Changing the power factor of a circuit when connected to transformer
This is a pictorial answer which may help your thinking.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. (a) 3 Ω resistor only. (b) 3 Ω resistor and 3 Ω inductance.

In Figure 1b we can see that the impedance of the inductor is at right angles to that of the resistor. Since the impedances are equal in magnitude we can show by Pythagoras Theorem that the combined impedance of the two elements has a magnitude of \$ \sqrt 2 R \$.

From \$ I = \frac {V}{Z} = \frac {V}{\sqrt{2}R} \$ we can see that the current is \$ \frac {1}{\sqrt{2}} \$ (about 70%) of that with the resistor on its own.

So, for your 48 V supply into 3 Ω we would expect 16 A as you calculated. For 3 - j3 Ω we would expect \$ I = \frac {16}{\sqrt{2}} = 11.3 \; A \$. (Someone correct me if I have the sign wrong.)