An inductor 20 mH, a capacitor 100μF  and a resistor 50 Ω 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 isZ=R2+XC-XL2; where XC= capacitive reactance                                                   and XL= inductive reactance.  Also XC=1ωC and XL=ωL∴Z=(50)2+1314×100×10-6-314×20×10-32      or   Z=56Ω The power loss in the circuit is Pav=VrmsZ2R∴Pav=10(2)562×50=0.79 W