A series AC circuit contains an inductor (20mH), a capacitor (100 $\mathrm{\mu }$F), a resistor (50 $\mathrm{\Omega }$) and an AC source of 12V, 50 Hz. Find the energy dissipated in the circuit in 1000 s.

Energy dissipated in the circuit:

$$\begin{gathered} \begin{array}{l}\mathrm{E}={\mathrm{I}}_{\mathrm{rms}}\times {\mathrm{V}}_{\mathrm{rms}}\cos \mathrm{\varphi } \mathrm{t}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{\mathrm{Z}}\times {\mathrm{V}}_{\mathrm{rms}}\times \frac{\mathrm{R}}{\mathrm{Z}}\times \mathrm{t}=\frac{{\mathrm{V}}_{\mathrm{rms}}^2\mathrm{Rt}}{{\mathrm{Z}}^2}\\ {\mathrm{Z}}^2={\mathrm{R}}^2+{\left(\mathrm{\omega L}-\frac{1}{\mathrm{\omega C}}\right)}^2\\ =(50{)}^2+{\left(2\mathrm{\pi }\times 50 \times 20\times {10}^{-3}-\frac{1}{2\mathrm{\pi }\times 50\times 100\times {10}^{-6}}\right)}^2\\ =2500+{\left(2\mathrm{\pi }-\frac{100}{\mathrm{\pi }}\right)}^2=3153.68\\ \end{array} \\ \therefore \mathrm{E} = \frac{{\mathrm{V}}_{\mathrm{rms}}^2\mathrm{Rt}}{{\mathrm{Z}}^2} \\ \Rightarrow \mathrm{E} = \frac{{12}^2 \times 50\times 1000}{3153.68} = \\ \Rightarrow \mathrm{E} = 2283.047 \mathrm{J} = 2.283 \mathrm{kJ} \end{gathered}$$