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