In the given circuit, initially switch `S_(1)` is closed, and `S_(2)` and `S_(3)` are open. After charging of capacitor, at `t = 0`, `S_(1)` is opened and `S_(2)` and `S_(3)` are closed. If the relation between inductance, capacitance and resistance is `L = 4 CR^(2)`, tehn find the time (in s) after which current passing through capacitor and inductor will be same (given `R = In2 m Omega,L = 2 mH`)

Answer

`(1)`

Explanation

After charging, charge on capacitor `= C epsilon`
Now, at `t = 0` two circuits are formed
1. Discharging of capacitor
`q = C epsilon e^(-t//tau_(C)) = Cepsilon e^(-t//RC)`
`i_(1) = - (dq)/(dt) = (epsilon)/(2R) e^(-t//RC)`
2. Growth of current in `L - R` circuit.
`i_(2) = (epsilon)/(2R) [1 - e^(-t//tau_(L))]`
Now, `i_(1) = i_(2)`
`(epsilon)./(2R) e^(-t//tau_(C)) = (epsilon)/(2R) [1 - e^(-t//tau_(L))]` ..(i)
Given `l = 4CR^(2)`
`(L)/(2R) = 2RC = (1)/(In 2) rArr tau_(C) = tau_(L) = (1)/(In 2)`
Solve to get `t = 1 s`