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

XC=1ωC=1340 x 50 x 10-6=58.8 Ω XL=ωL=340 x 20 x 10-3=6.8 Ω Z=R2+XC-XL2=402+58.8-6.82=4304 Ω P=i2rmsR=VrmsZ2R=10/243042 x 40=50 x 404304=0.47 WSo best answer( nearest answer) will be (1).