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)/(2.)q_(20)^(2)/(C)`or`U_(C)=(1)/(2)xx((7.5xx10^(-3)^(2))/((25xx10^(-6))=1.125J`
therefore Total energy` E=U_(L) +U_(C)=1.125 J` Att=1.2 ms
`U_(L) =(1)/(2) Li^(2) = (10xx10^(-3))(10.13)^(3)`
`U_(C)=E-U_(L) =1.125-0.5130 = 0.612 J`
Otherwise `U_(C)` can be calculated as`U_(C)=(1)/(2)(q^(2))/(C)=(1)/(2)xx(5.53xx10^(-3)/(25xx10-(-6)=0.612J`