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

Solution :
$$\eqalign{ & {\text{Given}}:L = 20\,mH;C = 50\,\mu F;R = 40\Omega \cr & V = 10\sin 340\,t \cr & \therefore {V_{{\text{runs}}}} = \frac{{10}}{{\sqrt 2 }} \cr & {X_C} = \frac{1}{{\omega C}} = \frac{1}{{340 \times 50 \times {{10}^{ - 6}}}} = 58.8\,\Omega \cr & {X_L} = \omega L = 340 \times 20 \times {10^{ - 3}} = 6.8\,\Omega \cr & {\text{Impedance, }}Z = \sqrt {{R^2} + {{\left( {{X_C} - {X_L}} \right)}^2}} \cr & = \sqrt {{{40}^2} + {{(58.8 - 6.8)}^2}} = \sqrt {4304} \,\Omega \cr} $$
Power loss in $$A.C.$$  circuit,
$$\eqalign{ & P = i_{rms}^2R = {\left( {\frac{{{V_{rms}}}}{Z}} \right)^2}R = {\left( {\frac{{\frac{{10}}{{\sqrt 2 }}}}{{\sqrt {4304} }}} \right)^2} \times 40 \cr & = \frac{{50 \times 40}}{{4304}} \simeq 0.51\,W \cr} $$