A capacitor of capacitance `25muF` is charged to `300V`. It is then connected across a `10mH` inductor. The resistance in the circuit is negligible.
a. Find the frequency of oscillation of the circuit.
b. Find the potential difference across capacitor and magnitude of circuit current `1.2ms` after the inductor and capacitor are connected.
c. Find the magnetic energy and electric energy at `t=0` and `t=1.2ms`.

Explanation

At t = 0 : Current in the circuit is zero. Hence,`U_(L) = 0` Charge on the capacitor is maximum
Hence, `U_(c)=1/2q_(0)^(2)/C`
or `" "U_(c)=1/2 times (7.5 times 10^(-3))^(2)/(25 times 10^(-6))=1.125J`
`therefore " "` Total energy `E=U_(L)+U_(C)=1.125J`
At t = 1.2 ms
`U_(L)=1/2Li^(2)=1/2(10 times 10^(-3))(10.13)^(2)=0.513J`
`U_(C)=E-U_(L)=1.125-0.513=0.612J`
Otherwise `U_(C)` can be calculated as,
`U_(C)=1/2q^(2)/C=1/2 times (5.53 times 10^(-3))^(2)/((25 times 10^(-6)))=0.612 J`