In This post you will get answer of Q.
The current (i) at time $t=0$ and $t=infty$ respectively for the given circuit is
Solution:
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} $
Solution:
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} $