#### Answer

(a) $I = 0$
(b) $I = 2.0~A$

#### Work Step by Step

(a) We can find the current immediately after the switch closes:
$I(t) = \frac{V}{R}(1-e^{-Rt/L})$
$I(t) = \frac{V}{R}(1-e^0)$
$I(t) = \frac{V}{R}(1-1)$
$I(t) = 0$
(b) We can find the current after the switch has been closed for a long time:
$I(t) = \frac{V}{R}(1-e^{-Rt/L})$
$I(t) = \frac{V}{R}(1-\frac{1}{e^{Rt/L}})$
$I(t) = \frac{V}{R}(1-0)$
$I(t) = \frac{V}{R}$
$I(t) = \frac{10~V}{5~\Omega}$
$I(t) = 2.0~A$