In the circuit shown in Fig. the battery has negligible internal resistance. Show that the current in the circuit through the battery rises instantly to its steady state value $E/R$ when the switch is closed, provided that the resistance $R$ is $\sqrt{L/C}$.

Let the current 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 branch; so
$I_L=\frac{E}{R}[1-e^{-(R/L)t}]$

And $I_C=\frac{E}{R}{e}^{-t/RC}$
Hence the current through the battery is
$I=\frac{E}{R}[1-e^{-(R/L)t}+e^{-t/RC}]$
If the current has to reach its final value $E/R$ instantaneously, the exponential terms must cancel out, i.e., 

$e^{-t/{\tau }_L}=e^{-t/{\tau }_C}$

which is possible if ${\tau }_L={\tau }_C$
$L/R=RC;R=\sqrt{(L/C)}$
For all $t>0$, this is the desired result.