An oscillating circuit consist of a capacitor with capacitance `C=4.0 mu F` and a coil with inductance `L=2.0 mu H` and active resistance `R=10 Omega`. Find the ration of the energy of the coil's magnetic field to that of the capacitor's electric field at the moment when the current has the maximum value .

Explanation

`I =I_(m) e^(-betat) sin omegat`
`beta=(R)/( 2L), omega_(0)=sqrt((1)/( LC)), omega=sqrt(omega_(0)^(2)-beta^(2))`
`I=-dot(q),q=` charge on the capacitor
Then `q=I_(m)e^(-betat) ( sin(omegat+delta))/( sqrt(omega^(2)+beta^(2))), tan delta=(omega)/( beta)`
Thus `W_(M)=(1)/(2) LI_(m)^(2) e^(-2 betat) sin ^(2) omegat`
` W_(E)=(I_(m)^(2))/( 2 C) (e^(-2betat) sin^(2) (omegat+ delta))/( omega^(2)+ beta^(2))=(LI_(m)^(2))/(2)e^(-2 betat)sin ^(2)( omegat + delta)`
Current is maximum when `(d)/(dt) e^(-betat)sinomegat=0`
Thus` - betasin omegat+ omega cos omegat=0`
or `tan omegat=(omega)/( beta)= tan delta`
i.e.` omega t=n pi + delta`
and hence `(W_(M))/( W_(E))=(sin^(2)(omegat))/( sin^(2)(omegat+ delta))=(sin^(2)delta)/( sin^(2) 2 delta)=(1)/( 4 cos ^(2) delta)`
`=(1)/( 4 beta^(2)//omega_(0)^(2))=(omega_(0)^(2))/(4 beta^(2))=(1)/(LC)xx(L^(2))/(R^(2))=(L)/(CR^(2))=5`.
`(W_(M)` is the magnetic energy of the inductance coil and `W_(E)` is the electric energy of the capacitor. `)`