An inductor 20 mH, a capacitor 50 μF and a resistor 40 $\mathrm{\Omega }$ are connected in series across a source of emf V = 10sin 340 t. The power loss in A.C. circuit is :

$$\begin{gathered} {\mathrm{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega C}}=\frac{1}{340 \mathrm{x} 50 \mathrm{x} {10}^{-6}}=58.8 \mathrm{\Omega } \\ {\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L}=340 \mathrm{x} 20 \mathrm{x} {10}^{-3}=6.8 \mathrm{\Omega } \\ \mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{C}}-{\mathrm{X}}_{\mathrm{L}}\right)}^2}=\sqrt{{40}^2+{\left(58.8-6.8\right)}^2}=\sqrt{4304} \mathrm{\Omega } \\ \mathrm{P}={{\mathrm{i}}^2}_{\mathrm{rms}}\mathrm{R}={\left(\frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{Z}}\right)}^2\mathrm{R}={\left(\frac{10/\sqrt{2}}{\sqrt{4304}}\right)}^2 \mathrm{x} 40=\frac{50 \mathrm{x} 40}{4304}=0.47 \mathrm{W} \end{gathered}$$

So best answer( nearest answer) will be (1).