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An inductor 20 mH, a capacitor 100 muF a...

An inductor `20 mH`, a capacitor `100 muF` and a resistor `50 Omega` are connected in series across a source of emf, `V=10 sin 314 t`. The power loss in the circuit is

A

2.74 W

B

0.43 W

C

0.79 W

D

1.13 W

Text Solution

Verified by Experts

The correct Answer is:
C
`P_(av)=((V_(RMS))/(Z))^(2)R`
`Z=sqrt(R^(2)+(omegaL-(1)/(omegaC))^(2))=56 Omega`
`therefore P_(av)=((10)/((sqrt(2))56))^(2)xx50=0.79 W`
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