This is just a partial answer, but I think the question is quite open ended...
If $K$ is a p-adic field there is only one non-zero primeideal! The totally ramified extensions are precisely the ones that come from adjoining a root of an Eisenstein polynomial. (This can be found in any standard text about local fields/alg.number theory, e.g. Neukirch)
Example: X^2-2 is Eisenstein in $2$, so $\mathbb{Q}_2(\sqrt{2})/\mathbb{Q}_2$ is totally ramified.
This can be used to construct extensions of number fields that are totally ramified over one prime, there is for example the following
Proposition: Let $K$ be a numberfield. If $P(X)\in\mathcal{O}_K$ is Eisenstein in a prime $p$ of $K$, then $P(X)$ is irreducible and $p$ is totally ramified in the extension $K(\alpha)/K$ ($\alpha$ a root of $P(X)$).
However this proposition doesn't tell you wether other primes ramify of not. Over $\mathbb{Q}$ the only other primes that can ramify are the ones that divide the discriminant of the polynomial.
If you are familiar with more advanced number theory you should have a look at Class field theory. This gives you for any number field $K$ the existence of an abelian extension $F/K$ which ramifies precisely at one prime (and much more!).