I'm going to use the transformer as an example.
A transformer will have leakage inductance i.e. "spare" inductance in each winding that isn't coupling through the core. This leakage inductance is a series component and can drop voltage across it when load currents are taken. This is a significant reason why regulation is not perfectly 100% in a transformer.
So, if the net leakage inductance is X and the load is (say) 50X and reactive, a simple potential divider is formed and the secondary voltage is: -
\$\dfrac{50}{50+1}\$ = 98%
But, if the same load current were taken with a resistor equivalent in impedance to 50X then the secondary voltage would be: -
\$\dfrac{50}{\sqrt{50^2+1^2}}\$ = 99.98%
If the copper losses were as significant as the leakage inductance in their effect, regulation would be the same for inductive and resistive loads.