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

Options

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

Answer

A

Explanation

`X_(C) = (1)/(omegaC) = (1)/(340 xx 50 xx 10^(-6)) = 58.8 Omega `
`X_(L) = omega L = 340 xx 20 xx 10 ^(-3) = 6.8 Omega `
` Z = sqrt(R^(2) + (X_(C) - X_(L))^(2))`
`= sqrt(40 ^(2) + (58.8 - 6.8)^(2)) = sqrt(4304) Omega `
`P = i_(rms)^(2) R = ((V_(rms))/(Z))^(2)R`
`= ((10 // sqrt(2))/(sqrt(4304)))^(2) xx 40 = (50 xx 40)/(4304) = 0.47 W ~~ 0.50 W`
So, the best answer (nearest answer) will be (a).