- #1
RedX
- 970
- 3
Say you touch the hot wire and a piece of plumbing simultaneously. Then current flows through the hot wire, through you, to the ground, and I assume (though I'm not sure) to where the ground wire is tied to the neutral wire (at the breaker panel). But is the current in the ground mainly through the resistance of the earth, or the capacitance? I've attached a picture to clarify what I mean:
In the picture a short has occurred, which puts a high voltage across the ground resistor and capacitor (the load is unaffected by this short and is irrelevant - I just drew it to add context to the picture). The current through the resistor is about 120/Rground - I do not know if this current is high because I have no idea what the resistance of the ground is. The capacitor is now charging up from 0 V to 120 V. The current through the capacitor would be charging an RC circuit, 120/Rwire[1-exp(-t/RwireCground)]
So the total current would be the sum:
I=120/Rground+120/Rwire[exp(-t/RwireCground)]
So which of the two terms would be bigger? I have no idea what the capacitance of the Earth is. I don't even know if I'm modelling it correctly. The idea is that the Earth in general is an infinite sink for charge since it's so big. However, charge won't flow from the hot wire to the Earth unless the same amount of charge can jump back from the Earth into the circuit, or else the powerline would start to lose electrons. So I'm modelling this as having one plate of a capacitor at the place where the hot wire shorts to earth, and the other plate of the capacitor at the place where the ground is tied to the neutral, so whenever electron flows into one plate, it flows out of the other plate.
For the situation I outlined at the beginning, you would add the resistance of the human to all resistances to get your current.
I guess I just have a hard time believing that the resistance of the ground is small, so I want to believe it's capacitance instead, but I'm not sure.
*Actually, to be more accurate, the current should be:
I=[120/Rground](1-Rwire[exp(-t/RwireCground)])+120/Rwire[exp(-t/RwireCground)]
but I'm more interested in if it's the steady state value of the resistor current, or capacitor current that is responsible for shock.
In the picture a short has occurred, which puts a high voltage across the ground resistor and capacitor (the load is unaffected by this short and is irrelevant - I just drew it to add context to the picture). The current through the resistor is about 120/Rground - I do not know if this current is high because I have no idea what the resistance of the ground is. The capacitor is now charging up from 0 V to 120 V. The current through the capacitor would be charging an RC circuit, 120/Rwire[1-exp(-t/RwireCground)]
So the total current would be the sum:
I=120/Rground+120/Rwire[exp(-t/RwireCground)]
So which of the two terms would be bigger? I have no idea what the capacitance of the Earth is. I don't even know if I'm modelling it correctly. The idea is that the Earth in general is an infinite sink for charge since it's so big. However, charge won't flow from the hot wire to the Earth unless the same amount of charge can jump back from the Earth into the circuit, or else the powerline would start to lose electrons. So I'm modelling this as having one plate of a capacitor at the place where the hot wire shorts to earth, and the other plate of the capacitor at the place where the ground is tied to the neutral, so whenever electron flows into one plate, it flows out of the other plate.
For the situation I outlined at the beginning, you would add the resistance of the human to all resistances to get your current.
I guess I just have a hard time believing that the resistance of the ground is small, so I want to believe it's capacitance instead, but I'm not sure.
*Actually, to be more accurate, the current should be:
I=[120/Rground](1-Rwire[exp(-t/RwireCground)])+120/Rwire[exp(-t/RwireCground)]
but I'm more interested in if it's the steady state value of the resistor current, or capacitor current that is responsible for shock.