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A capacitor having capacitance 2 muF is charged to a potential difference of 50 V. it is then diconnected from battery and connected to an inductor of inductance 5 mH. Peak current that flows through the inductor is

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Solution :(a) When entire electrical energy of capacitor gets CONVERTED into MAGNETIC energy of INDUCTOR then at that instant CURRENT in the circuit becomes maximum because energy stored in the inductor is proportional to the SQUARE of the current flowing .
`(1)/(2) Li_(0)^(2) = (1)/(2)CV^(2)`
`implies 5xx10^(-3)xxi_(0)^(2)=2xx10^(-6)xx2500`
`implies i_(0)^(2)= (2xx10^(-6)xx2500)/(5xx10^(-3))=1`
`implies i_(0)=1 A `


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