What happens if a motor draws more amps than a battery can provide?
I'll answer your first scenario only, because I think it's more interesting. Let's dissect:
the battery I'm using can only provide 20 amps
But what that really means? What would happen if you try to pull 21 amps? If it is a smart battery, some sort of overcurrent protection might kick in. If yes, then you have the answer - the motor will just not run.
OK, let's say it's a plain chemical battery — no fuses, no BMS. The 20 amps figure likely is printed on the battery, this is its rated current. It's important to realize that
- Ratings like that have certain requirements to be met, e.g. temperature, state-of-charge, etc.
- Such ratings are not hard limits, they can be exceeded given the right circumstances. But you need to know what you are doing.
By exceeding the rated current you'd run into some issues, which the manufacturers tried to guard you from. Likely the battery will heat more (power lost through internal resistance is proportional to the square of the current. Your example is at 125% of rated current, so heating power is 156% of the nominal that was deemed acceptable by the designers). Will the battery tolerate it is not clear cut. Heating takes some time (the battery has thermal inertia). For a short time — e.g. a minute — the heat rise is likely acceptable. Or you may compensate the extra heat by supplying ample cooling (a water cooling jacket if you really insist).
Also, consider that a battery — even at full rated current — still needs to be a decent voltage source, so its voltage cannot sag too deeply. Let's imagine a specific example: it's a 12V battery, and the designers decided 5% voltage sag at rated current is good. 5% of 12V is 0.6V; this is at 20 amps, so 0.03 ohms of internal resistance. If you were to short the battery terminals with a heavy copper bar, that 12V/0.03 ohms means 400 amps will flow (sadly the batteries don't behave that predictably, but the figures here aren't too wrong, so let's stick with them). This partially answers your #2 scenario - the battery will likely be able to start your motor.
I think the biggest problem in your #1 scenario is actually precisely with the starting current. By using an undersized battery you increase the chance that the battery could be incapable of supplying the starting current. If the motor has any load attached, it may fail to turn (just like how a car may not start due to its battery being too old). This would lead to a pretty bad scenario, in which the motor does not turn, but consumes a ton of current. If that situation is left unchecked, it can get really ugly. The possible outcomes I can think of:
- the mildest one. Everything gets hot, but no fire or melting, the battery just depletes very quickly. As we assumed it has no protection, it gets depleted down to 0V. For most chemistries (and especially Li-Ion) the battery will be ruined, by a chemical process which happens inside it below certain voltages. The motor in this case is unaffected.
- the motor windings get so hot that their insulation melt, or the windings break open. In this case the motor is ruined.
- the battery may get so hot that it bursts into flames or explodes. Pretty much worst case scenario.
So is it potentially dangerous? Yes. Could the battery be damaged? Yes.
Both scenarios are potentially hazardous. In the first case, the internal resistance in the battery will probably reduce the current sufficiently to prevent damage to the motor, but the battery may be damaged. In the second case, the internal battery resistance will not reduce the motor current sufficiently tp prevent damage to the motor. Some small motors may not draw so much current when they start, but those motors are not very efficient because their internal resistance is wasting power during normal operation.