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Assuming I have a grounded electrical device, for example a microwave. The left drawing shows that the phase wire is connected (due to a fault in the wire) to the microwave case, and an electrical path is created from the phase, to the case, and from there diverging to the human (displayed as a resistor) and to the grounding. The right image shows the equivalent circuit.

I couldn't understand however, how does the grounding helps reduce the current on the human? There is a voltage of Vphase on both the human and the resistor to the grounding (it is the equivalent resistance of the grounding wire). They are connected in parallel. It appears that the current on the human is Vphase/R_human whether the device is grounded or not. So how does the grounding help? Also, what is the potential of the case during regular times (I thought it it probably 0) and during faulty times (thought it is Vphase, beacuse now Vphase is directly connected to it) ?

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  • \$\begingroup\$ In such a fault scenario (with the MW's case connected to protective earth), the residual-current device would trigger right when the fault occurs. So there is no chance of toughing the live case. If there is no RCD, good luck... \$\endgroup\$ – Sim Son Nov 10 '20 at 14:41
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Your equivalent circuit ignores what you'd call Rphase: the series resistance of the phase supply wiring and the internal resistance of the supply itself. The supply cannot provide an infinite current. Rgrounding_wire_resistance will be relatively very small compared to that, so there is a potential divider you haven't shown.

Secondly, a real-world mains socket supply has unavoidable over-current protection in the consumer unit (RCDs/fuses) and (in UK) in the mains plug (fuse). This cuts the supply before something in the loop of the wiring and the short-circuit melts. Earthing is not expected to protect people just on its own.

So Rgrounding_wire_resistance is to be low enough to divert drop as much supply current as possible away from a user-contact area, leaving a low voltage on that area until the protection cuts the supply off.

The Rgrounding_wire_resistance has a minimum limit: what's practical to install (not really thick wires everywhere) and economical for widespread use.

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  • \$\begingroup\$ Ok, so if I denote Rs the resistance of the supply (as you suggested), Rg of the grounding wire, Rh of the human, and lets assume the supply is 220V, Rs is 10[ohm], Rg is 1[ohm], Rh is 1k[ohm], than we have I = 20[A] on the supply, and I_h = 20[mA] on the human (which is, as far as i know dangerous current). If we compare it to no grounding at all, than Ih = 220/(Rs+Rh)=0.21[A]. It appears that the grounding didnt help at all. how come? \$\endgroup\$ – Jonathan Nov 10 '20 at 14:29
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    \$\begingroup\$ @Jonathan - that 20A passing through the ground will trip the supply circuit breaker within a few milliseconds and disconnect the supply before it can do any harm. Either the circuit breaker/fuse will trip because of over-current or the RCD will detect the current passing to the ground (only needs 30mA to trip). \$\endgroup\$ – Kevin White Nov 10 '20 at 15:26
  • \$\begingroup\$ @Jonathan, it appears wrong because your assumptions and equation are wrong. Ih is not 220/(Rs+Rh) and Rs is not 10R. You can easily investigate this further yourself, reading and understanding the answer then using it as a guide for your investigation. \$\endgroup\$ – TonyM Jan 30 at 14:26
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There are two scenarios. The resistance of the phase is low or the resistance of the phase is high.

If the resistance of the phase is low, a ground connection will trip the protection device. Protection devices with a dead short are typically very fast.

In the case that the phase has enough resistance that the protection device does not trip things get a bit more complicated. Suppose the feed is 120V and the protection device is 15A. The resistance of the feed will need to be at minimum 8ohms to keep the protection device from tripping. A good ground resistance will be sub 1ohm. We can calculate the voltage drop using those numbers. 120*(1/(8+1)) = 13V. About 13V would appear on the case worst-case scenario.

Finally the human touching the case. The resistance of the human body varies but 1500ohms is a good worst-case scenario. If we put 1500ohms in parallel with 1ohm the current divider gives us about 10mA. That is considered a bad shock but not lethal. This is far reduced when the resistance of the ground is lower.

Something that I want to note; is that the case of the phase having enough resistance not trip the protection is highly unlikely. At the very least it would not go unnoticed for any extended period of time. There would be 1800 watts of power being dissipated and that would most likely catch something on fire.

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  • \$\begingroup\$ Ok, so if I denote Rs the resistance of the supply (as you suggested), Rg of the grounding wire, Rh of the human, and lets assume the supply is 220V, Rs is 10[ohm], Rg is 1[ohm], Rh is 1k[ohm], than we have I = 20[A] on the supply, and I_h = 20[mA] on the human (which is, as far as i know dangerous current). If we compare it to no grounding at all, than Ih = 220/(Rs+Rh)=0.21[A]. It appears that the grounding didnt help at all. how come? \$\endgroup\$ – Jonathan Nov 10 '20 at 15:31
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    \$\begingroup\$ @Jonathan If you want to compare Apples to Apple the Rg of 20A wiring is going to be less than the Rg of 15A wiring. As the ampacity of the wires increases so does its resistance decreases. \$\endgroup\$ – vini_i Nov 10 '20 at 17:09
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    \$\begingroup\$ @Jonathan I added a little something to my answer. In your case that would be 4000 watts of power being dissapated. \$\endgroup\$ – vini_i Nov 10 '20 at 17:56

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