Although in theory that works, it is not practical ... at all.
Let's say your load is 10 Ohms (Making this simple) and it is rated for 10V. This would yield a current of 1A. So let's say I have 10,000V available and want 1A to run through my load, I would add let's say a 10,000 Ohm resistor in series which yields approximately 1A.
Even though this work, now let's talk about power.
P = I*V = (I^2)*R$$P = I \cdot V = I^2R$$
For the 10 Ohm resistor, we would dissipate P = (1^2)*10 = 10 watts\$P = 1^2 \cdot 10 = 10 \;watts\$. Where in the 10,000 Ohm resistor we would dissipate P = (1^2)*10,000 = 10,000 watts\$P = 1^2 \cdot 10,000 = 10,000 \;watts\$.
Of course this is not desirable because our power transfer is terrible. We would like maximum power delivered to our load. Therefore we often use a transformer to step down the voltage to an appropriate level.
Of course motors are not resistors (Real and imaginary loads), but the demonstration works for this question. Motors are built with coils, behaving like an inductive load and can't be treated exactly like a resistor. Either way, it is best to deliver maximum power to your load or else you will be wasting valuable energy in your 10,000 Ohm resistor.
Hope that helps, please comment if you have further questions.
