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

Answer

951J, 0.52 cos 314 t.

Explanation

Here, time of 1 cycle `T=1//50 s`, we have to calculate the average energy as time gt gt T.
`=V_(rms) I_(rms) cos (phi)t =(I_M)/(sqrt(2)) xx (V_M)/(sqrt(2)) xx(R )/(Z) t`
`V=(V_(M)^(2)R)/(2Z^(2))t` (`:. I_(M)=(V_M)/(Z))`
now, `Z=sqrt(R^(2)+(omega L -(1)/(omega C)^(2))`
`=sqrt((50)^(2)+(314 xx 20 xx 10^(-3) -(1)/(314 xx 100 xx 10^(_6)))^(2))`
`=sqrt(3153.6) ~~ 56 Omega`
Energy consumed `=(10^(2)xx50xx20xx60)/(2xx3153.6) = 951 J`
When resistance is removed
`cos(phi)=(R )/(Z')'=(1)/(omega C) - (omega L') = (1)/(314xx10^(-4)) - 314xx40xx10^(_3) = 19.3`
[here`X_(C)gtX_(L)`, hence `phi=pi//2]`
`I=(V_M)/(Z') sin (omega t + phi) =(10)/(19.3) sin (314 t + (pi)/(2))`
`=0.52 cos 314 t`.