For the given circuit the current i through the battery when the key in closed

In steady state, inductor behaves as a conducting wire.

So, equivalent circuit becomes 

$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$

$\Rightarrow R_{\text {eq }}=1 \Omega$

$\Rightarrow$ Circuit becomes

$\Rightarrow \mathrm{i}=\frac{30}{3}=10 \mathrm{~A}$