impedance of the series LCR circuit is $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{40^2 + (58.82 - 6.8)^2} = \sqrt{40^2 + 52.02^2} \approx \sqrt{1600 + 2706} = \sqrt{4306} \approx 65.62 \, \Omega$. The RMS current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{V_m / \sqrt{2}}{Z} = \frac{10}{\sqrt{2} \times 65.62}$. The power loss in the circuit is dissipated only in the resistor and is given by $P = I_{rms}^2 R$ . $P = \left( \frac{10}{\sqrt{2} \times 65.62} \right)^2 \times 40 = \frac{100}{2 \times 4306} \times 40 = \frac{2000}{4306} \approx 0.46$ W.