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

`R=40 Omega, X_(L)=omegaL=340xx20xx10^(-3)=6.8 Omega`
`X_(c)=1/(omegac)=1/(340xx50xx10^(-6))=(10^(4))/170=58.8`
`Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))`
`=sqrt((40)^(2)+(6.8+58.8)^(2))=sqrt(4304)`
`bar(P)=1/2V_(o)I_(o)cos phi=1/2V_(o) (V_(o))/Z R/Z=(V_(o)^(2)R)/(2Z^(2))`
`=((10)^(2)xx40)/(2xx4304)~=0.46 W=0.51 W`