$\triangle ABC$的内角$A$,$B$,$C$的对边分别为$a$,$b$,$c$,若$a\cos B=b\sin A$,$C=\dfrac{\pi}{3}$,$c=\dfrac{3}{2}$,$b=$（ ）
$\text{A.}$ $\dfrac{\sqrt{6}}{2}$
$\text{B.}$ $\dfrac{3\sqrt{2}}{2}$
$\text{C.}$ $\dfrac{3\sqrt{2}+\sqrt{6}}{4}$
$\text{D.}$ $\dfrac{3\sqrt{3}-3}{2}$