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If a capacitor of 2.0muF is charged to 2...

If a capacitor of `2.0muF` is charged to 20V and then suddenly short-circuited by a coil of negligible resistance and of inductance `8.0muH` Calculate the maximum amplitude and the frequency of the resulting current oscillations.

Text Solution

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Here, `C=2.0xx10^-6F, V=20V,`
`L=8.0xx10^-6H`
The equation for the oscillatory discharge of
capacitor through inductance is given by
`q=q_0 sin omegat`, where `omega=1/(sqrt(LC))`
The resulting oscillatory current will be
`I=(dq)/(dt)=q_0omega cos omegat=I_0 cos omegat....(i)`
where `q_0omega=I_0`, amplitude of oscillatory current.
`:.` Amplitude of oscillatory current
`I_0=q_0omega=CVxx1//sqrt(LC)=Vsqrt(C/V)`
`=20xxsqrt((2.0xx10^-6)/(8.0xx10^-6))=10A`
Also frequency of oscillatory current
`v=1/(2pisqrt(LC))`
`=1/(2xx3.14sqrt((8.0xx10^-6)xx(2.0xx10^-6)))`
`=4.0xx10^4Hz`
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