Answer
$i_1 = 2.0~A$
Work Step by Step
When the switch is first closed, the inductor acts like a broken wire, so all the current will flow through $R_1$:
We can find $i_1$:
$i_1 = \frac{\mathscr{E}}{R_1}$
$i_1 = \frac{10~V}{5.0~\Omega}$
$i_1 = 2.0~A$
