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. `0.6 W`
B. `0.76 W`
C. `0.89 W`
D. `0.046 W`

Answer

D

Explanation

`L = 20 mH = 20 xx 10^(-3) H, C = 50 xx 10^(-6) F`,
`R = 40 Omega, V = 10 sin 340 t,` Power l oss `= ?`
`:. E_(0) 10 V, Omega = 340 rad//s`
`X_(L) = omega L = 340 xx 20 xx 10^(-3) = 6.8 Omega`
`X_(C ) = (1)/(omega C) = (1)/(340 xx 50 xx 10^(-6)) = (1000)/(17) Omega`
`= 58.8 Omega`
`Z = sqrt(R^(2) + (X_(C ) - X_(L))^(2))`
` sqrt(40^(2) + (58.8 = 6.8)^(2)) = 65.6 ohm`
`I_(upsilon) = (E_(upsilon))/(Z) = (E_(0))/(sqrt Z) = (10)/(1.414 xx 65.6) = 0.1078 A`
The power loss `= I_(upsilon)^(2) xx R = (0.1078)^(2) xx 40`
`= 0.046 W`
Choice (d) in closest.