I have a mini motor that's rated at 3 volts. Does this mean that the 3 is the number of volts that will produce the desired amperage for the motor? Could I technically make it work by plugging it into a 10,000 volt circuit and add resistance until it's at the same amperage as it would be for 3 volts?
No, motors don't work that way.
Generally speaking, the voltage that you apply to a motor determines the speed at which it runs, but the current that it draws depends on the mechanical load (torque) that it is driving.
If you applied 10,000 V to a 3-V motor, even through a resistor, it would try to spin far too fast, probably destroying itself.
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.
$$P = I \cdot V = I^2R$$
For the 10 Ohm resistor, we would dissipate \$P = 1^2 \cdot 10 = 10 \;watts\$. Where in the 10,000 Ohm resistor we would dissipate \$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.
Short answer: not in practice. Long answer:
Effects of voltage and current
Voltage and current ratings generally protect against independent failure mechanisms, in most cases respectively insulation breakdown (especially for semiconductors), and damage due to excessive Joule heating.
Some components are more flexible than others around their voltage and current rating, up to what is called "absolute maximum" ratings. DC motors are of that kind. Increase the voltage and the power of the motor will increase (meaning, either more speed for a given torque, or more torque for a given speed) - that is, assuming the motor can draw the current it needs. This means that the current is only limited by limiting the voltage across the motor (this is also true for other devices, it can be seen on current limiting power supplies).
You could run a 3V motor at probably 12V with a giant heatsink to make sure the heat gets away faster than it heats up the wires (otherwise they'll melt); however you cannot run such a motor at 10,000V because 1) the insulation will break, increasing drastically the current drawn by the motor 2) whether the insulation broke or not, the current will be so high the wires will melt. Note that depending on the shaft alignment and the bearings, too high a voltage may generate too much vibration and destroy the motor before the heat melts the coil.
Limiting via power resistor
Now, you suggest adding a device to drop the voltage from 10000V to 3V. That would make the motor happy: as long as it gets 3V across it, it will draw current up to its stall current and hopefully it has been designed to cope with that current. However, you are dropping 9997V. If any current is drawn through the device that drops that voltage, this is power that has to be dissipated somehow. 9997*I, even if I is as low as 50mA is 500W... To put that into perspective, 5W aluminium clad resistors are about as big as a finger. Note that power resistors will only drop a fixed voltage at a given current: change the mechanical load on the motor and the voltage across the motor will change, this will lead to unstability... Assuming a 200k resistor for a design current of 50mA, +/-15uA will have a +/-3V effect on the motor! Not to mention that 500W are wasted for 0.15W used.
Limiting via buck converter
You would need a converter with a better efficiency, like buck DC DC converters. Theoretically, they do not drop the voltage themselves, but switch between full voltage and no voltage to achieve the required voltage in average (in this case, the motor inductance filters the current, which filters the voltage via the motor resistance). However, this means switching ON 0.03% of the time - the motor inductance will give you how long full voltage can be applied for before the current is too high, and that will not be long. Therefore 0.03% of that time will be very short, chances are no converters will be able to. The vibrations of the motor due to the high current ripple would probably ruin the motor and its load anyway.
Whether a power resistor or a switching converter is considered, both will have to have 10,000V insulation, which means your product will be huge to respect the creepage distances and insulators thicknesses.
Unless you mean 10,000VAC from the supply grid to 3VDC? In that case, you will have to use a step down transformer and then a rectifier, but I genuinely hope that you know what you are doing.
Depends on what your goal is. If you want relatively constant speed with varying loads, then a constant voltage source (or even slightly negative series resistance) is best. The motor will be happy unless you load the shaft too heavily, provided you give it no more than the rated voltage. Too heavy a load and it will overheat.
If you want a constant torque with varying loads, then a constant current source is best. The motor will be happy unless you load the shaft too lightly, provided you give it no more than the rated run current. Too light a load and the RPM will wind up and it could damage itself (the motor voltage will exceed the rated voltage under those conditions). A large series resistor behaves more like a constant current source. A >10K source resistance source is easily simulated electronically without the very high voltage and insane resistor.
In reference to your original question, in neither case are you actually exceeding the rated voltage at the motor terminals (the only thing the motor cares about).
Motors can be damaged by exceeding any of the following maxima:
Mechanical force on components thereof (may cause instant failure)
Amount of magnetic flux (may cause instant failure)
Rotational speed, vibration, etc. (may greatly accelerate wear, or in some cases generate forces resulting in instant failure)
Amount of heat generation in the windings (intermittent operation may allow motor to tolerate more power (heat/second) than sustained)
Amount of heat generated in the bearings (intermittent operation may allow motor to tolerate power (heat/second) than sustained).
It may be possible for a "3-volt" motor to be operated safely at higher voltages under some conditions if those conditions reduce the stress factor which was limited operation to 3 volts and do not raise any other stress factor enough to cause trouble. What's important is not keeping the motor voltage below the rating, but ensuring that all of the other stress factors stay within their corresponding limits.
A motor, electrically speaking, is somewhat like an inductor (ignoring any motor loading effects). Inductors resist any change in current so when you apply a voltage to the inductor (read motor), the current will initially be zero but rise exponentially (see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indtra.html). This results in the full 10,000 volts (assuming you could apply this instantaneously) being applied across the motor windings for a short period of time regardless of what resistor value you use. The high voltage would likely breakdown the winding insulation and result in short circuits across the motor windings.
Practically speaking, a circuit using 10,000 V and a resistor to drive a 3 V motor would be hugely inefficient. The motor current will depend on how much the motor is being loaded. In other words if you selected a resistor that would give 3 V across the motor at no motor load, then added some mechanical load, the voltage across the motor would drop because its trying to draw more current. This would slow the motor down and it would try to draw more current reducing the voltage further and eventually stalling the motor at much less than its rated torque.