Every transformer has impedance in its own windings, and the nameplate states it as percent impedance, $\%Z$. Read the definition carefully, because everything else follows from it: $\%Z$ is the percentage of rated voltage you must apply to the primary, with the secondary shorted, to push RATED current through. A 5% unit needs 5% of rated volts to make 100% of rated amps.

At working loads that impedance costs you a few volts: $V_d = \dfrac{\%Z}{100}\cdot\dfrac{S_L}{S_R}\cdot V_R$, where $V_d$ is the drop inside the windings in volts, $S_L$ is the actual load and $S_R$ the transformer's rating (both kVA, so the ratio is bare), and $V_R$ is the rated secondary voltage. The ratio $S_L/S_R$ is simply how much of the nameplate you are using — at full load it is 1, and the drop is the full $\%Z$ of rated volts.

Measure that same droop instead of predicting it, and you have voltage regulation: $\%VR = \dfrac{100(V_{nl} - V_{fl})}{V_{fl}}$, where $V_{nl}$ is the terminal voltage with the load off and $V_{fl}$ the voltage at rated load, both in volts. The base is the FULL-LOAD reading — the loaded terminal, the one your equipment actually lives on.

Now short the secondary and read the same nameplate the other way. If $\%Z$ of rated voltage drives rated current, then full rated voltage drives $\dfrac{100}{\%Z}$ times rated current: $I_{SC} = \dfrac{100\, I_{FL}}{\%Z}$. A 5% transformer can deliver twenty times its rated current into a bolted fault. That is the number a panel's interrupting rating has to beat, and note the direction it runs — a LOWER impedance means a HIGHER fault current, which is why the stiffest transformers are the ones that need the most expensive gear behind them.