4 ms·

Potential difference between two otherwise not-connected systems is not well defined. In practice the system will not be totally ungrounded, but there will be s

by dfox 7y ago

Potential difference between two otherwise not-connected systems is not well defined. In practice the system will not be totally ungrounded, but there will be some high-impedance path to ground, and the impedance is hopefully large enough to not allow any significant fault current to flow.

nkurz 7y ago
I think this is getting close to the right answer, but I still don't understand the details. Let's say that we set up a large ground isolated system, and have it driving a resistance heater (20A at 120VAC). Standing barefoot on wet concrete, I tightly grab one of the leads to the heating element. With my other hand, I use a hammer smash the heating element so it's no longer a complete circuit. What happens and why? I suspect that I'll get a terrible shock and likely die, because some substantial current will start flowing through my body. You seem to be saying that I'll be fine. Why do you believe this? My guess is that the typical ground isolated system is safe not because of the isolation per se, but because the transformer used is small enough that the maximum current is not life threatening. When the circuit is intact, I'm unlikely to get anything other than a static shock. When the circuit is broken, it's current limited. My intuition is that this is the same as having a very small generator with only a few coils of wire. What am I missing?
- pjc50 7y ago
  > When the circuit is broken, it's current limited. When the circuit is broken .. by definition, there isn't any current flowing! You need a loop somehow from one of the terminals of the transformer through you to the other terminal in order to get a current flow. (There might be some transient effect caused by the inductance of the system within one half-cycle, which may give you an arc as you smash the heater, but that can't continue and is unlikely to pass through you)
- kragen 7y ago
  > because the transformer used is small enough that the maximum current is not life threatening. No, this is definitely not the case. 8mA is 1W at 120V, and although 8mA will not actually kill you, it will be a very nasty experience you will not want to repeat. 80mA, I wouldn't want to sell you life insurance. 160mA can definitely kill you. Efficient transformers are also not effective current-limiting devices; their coils do have some resistance, and their magnetics have some leakage inductance which limits the current, but these only produce a small voltage loss at their rated output. Say you have a transformer designed for a 50-watt razor (400mA at 120V, several times more than you can survive). If its coils are dissipating 10 watts, it's going to get pretty hot there inside the wall while you're shaving, so this means that its coil resistance is already impractically high for this application. (Think about how hot a 10-watt compact fluorescent or LED bulb gets, even though it's exposed to airflow.) Yet that's only 62.5Ω, which means that if you short out its output, you're going to get 1900 mA through the short, almost five times its rated output and 19 times as much current as is needed to kill you. Using a more realistic coil resistance gives even larger peak currents. (Also, using 20% of the total circuit power means that the razor sees 20% lower voltage, which may cause it to fail to work.) The bottom line is, you can't depend on the transformer burning itself up or otherwise limiting current to protect you. The correct answer is given in my other response above. Special transformers like neon-sign transformers and arc-welding transformers do have sufficiently high leakage inductance to not burn up if you short out their output, but such transformers are not used for general power application. Generally their maximum currents are still far too high to provide much protection for squishy human bodies.

by dfox 7y ago

nkurz 7y ago

I think this is getting close to the right answer, but I still don't understand the details. Let's say that we set up a large ground isolated system, and have it driving a resistance heater (20A at 120VAC). Standing barefoot on wet concrete, I tightly grab one of the leads to the heating element. With my other hand, I use a hammer smash the heating element so it's no longer a complete circuit. What happens and why? I suspect that I'll get a terrible shock and likely die, because some substantial current will start flowing through my body. You seem to be saying that I'll be fine. Why do you believe this? My guess is that the typical ground isolated system is safe not because of the isolation per se, but because the transformer used is small enough that the maximum current is not life threatening. When the circuit is intact, I'm unlikely to get anything other than a static shock. When the circuit is broken, it's current limited. My intuition is that this is the same as having a very small generator with only a few coils of wire. What am I missing?

pjc50 7y ago
> When the circuit is broken, it's current limited. When the circuit is broken .. by definition, there isn't any current flowing! You need a loop somehow from one of the terminals of the transformer through you to the other terminal in order to get a current flow. (There might be some transient effect caused by the inductance of the system within one half-cycle, which may give you an arc as you smash the heater, but that can't continue and is unlikely to pass through you)
kragen 7y ago
> because the transformer used is small enough that the maximum current is not life threatening. No, this is definitely not the case. 8mA is 1W at 120V, and although 8mA will not actually kill you, it will be a very nasty experience you will not want to repeat. 80mA, I wouldn't want to sell you life insurance. 160mA can definitely kill you. Efficient transformers are also not effective current-limiting devices; their coils do have some resistance, and their magnetics have some leakage inductance which limits the current, but these only produce a small voltage loss at their rated output. Say you have a transformer designed for a 50-watt razor (400mA at 120V, several times more than you can survive). If its coils are dissipating 10 watts, it's going to get pretty hot there inside the wall while you're shaving, so this means that its coil resistance is already impractically high for this application. (Think about how hot a 10-watt compact fluorescent or LED bulb gets, even though it's exposed to airflow.) Yet that's only 62.5Ω, which means that if you short out its output, you're going to get 1900 mA through the short, almost five times its rated output and 19 times as much current as is needed to kill you. Using a more realistic coil resistance gives even larger peak currents. (Also, using 20% of the total circuit power means that the razor sees 20% lower voltage, which may cause it to fail to work.) The bottom line is, you can't depend on the transformer burning itself up or otherwise limiting current to protect you. The correct answer is given in my other response above. Special transformers like neon-sign transformers and arc-welding transformers do have sufficiently high leakage inductance to not burn up if you short out their output, but such transformers are not used for general power application. Generally their maximum currents are still far too high to provide much protection for squishy human bodies.