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Initially capacitor 'A' is charged to a ...


Initially capacitor 'A' is charged to a potential drop `epsilon` and capacitor B is uncharged At t=0, switch S is closed, then the maximum current through the inductor is

A

`epsilonsqrt((C)/(2L))`

B

`epsilonsqrt((C)/(L))`

C

`2epsilon((C)/(L))`

D

`(epsilon)/(2)sqrt((C)/(L))`

Text Solution

Verified by Experts

The correct Answer is:
A

When the current through the inductor is maximum
`V_(L)=L(di)/(dt)=0`
Hence capacitors A and B will be in parallel now using conservation energy
`(1)/(2)Cepsilon^(2)=2xx(1)/(2)C(epsilon^(2))/(4)+(1)/(2)Li_(max)^(2)`
`(Cepsilon^(2))/(4)=(1)/(2)Li_(max)^(2)`
`i_(max)=epsilonsqrt((C)/(2L))`
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