Physics - EMI/AC Question with Solution | TestHub

At $t=0$, current through inductor is zero,

hence $R_{\mathrm{eq}}=\left(5+1\right)\Vert \left(5+4\right)=\frac{18}{5}$

$i_1=\frac{E}{18/5}=\frac{5E}{18}$

At $t=\infty ,$ inductor becomes a simple wire and now the circuit will be as shown in figure

hence $R_{\mathrm{eq}}=\left(5\parallel 5\right)+\left(4\parallel 1\right)=\frac{33}{10};$ ($\parallel \Rightarrow$ parallel)

$i_2=\frac{E}{33/10}=\frac{10E}{33}$