An inductor 20 mH, a capacitor 100$\mu F$  and a resistor 50 $\Omega$ are connected in series across a source of emf, V=10 sin 314 t. The power loss in the circuit is

Impedance  Z in an ac circuit is

$$\begin{gathered} Z=\sqrt{R^2+{\left(X_C-X_L\right)}^2};\text{ where }{X}_C=\text{ capacitive reactance } \\ \text{ and }{X}_L=\text{ inductive reactance. } \end{gathered}$$

$\text{ Also }{X}_C=\frac{1}{\omega C}\text{ and }{X}_L=\omega L$

$$\begin{gathered} \therefore Z=\sqrt{(50{)}^2+{\left(\frac{1}{314\times 100\times {10}^{-6}}-314\times 20\times {10}^{-3}\right)}^2} \\ \text{ or } Z=56\Omega \end{gathered}$$

$\text{ The power loss in the circuit is }{P}_{av}={\left(\frac{V_{rms}}{Z}\right)}^{2}R$

$\therefore P_{av}={\left(\frac{10}{\left(\sqrt{2}\right)56}\right)}^2\times 50=0.79 \mathrm{W}$