An inductor `20xx10^(-3)` a capacitor `100(mu)F` and a resistor `50 Omega` are connected in series across a source of emf `V=10 sin 314 t`.
Then the energy dissipated in the circuit in 20 mim is

Options

A. 960 J
B. 900 J
C. 250J
D. 500J

Answer

A

Explanation

`Z^(2)=(X_(C )-X_(L))^(2) + R^(2)=(31.85 - 6.28)^(2) + (50)^(2) = 3154`
`P=((E_(rms)^(2))/(Z^(2))) R= ((10//sqrt(2))^(2))/(3154) xx 50 = 0.8 W`
Heat produced in 20 min `=(0.8)(20xx60)=960 J`
`X_(C )-X_(L) = 31.85 - 2(6.28)=19.29`
`I_(m)=(10)/(19.29)=0.52`
Hence, `I=0.52 sin (314t+(pi)//(2))=0.52 cos 314 t`.