# How does higher amperes NOT affect a DC motor for example?

I'm shocked that no one has asked this question on the whole internet. So the question is mindblowing and frustrating.

If I have a 12V DC supply with say 500 milliampere, I can for example short circuit the supply without any issue.

But if I use my car battery with 12V that has incredibly high amount of amperes and short circuit it, it would generate an insane amount of sparks and could cause fire. I get why this is happening, so here on everything is clear in my head. But these two points are important to keep in mind for the question itself.

But now for the mindblowing question.

How can a 12V DC motor run just as well on both the weak power supply and the car battery that's incredibly powerful? I get that "a motor/component only draws as much current as it needs" and "ampere is the size of the water pipe" concept. But, whatever you put on the circuit. once that something is connected you are essentially going from open to closed circuit, which is similar to a short circuit with a component/resistance/DC motor in middle, so if we have the car battery with super high amps and connect it with the DC motor in the circuit why doesn't it burn up everything from the battery terminals to the dc motor?

Essentially we ARE short circuiting it with just a resistance (DC motor) in the middle.

• Explain why you think there is a short circuit (there isn't of course). Commented Sep 3, 2020 at 9:06
• "Im shocked that noone has asked this question in the whole internet." This should be your first indication that something is wrong with your reasoning and that there is some detail you have overlooked. Commented Sep 3, 2020 at 9:30
• Does it also shock you that a small engine can run on both a small and large gas tank? Because it's the same thing. Commented Sep 3, 2020 at 13:36
• A short circuit means there is nothing in the middle except a wire. If there's a motor in the middle then it's not a short circuit. Learn about Ohm's Law. Commented Sep 3, 2020 at 14:23
• Because the motor is not a short circuit (though it gets pretty close if you hold it stalled)
– user16324
Commented Sep 3, 2020 at 17:42

## 5 Answers

Despite your protest that you understand that "a motor/component only draws as much current as it needs," you don't understand that "a motor/component only draws as much current as it needs."

A motor isn't simply a wire. A wire connected to the terminals of a car battery will do what you describe: showers of sparks, wire gets hot, melts, burns, etc.

That a battery can supply insane amounts of current doesn't mean that it must supply insane amounts of current.

I'm going to skip motors because they are complicated to explain.

Light bulbs are conceptually easier, so I'll use those.

Take a headlight from your car. If you connect it to a car battery, something like 5 amperes of current will flow through the headlight. Headlights operate on 12 volts, and consume around 60 watts of power - that's about 5 amperes of current at 12 volts.

Now take a tail light from your car. It also operates on 12 volts. If you connect it to your car battery, it will draw about 0.42 amperes of current. Tail lights are rated at about 5 watts, which is about 0.42 amperes at 12 volts.

One light draws more current than the other, and neither explodes when connected to your car battery.

Why?

Take a close look at the filament in each bulb.

Here's an H3 headlight:

Here's a 5 watt tail light:

The headlight has a short, thick wire as a filament. The 5 watt tail light has a long, skinny wire as a filament.

Each filament has a property called "resistance."

A short, thick wire has less resistance than a long skinny wire.

Different metals have different resistances. Given copper wire and a tungsten wire of the same length and diameter, the copper wire will have lower resistance.

For a given voltage and a given resistance, you can calculate the current:

$$I =\frac {E}{R}$$

Where:

• $$\I\$$ is current in amperes
• $$\E\$$ is voltage in volts
• $$\R\$$ is resistance in ohms

This is known as "Ohm's law." This is why neither the headlight or the taillight bulbs draw all of the thousand (!) or so amperes that a car battery can deliver.

Take the headlight. Given that it draws around 5 amperes from a 12 volt battery, Ohm's law says it has a resistance of about 2.4 ohms.

For the 5 watt tail light,the resistance works out to about 29 ohms.

For a piece of copper wire, the resistance is far lower. Copper is one of the best conducting materials available.

If you connect the terminals of your car battery with a 2 foot long piece of 22AWG wire, the resistance would be about 0.032 ohms, and the current would be about 375 amperes. You'd get the "sparking, melting, and catching fire" that you expected.

Your assumption that connecting a component to a battery is the same as a short circuit is wrong.

Components have resistance (and other properties) that make them different from a short circuit through a piece of wire.

For an analogy from real life, consider a garden hose without a nozzle.

It can deliver maybe 12 gallons (45 liters) per minute. That's its "short circuit current" - like the maximum current a car battery can deliver.

If you bend the hose over so that it is kinked and partially squeezed shut, then you won't get that maximum amount of water per minute, you'll get something less. That's the resistance of your headlight. If you squeeze the hose really hard, the water will just dribble out - that's the resistance of the taillight.

More kink = less water flow.

More resistance = less current flow.

To take the analogy further, if you connect the garden hose to a high pressure water main pipe (like the ones feeding all the houses in your neighborhood,) then the hose will thrash around and slap you, and probably burst if it gets kinked or anything blocks its outlet. That's like short circuiting your car battery.

• I like the last analogy about high pressure water main pipe and trying to stop it with a small water hose, isn't that what we are essentially doing when connecting a car battery with alot of amperes into a tiny motor/lamp? Shouldn't it just burst through the motor/lamp the same way high pressure main would burst through our small garden hose since there is sooo much current and very little resistance? Commented Sep 3, 2020 at 20:48
• @CoffeDev These are good questions to be asking. The water pressure is analogous to voltage and the water flow rate is analogous to the current. A car battery is low voltage (low water pressure) and capable of high current (high water flow rate). However it doesn't matter that the car battery is capable of delivering high current it lacks the pressure to "push" high current through. Something else to note is the small hose would be a high resistance as it takes higher pressures for a given flow rate. Commented Sep 8, 2020 at 13:50
• @Clipboard_Waving_Enginerd Thanks alot! now i fully understand how it works! :D Commented Sep 8, 2020 at 16:39

Because the motor has an internal voltage source that is in anti-series to the external source as it runs. It's called counter-electromotive force and is a result of the magnetic fields in a running motor.

simulate this circuit – Schematic created using CircuitLab

In result, there's not 12V across the winding resistance but less than 1V. The actual number depends on the speed of the rotor. That's why you shouldn't power a motor that is mechanically locked. It cannot build up that counter voltage and will burn out quickly.

If you put external torque on the running rotor, the speed decreases and so does the counter-voltage, which leads to more current running through the rotor which creates a counter-torque. As soon external torque and internal counter-torque have balanced each other again, there's a new equilibrium at a lower running speed.

So, why does the very same motor burn out then if you lock it mechanically and connect it to a car battery, but stays intact on the 500mA power supply?

Because it's e.g. a windscreen wiper motor designed for a current of 1.5A. If it's hold at standstill on the 500mA power supply, it will likely want to draw 20A. But it can't. Because the power supply can't deliver. And it can stand the 500mA running through it at standstill as it's designed for 1.5A in the first place.

• In reality, there is no voltage source, and no winding resistance. There is just a winding... Commented Sep 3, 2020 at 14:25
• It's pretty obvious the OP wanted to know about the equivalent circuit. He even brought up the winding resistance himself. Commented Sep 3, 2020 at 16:33

In fact, the circuit DOES pull only as much current as it wants. The question is in load resistance. When you connect your motor, you only draw a certain amount of current, because motor has resistance. So the current is 12V/Rload=Iload (or any other load). So your current very much finite and limited. When you just short + with -, you connect close to zero resistance, which causes very high current.

I don't know much about DC motors specifically, never needed to work with them, but @Janka gave some explanation about it (btw interesting for me too when it comes to burning out not moving motors, thx).

As for power supply and battery, first, the battery may have no short circuit protection, while power supplies can have chips that cut the power in case of short circuit or can somehow take it. Second of all, the more current you pull from the battery, the more its voltage falls immediately (there is an internal resistance, see it as a series resistor on the + terminal of the battery, but inside it, so you lose voltage there proportionally to current, roughly).

Inductance can also cause sparks and/or high voltage spikes (60V voltage spike in 5V circuit is real) when connecting/disconnecting inductive load, this is usually compensated with flyback diode (google this name, nothing hard about it)

Closed circuit is any functioning circuit, it's the resistance that matters, it limits the current.

I'm shocked that no one has asked this question on the whole internet. So the question is mindblowing and frustrating.

There have been very many questions on this site by people who don't understand that a load only draws the current required, not whatever the supply limit is.

If I have a 12V DC supply with say 500 milliampere, I can for example short circuit the supply without any issue.

Not necessarily. If the PSU does not have short-circuit protection it may

• burn out some components and turn off.
• blow it's internal fuse.
• catch fire.

But if I use my car battery with 12V that has incredibly high amount of amperes ...

They're credibly high. Believe the values written on the datasheet (written by credible manufacturers).

Once that something is connected you are essentially going from open to closed circuit, which is similar to a short circuit with a component/resistance/DC motor in middle,

No. A short-circuit implies that the current hasn't traveled the correct route through the whole circuit. In the situation you have described all the current will flow through the motor as intended.

... so if we have the car battery with super high amps and connect it with the DC motor in the circuit why doesn't it burn up everything from the battery terminals to the dc motor?

Because the motor has resistance which limits the current. In addition, as the motor accelerates to operational speed the back-EMF reduces the current drawn.

Essentially we ARE short circuiting it with just a resistance (DC motor) in the middle.

No we ARE NOT short-circuiting the battery. We are closing the circuit with a known load.

My standard answer to this question is, "Why when you plug in a lightbulb in your house does it not take all the power from your national grid?" Can you think of the answer?

Blockquote If I have a 12V DC supply with say 500 milliampere, I can for example short circuit the supply without any issue.

Not really. Depending on how the 12V supply was designed, it can

1. just burn
2. stop delivering any current until the short circuit situation is resolved and you replace the internal fuse
3. stop delivering any current until the short circuit situation is resolved, then resume automatically.
4. deliver exactly 500 mA and no more

Case 3 is the most common in commercial supplies. Case 1 would be a very bad and dangerous one. But maybe yours is case 4. Case 2 is what you should do when testing such stuffs: Adding a replaceable fuse between the source and your DUT (device under test).

My point is a short circuit is never a normal situation and it's only because there is a current limiting system or a fuse that it doesn't cause sparks, melting and fire.

In a car, you see sparks because the starter needs as much amperes as possible to start the engine and there isn't so much protection. If you are creating a short directly from one battery pole to the other, of course, then there is no protection at all and you see what happens.