Power factor correction of a linear inductive load is exactly the same as tuning a parallel circuit of a capacitor and inductor. You pick a capacitor that works with the value of the transformer's magnetizing inductance to satisfy this: -
60Hz = \$\dfrac{1}{2\Pi\sqrt{L_M.C}}\$
Where \$L_M\$ is the transformer's magnetizing inductance and C is the capacitor chosen to "neutralize" the current taken by the coil.
For a pure lossless inductance and capacitance, the resultant current taken from the supply is zero.
If your transformer primary indicates an inductive reactance at 60Hz of 300 ohms, the magnetizing inductance value is this divided by \$2\Pi\times 60\$ = 0.8 henries.
The capacitive reactance required is 300 ohms and this is 8.8 uF. As a sanity check: -
F = \$\dfrac{1}{2\Pi\sqrt{0.8\times 8.8\times 10^{-6}}}\$ = 59.98Hz
So my question is does inductive motors
I think you mean induction motor.
Electric motors are energy conversion devices and an ideal electric motor converts 100% of the electrical power it receives from the circuit to mechanical power - the ideal motor itself does not consume power but, rather, changes its form.
Of course, for real motors, there are loses due to e.g., winding resistance, friction, etc.
Now an inductor and a capacitor are energy storage circuit elements - they alternately store and then release energy from and to the electric circuit to which they are attached.
So, for ideal inductors and capacitors, there is no loss - on average, their associated electric power is zero. The instantaneous power associated with an ideal reactive element is alternately positive and negative.
However, for a resistance, the electric power from the attached circuit is dissipated - converted to heat. The instantaneous power is always positive.
Best Answer
It's just as easy both ways theoretically but consider "size" as the main practical limitation. A 10 uF non-polarized capacitor running on 230 V AC will have an impedance of 318 ohms at 50 Hz and you should be able to hold a couple of them in your hand (discharged of course) very easily. They will also be less than a dollar each.
The equivalent inductance is about 1 henry and will take a current of about 720 mA and, to avoid saturation problems will weigh more than a couple of kilogrammes and not that easily fit in your hand. Price probably about $10+.
But, the bottom line is that most pieces of equipment that cause non ideal power factors are motors and these are corrected by capacitors.
It's not anyone's preference, it's practicality.