For a strongly regular simple undirected graph to be disconnected, does each of its component has to be complete (aka $\mu=0$)?
My reasoning is that two vertices disconnected components have 0 common neighbours, thus that $\mu$ has to be zero. Then by $${\displaystyle (v-k-1)\mu =k(k-\lambda -1)}$$ either $k=0$ (disjoint vertices) or $\lambda =k-1$. The second case implies for a vertex $u$, a neighbour $v$ can only connect to $u$ and all other neighbours of $u$, which seems to me that it can only be a clique of size $k+1$.
Edit: The proof was found in the book Algebraic Graph Theoryproof.
