# Given an non-deterministic finite automaton, will its determinization always have unreachable states?

Given an NFA that accepts the regular language L, will its equivalent DFA which accepts the same language L always have unreachable states. If it does, why?

No, there aren't always unreachable states. Consider the NFA with one state, $$q$$, and no transitions. (It accepts the language $$\{\epsilon\}$$ if $$q$$ is accepting, and accepts $$\emptyset$$, otherwise.)
If you determinize this automaton, you end up with a two-state DFA with a transition from the start state $$\{q\}$$ to the other state, $$\emptyset$$, so both states are reachable.
The powerset construction shows that for every NFA with $$n$$ states there is an equivalent DFA with $$2^n$$ states. Every example in which this is tight – that is, the minimal DFA has $$2^n$$ states – is a counterexample to your claim. You can find such an example, for arbitrary $$n$$ and over a binary alphabet, in this answer.