An oscillating circuit consisting of a capacitor with capacitance `C` and a coil of inductance `L` maintans free undamped oscillations with voltage amplitude across the capacitor equal to `V_(m)`. For an arbitrary moment of time find the relation between the current `I` in the circuit and the voltage `V` acroos the capacitor. Solve this problem usdini `Omega`'s law and then the energy conservation law.

Explanation

Since there are no sources of emf in the circuit, Omega's 1 law reads
`(q)/(c)=-L(dI)/(dt)`
where `q=` change on the capacitor, `I=(dq)/ (dt)=` current through the coil. Then
`(d^(2)q)/(dt^(2))+ omega_(0)^(2) q==0, omega_(0)^(2)=(1)/(LC).`
`q=q_(m) cos ( omega_(0) t + alpha)`
The solution for this equation is
`I=-omega_(0) C V _(m) sin ( omega_(0)t+ alpha)`
and `V=V_(m) cos ( omega_(0)t- alpha)`
`V^(2)+(I^(2))/( omega_(0)^(2)C^(2))=V_(m)^(2)`
or ` V^(2)+(LI^(2))/( C)=V_(m)^(2)`
By energy conservation `(1)/(2) LI^(2)+(q^(2))/( 2 C)=` constant ltbr. When the `P.D.` across the capacitor takes its maximum value `V_(m)` the current `I` must be zero.
Thus` "` constant `"` `=(1)/(2) CV_(m)^(2)`
Hence `(LI^(2))/( C)+V^(2)=V_(m)^(2)` once again.