For the Jacobian criterion it is fundamental to consider $I(V)$ and not just an ideal $I$ such that $V = V(I)$. For instance, take $I = ((x-1)^2)$. Then $V(I) = \{p\}\subset\mathbb{A}^1$. We have $rank (J_p(I)) = 0$. However, $dim(V) = 0\neq dim(\mathbb{A}^1)-rank (J_p(I))$.

For the Jacobian criterion it is fundamental to consider $I(V)$ and not just an ideal $I$ such that $V = V(I)$. For instance, take $I = ((x-1)^2)$. Then $V(I) = \{p\}\subset\mathbb{A}^1$. We have $rank (J_p(I)) = 0$. However, $dim(V) = 0\neq dim(\mathbb{A}^1)-rank (J_p(I))$.

The answer is negative also if the polynomials generating $I$ are irreducible. For instance, consider the quadric surface $Q$ given by $$ \det \left(\begin{matrix} x & y \\ y & z \end{matrix}\right) =0 $$ and the cubic surface $S$ given by $$ \det \left(\begin{matrix} x & y & z \\ y & z & w \\ z & w & x \end{matrix}\right) =0 $$ On a general point $p = [u^3:u^2v:uv^2:v^3]\in Y$ we have $Jac(Q)(p) = (uv^2,-2u^2v,u^3,0)$ and $Jac(S)(p) = (v^2(u^4-v^4),-2uv(u^4-v^4),u^2(u^4-v^4),0)$. Therefore, $\mathbb{T}_pQ = \mathbb{T}_pS$ for a general point $p\in Y$. This means that, if $Y\subset\mathbb{P}^3$ is the twisted cubic then $Q\cap S = Y$ set-theoretically. However, scheme-theoretically $Q$ and $S$ cut $Y$ twice.

Now, take the affine chart $w\neq 0$. Then, if $I =(xz-y^2,x^2z-x-y^2x+2yz-z^3)$ we have that $V(I)$ is the affine twisted cubic $C$. However, $rank(Jac_{(0,0,0)}(I)) = 1$. Then $$dim(\mathbb{A}^3)- rank(Jac_{(0,0,0)}(I)) = 2 \neq dim(C) = 1.$$

A simpler example: take the two curves $C = \{y-x^2 = 0\}$ and $L = \{y = 0\}$ in $\mathbb{A}^2$. Note that $L$ is the tangent line of $C$ in $(0,0)$. Both $C$ and $L$ are irreducible. Now, if $I = (y-x^2,y)$ then $V(I) = \{(0,0)\}$. On the other hand the scheme defined by $I$ is non-reduced. Indeed it is the origin with multiplicity two. In this case $rank(Jac_{(0,0)}(I)) = 1$, and $$dim(\mathbb{A}^2)- rank(Jac_{(0,0,0)}(I)) = 1 \neq dim(V(I)) = 0.$$