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# An inductor $20mH,$a capacitor and a resistor $40\Omega$ are connected in series across a source of emf The power loss in A.C. circuit is

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a
0.76 W
b
0.89 W
c
0.46 W
d
0.67 W

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detailed solution

Correct option is C

Here, L=20mH=20×10-3H,C=50μF=50×10-6 FR=40Ω,V=10sin340t=V0sinωtω=340rads-1,V0=10 VXL=ωL=340×20×10-3=6.8ΩXC=1ωC=1340×50×10-6=10434×5=58.82ΩZ=R2+XC-XL2=(40)2+(58.82-6.8)2=(40)2+(52.02)2=65.62ΩThe peak current in the circuit is I0=V0Z=1065.62 A,cosϕ=RZ=4065.62Power loss in A.C. circuit =VrmsIrmscosϕ=12V0I0cosϕ=12×10×1065.62×4065.62=0.46 W

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