# Amps kills but not USB?

What am I missing here? I know USBs don't kill you so I know I've got something wrong.

Google tells me that around 0.1A has a really good chance of killing you; and everything I read says it's the amps that kill you not the volts. So I know van der grafs and static shocks won't kill you.

My USB charger puts out 5v DC at 2.1A. But it aint going to kill me.

Things I think I know (dangerous - I know)
1. So I can see the volts are much lower, but if 50,000 volts won't kill you from a van der graf then I don't think the difference between 5v and 110v is the key

2. I'm missing resistance. I don't really know what "5V at 2.1A" means. I though I did, but if V = IR and I put the positive and negative ends of the USB together so resistance was close to zero; 5 = 2 x 0 ??? I got that wrong somewhere!

3. Google tells me that the body has a resistance of anywhere between 100,000 ohms and 100 ohms (if you are wet). It also tells me "For currents above 10 milliamps, muscular contractions are so strong that the victim cannot let go of the wire that is shocking him. At values as low as 20 milliamps, breathing becomes labored, finally ceasing completely even at values below 75 milliamps"

So if I stick that in the V=IR equation, Amps = 5/100 = 0.05; which suggests my USB could kill me.

What am I missing?

Related Other Physics Topics News on Phys.org
nsaspook

Ignore the 'Fux' stuff for some good information.

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DrClaude, nasu, russ_watters and 1 other person
2. I'm missing resistance. I don't really know what "5V at 2.1A" means. I though I did, but if V = IR and I put the positive and negative ends of the USB together so resistance was close to zero; 5 = 2 x 0 ??? I got that wrong somewhere!
Basically all power sources you can buy separately, are voltage sources. A voltage source will always produce the rated voltage, but the current rating is only a maximum. As long as the current rating is not exceeded, the current will be determined by the voltage of the voltage source and the resistance of the load.
If the resistance of the load is too low, the current rating will be exceeded and the following may then happen.
- the voltage source shuts down because of a fuse or an electronic safety circuit.
- the voltage is lowered.
- the voltage source may get damaged.

Current sources also exist, but normally not as separate power supplies, but as a part of a circuit.
A current source will produce a constant current, whatever the load is, up to a maximum voltage.

davenn
Gold Member
2019 Award
What am I missing here? I know USBs don't kill you so I know I've got something wrong.

Google tells me that around 0.1A has a really good chance of killing you; and everything I read says it's the amps that kill you not the volts. So I know van der grafs and static shocks won't kill you.

My USB charger puts out 5v DC at 2.1A. But it aint going to kill me.

Things I think I know (dangerous - I know)
1. So I can see the volts are much lower, but if 50,000 volts won't kill you from a van der graf then I don't think the difference between 5v and 110v is the key

2. I'm missing resistance. I don't really know what "5V at 2.1A" means. I though I did, but if V = IR and I put the positive and negative ends of the USB together so resistance was close to zero; 5 = 2 x 0 ??? I got that wrong somewhere!

3. Google tells me that the body has a resistance of anywhere between 100,000 ohms and 100 ohms (if you are wet). It also tells me "For currents above 10 milliamps, muscular contractions are so strong that the victim cannot let go of the wire that is shocking him. At values as low as 20 milliamps, breathing becomes labored, finally ceasing completely even at values below 75 milliamps"

So if I stick that in the V=IR equation, Amps = 5/100 = 0.05; which suggests my USB could kill me.

What am I missing?
you are missing the fact that the current needs a voltage to "push" it through a given resistance

My USB charger puts out 5v DC at 2.1A. But it aint going to kill me.
that's right because the 5V isn't enough to overcome your skin resistance ... up that to 50 or more volts and then you will start feeling the current flowing through you

Things I think I know (dangerous - I know)
1. So I can see the volts are much lower, but if 50,000 volts won't kill you from a van der graf then I don't think the difference between 5v and 110v is the key

tho the voltage of a VDG generator is very high, many 1000's of volts, the current is extremely low, micro to a few milliAmps

2. I'm missing resistance. I don't really know what "5V at 2.1A" means. I though I did, but if V = IR and I put the positive and negative ends of the USB together so resistance was close to zero; 5 = 2 x 0 ??? I got that wrong somewhere!
shorting the USB supply lines is no different to shorting battery terminals etc ... minimum resistance will = maximum current flow ( limited by what the supply can physically provide 1A 10A 100A whatever the supply capability

3. Google tells me that the body has a resistance of anywhere between 100,000 ohms and 100 ohms (if you are wet). It also tells me "For currents above 10 milliamps, muscular contractions are so strong that the victim cannot let go of the wire that is shocking him. At values as low as 20 milliamps, breathing becomes labored, finally ceasing completely even at values below 75 milliamps"

Yes dry skin is quite high resistance and it needs considerable voltage to "push" some current through that resistance
again .. 50V up can start doing that. 115V or 240V AC mains --- no problem at all and electrocution can occur

So if I stick that in the V=IR equation, Amps = 5/100 = 0.05; which suggests my USB could kill me.
NO, for all the reasons I have already given ( unless you put the electrodes directly across the heart you might well get fibrillation... ie... no skin or other body resistance in the way)

Dave

Thank you everyone - that's made so much sense. Loved the video - watched almost every video on his channel.

nsaspook
Thank you everyone - that's made so much sense. Loved the video - watched almost every video on his channel.
He's a great teacher.
He is also a professional engineer showing the dangers, don't do this at home.

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Hello guys. I have a question regarding this topic. A power source has a label with both the output voltage and current. How can I know what's the maximum current it can give if it's not the one in the label? For instance, if I got a zero resistance load, how can I know which current is gonna flow through it?
Thank you

phinds
Gold Member
2019 Award
Hello guys. I have a question regarding this topic. A power source has a label with both the output voltage and current. How can I know what's the maximum current it can give if it's not the one in the label? For instance, if I got a zero resistance load, how can I know which current is gonna flow through it?
Thank you
The listed current is the maximum current that the source is able to supply. If you short it, as you said, it will overheat and likely be damaged unless it has a fuse in it.

So if my source's label says its output current is 2 Amps then I'll get those two Amps through the device whenever I short its terminals? I ask this because mathematically you would get infinite amps when you get any voltage divided by zero resistance.

phinds
Gold Member
2019 Award
So if my source's label says its output current is 2 Amps then I'll get those two Amps through the device whenever I short its terminals? I ask this because mathematically you would get infinite amps when you get any voltage divided by zero resistance.
First of all, you don't HAVE zero resistance, you just have a very low resistance. Second, maybe, maybe not. You may not get the two amps at all other than for a very brief time before the fuse blows or in the absence of a fuse, the supply stops working at all properly because you are preventing it from maintain the output voltage for which it is designed.

I think you are getting hung up on the correct definitions surrounding ideal sources ideal shorts and forgetting that those ideal sources and ideal shorts don't exist in the real world.

There is no such thing as zero resistance unless you are talking about superconductors, and even those are not quite zero resistance , just very close to it.
If you short the power supply at room temperature with say copper wire, there still will be a small but significant resistance.
The result will be a very hot and possibly melted wire, or instead the power supply will get damaged unless it has some kind of overload protection built in.

Ok, I do understand the security measures to avoid damaging my source. I know some of them are fused and if not, they can be really harmed because of high currents and may remain useless forever. I know there's nothing with zero resistance. I'm not shorting any terminals, I promess. My question is: is the output current written on my source's label the maximum the device can give?
I understand it may not reach the exact value. I know for sure things in practice aren't ideal, I referred to the to the two Amps as the maximum current.

When you see a power supply rated at 'x' amps, this generally means that the device is guaranteed to provide that amount of current safely.
Most modern power supplies do in fact have circuitry built in, or a fuse at least, which prevents damage if a short circuit happens and the device operating is outside of it's safe design.
The more sophisticated (= more expensive) kind of power supply have circuitry using power transistor which strictly limits the amount of current that can be delivered.

Oh ok, got it got it. So for instance:

Output: 3,75 V/340 mA. It means the device can provide 340 mA without any risk of burning/melting etc.?

russ_watters
Mentor
My question is: is the output current written on my source's label the maximum the device can give?
No. The actual maximum it can give will depend on things like ambient temperature and ventilation, which can prevent overheating. Not to mention the safety factor the power supply was designed for and how long you are drawing the current for. You might draw 5 amps for a few seconds and it won't have time to burn-up. Or you might draw 2.5 amps for 5 minutes and it will. It is tough to know for sure.

Bottom line, the 2 amps is a rating, which means it is guaranteed to be capable of 2 amps under a certain set of conditions.
Output: 3,75 V/340 mA. It means the device can provide 340 mA without any risk of burning/melting etc.?
Assuming it isn't being used in some other unusual way, yes (in an oven, under water, etc.).

I know this is a dumb question but is the rating on the label? Like the picture above for instance?

Ah ok, didn't see your last message below haha. Thank you guys! It's clear now!