Question

How long after closing the switch will the current through the inductor reach one-half of its maximum value?

Answer 1

After one half-life
$t=t_{1/2}=\left(\ln 2\right){\tau }_L=0.693\frac{L}{R}=\frac{\left(0.693\right)(1.25\times {10}^{-3})}{50}$
$=1.73\times {10}^{-5}s$