In the circuit shows in Fig. the battery has negligible internal resistance. Show that the current in the circuit through the battery rises instanlty to its steady state value `E//R` when the switch is closed, provided that resistance `R` is `sqrt(L//C)`.

Explanation

Let the currents through inductive branch and capacitive branch be `I_(L)` and `I_(C)`, respectively. Then the current through the battery, from `KCL`, is `I = I_(L) + I_(C)`.
Since the battery is connected in parallel to the `RL` and `RC` branches of the circuit, the current in the `RL` branch is unaffected by the presence of the `RC` branch, so
`I_(L) = (E)/(R) [1 -e^(-(R//L)t)]`and `I_(C) = (E)/(R) e^(-(t//RC))`
Hence, the current through battery is `I = (E)/(R) [1 -e^(-(R//L)t) + e^(-(t//RC))]`
If the current has to reach its final value `E//R` instananeously, the exponitial terms must cancel out, i.e.,
`e^(-(t//tauL)) = e^(-(t//tauC))`, which is possible if `tau_(L) = tau_(C)`.
`L//R = RC,R = sqrt((L//C))`
For at `t gt 0`, this is the desired result.