A capacitor of capacity `2muF` is changed to a potential different of `12V` . It is then connected across an inductor of inductance `0.6mH` What is the current in the circuit at a time when the potential difference across the capacitor is `6.0V` ?

Explanation

As the capacitor is charged to a p.d of 12 V the initial charge on the capacitor is
`q_(0)=CV_(0)=2xx10^(-6)xx12` Coul. `" " cdots(1)`
At any instant as the capacitor discharges through the inductor ( LC circuit ) the
`q = q_(0) cos omega t " " cdots(2)` [ because at t=0, q=`q_(0)`]
But q = CV
where V is the p.d. at the instant .t.
From (1) and (3) we obtain `(q)/(q_(0))=(V)/(V_(0))`
Putting the value of V and `V_(0)` we obtain
`(q)/(q_(0))=(1)/(2) implies cos omega =(1)/(2)`
`implies omega t = cos^(-1)((1)/(2))`
`implies omega t= pi//3` rad . `" " cdots(4)`
Here `omega = (1)/(sqrt(LC))= (1)/((0.6xx10^(-3)xx2xx10^(-6))^((1)/(2)))`
`implies omega = (10^(5))/(2sqrt(3))` rad /sec `" " cdots (5)`
The current through the circuit at that instant is given by .
`i=(dq)/(dt)=(d)/(dt)[q_(0)cos omega t]`
`implies i = -q_(0) omega sin omega t `
`implies |i| = q_(0)omega sin omega t `
Putting the value of `q_(0)` from (1) `omega` from (5) and `omega t` from (4) we obtain.
(i) = `2xx10^(-6)xx12xx(10^(5))/(2sqrt(3)) "sin" (pi//3)`
`implies [i] = 0.69 A. `