An oscillating circuit consists of a capacitor with capacitance `C`, a coil of inductance `L` with negligible resistance, and aswitch. With the switch disconnected , the capacitor was charged to a voltage `V_(m)` and then at the moment `t=0` the switch was closed . Find `:`
`(a)` the current `(I(t))` in the circuit as a function of time,
`(b)` the emf of self- inductance in the coil at the moments when the electric energy of the capacitor is equal to that of the current in the coil.

Explanation

After the switch was closed, the circuit satisfies
`-L(dI)/( dt)=(q)/(C)`
or `(d^(2)q)/(dt^(2))+ omega_(0)^(2)q=0implies q=CV_(m) cos omega_(0)t`
where we have used the fact that when the switch is closed we must have
`V=(q)/(C)=V_(m), I=(dq)/( dt)=0` at `t=0`
Thus `(a)` `I=(aq)/( dt)=-CV_(m)omega_(0) sin omega_(0)t`
`=-V_(m) sqrt((C)/(L)) sin omega_(0)t`
`(b)` The electrical energy of the capacitor is `(q^(2))/( 2 C) alpha cos^(2) omega_(0)t` and of the inductor is
`(1)/(2) L I^(2) alphasin^(2) omega_(0)t.`
The two are equal when ltbr` ` omega_(0)t=(pi)/( 4)`
At that instant the emf of the self `-` inductance is
`-L(di)/(dt)=V_(m) cos omega_(0)t =V_(m) //sqrt(2)`