Say a person touches the live wire as shown in the first attachment, the current will flow through him and dissipates into the ground as shown in the second attachment right? the fault current through a person does not really “return” to the source as drawn in pencil correct? Also is it possible that the return current is less than the outgoing current? 2/3A vs 1/3A , assuming that the resistance of a person is 10 ohms . Much appreciated!enter image description here

Sorry the second picture was not able to be attached , it just shows current dissipating into the ground in ring shaped waves propagating away from a grounding rod in the earth


The current DOES return to the source. Current has to flow in a closed loop It will come 'out' of the ground back to the bottom connection of your 10V source. I have annotated your bottom picture. Thus the return path will again have 1/3 + 1/3 = 2/3 Amps.

enter image description here

If the 1/3 Amp going through your body would somehow 'disappear' into the ground the source would have 2/3 Amps leaving at the top and 1/3 Amp returning at the bottom. This is not possible.

  • \$\begingroup\$ Conversely, if the two grounds were not connected together, for example one was a local ground, the isolated equipment chassis, and the other the floor of the room, then no return path exists and no current would flow through the person. \$\endgroup\$ – Peter Jennings Jul 30 at 10:51
  • \$\begingroup\$ Since current dissipates into the ground in waves of expanding rings around the persons feet how exactly does the current return to the source through the earth , say the source grounding rod is 2 km away . \$\endgroup\$ – Jason Jul 30 at 12:37
  • \$\begingroup\$ Ok is it more like the source grounding rod “ sucking up” electrons/current from the earth? \$\endgroup\$ – Jason Jul 30 at 12:48
  • \$\begingroup\$ The current won't flow in perfect "rings". Some current might initially propagate away from the guy's foot in a direction away from the source, but it will eventually loop back to the source. You can solve for the actual current distribution numerically using a finite element method (FEM). \$\endgroup\$ – The Photon Jul 30 at 16:09

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