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 $$I_L=dfrac{E}{R}[1-e^{-(R/L)t}]$$ and $$ I_C=dfrac{E}{R}e^{-(t/RC)}$$ Hence,the current through battery is $$I=dfrac{E}{R}[1-e^{-(R/L)t+e^{-(t/RC)}}]$$ If the current has to reach its final value E/R instantaneously, the exponential terms must cancel out, i.e $$e^{-t/ au_L}$$,which is possible if $$ au_L= au_C$$ $$L/R=RC,R=sqrt{(L/C)}$$ For all t > 0 this is the desired result.