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

a. The frequency of oscillation of the circuit is
`f=1/(2pisqrt(LC))`
Substituting the given values, we have
`f=1/(2pisqrt((10xx10^-3)(25xx10^-6)))=318.3Hz`
b. Charge across the capacitor at time `t` will be
`q=q_0cosomegat` and `i=-q_0omegasinomegat`
Here `q_0=CV_0=(25xx10^-6)(300)=7.5xx10^-3C`
Now, charge in the capacitor after `t=1.2xx10^-3s` is
`q=(7.5xx10^-3)cos(2pixx318.3)(1.2xx10^-3)C`
`=-5.53xx10^-3C`
`:.PD` acorss capacitor `V=(|q|)/C=(5.53xx10^-3)/(25xx10^-6)=221.2`volt
The magnitude of current in the circuit at `t=1.2xx10^-3s` is
`|i|=q_0omegasinomegat`
`=(7.5xx10^-3)(2pi)(318.3)sin(2pixx318.3)(1.2xx10^-3)A`
`=10.13A`
c. At `t=0` Curren in the circuit is zero.
Hence `U_L=0`
Charge in the capacitor is maximum
Hence `U_C=1/2q_0^2/C`
or `U_C=1/2xx((7.5xx10^-3)^2)/((25xx10^-6))`
`:.` Total energy , `E=U_L+U_C=1.125J`
At `t=1.2ms`
`U_L=1/2Li^2`
`=1/2(10xx10^-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`
`=1/2xx((5.53xx10^-3)^2)/((25xx10^-6))`
`=0.612J`