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A capacitor of capacitance 25muF is char...

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`.

Text Solution

Verified by Experts

Charge across the capacitor at time I will be,
`q=q^(0)cosomegat`
`i=-q^(0)omegasinomegat`
`q^(0)=CV_(0)=(25xx10^(6))(300)=7.5xx10^(-3)C` Now, charge in the capacitor after `t=1.2xx10^(-3)s is`
`q(7.5xx10^(-3)cos(2pixx318.3)(1.2xx10^(-3)C=5.53xx10^9-3)C`
.therefore P. D. across capacitor`V=|(q)/(C)=(5.5xx10_(3))/(25xx10_(6))=221.2volt`
The magnitude of current in the circuit at `t=1.2xx10^9-3`s is
`|i|=q_(0)omegasinomegat=(7.5xx10^(-3)(2pi)(318.3)sin(2pixx318.3)(1.2xx10^(-3))A=10.13A`
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