I have always wondered about the relations between the application and the battery source. Please forgive me if my question sounds so simple.
Let's say for example:
I have a large battery 12v, 350amp as a source. I have a lightbulb that operates at 12v and draw 2amp.
Will my bulb pop because the amp from the source is too high or will it be okay?
If I understand this correctly, the 350amp from the source is like a max "threshold" that an application cannot draw past, and since the lightbulb is only drawing 2amps, it should be okay.

 

The load only draws/consumes the amount of current it demands (2amps), IOW the 350amp is the full capacity the battery is capable of, albeit just for a few seconds!
Max..

 

So the source isn't going to dump 350amp to the load, correct?

 

Correct.

 

The load current obeys Ohm's law, which is current = voltage / load-resistance, so it makes no difference how much current the source can supply, the current is determined by the source voltage and the load resistance only.

 

Thanks for the clarifications, everyone!
It will become a problem if the load is 5v and the source is 12v correct? In this case, the load will fry?

 

It will become a problem if the load is 5v and the source is 12v correct? In this case, the load will fry?

Or it could just saute a little!
Max.

 

Thanks for the clarifications, everyone!
It will become a problem if the load is 5v and the source is 12v correct? In this case, the load will fry?

Yes.. voltage must match.. Current rating of the supply can be higher (and for safety reasons I always recommend at least 125% higher than the maximum continuous load)

Note: this whole conversation does not apply to certain devices.. For example.. LEDs.. They need to be fed the proper amount of current or all will pass right through them and they will go up in smoke.
Typically LED circuits incorporate a resistor which limits the current through the LED or whats called a "constant current" power supply which is different from a "constant voltage" power supply which is the discussion topic here..

A constant voltage supply will maintain a constant voltage no matter the current draw.. While a constant current supply will maintain a constant current for devices like LEDs which will NOT only draw what they need..

 

If the load is 12V/350A=0.0343ohms then it will draw 350amps.

Your headline question was can a motor be burnt out. It is a more complex question with the motor as a load. When it is running at full speed and no mechanical load the back emf balances the supply and even though the battery can deliver 350A it may only deliver 1A. However if you stall the motor by applying a mechanical brake then there is no back emf and the 12V is applied to a winding that may be only 0.5ohms in which case 24A will flow from your massive battery and may well fry the motor if it is rated at 5A maximum load. A motor drive usually has a current limit circuit to prevent over current during stall conditions. So if you have a small motor and you run it from a car battery you could blow it up if it stalls.

 

It will become a problem if the load is 5v and the source is 12v correct? In this case, the load will fry?

Well, clearly, power consumption would increase by a factor of 5.76 (i.e. 144/25) -- Of course more 'complex' loads would likewise be subject to undue dielectric stresses, switch/brush arcing, etc...

Best regards
HP