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

$\text{ Here, }L=20\mathrm{mH}=20\times {10}^{-3}\mathrm{H},C=50\mu \mathrm{F}=50\times {10}^{-6} \mathrm{F}$

$\begin{array}{l}R=40\Omega ,V=10\sin 340t=V_0\sin \omega t\\ \omega =340rad{s}^{-1},V_0=10 \mathrm{V}\\ X_L=\omega L=340\times 20\times {10}^{-3}=6.8\Omega \end{array}$

$\begin{array}{l}{X}_C=\frac{1}{\omega C}=\frac{1}{340\times 50\times {10}^{-6}}=\frac{{10}^4}{34\times 5}=58.82\Omega \\ Z=\sqrt{R^2+{\left(X_C-X_L\right)}^2}=\sqrt{(40{)}^2+(58.82-6.8{)}^2}\\ =\sqrt{(40{)}^2+(52.02{)}^2}=65.62\Omega \end{array}$

The peak current in the circuit is 

$I_0=\frac{V_0}{Z}=\frac{10}{65.62} \mathrm{A},$

$\cos \varphi =\frac{R}{Z}=\left(\frac{40}{65.62}\right)$

Power loss in A.C. circuit 

$\begin{array}{l}=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi =\frac{1}{2}{V}_0{I}_0\cos \varphi \\ =\frac{1}{2}\times 10\times \frac{10}{65.62}\times \frac{40}{65.62}=0.46 \mathrm{W}\end{array}$