Originally Posted by DavidW
Originally Posted by dasanman69
1 amp doesn’t sound like much but it is. You can have 100 gallons of water (high voltage) come out of a pipe at a slow rate (low amps, milli amps) or 20 gallons of water (low voltage) come out of the same pipe at a very fast rate (high amps, 1), which would most likely be deadly?
I understand that it's amperage that kills and it doesn't take a lot of it to kill. But voltage does matter. 1 amp at 5V will not kill you. Otherwise there would be more people being electrocuted by the mishandling of the end of a USB cable that is plugged into a 5V USB power source. But as little as .1 amp at 110V will. Voltage is not volume, it's pressure. 100 gallons of water (Amps) coming out at 100 PSI (high voltage) is more deadly than 100 gallons of water (Amps) coming out at 1 PSI (low voltage). And if the water pipe (wire) can't handle the pressure, it burst.
At least two things had to fail in this charger for someone to get electrocuted. The first is that part of the rectrifier circuit fails and sends 220 VAC into the 5V line. The second is that the device that limits the amperage at 1A at 5V failed. This device, be it a fuse, fusable link or some sort of thermister circuit that shunts the voltage away from the 5V line, all works off Wattage. (Volts X Amps) In other words, when too much heat builds up, the voltage is cut off. The voltage cut off limit should be around 5W. (5V X 1A) If 200V when through, then this device should have cut off the voltage at around .025A (25 milliamps) That's because 200V X .025A = 5W. And the voltage should have been cut off instantly as the person was lying on a cement ground, no shirt and most likely wet from sweat because he was essentially ground for the current. In other words, almost a short.
Now if this device failed, then the USB cable should have shown signs of damage as the current wasn't going to stop just because the person may have already been dead with the first few 10ths of an amp. If he was on top of the iPhone, then the current should have continued until the USB cable melted or some other plastic parts of the charger. Even at 1 amp at 200V, that's 200W. So if this device failed or wasn't present at all, then the USB cable (or other plastic parts) should show signs of damage. Now if this device was present and didn't fail, then the cable would still be intact, but the person mostly likely would have ended up with a bad shock as only about 25 milliamp got through for what should have been a very, very brief time. But for sure, lying on a cement floor, no shirt, sweaty, clutching the phone (and most likely in his left hand) and having voltage up to 220VAC , when the charger failed, didn't help in this case.
Another way this charger can fail is that some how 220VAC is sent through the negative (ground) line.(-5V) This would bypass the fuse (or other current limiting device) as most fuses are only on the positive line.(+5V). In this case the outside metal covering the pins of a USB connector is becomes hot. Which means the metal on the outside of the iPhone 4s is now hot when it's connected by the 30 pin Apple connector because the metal shield covering it's pins is now hot. (I don't think the lightning connector has a metal shield around it.) But the amount of current (amps) that will pass through in this case would have surely melted the charger or cable and the most likely parts of the iPhone itself, in a very short time.
I agree with you that dasanman69's water analogy was not quite right, and I'm impressed that you went to the trouble to try to explain it better. However, I'm afraid that you have also misunderstood the fundamental principles involved here.
Firstly, it is current that kills. Current is all that matters physiologically. That does not mean that voltage is unimportant, but only because the voltage is what drives the current, as described by Ohm's Law (V = IR for a purely resistive load). A functional USB supply will not kill you, not because it cannot supply enough current (less than 50 mA AC or 500 mA DC can cause fibrillation and death), but because the resistance of the human body (specifically due to skin resistance), even when wet, exceeds 1000 Ohms which, at 5 V will only sink 5 mA. That's DC current too, which is much less effective as noted above.
Secondly, you seem to have misunderstood how a fuse works, and its implications for circuit protection. In a simple series circuit, the conserved quantity is current. Here the water analogy works well - current = flow rate (e.g. gallons per minute) - and, just like water flowing through a set of pipes of varying diameter and length ( equivalent to different resistances), the flow rate is the same at all points in the system. Voltage, on the other hand, varies through the circuit according to the resistance of the circuit components and the current - again according to V = IR. In a simple circuit with a supply, two conductors and a load, most of the voltage drop is across the load (high resistance), with very little across the conductors (low resistance). A fuse is a simple component that is designed to fail when the current through it exceeds a pre-determined value (and technically for a pre-determined time). The fuse knows nothing about the overall supply voltage, except in the case of a total short of the load, at which point it will become the highest resistance in the circuit, nor does it know anything about the total power being dissipated elsewhere in the circuit.
Power dissipation is, in general, different in each circuit component (including the conductors) since (again in a purely resistive circuit) it is given by the product of voltage (across that component) and current (the same everywhere in the circuit). So P = VI ( and also, trivially P = I²R = V²/R by substitution), but that V is not the supply voltage, except approximately for the load in a good circuit in which the conductor resistance (and thus voltage drop) is low. Thus, in a well designed circuit, most of the power is dissipated in the intended load.
So, even when the switching voltage regulator in the power supply fails, and 220 V AC is fed into the output down the USB cable, the cable will still happily transport 1 A. It's a low resistance conductor, and it is not dissipating 220 W - that would be dissipating in whatever load (which would need to be around 220 Ohms to draw 1 A) it was feeding. Short it out, and then far more than 1 A will flow, but a human body is not a short - it's typically between 1000 Ohms (wet), which would give around 220 mA AC (fatal) and 100,000 Ohms (very dry) which would draw about 2 mA AC (not fatal).
That is why a fuse cannot protect against electrocution in any circuit designed to carry more than about 20 mA AC or 200 mA DC - it cannot distinguish a fault mode dumping current into a human body from design mode into the intended load. Hence the use of ground fault circuit interrupters, which detect when the current flowing from the power supply (start of the circuit) is different from the current flowing back to the power supply (end of the circuit). The assumption is that the missing current is flowing to ground somewhere that it should not (maybe through a person), and the circuit is rapidly interrupted. These devices detect a current mismatch of as little as 5 mA - well below the AC or DC threshold for electrocution.