An inductor `20xx10^(-3)H` a capacitor `100 (mu)F`, and a resistor `50 Omega` are connected in series across a source of emf `V=10 sin 341 t`. Find the energy dissopated in the circuit in 20 min. If the resistance is removed from the circuit and the value of inductance is doubled, then find the variation of current with time in the new circuit.

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`.