In the circuit diagram shown, intially there is no energy in the inductor and the capacitor. The switch is closed at `t=0` . Find the current `I` as a function of time `t` .

Explanation

At any time `t`
`i_(1)=(E)/(2R)(1-e^((-2Rt)/(L)))`
Change on capacitor
`q=CE(1-e^(-t//RC))`
`i_(2)=(dq)/(dt)=CE[0-e^((-t)/(RC))(-1/(RC))]`
`i_(2)=(dq)/(dt)=CE[0-e^(-(t)/(RC)((-(1)/(RC))]`
`=(E)/(R)e^((t)/(RC))`
`i=i_(1)+i_(2)`
`=(E)/(2R)(1-e^(-(2Rt)/(L)))+(E)/(R)e^((-t)/(RC))`