At $t =0$, current through inductor is zero,
hence $R _{ eq }=(5+1) \|(5+4)=\frac{18}{5}$
$i _{1}=\frac{ E }{18 / 5}=\frac{5 E }{18}$
At $t=\infty,$ inductor becomes a simple wire and now the circuit will be as shown in figure
hence $R _{ eq }=(5 \| 5)+(4 \| 1)=\frac{33}{10} ; \quad(\| \Rightarrow $ parallel $)$
$i _{2}=\frac{ E }{33 / 10}=\frac{10 E }{33} $
