jmc

Let $a=\sqrt{t-3}.$ Then $a^2=t-3,$ so $t=a^2+3.$ Then $$\frac{t}{\sqrt{t-3}}=\frac{a^2+3}{a}=a+\frac{3}{a}.$$By AM-GM, $$a+\frac{3}{a}\ge 2\sqrt{3}.$$Equality occurs when $a=\sqrt{3},$ or $t=6,$ so the minimum value is $\boxed{2\sqrt{3}}.$