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

Options

A. `0.89 W`
B. `0.51 W`
C. `0.67 W`
D. `0.76 W`

Answer

B

Explanation

`X_(C)=1/(omegaC)=1/(340xx50xx10^(-6))=58.8 Omega`
`X_(L)=omegaL=340xx20xx10^(-3)=6.8 Omega`
`Z=sqrt(R^(2)+(X_(C)-X_(L))^(2))`
`=sqrt(40^(2)+(58.8-6.8)^(2))=sqrt(43.04)Omega`
`P=i_(rms)^(2)R=((V_(rms))/Z)^(2) R`
`=((10//sqrt(2))/(sqrt(4304)))^(2)xx40=(50xx40)/4304=0.47 W` So best answer (nearest answer) will be (b)