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I always understood that one of the two purposes of protective earthing is keeping the electric potential difference between the chassis of a devide and the ground (the ground I am standing uppon while using this device) at a low rate, preferably lower than what is considered harmful to the human body. However, looking at this particular realisation of the protective grounding (TN-Net) enter image description here there is something I don't understand:

The PE-Connection will keep the chassis at the same Potential as the ground located at the star point of the system. However (as this question indicates), the ground I am standing on in the moment I am touching the device may have a different potential than the one that exists at the point where the network is grounded. How will this help me not being shocked?

EDIT: I know that there are other protective measures, like the RCD shutting down when it detects a difference between the L1 current and the N current. I'd still like to know wether the difference in potentials do have an effect (or why they don't have an effect).

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  • \$\begingroup\$ What do your local wiring regulations insist on the earth resistance is and, how much current might have to flow to produce (maybe) enough voltage to give you a slight shock AND with that current flow, won't the breakers trip rendering the installation harmless (irrespective of whehter you have supplementary protection such as RCDs or GFCIs). \$\endgroup\$ – Andy aka Mar 20 at 17:26
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The thing that is not being drawn in all the diagrams is the parasitic resistance and inductance. Wires have resistance (and even earth does also), so if there is any current it will generate a voltage, but not much if the ground wire is sufficiently large.

If the ground cable is a few ohms, the amount of current the ground wire could support before a dangerous voltage 60V+ could build on the chassis would be about ~60A. The neutral wire would also take current away before you have much current on the ground wire in the event of a short\fault.

Not to mention that most systems like this would also have circuit protection like circuit breakers, so as soon as there is a short, the system shuts down.

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  • \$\begingroup\$ You argue that the for a high current of 60 A the voltage would be about 60 V. But it's 60 V with respect to the ground at the location of the star point, not 60 V with respect to the ground that I am standing on when I am touching the faulty chassis. The Potential difference between chassis and the earth I am standing at could be different. \$\endgroup\$ – Quantumwhisp Mar 21 at 6:27

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